Chapter 11: Adsorption and Colloids

In the early 1860s, the Scottish chemist Thomas Graham was doing something that sounds almost dull. He was measuring how fast dissolved substances travel through a sheet of parchment. Salt, sugar and urea slipped through without much trouble. Gelatin, gum and egg albumin crawled — and no matter how carefully he tried, they refused to form crystals.

Graham decided he was looking at two families of matter. The fast, crystal-forming substances he called crystalloids. The slow, glue-like ones he named colloids, from the Greek word kolla, meaning glue. The apparatus he built to separate one from the other he called a dialyser, and the process, dialysis — words a hospital still uses today.

Graham thought he had found a difference of kind. Later chemists showed it was really a difference of size. Colloidal particles are simply far larger than an ion or a small molecule, yet far too small to see or to settle out. They are big enough to be trapped by a membrane, and big enough to scatter light — which is why milk is white and a torch beam is visible in fog.

That size has one more consequence. Break matter into extremely fine particles and you create an enormous amount of surface. And surfaces, as this chapter will show, are where a great deal of chemistry quietly happens.

Diagram

Fig. 11.0 – Split opening illustration: (left) a parchment bag of gelatin solution suspended in water, small ions passing out through the membrane while large colloidal particles stay inside; (right) a torch beam passing through two beakers — invisible in a clear salt solution, visible as a bright cone in a milky colloidal dispersion. Label “crystalloid”, “colloid”, “membrane”, “scattered light”.

NoteCuriosity Corner

Q1. The same mass of charcoal cleans up far more dye from a solution when it is a fine powder than when it is a single lump. Why should merely crushing it make such a difference? Q2. Cooling a gas makes it stick to a solid surface more strongly — yet one type of adsorption actually increases when you heat the system. How can both of these be true? Q3. Milk looks white, and fog makes the beam of a headlight visible in the air, but a clear salt solution does neither. What must be different about the particles inside them? Q4. A pinch of alum is enough to clear a bucket of muddy water, but a gelatin solution needs a large amount of salt before anything separates out. Why does one need so little and the other so much? Q5. A catalyst speeds up a reaction, and yet at the end of it all we recover the catalyst unchanged in mass and in chemical composition. If it took part in the reaction, how does it come back untouched?

Keep these five questions in mind as you read. The chapter gives you everything you need to reason out each one, and you will meet the answers along the way.

11.1 Introduction

Most of the chemistry you have studied so far happens inside a substance — inside a beaker, inside a reacting mixture, inside a cell. But a great deal of chemistry happens only at the outermost layer of matter, where one substance meets another. This chapter begins there, at the surface.

Two Everyday Observations

Dip a metal spoon in milk and take it out. You will notice that a film of milk particles covers the surface of the spoon. The milk has not soaked into the metal; it is clinging to the outside.

Now take a cold water bottle out of the refrigerator and keep it on a table for a while. Water vapour is seen to condense on the outer surface of the bottle, forming droplets or a film. Again, nothing has entered the glass or the plastic — the water molecules have simply gathered on the surface.

In both cases the same thing has happened. The milk particles in the first case, and the water molecules from the air in the second, get adsorbed on the surface of the spoon and of the bottle.

The Invisible Film on Everything

The surfaces of all the objects around us are exposed to the atmosphere. Water molecules, as well as other gas molecules from the air such as \(\mathrm{N_2}\) and \(\mathrm{O_2}\), form an invisible multi molecular film on these objects — that is, a film built of several layers of molecules rather than a single one. Your desk, your window pane and the page you are reading all carry such a film right now.

This gathering of a substance on a surface is known as the phenomenon of adsorption. Notice carefully what is being claimed: the material collects at the surface and stays there, instead of spreading uniformly into the interior. That one distinction runs through the whole of the first half of this chapter, and getting it right early saves a great deal of confusion later.

NoteDefinition

Adsorption is the accumulation of molecules of a substance on the surface of a solid or a liquid, the substance staying at the surface instead of spreading uniformly into the bulk.

The film on the chilled bottle is easy to picture if we magnify a small patch of its surface, as drawn in Fig. 11.12.

Diagram

Fig. 11.12 – A chilled water bottle on a table, with a magnified circular inset of a small patch of its outer surface showing a film several molecules thick made up of H₂O, N₂ and O₂ molecules resting on the surface. Label “adsorbed multi molecular film”, “surface of bottle”, “molecules from air”.

Q. If a film of gas molecules is sitting on every object around us, why can we not see it?

A. Because it is only a few molecules thick — far thinner than the wavelength of visible light — so it does not change the way the object looks. We notice it only when it grows thick enough to become a visible layer, as when water vapour builds up into droplets on a chilled bottle.

The word adsorption was coined in the nineteenth century from the Latin ad, meaning “to” or “at”, joined to sorbere, meaning “to suck in”. The prefix was chosen deliberately, to set the word apart from absorption, in which a substance is drawn into the interior of a material. The two words are therefore not careless spellings of each other: that single letter carries the whole scientific difference between a surface process and a bulk process, and examiners rely on students knowing it.

This invisible film has a practical consequence in the laboratory. A clean glass surface always carries an adsorbed layer of moisture from the air, and a finely powdered or hygroscopic solid carries a great deal more. That is why a solid meant for accurate weighing is dried first and then stored in a desiccator, and why a crucible or weighing bottle is lifted with tongs or held through a clean tissue rather than with bare fingers — the moisture and skin oils left by fingertips are adsorbed on the glass and register as extra mass on a sensitive balance. If a sample appears to gain a little mass each time it is weighed, adsorption from the air is usually the reason.

(question only — no solution)

Two identical clean glass slides are left in the same room. Slide A has been lying on the bench all morning; slide B has just been taken out of a freezer. Only slide B develops a visible film of moisture. State whether slide A also carries an adsorbed layer of molecules from the air, and explain why the film becomes visible on only one of the two slides.

11.2 Adsorption

Adsorption is not an accident of dirty surfaces. It happens because a surface is, in a real physical sense, an incomplete place. To see why, we have to look at what a molecule sitting at a surface experiences compared with one buried deep inside.

11.2.1 Unbalanced Forces

Consider a surface of a liquid or a solid. The molecular forces at the surface of a liquid are unbalanced, or in an unsaturation state. In solids, the ions or molecules at the surface of a crystal do not have their forces satisfied by close contact with other particles.

The reason is a matter of simple geometry. A molecule in the interior of the liquid or solid is completely surrounded by neighbours, so the attractions pulling on it from every direction cancel out. A molecule at the surface has neighbours below it and beside it, but nothing above — so the pull on it does not cancel. There is a net inward force, and a set of attractive forces left unused. This contrast is drawn in Fig. 11.1, where molecule A is a bulk molecule and molecule B is a surface molecule.

Diagram

Fig. 11.1 – Unbalanced forces: a container of liquid with molecule A deep inside the bulk, arrows radiating equally in all directions around it (forces balanced), and molecule B at the surface with arrows drawn sideways and downwards only, giving a net inward pull. Label “A – bulk molecule”, “B – surface molecule”.

Because of this unsaturation, solid and liquid surfaces tend to attract gases or dissolved substances with which they come in close contact. Thus the substance accumulates on the surface of the solid or liquid. In everyday language, a surface is “hungry”, and adsorption is how it satisfies that hunger.

Q. Why does a substance accumulate only on the surface, and not equally throughout the solid?

A. Because only the surface particles have unsatisfied forces. The particles inside are already surrounded on all sides by their own neighbours, so they have nothing left over with which to hold an incoming molecule.

What is adsorption?

11.2.2 Why Does Adsorption Occur?

The adsorption phenomenon is caused by London dispersion forces or van der Waals forces. These are short range and additive. Short range means their strength falls off sharply as the distance between particles increases, so a molecule must come very close to the surface before it feels them. Additive means the individual attractions add up: the adsorption force is the sum of all the interactions between all the atoms involved, which is why a surface made of a great many atoms can hold an incoming molecule quite firmly even though each single attraction is weak.

The pulling interactions cause the surface of a liquid to tighten like an elastic film. A measure of the elastic force at the surface of a liquid is called surface tension (refer to Chapter 10). This is the same property that lets a needle float on water and pulls a small drop into a sphere.

Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area. Read that carefully: to make more surface, you must spend energy. It follows that a system with a large surface is carrying extra energy that it would rather not have.

NoteDefinition

Surface tension is the amount of energy required to stretch or increase the surface of a liquid by unit area.

surface tension (SI unit: \(\text{N m}^{-1}\), which is the same as \(\text{J m}^{-2}\) — force per unit length is numerically the same as energy per unit area. It is always a positive quantity; a larger value means more energy is needed to create fresh surface.)

There is a tendency to have minimum surface tension, that is, a decrease of free energy, which leads to adsorption. The idea behind “free energy” is one you will meet formally in thermodynamics: a system left to itself moves towards the state of lowest free energy, and any change that lowers free energy tends to happen on its own. A bare surface carries unsatisfied forces and therefore extra free energy. When gas molecules or dissolved particles settle on it, some of those unsatisfied forces are used up, the surface energy falls, and the system becomes more stable. Adsorption is simply a surface taking the cheapest route available to a lower energy.

Terms Used in Adsorption

Two terms are needed constantly from here on, and students lose marks by mixing them up.

NoteDefinition

Adsorbent: the material or substance present in the bulk, on the surface of which adsorption takes place is called adsorbent.

NoteDefinition

Adsorbate: the substance getting adsorbed on the adsorbent is called adsorbate.

A quick way to keep them straight is to ask which one does the holding. The adsorbent is the host — it provides the surface and stays put. The adsorbate is the guest — it arrives and settles on that surface. When hydrogen gas collects on finely divided nickel, nickel is the adsorbent and hydrogen is the adsorbate.

11.2.3 Examples of Adsorption and Absorption

You know that when cotton is dipped in water, the cotton becomes wet with water, which is due to absorption. Here the water does not stop at the surface of the fibres — it travels right into the body of the cotton and is distributed throughout it. That is the essential mark of absorption, and it is exactly what adsorption is not.

Some more examples of adsorption:

  1. Adsorption of gases like hydrogen and oxygen by finely divided metals, namely platinum, palladium, copper, nickel, etc.

  2. Adsorption of gases like nitrogen and carbon dioxide by activated charcoal.

  3. Removal of colouring matter like an organic dye, for example methylene blue. When charcoal is added to methylene blue solution and shaken, it becomes colourless after some time, as dye molecules accumulate on the surface of charcoal.

The third example is worth pausing on, because the evidence is visible to the eye. The dye has not been destroyed and it has not soaked into the interior of the charcoal grains — it has been pulled out of the solution and parked on the charcoal’s surface. If that charcoal is later filtered off and treated suitably, the dye can be recovered.

Table 11.1 shows a comparison between adsorption and absorption.

Table 11.1: Comparison between Adsorption and Absorption

Adsorption Absorption
1. Adsorbed matter is concentrated only at the surface and does not penetrate through the surface to the bulk of adsorbent. Adsorption is a surface phenomenon. Absorbed matter is uniformly distributed inside as well as on the surface of the bulk of substance. Absorption is a bulk phenomenon.
2. Concentration of the adsorbate is high only at the surface of the adsorbent. Concentration of the absorbate is uniform throughout the bulk of the absorbent.
3. It is dependent on temperature and pressure. It is independent of temperature and pressure.
4. It is accompanied by evolution of heat known as heat of adsorption. It is not accompanied by evolution or absorption of heat.
5. It depends on surface area. It is independent of surface area.
6. Example: Adsorption of a gas or liquid like acetic acid by activated charcoal. Example: Absorption of water by cotton. Absorption of ink by blotting paper.

The forces responsible for adsorption carry two names for a good reason. Johannes Diderik van der Waals, the Dutch physicist, argued in the nineteenth century that real gas molecules must attract one another — an idea built into the equation of state that still bears his name, and one that won him the Nobel Prize in Physics in 1910. What he could not say was why neutral, non-polar atoms should attract at all. That explanation came decades later from the physicist Fritz London, who showed that the constant shifting of electron clouds creates fleeting, momentary dipoles which pull on neighbouring atoms. This is why the weakest of these attractions is called the London dispersion force, while “van der Waals forces” survives as the broader family name.

Hospitals use adsorption as an emergency treatment. A patient who has swallowed certain poisons or taken an overdose of some medicines may be given activated charcoal to drink as a slurry. The charcoal is not digested and does not enter the bloodstream; it travels through the gut, and the toxic molecules are adsorbed onto its huge internal surface and carried out of the body instead of being absorbed into the blood. Note how precisely the two words divide the work here — the treatment succeeds because adsorption onto charcoal wins the race against absorption through the intestinal wall. It is not a universal antidote, which is why it is given under medical supervision and only for poisons that charcoal is known to hold.

11.2.4 Desorption

Adsorption can also be reversed. The process of removal of an adsorbed substance from a surface on which it was adsorbed is called desorption.

NoteDefinition

Desorption is the process of removal of an adsorbed substance from the surface on which it was adsorbed.

Desorption is usually brought about by heating the adsorbent or by lowering the pressure of the gas above it. This matters industrially, because an adsorbent that can be cleaned by desorption can be used again and again instead of being discarded.

11.2.5 Sorption

Sometimes both processes go on at once. When a chalk is dipped in ink, ink molecules are adsorbed at the surface of the chalk and the surface becomes coloured, while the solvent of the ink goes deeper into the chalk due to absorption. When both adsorption and absorption occur simultaneously, it is known as sorption.

NoteDefinition

Sorption is the phenomenon in which adsorption and absorption take place simultaneously.

The chalk-and-ink observation is a neat one because a single experiment separates the two processes: the coloured dye stops at the surface, while the liquid it was dissolved in keeps travelling inward. The two paths are set out side by side in FC 11.1.

Try this: Dip a chalk in ink. What do you observe?

Flowchart

FC 11.1 – Chalk dipped in ink, showing sorption as two simultaneous paths: chalk + ink → branch 1: dye molecules held at the surface → surface becomes coloured → adsorption; branch 2: solvent of the ink travels into the interior of the chalk → distributed through the bulk → absorption; both branches together → sorption.

Adsorption vs Absorption vs Sorption | Feature | Adsorption | Absorption | Sorption | |—|—|—|—| | Where the substance goes | Stays at the surface only | Spreads uniformly through the bulk | Both, at the same time | | Type of phenomenon | Surface phenomenon | Bulk phenomenon | Combination of the two | | Concentration | High at the surface, low inside | Same everywhere inside | High at the surface and also present inside | | Effect of surface area | Depends strongly on surface area | Independent of surface area | Surface area affects the adsorption part | | Example | Methylene blue held on charcoal | Water taken up by cotton | Ink on a piece of chalk |

The three cases are easiest to remember as pictures rather than as words, and Fig. 11.13 places them together.

Diagram

Fig. 11.13 – Three side-by-side panels: (a) adsorption — gas molecules sitting in a layer on the outer surface of a charcoal grain, interior empty; (b) absorption — water molecules spread evenly throughout a piece of cotton; (c) sorption — a stick of chalk in ink, dye molecules shown only on the outer surface while solvent molecules are shown penetrating inward. Label each panel and mark “surface only”, “throughout bulk”, “both”.

Classify each of the following as adsorption, absorption or sorption, and in the cases of adsorption identify the adsorbent and the adsorbate.

(a) Water vapour collecting on the outside of a chilled bottle.

(b) Ink spreading through blotting paper.

(c) Nitrogen gas taken up by activated charcoal.

(d) Ink on a stick of chalk.

Solution.

  1. The water molecules gather only on the outer surface of the bottle and do not enter the material of the bottle. This is adsorption. The bottle surface is the adsorbent and water is the adsorbate.

  2. The ink is drawn into the body of the paper and is distributed uniformly through it, not held at the surface. This is absorption.

  3. Nitrogen molecules accumulate on the surface of the charcoal, including the surfaces of its pores, and are not distributed uniformly through the solid carbon. This is adsorption, with activated charcoal as the adsorbent and \(\mathrm{N_2}\) as the adsorbate.

  4. The dye stays on the surface while the solvent travels inward, so adsorption and absorption occur together. This is sorption.

The test to apply in every such question is a single one: ask whether the concentration of the added substance is high only at the surface, or the same throughout the bulk.

(question only — no solution)

For each of the following, name the adsorbent and the adsorbate, and state one reason for your choice:

(a) hydrogen gas held by finely divided palladium,

(b) acetic acid taken up from solution by activated charcoal,

(c) methylene blue removed from water by charcoal.

11.3 Types of Adsorption

There are mainly two types of adsorption phenomenon, depending on the nature of the forces involved. Either the adsorbate is held to the surface by weak physical attractions, or it is held by actual chemical bonds. Almost everything else that distinguishes the two types — the heat released, the effect of temperature, the number of layers formed, whether the process can be reversed — follows from that one difference in the nature of the force.

11.3.1 Physical Adsorption or Physisorption

When the adsorbate, such as gas molecules, accumulates at the surface of a solid on account of weak van der Waals forces, the adsorption is termed physical adsorption or physisorption. No new chemical bond is formed; the molecule simply sticks.

NoteDefinition

Physisorption is the adsorption in which the adsorbate is held on the surface of the adsorbent by weak van der Waals forces, with no chemical bond being formed.

The van der Waals forces involved are similar to the forces causing condensation of a gas into a liquid. Thus, heat is released in physisorption, and the heat released during physisorption is of the same order of magnitude as the heat of condensation. That comparison is worth holding on to: physisorption is very nearly a gas condensing, except that it condenses onto a surface instead of into a bulk liquid.

Key Features of Physisorption

Due to the weak nature of van der Waals forces, physisorption is also weak in nature. Several consequences follow, and each one is a favourite in examinations.

The adsorbed gas forms several layers of molecules at high pressures. Once the first layer of molecules is in place, incoming molecules can be held by van der Waals attraction to the molecules already there, so the layer can grow into a multimolecular one.

The extent of adsorption is large at low temperatures. Since the attraction is weak, a molecule moving too fast simply bounces away; cooling the gas slows its molecules down so that the weak forces are able to hold them.

The equilibrium is attained rapidly. There is no energy barrier to be crossed, so a molecule that reaches the surface is held almost at once.

The physisorption is readily reversed by lowering the pressure of the gas or by raising the temperature. Lowering the pressure means fewer molecules arrive to replace those that leave; raising the temperature gives the adsorbed molecules enough thermal energy to break free of the weak hold.

Q. Why can physisorption build up many layers when chemisorption forms only one?

A. Van der Waals forces act between the adsorbate molecules themselves as well as between adsorbate and surface, so fresh molecules can settle on the layer already present. A chemical bond, by contrast, can form only with the atoms of the surface. Once every surface site has bonded, there is nothing left for a second layer to bond to.

11.3.2 Chemical Adsorption or Chemisorption

Chemisorption was first investigated in 1916 by the American chemist Irving Langmuir (1881–1957). When the gas molecules accumulate on the surface of a solid adsorbent by means of chemical bonds, be they covalent or ionic, the adsorption is called chemical adsorption, or chemisorption.

NoteDefinition

Chemisorption is the adsorption in which the adsorbate is held on the surface of the adsorbent by chemical bonds, covalent or ionic.

Chemisorption is specific in nature. This follows directly from the definition: a chemical bond can form only between particular partners, so a given surface will chemisorb only those gases with which it can actually bond. Oxygen chemisorbs on tungsten and hydrogen on nickel, but no surface chemisorbs everything.

Chemisorption involving the gas–solid system as adsorbate and adsorbent is usually exothermic. It means that heat is released during this process, which is what we expect whenever bonds are formed. (Exception: the adsorption of hydrogen on glass is endothermic — heat is absorbed during the process. This is due to the dissociation of hydrogen, since the \(\mathrm{H\!-\!H}\) bond must be broken first, and breaking a bond costs energy.)

The heat evolved in chemisorption per mole of adsorbate is nearly of the same order of magnitude as that accompanying chemical bonding. Compare this with physisorption, where the heat released was of the order of a heat of condensation. The heat of adsorption is therefore a piece of experimental evidence that tells us which type of adsorption we are dealing with.

heat of adsorption (Usual unit: \(\text{kJ mol}^{-1}\). The values quoted in data tables — such as 20 – 40 for physisorption and 40 – 200 for chemisorption — are magnitudes of the heat released. If the same quantity is written as an enthalpy change, \(\Delta H\), it carries a negative sign for an exothermic adsorption. A common error is to compare a positive magnitude taken from a table with a negative \(\Delta H\) value and draw the wrong conclusion about which process releases more heat.)

Why Chemisorption Needs Activation Energy

Chemisorption involves a large energy of activation and is referred to as activated adsorption. Activation energy is the minimum energy a molecule must possess before a chemical change can take place — here, before old bonds can be loosened and a new bond to the surface can be made. A molecule that arrives with less than this energy is turned back, even though the final chemisorbed state would be the more stable one. The contrast between the two types is drawn as an energy profile in Fig. 11.14.

Diagram

Fig. 11.14 – Potential energy plotted against distance of the molecule from the surface, two curves on one set of axes: a shallow physisorption well with no barrier, and a deeper chemisorption well that can be reached only after crossing an energy hump. Label “physisorption (shallow well, no barrier)”, “chemisorption (deep well)”, “activation energy”, “heat of adsorption”.

Chemisorption increases with increase in temperature in the beginning, as a greater number of molecules can have the activation energy. But after a certain temperature, chemisorption decreases with an increase in temperature, as the chemical bonds break. A plot of the extent of chemisorption against temperature therefore rises to a maximum and then falls away, as sketched in Fig. 11.15.

Diagram

Fig. 11.15 – Extent of chemisorption (y-axis) plotted against temperature (x-axis): the curve rises to a maximum and then falls. Mark the rising part “more molecules acquire activation energy”, the maximum, and the falling part “chemical bonds break”.

the curve of extent of chemisorption against temperature in Fig. 11.15 (Study the shape of the curve and answer: (i) In which region does a rise in temperature help adsorption, and what is happening to the molecules there? (ii) In which region does a rise in temperature hinder it, and what is happening to the bonds there? (iii) Sketch on the same axes the shape you would expect for physisorption of the same gas, and state how the two curves differ at the low-temperature end. — no answer given here)

Q2. Cooling a gas makes it stick to a solid surface more strongly — yet one type of adsorption actually increases when you heat the system. How can both of these be true?

A. Because the two statements describe two different kinds of holding force. Physisorption depends on weak van der Waals attraction, with no barrier to cross, so the only thing temperature does is give the molecules enough energy to escape — cooling therefore increases physisorption. Chemisorption requires a chemical bond to form, and bond formation needs activation energy. Heating supplies that energy, so more molecules can react with the surface and chemisorption at first increases. The two effects run in opposite directions because in one case temperature is only helping molecules leave, while in the other it is also helping them react. Even chemisorption, though, eventually falls off at very high temperature, once the heat begins to break the very bonds it helped to form.

From Physisorption to Chemisorption

Sometimes at low temperature, physisorption occurs, which passes into chemisorption as the temperature is raised. The same gas on the same solid can therefore show both behaviours — held loosely by van der Waals forces when cold, then bonded chemically once it has enough energy to react. Nitrogen on iron is the standard illustration, and it appears in Table 11.2 as an example of both types.

Besides, chemisorption is dependent on the surface area of the adsorbent — as indeed physisorption is, since both processes need surface on which to happen. A full comparison of the two types is set out in Table 11.2.

Table 11.2: Comparison of physisorption and chemisorption

Physisorption Chemisorption
1. The forces operating are weak van der Waals forces. The forces operating are chemical in nature (covalent or ionic bonds).
2. Not specific in nature. All gases adsorb on all solids. For example, all gases adsorb on charcoal. Highly specific and occurs only when chemical bond formation is possible between adsorbent and adsorbate. For example, adsorption of oxygen on tungsten, hydrogen on nickel, etc.
3. The heat of adsorption is low and lies in the range 20 – 40 kJ mol\(^{-1}\). Higher heat of adsorption and lies in the range 40 – 200 kJ mol\(^{-1}\).
4. Occurs at low temperature and decreases with an increase of temperature. Favoured at high temperature; the extent of chemical adsorption is lowered at very high temperature, due to bond breaking.
5. For example: at low temperature \(\mathrm{N_2}\) gas is physically adsorbed on iron. For example, \(\mathrm{N_2}\) gas chemically adsorbed on iron at high temperature forms a layer of iron nitride, which desorbs at very high temperature.
6. Reversible. Irreversible.
7. Physisorbed layer may be a multimolecular layer of adsorbed particles under high pressure. Chemisorption forms a monomolecular layer of adsorbed particles.

Irving Langmuir came to surface chemistry through a very practical problem. Working in the research laboratory of an American electrical company, he was trying to understand why tungsten filament lamps blackened and failed — which meant studying how gases behave on hot metal surfaces. Out of that work came the idea that a surface offers a fixed number of sites, each able to hold one molecule and no more: exactly the monomolecular layer that Table 11.2 records as a mark of chemisorption. Langmuir was awarded the Nobel Prize in Chemistry in 1932 for his discoveries in surface chemistry, and he remains one of the few industrial chemists ever to have won it.

The distinction between the two types of adsorption sits at the centre of the problem of storing hydrogen as a fuel. Hydrogen carries a great deal of energy for its mass and burns to give only water, but as a gas it occupies an inconvenient volume. One approach holds it on solids of very large surface area, where physisorption stores it and a gentle warming or a drop in pressure releases it again — convenient, but the weak forces mean low temperatures are usually needed. The other approach uses metals and alloys that chemisorb hydrogen and take it up as a hydride, packing far more into the same volume, but now the bonds are strong and heat must be supplied to get the hydrogen back. The engineering problem is thus a chemical one: a store that binds hydrogen too weakly will not hold it, and one that binds it too strongly will not give it up.

A gas is adsorbed on a solid surface and the following observations are made. The heat of adsorption is \(160\ \text{kJ mol}^{-1}\); the adsorption is very slow at room temperature but becomes rapid on heating; only one layer of adsorbed molecules is formed however high the pressure; and the gas cannot be recovered by simply reducing the pressure. Identify the type of adsorption and justify each observation.

Solution.

The type is chemisorption. Each observation points the same way.

  1. A heat of adsorption of \(160\ \text{kJ mol}^{-1}\) lies within the range 40 – 200 kJ mol\(^{-1}\), which is the range associated with chemical bond formation, and well above the 20 – 40 kJ mol\(^{-1}\) typical of van der Waals attraction.

  2. Adsorption being slow when cold and fast on heating shows that an activation energy has to be supplied. Physisorption needs no such energy and reaches equilibrium rapidly even at low temperature.

  3. Formation of only a single layer, even at high pressure, shows that the adsorbate is bonded directly to the surface atoms. Once every site is occupied no further bonding is possible, so the layer stays monomolecular.

  4. The process not being reversed on lowering the pressure confirms that strong chemical bonds, and not weak physical attractions, are holding the molecules in place.

All four observations agree, which is how a question of this kind should always be answered — the type is decided by the set of properties taken together, not by any single one of them.

(question only — no solution)

Classify each of the following as physisorption or chemisorption, giving one reason in each case:

(a) the heat of adsorption of a gas on a solid is \(28\ \text{kJ mol}^{-1}\);

(b) oxygen is taken up by tungsten;

(c) the adsorbed gas is completely released when the pressure above the solid is reduced at constant temperature;

(d) \(\mathrm{N_2}\) held on iron at high temperature forms a layer of iron nitride.

11.4 Factors Affecting Adsorption of Gases on Solids

All solids adsorb gases to some extent. The extent of adsorption depends upon a number of factors, five of which are discussed in this section. Notice as you read that each factor is really the same question asked in a different way: how much surface is available, and how strongly can that surface hold on to the gas that reaches it? The five factors are collected in FC 11.2.

Flowchart

FC 11.2 – Extent of adsorption of a gas on a solid, branching into five factors: (1) nature of the adsorbate → higher critical temperature → more adsorption; (2) nature of the adsorbent → more porous → more adsorption; (3) surface area of adsorbent → larger area → more adsorption; (4) temperature → higher temperature → less adsorption; (5) pressure of gas → higher pressure → more adsorption, until saturation is reached.

Nature of the Adsorbate (Gas)

The amount of gas adsorbed by a solid depends on the nature of the gas. Gases having high critical temperature liquify easily and can readily be adsorbed (refer to Chapter 10).

Critical temperature is the temperature above which a gas cannot be liquefied, no matter how much pressure is applied to it. A high critical temperature therefore tells us that the molecules of that gas attract one another strongly, since they can be pulled into the liquid state even when the temperature is fairly high. Those same strong attractions are what let the molecules be held on a solid surface, which is why the two properties go together.

The gases such as \(\mathrm{SO_2}\), \(\mathrm{Cl_2}\) and \(\mathrm{NH_3}\), which are easily liquifiable, are adsorbed to a large extent compared to \(\mathrm{N_2}\), \(\mathrm{O_2}\), \(\mathrm{H_2}\), etc., which are difficult to liquify. Table 11.3 sets the two quantities side by side, and the pattern is unmistakable.

Table 11.3: Critical temperature of gases and volume adsorbed

Gas Critical temperature/K Volume adsorbed/cm\(^3\)
\(\mathrm{N_2}\) 126 08
\(\mathrm{HCl}\) 324 72
\(\mathrm{NH_3}\) 406 181
\(\mathrm{Cl_2}\) 417 235
\(\mathrm{SO_2}\) 430 380

extent of adsorption along a series of gases arranged by critical temperature (Along the series \(\mathrm{N_2} \rightarrow \mathrm{HCl} \rightarrow \mathrm{NH_3} \rightarrow \mathrm{Cl_2} \rightarrow \mathrm{SO_2}\), critical temperature rises from 126 K to 430 K and the volume adsorbed rises from 8 cm\(^3\) to 380 cm\(^3\) — because a higher critical temperature means stronger intermolecular attraction, and a gas whose molecules attract each other strongly is also held strongly by a surface. Read backwards, the trend is just as useful: a gas that is hard to liquefy is hard to adsorb.)

Diagram

Fig. 11.16 – Scatter plot of volume adsorbed (cm³, y-axis) against critical temperature (K, x-axis) using the five data pairs of Table 11.3, points labelled N₂, HCl, NH₃, Cl₂ and SO₂, with a smooth rising trend line drawn through them.

the plot of volume adsorbed against critical temperature in Fig. 11.16 (Using the plotted points, answer: (i) Is the relationship between the two quantities direct or inverse? (ii) Between which two neighbouring gases in the series is the increase in critical temperature small while the increase in volume adsorbed is large, and what does that tell you about how carefully critical temperature alone can be used to predict adsorption? (iii) The critical temperature of \(\mathrm{O_2}\) is lower than that of \(\mathrm{N_2}\). From the trend of the graph, predict whether the volume of \(\mathrm{O_2}\) adsorbed would be greater or smaller than 8 cm\(^3\). — no answer given here)

The idea of a critical temperature came from careful work on carbon dioxide in the 1860s by the Irish chemist Thomas Andrews. He compressed the gas at a series of fixed temperatures and found something unexpected: below a certain temperature, pressure produced a liquid with a visible surface, and above it no amount of pressure would do so. The gas simply became denser and denser with no liquid ever appearing. Andrews had found what we now call the critical point, and the temperature he identified for carbon dioxide is the reason a \(\mathrm{CO_2}\) cylinder at room temperature holds liquid while an oxygen cylinder cannot.

Nature of the Adsorbent

Substances which provide a large surface area for a given mass are effective as adsorbents and adsorb appreciable volumes of gases. Silica gel and charcoal are effective adsorbents due to their porous nature.

The word porous is doing the real work here. A porous solid is riddled with fine channels and cavities, and the walls of every one of those channels are surface. So a single small grain of charcoal offers an internal surface enormously larger than the outer surface you can see, which is why so modest a quantity of it can hold so much gas.

Surface Area of Adsorbent

Adsorption is a surface phenomenon. Hence the extent of adsorption increases with an increase in the surface area of the adsorbent. Finely divided substances, rough surfaces and colloidal substances are good adsorbents, as they provide a larger surface area for a given mass.

Note that this factor concerns the same material in different states of division, whereas the previous factor concerned different materials. Grinding a solid does not change its chemistry at all, yet it can multiply its adsorbing power many times over, simply by converting hidden interior into exposed surface.

Q1. The same mass of charcoal cleans up far more dye from a solution when it is a fine powder than when it is a single lump. Why should merely crushing it make such a difference?

A. Because adsorption happens only at the surface, and crushing creates surface. In a lump, almost all the carbon atoms are buried in the interior, where their forces are already satisfied by their neighbours and no dye molecule can reach them. Every cut we make through the lump turns interior into two fresh faces of exposed surface, and the finer the powder, the more faces there are. The mass of charcoal, and its chemical nature, are exactly the same before and after — only the area available for dye molecules to settle on has increased, and it is the area, not the mass, that sets how much dye can be held.

A solid cube of edge length 1 cm is powdered into small cubes, each of edge length 0.01 cm. Calculate the total surface area of the powder, and compare it with the surface area of the original cube. Does the volume change?

Solution.

Surface area of the original cube. A cube has six square faces, so

\[A_1 = 6\,l^2 = 6 \times (1\ \text{cm})^2 = 6\ \text{cm}^2\]

Number of small cubes. The volume of the original cube is $ (1 )^3 = 1 ^3 $, and the volume of each small cube is $ (0.01 )^3 = 1{-6} 3 $. Therefore

\[N = \frac{1\ \text{cm}^3}{1\times10^{-6}\ \text{cm}^3} = 1\times10^{6}\]

Total surface area of the powder. Each small cube has surface area

\[6 \times (0.01\ \text{cm})^2 = 6\times10^{-4}\ \text{cm}^2\]

so for a million such cubes,

\[A_2 = 1\times10^{6} \times 6\times10^{-4}\ \text{cm}^2 = 600\ \text{cm}^2\]

Comparison. \(A_2 / A_1 = 600/6 = 100\). The surface area has become 100 times larger.

Volume. No matter has been added or removed, so the total volume stays at \(1\ \text{cm}^3\).

This is the whole argument for using finely divided adsorbents in one calculation: the same mass and the same volume of solid, but a hundred times the surface on which adsorption can occur.

(question only — no solution)

A solid cube of edge length 2 cm is ground into cubes of edge length 0.02 cm. Calculate (a) the number of small cubes formed, (b) the total surface area of the powder, and (c) the factor by which the surface area has increased.

Temperature

Adsorption is an exothermic process. According to Le Chatelier’s principle (Chapter 12), it is therefore favoured at low temperature. The amount of gas adsorbed is thus inversely proportional to temperature.

To see why, it helps to write adsorption as an equilibrium, with heat shown as one of the products:

\[\text{Gas} + \text{Solid surface} \rightleftharpoons \text{Gas adsorbed on surface} + \text{Heat}\]

NotePrinciple / Law

Le Chatelier’s principle: when a system at equilibrium is disturbed by a change in temperature, pressure or concentration, the equilibrium shifts in the direction that opposes the change.

Raising the temperature is a way of adding heat to this system. The equilibrium then shifts in the direction that consumes heat — that is, backwards, towards desorption — so less gas remains adsorbed. Cooling has the opposite effect and favours adsorption. This single argument explains the temperature behaviour of physisorption completely, and you have already seen in the previous section why chemisorption, which needs activation energy, is the exception that behaves differently at first.

Figure 11.2 shows plots of the volume of \(\mathrm{N_2}\) adsorbed per unit mass of adsorbent against the pressure of the gas at different temperatures. As the temperature increases from 193 K to 273 K at a constant pressure \(P\), the amount of gas adsorbed decreases.

Diagram

Fig. 11.2 – Variation of adsorption with temperature and pressure: three curves of x/m (y-axis) against P (x-axis), each rising steeply at first and then flattening, labelled 195 K, 244 K and 273 K from the topmost curve downwards. Mark a single constant pressure P with a vertical dashed line cutting all three curves, showing x/m falling as the temperature rises.

Read Fig. 11.2 the way an examiner will ask you to read it. Along any one curve, we hold the temperature fixed and watch the effect of pressure. Moving between curves at a fixed pressure, we hold the pressure fixed and watch the effect of temperature. Each curve is a separate experiment carried out at its own constant temperature — which is exactly what the next section will call an isotherm.

Pressure of Gas

At any temperature, the extent of gas adsorbed increases with an increase of pressure. The extent of adsorption is directly proportional to the pressure of the gas.

At high pressures, the extent of adsorption becomes independent of the pressure. The surface of the adsorbent is then almost fully covered by adsorbed gaseous molecules. This is why every curve in Fig. 11.2 flattens out towards the right instead of rising for ever.

Q. Why does raising the pressure eventually stop increasing the amount of gas adsorbed?

A. Because a surface offers only a limited number of places for molecules to occupy. Once nearly every site is filled, an incoming molecule finds nowhere to settle, and pushing more gas into the vessel cannot create fresh surface. The adsorbent is then said to be saturated.

An oxygen concentrator of the kind used in hospitals and homes is built entirely on the factors in this section. Air is pumped at raised pressure through a column packed with a porous adsorbent that holds nitrogen far more strongly than oxygen, so the gas leaving the column is greatly enriched in oxygen. The pressure is then dropped, the adsorbed nitrogen is released — desorption, exactly as described earlier — and the cleaned column is ready for the next cycle. Two columns are usually worked alternately, one adsorbing while the other is being regenerated, so the supply to the patient never stops. Notice that no chemical is consumed and nothing is stored: the machine works by repeatedly raising and lowering a pressure, and the pressure dependence of adsorption does the rest.

(question only — no solution)

Equal masses of the same activated charcoal are placed in two identical vessels at the same temperature and pressure. One vessel is filled with \(\mathrm{Cl_2}\) and the other with \(\mathrm{H_2}\). Predict in which vessel the greater volume of gas will be adsorbed, and justify your prediction using the data of Table 11.3 and the idea of critical temperature.

11.5 Adsorption Isotherm

In the previous section we changed the temperature and the pressure together and watched the amount of adsorbed gas change. To study adsorption properly, a chemist does something more disciplined: the temperature is held fixed, and only the pressure of the gas (or the concentration of the solution) is varied. The curve obtained from such an experiment is called an adsorption isotherm.

The relationship between the amount of a substance adsorbed per unit mass of adsorbent and the equilibrium pressure (in the case of a gas) or the concentration (in the case of a solution) at a given constant temperature is called an adsorption isotherm. Various adsorption isotherms are known.

NoteDefinition

An adsorption isotherm is the relationship between the amount of a substance adsorbed per unit mass of the adsorbent and the equilibrium pressure of the gas, or the concentration of the solution, at a given constant temperature.

The word breaks up helpfully: iso- means equal and -therm refers to temperature, so an isotherm is a curve along which the temperature is the same everywhere. Each of the three curves you met in Fig. 11.2 was therefore a separate isotherm, since each was recorded at its own fixed temperature.

Freundlich Adsorption Isotherm

Freundlich proposed an empirical equation for the adsorption of a gas on a solid.

\[\frac{x}{m} = k\,P^{1/n} \qquad (n > 1) \qquad \ldots (11.1)\]

Where \(x\) = mass of the gas adsorbed, \(m\) = mass of the adsorbent at pressure \(P\), so that \(\dfrac{x}{m}\) = mass of gas adsorbed per unit mass of adsorbent, and \(P\) = equilibrium pressure.

\(k\) and \(n\) are constants which depend on the nature of the adsorbate, the nature of the adsorbent and the temperature. Because they depend on temperature, a fresh pair of values belongs to every isotherm; you cannot use \(k\) and \(n\) measured at one temperature to predict adsorption at another.

Equilibrium pressure means the steady pressure that remains once adsorption and desorption are going on at equal rates, so that the amount adsorbed no longer changes with time. Until that steady state is reached, the reading would still be drifting and the measurement would be meaningless.

NotePrinciple / Law

Freundlich adsorption isotherm: at a constant temperature, the mass of gas adsorbed per unit mass of adsorbent varies as the equilibrium pressure raised to the power \(1/n\), that is \(\dfrac{x}{m} = k\,P^{1/n}\) with \(n > 1\). The relation is empirical and holds good only over a limited range of pressures.

\(x/m\), \(k\) and \(1/n\) (\(x/m\) is a mass of adsorbate per unit mass of adsorbent, so its unit is \(\text{g g}^{-1}\); it is sometimes reported instead as a volume of gas adsorbed per gram, as in Table 11.3. The factor \(1/n\) is a pure number with no unit, lying between 0 and 1. The constant \(k\) is not dimensionless — its numerical value depends on the units chosen for pressure or concentration, so always quote the units used along with it.)

Q. What does it mean to call Freundlich’s equation “empirical”?

A. It means the equation was written to fit experimental curves rather than derived from a theory of what happens on the surface. That is why it carries no explanation of \(k\) and \(n\) in terms of surface sites, and why it fails outside a limited pressure range — an equation that only copies the data cannot be trusted where no data were taken.

Adsorption from Solution

In the case of a solution, \(P\) in Eq. (11.1) is replaced by the concentration, and thus

\[\frac{x}{m} = k\,C^{1/n} \qquad \ldots (11.2)\]

This substitution makes physical sense. What drives a molecule onto the surface is how crowded the surrounding fluid is with those molecules, and in a gas that crowding is measured by pressure while in a solution it is measured by concentration.

Graphical Representation

A graphical representation of the Freundlich equation is shown in Fig. 11.3. The curve rises steeply at first, then bends over, and finally becomes almost horizontal.

Diagram

Fig. 11.3 – Adsorption isotherm: x/m plotted against P, a single curve rising steeply from the origin, then bending and flattening to a nearly horizontal line at high P. Mark three regions along the curve — “low pressure: x/m ∝ P”, “intermediate pressure: x/m ∝ P^(1/n)”, “high pressure: x/m independent of P”.

the three regions of the adsorption isotherm in Fig. 11.3 (From the shape of the curve alone, answer: (i) In which region is the graph a straight line through the origin, and what value of \(1/n\) does that correspond to? (ii) In which region is the graph horizontal, and what value does \(1/n\) approach there? (iii) What is physically happening to the surface of the adsorbent in the flat region, and why can raising the pressure further not help? — no answer given here)

A curve is awkward to work with, because reading two constants off a bend is difficult. The standard trick is to turn the curve into a straight line, and logarithms do exactly that.

After taking the logarithm of the above Eq. 11.2, we get

\[\log \frac{x}{m} = \log k + \frac{1}{n}\log C \qquad \ldots (11.3)\]

The two steps in between are worth writing out, since this manipulation appears again and again in physical chemistry. Taking logarithms of both sides of Eq. (11.2),

\[\log \frac{x}{m} = \log \left( k\,C^{1/n} \right)\]

The logarithm of a product is the sum of the logarithms, so the right-hand side becomes \(\log k + \log C^{1/n}\). The logarithm of a power brings the index out in front, so \(\log C^{1/n} = \dfrac{1}{n}\log C\), which gives Eq. (11.3). A multiplication and a power have been converted into an addition and a multiplication — and that is precisely what turns a curve into a line.

Plotting \(\log \dfrac{x}{m}\) against \(\log C\), or against \(\log P\) for a gas, a straight line is obtained, which is shown in Fig. 11.4.

Diagram

Fig. 11.4 – Freundlich isotherm: log (x/m) on the y-axis plotted against log P on the x-axis, giving a straight line of positive slope. Mark “slope = 1/n” on the line and “intercept = log k” where the line meets the y-axis.

Compare Eq. (11.3) with the equation of a straight line, \(y = mx + c\). Here \(\log \dfrac{x}{m}\) plays the part of \(y\), \(\log C\) (or \(\log P\)) plays the part of \(x\), the gradient is \(\dfrac{1}{n}\) and the intercept is \(\log k\). So the slope of the straight line gives \(\dfrac{1}{n}\), while the intercept on the y-axis gives \(\log k\). Two constants that were buried inside a curved graph can now simply be measured with a ruler.

The factor \(\dfrac{1}{n}\) ranges from 0 to 1. Equation (11.3) holds good over a limited range of pressures, which is the price paid for using an empirical relation.

Two Limiting Cases

The value of \(\dfrac{1}{n}\) tells us at a glance how the adsorbent is behaving, and the two extremes are the cases most often asked about.

When \(\dfrac{1}{n} \to 0\), \(\dfrac{x}{m} \to\) constant, and the adsorption then is independent of pressure. Any quantity raised to the power zero equals one, so \(P^{1/n}\) becomes 1 and \(\dfrac{x}{m}\) reduces to \(k\). This is the flat, saturated region at the right of Fig. 11.3.

When \(\dfrac{1}{n} = 1\), \(\dfrac{x}{m} = kP\), that is \(\dfrac{x}{m} \propto P\), and the adsorption varies directly with pressure. Doubling the pressure now doubles the amount adsorbed, which is the behaviour seen in the initial straight portion of the isotherm at low pressure.

The experimental isotherms, as in Fig. 11.3, tend to saturate at high pressure. Freundlich’s equation, on the other hand, keeps rising for ever as long as \(1/n\) stays greater than zero — another reminder that it is a useful fit over a limited range and not a complete theory of the surface.

the value of 1/n and the behaviour of the isotherm | Value of \(1/n\) | Freundlich equation becomes | Dependence on pressure | Region of Fig. 11.3 | |—|—|—|—| | \(1/n = 1\) | \(x/m = kP\) | Directly proportional to \(P\) | Initial straight portion, low pressure | | \(0 < 1/n < 1\) | \(x/m = kP^{1/n}\) | Increases with \(P\), but less than proportionally | Curved middle portion | | \(1/n \to 0\) | \(x/m = k\) (a constant) | Independent of \(P\) | Flat portion, high pressure |

The equation is named after Herbert Freundlich, a German chemist who worked extensively on colloids and on the chemistry of surfaces in the early twentieth century and wrote one of the first systematic books on the subject. His isotherm was arrived at by fitting curves to measured data, at a time when no one yet had a picture of how molecules sat on a surface. The theoretical account came a little later, from the study of surfaces as a fixed set of adsorption sites. The order of events is a fair picture of how physical chemistry often advances: someone finds an equation that works, and someone else later explains why.

Adsorption isotherms are not a textbook curiosity — they are how an engineer sizes an activated-carbon filter. Suppose a factory has to strip an organic pollutant out of its waste water before discharge. A laboratory technician shakes measured masses of carbon with samples of the effluent, measures the concentration left in solution at equilibrium, and plots \(\log(x/m)\) against \(\log C\). From the slope and intercept come \(1/n\) and \(k\) for that particular pollutant on that particular carbon, and from those two numbers the mass of carbon needed to bring the effluent down to the permitted concentration can be calculated before a single tonne is bought. Since the spent carbon is usually regenerated by heating and used again, the design also decides how much waste the plant creates — which is why getting the isotherm right is a green-chemistry question as much as an economic one.

For the adsorption of a gas on activated charcoal at a constant temperature, a plot of \(\log \dfrac{x}{m}\) against \(\log P\) is a straight line of slope 0.5 with an intercept of 0.30 on the \(\log \dfrac{x}{m}\) axis, where \(P\) is measured in kPa.

(a) Calculate the values of \(n\) and \(k\).

(b) Calculate \(\dfrac{x}{m}\) when the equilibrium pressure of the gas is 100 kPa.

(c) State whether the adsorption in this pressure range is directly proportional to pressure.

Solution.

  1. Comparing the plotted line with Eq. (11.3), \(\log \dfrac{x}{m} = \log k + \dfrac{1}{n}\log P\), the slope of the line is \(\dfrac{1}{n}\) and the intercept is \(\log k\).

\[\frac{1}{n} = 0.5 \quad \Rightarrow \quad n = \frac{1}{0.5} = 2\]

\[\log k = 0.30 \quad \Rightarrow \quad k = 10^{0.30} = 2.0\]

(Recall that \(\log 2 = 0.3010\), so an intercept of 0.30 corresponds to \(k = 2.0\), with \(P\) in kPa.)

  1. Substituting in Eq. (11.1),

\[\frac{x}{m} = k\,P^{1/n} = 2.0 \times (100)^{0.5} = 2.0 \times 10 = 20\]

So \(\dfrac{x}{m} = 20\ \text{g g}^{-1}\); that is, 20 g of the gas is adsorbed per gram of charcoal at 100 kPa.

  1. No. Direct proportionality requires \(\dfrac{1}{n} = 1\), whereas here \(\dfrac{1}{n} = 0.5\). The amount adsorbed increases with pressure, but only as the square root of the pressure — so raising the pressure four times would increase \(\dfrac{x}{m}\) only twice.

(question only — no solution)

At a constant temperature, 1.0 g of an adsorbent adsorbs 0.60 g of a gas when the equilibrium pressure is 40 kPa. For this system \(\dfrac{1}{n} = 0.5\). Calculate the mass of the gas adsorbed by 1.0 g of the same adsorbent at an equilibrium pressure of 160 kPa, the temperature being unchanged.

(question only — no solution)

For a certain gas–solid system, the measured value of \(\dfrac{1}{n}\) at high pressure approaches zero. State what the Freundlich equation predicts about the effect of pressure on \(\dfrac{x}{m}\) in this range, sketch the shape of the \(\dfrac{x}{m}\) versus \(P\) graph you would expect, and explain the prediction in terms of the surface of the adsorbent.

11.6 Applications of Adsorption

Adsorption finds a large number of applications, which are illustrated here. As you go through them, notice how often the same two ideas from the previous sections are being put to work: a solid with a very large surface area, and the fact that different substances are adsorbed to different extents on the same surface.

Heterogeneous Catalysis

Solid catalysts are used in many industrial manufacturing processes. For example, iron is used as the catalyst in the manufacturing of ammonia; platinum in the manufacturing of sulphuric acid, \(\mathrm{H_2SO_4}\) (by the contact process). In the hydrogenation of oils, finely divided nickel is employed as catalyst.

In each case the reacting gases or liquids are first adsorbed on the surface of the solid, and it is on that surface that the reaction takes place. This is why catalysis belongs in a chapter on adsorption at all, and the connection is taken up properly in Sections 11.7 and 11.8.

Gas Masks

A gas mask is a device which consists of activated charcoal or a mixture of adsorbents. It is used for breathing in coal mines to avoid inhaling poisonous gases. The air drawn in by the wearer passes through the adsorbent, on whose surface the harmful gas molecules are held back while oxygen and nitrogen pass through largely untouched.

Control of Humidity

Silica and alumina gels are good adsorbents of moisture. Water vapour from the surrounding air is adsorbed on the vast internal surface of the gel, which is why small sachets of these solids are packed with medicines, electronic goods and leather articles to keep them dry.

Production of High Vacuum

Lowering the temperature at a given pressure increases the rate of adsorption of gases on charcoal powder. By using this principle, a high vacuum can be attained by adsorption.

A vessel evacuated by a vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum. The arrangement is drawn in Fig. 11.17.

Diagram

Fig. 11.17 – Production of high vacuum: a vessel to be evacuated connected by tubing through a stopcock to a second vessel packed with coconut charcoal, the second vessel immersed in a bath of liquid air. Arrows show residual gas molecules moving from the first vessel to the charcoal. Label “vessel evacuated by pump”, “coconut charcoal”, “liquid air”.

A mechanical pump can remove most of the gas from a vessel, but the last traces are hardest to pull out, since so few molecules remain to be pushed along. The cooled charcoal solves the problem in a different way: instead of pushing the molecules out, it holds them.

Q. Why must the charcoal be cooled by liquid air rather than simply left at room temperature?

A. Because adsorption is exothermic and is therefore favoured at low temperature, as Le Chatelier’s principle predicts. At the temperature of liquid air the charcoal holds far more gas, and holds it far more firmly, so even the last traces of air and moisture are taken up instead of being released again.

Adsorption Indicators

Adsorption is used to detect the end point of precipitation titrations. Dyes such as eosin and fluorescein are used as indicators.

For example, a solution of sodium chloride containing a small amount of fluorescein is titrated against silver nitrate solution.

\[\mathrm{NaCl} + \mathrm{AgNO_3} \rightarrow \underset{\text{white ppt}}{\mathrm{AgCl}} + \mathrm{NaNO_3}\]

A precipitation titration is one in which the added reagent throws the substance being estimated out of solution as an insoluble solid, and the end point is the moment at which exactly enough reagent has been added. Here every drop of silver nitrate removes chloride ions as a white precipitate of silver chloride.

When chloride ions are over, fluorescein is adsorbed on the white silver chloride precipitate and a red colour is developed. Thus a colour change from pale yellow to reddish pink is observed at the end point.

The reason the dye waits its turn is instructive. As long as chloride ions remain in solution, the precipitate adsorbs them in preference to the bulky dye anion. Once the chloride is exhausted, the precipitate surface is free to adsorb fluorescein instead, and the adsorbed dye has a different colour from the dye in solution. The signal, in other words, is produced by adsorption and not by any reaction of the dye.

fluorescein (and eosin) as adsorption indicators (What it does: an anionic dye that is pale yellow in solution but appears reddish pink once adsorbed on a fresh precipitate surface → Conditions it needs: a precipitation titration in which the precipitate is present in the vessel, and a solution not so acidic that the dye anion is protonated → Typical use: locating the end point in the titration of a halide solution against silver nitrate, where the dye is adsorbed only after the halide ions have been used up.)

Indicators of this kind are known as Fajans indicators, after the Polish-born chemist Kazimierz Fajans, who showed how the preferential adsorption of ions on the surface of a fresh precipitate could be used to signal an end point. The underlying idea is the one you met in Section 11.2 — a surface with unsatisfied forces will hold whichever available ion it can hold most strongly — put to work as an analytical tool.

25.0 cm\(^3\) of a sodium chloride solution containing a little fluorescein required 20.0 cm\(^3\) of 0.100 M silver nitrate solution to reach the end point, at which the solution turned reddish pink. Calculate (a) the molarity of the sodium chloride solution and (b) the mass of silver chloride precipitated. (Molar mass of \(\mathrm{AgCl} = 143.5\ \text{g mol}^{-1}\).)

Solution.

Step 1: moles of silver nitrate used. Molarity is moles per litre of solution, so

\[n(\mathrm{AgNO_3}) = 0.100\ \text{mol L}^{-1} \times 0.0200\ \text{L} = 2.00\times10^{-3}\ \text{mol}\]

Step 2: use the stoichiometry. From the balanced equation, \(\mathrm{NaCl}\) and \(\mathrm{AgNO_3}\) react in a 1 : 1 ratio, so

\[n(\mathrm{NaCl}) = 2.00\times10^{-3}\ \text{mol}\]

Step 3: molarity of the sodium chloride solution.

\[c(\mathrm{NaCl}) = \frac{2.00\times10^{-3}\ \text{mol}}{0.0250\ \text{L}} = 0.0800\ \text{mol L}^{-1}\]

Step 4: mass of the precipitate. Again by the 1 : 1 stoichiometry, \(n(\mathrm{AgCl}) = 2.00\times10^{-3}\ \text{mol}\), so

\[m(\mathrm{AgCl}) = 2.00\times10^{-3}\ \text{mol} \times 143.5\ \text{g mol}^{-1} = 0.287\ \text{g}\]

The molarity of the sodium chloride solution is \(0.0800\ \text{M}\) and the mass of silver chloride formed is \(0.287\ \text{g}\). Note that the indicator plays no part in the calculation at all — it only tells us when to stop adding silver nitrate.

(question only — no solution)

20.0 cm\(^3\) of 0.050 M potassium chloride solution containing a little fluorescein is titrated against 0.100 M silver nitrate solution. Calculate (a) the volume of silver nitrate solution required to reach the end point and (b) the mass of silver chloride precipitated. (Molar mass of \(\mathrm{AgCl} = 143.5\ \text{g mol}^{-1}\).)

Separation of Inert Gases

In a mixture of noble gases, different gases adsorb to different extents. Due to the principle of selective adsorption, the gases can be separated on coconut charcoal. The charcoal is cooled, the mixture is passed over it, and the more strongly adsorbed gases are held back while the more weakly adsorbed ones travel on — so that by controlling the temperature the members of the mixture can be taken off one by one.

(question only — no solution)

A mixture of helium and xenon is passed over coconut charcoal cooled to a low temperature. Predict which of the two gases will be adsorbed to the greater extent, and justify your answer using the idea of critical temperature introduced in Section 11.4.

Froth Flotation Process

A low grade sulfide ore is concentrated by separating it from silica and other earthy matter using pine oil as a frothing agent. Hydrophobic pine oil preferentially wets (adsorbs on) the sulfide ore, which is taken up in the froth.

The word hydrophobic means water-repelling. The finely powdered ore is stirred with water, pine oil and a stream of air. The oil is adsorbed only on the sulphide particles, giving them a water-repellent coating to which air bubbles readily cling, so those particles are carried upward into the froth and skimmed off. The silica and earthy matter, which the oil does not coat, stay wetted by water and settle at the bottom. The sequence is set out in FC 11.3 and the cell is drawn in Fig. 11.18.

Flowchart

FC 11.3 – Froth flotation process: powdered low grade sulphide ore + water + pine oil → agitation with a current of air → branch 1: pine oil adsorbed on sulphide particles → particles become hydrophobic → air bubbles attach → particles rise into the froth → froth skimmed off → concentrated ore; branch 2: silica and earthy matter remain wetted by water → not lifted by bubbles → settle at the bottom → gangue removed.

Diagram

Fig. 11.18 – Froth flotation cell: a tank of water containing powdered ore, with pine oil added, an air inlet with a stirrer at the bottom producing bubbles, a layer of mineral-laden froth at the top being skimmed off into a launder, and heavier gangue settling at the base. Label “froth with concentrated sulphide ore”, “air inlet”, “stirrer”, “gangue”.

Notice that nothing chemical has happened to the ore. Adsorption of a single thin layer of oil has changed only how the particle surface behaves towards water — and that is enough to separate a valuable mineral from worthless rock on an industrial scale.

Chromatographic Analysis

Chromatographic analysis is based on the selective adsorption of ions from solution using powdered adsorbents such as silica or alumina gel. It has several industrial and analytical applications.

The principle is a race. A mixture is allowed to travel through or over a bed of adsorbent, carried along by a solvent. A component that is strongly adsorbed spends most of its time held on the solid and moves slowly; one that is weakly adsorbed spends most of its time in the moving solvent and travels quickly. Given enough length of adsorbent, the components arrive at different places and the mixture is separated, as shown in Fig. 11.19.

Diagram

Fig. 11.19 – Column chromatography: a vertical glass column packed with alumina or silica gel, solvent being added at the top from a reservoir, a mixture separating into three distinct coloured bands at different heights inside the packing, and the separated component collected drop by drop in a beaker at the bottom. Label “solvent”, “adsorbent packing”, “separated bands”, “collected fraction”.

Chromatography began with plant pigments. In the early years of the twentieth century the Russian botanist Mikhail Tsvet passed a solution of leaf extract down a column packed with powdered calcium carbonate and watched the extract resolve itself into separate coloured bands, each a different pigment held with a different strength by the powder. He named the technique from the Greek chroma, colour, and graphein, to write — colour-writing — and the name has stuck even though most modern separations involve nothing coloured at all. Today the same principle, differential adsorption, separates drug samples, tests athletes for banned substances and checks the purity of industrial products.

Other Applications

Other applications include surface area determination and the purification of water. The first of these is a neat inversion of everything in this chapter: instead of using a known surface to adsorb an unknown gas, a known gas is adsorbed on an unknown solid, and the amount taken up is used to calculate how much surface that solid must have.

Adsorption is one of the main defences against dissolved contaminants in groundwater. In several parts of India, groundwater carries fluoride at concentrations high enough to damage teeth and bone, and in other regions the problem is arsenic. Beds of activated alumina are used to treat such water: the contaminant ions are adsorbed on the alumina surface as the water percolates through, and the treated water leaves fit to drink. When the bed is exhausted it can be regenerated chemically and returned to service, so the same alumina serves for many cycles. Two points from earlier sections decide whether such a unit works in practice — the surface area of the adsorbent, and the fact that a saturated adsorbent stops working even though nothing about it looks different, which is why treated water must be tested regularly rather than trusted indefinitely.

11.7 Catalysis

Catalysts are of importance in the chemical industry and in living organisms. A large number of the chemicals manufactured in industries make use of catalysts to obtain specific products. The use of a catalyst lowers the reaction temperature and energy costs significantly.

That last point deserves attention, because it explains why industry cares so much about catalysis. A reaction that would need a furnace at very high temperature may run acceptably fast at a much lower temperature once a suitable catalyst is present. Less heating means less fuel, cheaper plant construction and fewer unwanted side products — a saving repeated over every tonne manufactured.

A catalyst thus can be defined as a substance which, when added to a reacting system, increases the rate of a reaction without itself undergoing any permanent chemical change.

NoteDefinition

A catalyst is a substance which, when added to a reacting system, increases the rate of the reaction without itself undergoing any permanent chemical change.

Read the words permanent chemical change carefully. A catalyst may well take part in the reaction — it may be attacked, bonded to, or temporarily converted into something else — but by the end of the reaction it is returned to its original chemical form. That is why a small quantity can serve for a very long run, and why the catalyst never appears in the balanced overall equation.

How a Catalyst Speeds Up a Reaction

A catalyst does not push the molecules harder. It offers them an easier route. Every reaction has to cross an energy barrier, the activation energy you met in Section 11.3, and only those molecules that carry at least that much energy can react. A catalyst provides an alternative path over a lower barrier, so at the same temperature a far greater fraction of molecules can get across. The two paths are compared in Fig. 11.20.

Diagram

Fig. 11.20 – Reaction-coordinate diagram: potential energy on the y-axis against progress of reaction on the x-axis, with reactants and products at the same two energy levels in both cases; a tall barrier for the uncatalysed path and a lower barrier for the catalysed path drawn on the same axes. Label “reactants”, “products”, “Ea (without catalyst)”, “Ea (with catalyst)”. Mark that the energy levels of reactants and products are unchanged.

Notice one thing the diagram makes clear. The catalyst lowers the barrier, but it does not shift the energy levels of the reactants or the products. A catalyst therefore changes how fast a reaction reaches equilibrium; it does not change how much product is possible at equilibrium.

Catalysis is of two types, namely homogeneous and heterogeneous catalysis. The word phase is the key to telling them apart: a phase is a physically distinct, uniform part of a system, such as a gas mixture, a single solution, or a solid.

The word catalysis was introduced in 1835 by the Swedish chemist Jöns Jacob Berzelius, who had been collecting reports of reactions that ran only in the presence of some substance that seemed to survive the process untouched. He built the name from Greek roots meaning, roughly, “loosening down” — his image was of a substance that loosens the bonds holding the reactants back. It was a good name chosen for a slightly wrong reason, since we now describe catalysis in terms of a lowered energy barrier rather than a loosening force, but the word has outlasted the picture behind it.

11.7.1 Homogeneous Catalysis

When the reactants and the catalyst are in the same phase, it is said to be homogeneous catalysis.

NoteDefinition

Homogeneous catalysis is catalysis in which the catalyst and the reactants are present in the same phase.

Examples of Homogeneous Catalysis
  1. Iodide ion (\(\mathrm{I^{-}}\)) finds use as a homogeneous catalyst in the decomposition of aqueous hydrogen peroxide. Both \(\mathrm{I^{-}}\) and \(\mathrm{H_2O_2}\) are present in the same aqueous phase.

  2. Oxidation of sulfur dioxide to sulfur trioxide with dioxygen (\(\mathrm{O_2}\)) in the presence of nitric oxide as catalyst (lead chamber process).

\[2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \mathrm{NO}\text{(g)}\ } 2\,\mathrm{SO_3}\text{(g)}\]

Here all three species — the two reactant gases and the catalyst — are gases, so the system is a single gaseous phase.

  1. Hydrolysis of sugar is catalysed by the \(\mathrm{H^{+}}\) ion furnished by sulphuric acid.

\[\underset{\text{sucrose solution}}{\mathrm{C_{12}H_{22}O_{11}}\text{(aq)}} + \mathrm{H_2O}\text{(l)} \xrightarrow{\ \mathrm{H_2SO_4}\ } \underset{\text{glucose}}{\mathrm{C_6H_{12}O_6}\text{(aq)}} + \underset{\text{fructose}}{\mathrm{C_6H_{12}O_6}\text{(aq)}}\]

All the reactants and the catalyst are in the same solution phase. Note that the two products share the same molecular formula but are different sugars — glucose and fructose differ in the arrangement of their atoms, not in their composition.

  1. Enzyme catalysis is also an important type of homogeneous catalysis. Enzymes are the catalysts of living systems, dissolved in the same aqueous surroundings as the substances they act on.

Enzyme catalysis has quietly changed household chemistry. Older washing powders relied on hot water and harsh alkali to shift protein and fat stains. Modern detergents instead carry small quantities of enzymes that break those stains down, and because the enzymes work at moderate temperatures, clothes can be washed in cool or lukewarm water. The chemistry saves a great deal of electricity across millions of households — the heating of water, not the washing itself, is where most of a machine’s energy goes. The same logic applies in industry: replacing a high-temperature step run over a metal catalyst with a mild, enzyme-catalysed one in water cuts both energy use and solvent waste, which is why enzyme processes are a standard example of green chemistry in practice.

11.7.2 Heterogeneous Catalysis

When the reactant and the catalyst are in different phases, it is said to be heterogeneous catalysis.

NoteDefinition

Heterogeneous catalysis is catalysis in which the catalyst and the reactants are present in different phases.

The heterogeneous catalyst is generally a solid, and the reactants may either be gases or liquids. The solid catalyst is added to the reaction mixture. It does not dissolve in the reacting system. The reaction occurs on the surface of the solid catalyst — which is exactly why this section belongs in a chapter on adsorption, and why finely divided catalysts with large surface areas are preferred.

Examples of Heterogeneous Catalysis
  1. Dinitrogen (\(\mathrm{N_2}\)) and dihydrogen (\(\mathrm{H_2}\)) combine to form ammonia in the Haber process in the presence of finely divided iron along with \(\mathrm{K_2O}\) and \(\mathrm{Al_2O_3}\).

\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \xrightarrow[\ \mathrm{K_2O},\ \mathrm{Al_2O_3}\ ]{\ \mathrm{Fe}\text{(s)}\ } 2\,\mathrm{NH_3}\text{(g)}\]

Here \(\mathrm{Al_2O_3}\) and \(\mathrm{K_2O}\) are promoters of the \(\mathrm{Fe}\) catalyst. \(\mathrm{Al_2O_3}\) is added to prevent the fusion of \(\mathrm{Fe}\) particles. \(\mathrm{K_2O}\) causes chemisorption of nitrogen atoms. Molybdenum is also used as promoter.

  1. The hydrogenation reaction of vegetable oils to produce solid fat is used in the food industry. The reaction is catalysed by finely divided metals like Ni, Pd or Pt. Vegetable oil contains one or more carbon–carbon double bonds (\(\mathrm{C=C}\)) in its structure. On hydrogenation a solid product, which contains only carbon–carbon single bonds, is formed. It is called vanaspati ghee.

\[\underset{(\mathrm{C=C})}{\text{Vegetable oil (l)}} + \mathrm{H_2}\text{(g)} \xrightarrow{\ \mathrm{Ni}\text{(s)}\ } \underset{(\mathrm{C\!-\!C})}{\text{Vegetable ghee (s)}}\]

The catalyst here is a solid while the oil is a liquid and the hydrogen a gas, so three phases meet at the metal surface — as clear a case of heterogeneous catalysis as one could ask for.

The name vanaspati comes from Sanskrit, where it means simply “of plants” or “vegetation” — a label recording that the fat began life in a plant oil rather than in animal milk fat.

  1. Another important application of heterogeneous catalysts is in automobile catalytic converters. In automobile exhaust, a large number of air pollutants such as carbon monoxide, nitric oxide, etc. are present. The catalytic converter transforms the air pollutants into carbon dioxide, water, nitrogen and oxygen. The catalyst is poisoned by the adsorption of Pb (lead). Automobiles with catalytic converter require unleaded petrol. A cut-away view of a converter is given in Fig. 11.21.

Diagram

Fig. 11.21 – Cut-away view of an automobile catalytic converter: exhaust gases entering from the engine side, passing through a honeycomb ceramic support coated with finely divided metal catalyst, and cleaner gases leaving at the tailpipe end. Arrows in show CO, NO and unburnt hydrocarbons; arrows out show CO₂, N₂, O₂ and H₂O. Label “honeycomb support (large surface area)”, “metal catalyst coating”.

The honeycomb shape is not decoration. A converter must expose the largest possible catalyst surface to a fast-moving stream of gas without blocking its flow, and a honeycomb does both.

Promoters and Catalyst Poisons

Two terms used in the examples above need to be pinned down, since NCERT introduces them in passing and examiners ask for them directly.

NoteDefinition

A promoter is a substance which is not itself a catalyst but which, when added along with a catalyst, increases the activity of that catalyst.

NoteDefinition

A catalyst poison is a substance which destroys or greatly reduces the activity of a catalyst, usually by being strongly adsorbed on the catalyst surface.

The two definitions are mirror images, and both work through adsorption. A promoter such as \(\mathrm{K_2O}\) improves the surface — in this case by helping nitrogen atoms chemisorb — while \(\mathrm{Al_2O_3}\) protects it by keeping the iron particles from fusing together into larger lumps with less surface. A poison such as lead does the opposite: it is adsorbed so strongly on the active sites that the reactant molecules can no longer reach them, and since the poison does not leave, the damage is permanent. This is the whole reason leaded petrol had to be given up before catalytic converters could be fitted.

Homogeneous vs Heterogeneous Catalysis | Feature | Homogeneous catalysis | Heterogeneous catalysis | |—|—|—| | Phase of catalyst and reactants | Same phase | Different phases | | Usual physical state of catalyst | Gas or dissolved species | Generally a solid | | Where the reaction occurs | Throughout the bulk of the mixture | On the surface of the catalyst | | Role of surface area | Not important | Very important; finely divided catalysts are used | | Separating the catalyst afterwards | Difficult, as it is mixed with the products | Easy, as the solid can be filtered off | | Example | Nitric oxide in the lead chamber process | Finely divided iron in the Haber process |

nickel, palladium and platinum as hydrogenation catalysts (What they do: adsorb \(\mathrm{H_2}\) on their surface and dissociate it into reactive hydrogen atoms, which are then added across a \(\mathrm{C=C}\) double bond → Conditions they need: the metal in a finely divided form to give a large surface, hydrogen gas, and usually raised temperature and pressure; nickel is the cheap workhorse used industrially, while palladium and platinum are more active and used where milder conditions matter → Typical use: converting liquid vegetable oils containing \(\mathrm{C=C}\) bonds into solid fats, and hydrogenating unsaturated organic compounds generally.)

Classify each of the following as homogeneous or heterogeneous catalysis, giving a reason based on phases in each case.

(a) Decomposition of aqueous \(\mathrm{H_2O_2}\) in the presence of \(\mathrm{I^{-}}\) ions.

(b) Manufacture of ammonia from \(\mathrm{N_2}\) and \(\mathrm{H_2}\) over finely divided iron.

(c) Hydrolysis of sucrose solution in the presence of sulphuric acid.

(d) Removal of \(\mathrm{CO}\) and \(\mathrm{NO}\) from exhaust gases in a catalytic converter.

Solution.

  1. Homogeneous. The iodide ions and the hydrogen peroxide are both dissolved in water, so catalyst and reactant occupy one aqueous phase.

  2. Heterogeneous. The reactants \(\mathrm{N_2}\) and \(\mathrm{H_2}\) are gases while the iron catalyst is a solid; the reaction takes place on the iron surface.

  3. Homogeneous. The \(\mathrm{H^{+}}\) ions supplied by sulphuric acid, the sucrose and the water are all in the same solution phase.

  4. Heterogeneous. The pollutant gases pass over a solid catalyst, so catalyst and reactants are in different phases.

The test in every case is a single question — are the catalyst and the reactants in one phase or in two?

A sample of vegetable oil contains 0.50 mol of carbon–carbon double bonds. Calculate the volume of hydrogen gas, measured at STP, required for its complete hydrogenation to vegetable ghee. (Molar volume of a gas at STP \(= 22.4\ \text{L mol}^{-1}\).)

Solution.

Step 1: the stoichiometry. In hydrogenation, one molecule of \(\mathrm{H_2}\) is added across each \(\mathrm{C=C}\) double bond, converting it into a \(\mathrm{C\!-\!C}\) single bond. So

\[n(\mathrm{H_2}) = n(\mathrm{C=C}) = 0.50\ \text{mol}\]

Step 2: convert moles to volume at STP.

\[V = n \times 22.4\ \text{L mol}^{-1} = 0.50\ \text{mol} \times 22.4\ \text{L mol}^{-1} = 11.2\ \text{L}\]

So \(11.2\ \text{L}\) of hydrogen at STP is needed. The nickel catalyst does not enter the calculation at all, because it is not consumed — a useful reminder of what the definition of a catalyst actually means for stoichiometry.

(question only — no solution)

An oil molecule contains three carbon–carbon double bonds. Calculate (a) the number of moles of \(\mathrm{H_2}\) needed to hydrogenate 1.0 mol of this oil completely, (b) the mass of hydrogen required, and (c) the volume this hydrogen would occupy at STP. (Molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)

(question only — no solution)

In the Haber process, iron is used along with \(\mathrm{Al_2O_3}\) and \(\mathrm{K_2O}\). State the specific role of each of the two added oxides, and explain why they are described as promoters rather than as catalysts.

11.7.3 Inhibitors

Inhibitors are substances those decrease the rate of a chemical reaction. They are the opposite of catalysts in effect, and are sometimes called negative catalysts.

NoteDefinition

An inhibitor is a substance which decreases the rate of a chemical reaction.

Examples,

  1. Chloroform was once used as an anaesthetic. It forms a poisonous substance, carbonyl chloride, by air oxidation.

\[4\,\mathrm{CHCl_3}\text{(l)} + 3\,\mathrm{O_2}\text{(g)} \rightarrow 4\,\mathrm{COCl_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)} + 2\,\mathrm{Cl_2}\text{(g)}\]

When 2% ethanol is added to chloroform, the formation of \(\mathrm{COCl_2}\) is suppressed. Ethanol acts as an inhibitor and retards the above reaction. This is a genuine matter of safety rather than a laboratory curiosity: carbonyl chloride, better known as phosgene, is severely toxic, so bottles of chloroform are stabilised with a little ethanol and kept in dark bottles away from air.

  1. Hydrogen peroxide decomposes as

\[2\,\mathrm{H_2O_2}\text{(l)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]

The reaction is inhibited by the addition of dilute acid or glycerol acting as inhibitor. This is why a bottle of hydrogen peroxide from a chemist’s shop keeps for months instead of quietly fizzing itself away.

Q. Iodide ions catalyse the decomposition of hydrogen peroxide, yet dilute acid inhibits the very same reaction. Is that a contradiction?

A. No. A catalyst offers the reaction a new and easier path, while an inhibitor interferes with the path the reaction would otherwise take. The same reaction can therefore have both a catalyst and an inhibitor, and which one you add depends on whether you want the peroxide to decompose quickly, as in a laboratory demonstration, or to sit unchanged on a shelf.

11.8 Adsorption Theory of Heterogeneous Catalysis

We have said several times that a solid catalyst works on its surface. This section says exactly how, and in doing so it explains the most puzzling thing about catalysts — that they take part in a reaction and yet come out of it unchanged.

The catalytic action occurs on the surface of a catalyst. The mechanism involves five steps.

  1. Diffusion of reactants toward the surface of the catalyst.

  2. Adsorption of reactant molecules on the surface of the catalyst.

  3. Occurrence of chemical reaction on the catalyst surface and formation of an intermediate.

  4. Formation of products.

  5. Desorption of reaction products from the catalyst surface. Products leave the catalyst surface.

Two of these words need unpacking. Diffusion is simply the movement of molecules from a region where they are crowded to one where they are less crowded, driven by their own random motion; it is how a reactant molecule in the bulk of the gas finds its way to the catalyst in the first place. An intermediate is a short-lived species formed on the way from reactants to products — here, reactant fragments bonded to the surface — which exists during the reaction but is not present at the beginning or the end.

The steps involved can be shown as a cycle, given in FC 11.4, and the same five steps are drawn on an actual catalyst surface in Fig. 11.22.

Flowchart

FC 11.4 – Adsorption theory of heterogeneous catalysis, drawn as a closed cycle: diffusion of reactants to the surface → adsorption of reactant molecules on the surface → chemical reaction on the surface and formation of an intermediate → formation of products → desorption of products from the surface → arrow returning to the first step, showing fresh reactant molecules replacing the products and the cycle beginning again.

Diagram

Fig. 11.22 – A flat block of solid catalyst shown in five successive panels: (1) reactant molecules approaching the surface from the gas; (2) the same molecules held on the surface; (3) bonds within the adsorbed molecules breaking and new bonds forming, an intermediate shown attached to the surface; (4) product molecules formed but still attached; (5) product molecules leaving the surface while fresh reactant molecules arrive. Label “diffusion”, “adsorption”, “intermediate on surface”, “product formation”, “desorption”.

Fresh reactant molecules can replace the products to start the cycle again, as in step (i). This explains why the catalyst remains unchanged in mass and chemical composition at the end of the reaction.

Q5. A catalyst speeds up a reaction, and yet at the end of it all we recover the catalyst unchanged in mass and in chemical composition. If it took part in the reaction, how does it come back untouched?

A. Because the catalyst’s involvement is a loop, not a one-way change. It does genuinely take part: reactant molecules are adsorbed onto its surface and bonded to it, and it is that bonding which weakens the old bonds and lets the reaction happen over a lower energy barrier. But the very next step hands the surface back. The products formed on the surface are desorbed and leave, and every site they vacate is exactly as it was before they arrived, ready for fresh reactant molecules. So at any given instant part of the catalyst surface is chemically altered — that is the whole point — while over a complete cycle nothing is consumed. The catalyst is not hiding during the reaction; it is being used and returned, thousands of times over.

Q. If the surface is handed back at the end of every cycle, why do industrial catalysts ever have to be replaced?

A. Because processes outside the cycle slowly spoil the surface. A poison may be adsorbed so strongly that a site is never handed back, carbon deposits may block the pores, and prolonged heating can make fine particles fuse into larger ones with less surface. The catalytic cycle itself is endlessly repeatable; it is these side effects that end a catalyst’s working life.

11.8.1 Important Features of Solid Catalysts

Catalytic Activity

The activity of a catalyst depends on the strength of chemisorption. If a large number of reactant molecules (gas or liquid) are strongly adsorbed on the surface of the solid catalyst, the catalyst is said to be active.

However, the adsorption of reactant molecules on the surface — that is, the bond formed between the adsorbate and the adsorbent surface — should not be very strong, so that they are not immobilized. To be immobilized means to be pinned in place, unable to move or to rearrange. This is the balance on which catalytic activity depends, and it is worth stating as a rule.

If the bond to the surface is too weak, the reactant molecule bounces away before anything can happen to it. If the bond is too strong, the molecule is held so tightly that it can neither rearrange nor leave, the products cannot be desorbed, and the cycle in FC 11.4 stalls at step (iii). A good catalyst holds the reactant firmly enough to weaken its bonds, and loosely enough to let go afterwards. The dependence is sketched in Fig. 11.24.

Diagram

Fig. 11.24 – Catalytic activity (y-axis) plotted against strength of chemisorption of the reactant on the catalyst (x-axis): the curve rises to a maximum and then falls. Mark the left-hand region “adsorption too weak — reactant not held”, the peak “optimum strength”, and the right-hand region “adsorption too strong — molecules immobilized, products not desorbed”.

the curve of catalytic activity against strength of chemisorption in Fig. 11.24 (From the shape of the curve, answer: (i) Why does activity increase in the left-hand region as the adsorption becomes stronger? (ii) Which of the five steps of FC 11.4 fails in the right-hand region, and why? (iii) Two metals adsorb the same reactant, one very weakly and one very strongly. Predict which will be the better catalyst and justify your choice. — no answer given here)

strength of adsorption and catalytic behaviour | Strength of chemisorption | What happens to the reactant | Effect on the catalytic cycle | Catalytic activity | |—|—|—|—| | Too weak | Not held on the surface long enough | Cycle rarely gets past adsorption | Low | | Optimum | Held firmly, its bonds weakened, still able to rearrange | All five steps proceed | High | | Too strong | Immobilized on the surface | Products cannot be desorbed; sites stay blocked | Low |

It has been found that d-block metals such as Fe, V and Cr tend to be strongly active toward \(\mathrm{O_2}\), \(\mathrm{C_2H_2}\), \(\mathrm{C_2H_4}\), \(\mathrm{CO}\), \(\mathrm{H_2}\), \(\mathrm{CO_2}\), \(\mathrm{N_2}\), etc.; Mn and Cu are unable to adsorb \(\mathrm{N_2}\) and \(\mathrm{CO_2}\). The metals Mg and Li adsorb \(\mathrm{O_2}\) selectively.

Notice that this is a statement about which gases a metal can chemisorb at all — and since chemisorption is specific, as Section 11.3 established, a metal that cannot chemisorb a particular reactant simply cannot catalyse its reactions, however large its surface area.

The requirement that a catalyst surface stay clean has an expensive industrial consequence. Sulphur compounds are chemisorbed on metal catalysts so strongly that they are never released, and even a few parts per million of them in the feedstock can put a nickel catalyst permanently out of action. Refineries and fertiliser plants therefore run a desulphurisation step before the gas ever reaches the catalyst bed, purely to protect the catalyst downstream. The economics are striking: a plant will spend heavily on removing an impurity that plays no part in the reaction at all, because the alternative is replacing a catalyst charge worth far more. It is also a reminder for the laboratory — a catalyst that has stopped working has usually not been “used up”; something has been adsorbed on it.

Catalytic Selectivity

Some solid catalysts are selective in their action. The same gaseous reactants produce different products when different catalysts are used. This is a striking idea: the products of a reaction are decided not only by what you put in, but by the surface you run it over.

For example,

  1. The gaseous ethylene and \(\mathrm{O_2}\) react to produce different products with different catalysts.

\[2\,\mathrm{C_2H_4}\text{(g)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \mathrm{Pd/Al_2O_3}\ } 2\,\mathrm{CH_3CHO}\text{(g)}\]

\[2\,\mathrm{C_2H_4}\text{(g)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \mathrm{Ag/Al_2O_3}\ } 2\,\mathrm{C_2H_4O}\text{(g)}\]

The first product, \(\mathrm{CH_3CHO}\), is acetaldehyde. The second, \(\mathrm{C_2H_4O}\), is ethylene oxide, in which the oxygen atom bridges the two carbon atoms to give a strained three-membered ring; the two structures are drawn in Fig. 11.25. Both reactions consume exactly the same reactants in the same ratio, and the choice of catalyst decides which one you get.

Diagram

Fig. 11.25 – Two structural formulae side by side: (a) acetaldehyde, CH₃–CHO, showing the C=O double bond and the aldehydic hydrogen; (b) ethylene oxide, a three-membered ring of two carbon atoms and one oxygen atom, each carbon carrying two hydrogen atoms. Label each with its name and the catalyst that gives it.

  1. The gaseous carbon monoxide and \(\mathrm{H_2}\) produce different products by using different catalysts.

\[\mathrm{CO}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \xrightarrow{\ \mathrm{Ni}\ } \mathrm{CH_4}\text{(g)} + \mathrm{H_2O}\text{(g)}\]

\[\mathrm{CO}\text{(g)} + 2\,\mathrm{H_2}\text{(g)} \xrightarrow{\ \mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\ } \mathrm{CH_3OH}\text{(g)}\]

Here the very ratio in which the reactants are consumed changes with the catalyst — three moles of hydrogen per mole of carbon monoxide over nickel, but only two over the copper–zinc oxide–chromium oxide catalyst. This makes catalytic selectivity a favourite topic for find-the-product questions, so learn the pairs of catalyst and product together rather than separately.

the same reactants over different catalysts | Reactants | Catalyst | Product | Type of compound formed | |—|—|—|—| | \(\mathrm{C_2H_4} + \mathrm{O_2}\) | \(\mathrm{Pd/Al_2O_3}\) | \(\mathrm{CH_3CHO}\) | Aldehyde (acetaldehyde) | | \(\mathrm{C_2H_4} + \mathrm{O_2}\) | \(\mathrm{Ag/Al_2O_3}\) | \(\mathrm{C_2H_4O}\) | Cyclic ether (ethylene oxide) | | \(\mathrm{CO} + 3\,\mathrm{H_2}\) | \(\mathrm{Ni}\) | \(\mathrm{CH_4} + \mathrm{H_2O}\) | Alkane and water | | \(\mathrm{CO} + 2\,\mathrm{H_2}\) | \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\) | \(\mathrm{CH_3OH}\) | Alcohol (methanol) |

Predict the product formed in each of the following, and write the balanced equation.

(a) Ethylene and dioxygen passed over \(\mathrm{Ag/Al_2O_3}\).

(b) Carbon monoxide and dihydrogen passed over nickel.

(c) Carbon monoxide and dihydrogen passed over \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\).

Solution.

  1. Silver on alumina directs the oxidation of ethylene to ethylene oxide:

\[2\,\mathrm{C_2H_4}\text{(g)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \mathrm{Ag/Al_2O_3}\ } 2\,\mathrm{C_2H_4O}\text{(g)}\]

  1. Nickel directs the reaction to methane and water, consuming three moles of hydrogen for every mole of carbon monoxide:

\[\mathrm{CO}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \xrightarrow{\ \mathrm{Ni}\ } \mathrm{CH_4}\text{(g)} + \mathrm{H_2O}\text{(g)}\]

  1. The copper–zinc oxide–chromium oxide catalyst gives methanol, using only two moles of hydrogen per mole of carbon monoxide:

\[\mathrm{CO}\text{(g)} + 2\,\mathrm{H_2}\text{(g)} \xrightarrow{\ \mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\ } \mathrm{CH_3OH}\text{(g)}\]

In every part the reactants are the same; only the surface has changed. In such questions the catalyst written over the arrow is the piece of information that decides the answer.

56 g of carbon monoxide is completely converted to methanol by reaction with hydrogen over a \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\) catalyst. Calculate (a) the volume of hydrogen required at STP and (b) the mass of methanol formed. (Molar masses: \(\mathrm{CO} = 28\ \text{g mol}^{-1}\), \(\mathrm{CH_3OH} = 32\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)

Solution.

Step 1: moles of carbon monoxide.

\[n(\mathrm{CO}) = \frac{56\ \text{g}}{28\ \text{g mol}^{-1}} = 2.0\ \text{mol}\]

Step 2: use the balanced equation. From \(\mathrm{CO} + 2\,\mathrm{H_2} \rightarrow \mathrm{CH_3OH}\), each mole of \(\mathrm{CO}\) needs 2 mol of \(\mathrm{H_2}\) and gives 1 mol of \(\mathrm{CH_3OH}\). Therefore

\[n(\mathrm{H_2}) = 2 \times 2.0 = 4.0\ \text{mol}, \qquad n(\mathrm{CH_3OH}) = 2.0\ \text{mol}\]

Step 3: volume of hydrogen at STP.

\[V(\mathrm{H_2}) = 4.0\ \text{mol} \times 22.4\ \text{L mol}^{-1} = 89.6\ \text{L}\]

Step 4: mass of methanol.

\[m(\mathrm{CH_3OH}) = 2.0\ \text{mol} \times 32\ \text{g mol}^{-1} = 64\ \text{g}\]

So 89.6 L of hydrogen at STP is needed and 64 g of methanol is obtained. Note that the answer would be different over a nickel catalyst, where the stoichiometry is 1 : 3 and the product is methane — which is precisely what catalytic selectivity means.

(question only — no solution)

28 g of carbon monoxide is passed over a nickel catalyst with excess hydrogen so that the reaction goes to completion. Calculate (a) the volume of hydrogen consumed at STP and (b) the mass of methane formed. (Molar masses: \(\mathrm{CO} = 28\ \text{g mol}^{-1}\), \(\mathrm{CH_4} = 16\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).)

(question only — no solution)

A metal chemisorbs a reactant gas so strongly that the adsorbed molecules cannot move on its surface. Explain, with reference to the five steps of the catalytic cycle, why this metal is a poor catalyst even though it adsorbs the reactant very effectively.

Shape Selective Catalysis by Zeolites

Zeolites are alumino silicates with a three-dimensional network of silicates. Some silicon atoms in this network are replaced by aluminium atoms, giving an Al–O–Si framework. This results in a microporous structure — a solid honeycombed with channels and cavities of molecular dimensions.

The reactions in zeolites are dependent on the size and shape of the reactant or products, and also on the pores and cavities of the zeolites. Zeolites, therefore, are shape selective catalysts.

NoteDefinition

Shape selective catalysis is catalysis in which the reaction depends on the size and shape of the reactant and product molecules relative to the size and shape of the pores and cavities of the catalyst.

The idea is easiest to grasp as a sieve that also happens to be a catalyst, as shown in Fig. 11.23. A molecule small enough and of the right shape to enter a channel reaches the active sites inside and reacts; a molecule too large or too branched to enter never gets the chance. The selectivity here comes from geometry rather than from chemistry — which is exactly what distinguishes it from the catalytic selectivity of the previous subsection.

Diagram

Fig. 11.23 – Shape selective catalysis: a block of zeolite drawn in section with a regular array of uniform channels; a small straight-chain molecule shown entering a channel and reacting at a site inside, while a bulky branched molecule is shown stopped at the mouth of the channel, too large to enter. Label “Al–O–Si framework”, “pore of molecular size”, “molecule admitted”, “molecule excluded”.

In the petroleum industry, the zeolite catalyst ZSM-5 converts alcohols directly to gasoline (petrol) by dehydration, which gives a mixture of hydrocarbons. Dehydration here means the removal of water from the alcohol molecules; the fragments left behind join up inside the zeolite channels, and because those channels only permit hydrocarbons of a certain size to form and escape, the mixture obtained is in the size range useful as petrol.

zeolites, including ZSM-5 (What they do: act as solid acid catalysts inside pores of molecular size, so that only reactants and products of a suitable size and shape can take part → Conditions they need: the reactants in the gas or vapour phase, raised temperature, and a framework whose pore size matches the desired molecules; the framework must be kept free of deposits that would block the channels → Typical use: shape selective conversions in the petroleum industry, such as ZSM-5 converting alcohols to a petrol-range hydrocarbon mixture by dehydration.)

The name zeolite was given in the eighteenth century by the Swedish mineralogist Axel Fredrik Cronstedt, who heated a newly found mineral and watched it froth and give off steam as though it were boiling. He named it from Greek zein, to boil, and lithos, stone — a boiling stone. What he had actually seen was water escaping from the cavities in the framework, so the name records, quite accidentally, the very property that makes zeolites useful: they are stones full of holes.

11.9 Colloids

A number of substances we use in our day to day life are colloids. For example, milk, butter, jelly, whipped cream and mayonnaise. Colloid chemistry is the chemistry of everyday life.

Knowledge of colloid chemistry is essential for understanding many useful materials like cement, bricks, pottery, porcelain, glass and enamels; oils and lacquers; rubber, celluloid and other plastics; leather, paper, textiles, filaments, crayons, inks and road construction material. In many daily processes like cooking, washing, dyeing, painting, ore floatation, water purification, sewage disposal, smoke prevention, photography and pharmacy, the use of colloids is important.

Colloids are heterogeneous mixtures. The component of a colloid present in the largest proportion is called the dispersion medium, and the other components are called the dispersed phase. The particles of the dispersed phase are larger than the size of a molecule and smaller than particles which we can see with the naked eye.

NoteDefinition

A colloid is a heterogeneous mixture in which particles of one substance, the dispersed phase, are distributed through a second substance, the dispersion medium, the particles being larger than ordinary molecules but too small to be seen with the naked eye.

Between a Solution and a Suspension

Observe the formation of a solution of salt and water. Salt dissolves completely in water and forms a homogeneous system. On the other hand, ground coffee or tea leaves with milk form a suspension — the particles are large, and left alone they settle out.

Between the two extremes of solution and suspension, we observe a large group of systems called colloidal dispersions, or simply colloids. A colloid is therefore not a new kind of matter but a middle ground defined by particle size.

The essential difference between a solution and a colloid is particle size. Solutions contain solute particles with diameter in the range 0.1 to 2 nm, the size of a typical ion or small molecule. Solutions are transparent and may be coloured, and do not separate on standing.

On the other hand, colloids such as milk and fog contain particles of dispersed phase with diameters in the range 2 to 500 nm. They are translucent to light and do not separate on standing. The three kinds of mixture are placed on a size scale in Fig. 11.26.

Diagram

Fig. 11.26 – Horizontal particle-size scale in nanometres from 0.1 nm to beyond 1000 nm, divided into three labelled regions: “solution, 0.1 – 2 nm (transparent)”, “colloid, 2 – 500 nm (translucent, does not settle)” and “suspension (particles visible, settles on standing)”. Under each region, a small beaker sketch with a labelled example — salt solution, milk, and muddy water.

Solution vs Colloid vs Suspension | Feature | True solution | Colloid | Suspension | |—|—|—|—| | Diameter of particles | 0.1 to 2 nm | 2 to 500 nm | Larger still; visible to the naked eye | | Nature of the system | Homogeneous | Heterogeneous | Heterogeneous | | Appearance to light | Transparent | Translucent | Opaque | | Settling on standing | Does not separate | Does not separate | Particles settle out | | Example | Salt in water | Milk, fog | Tea leaves in milk, muddy water |

Note carefully that “does not separate on standing” is shared by solutions and colloids. This is why a colloid can be mistaken for a solution by eye, and why the optical test described in Section 11.9.5 is needed to tell them apart.

11.9.1 Examples of Colloids

Some examples of phenomena observed in our daily life which are understood in terms of colloids are as follows.

Blue colour of the sky. The sky appears blue to us because minute dust particles, along with minute water droplets dispersed in air, scatter blue light which reaches our eyes.

Blood. It is a colloidal dispersion of plasma proteins and antibodies in water. At the same time, blood is also a suspension of blood cells and platelets in water.

Soils. Fertile soils are colloidal in nature, where humus acts as a protective colloid. Soil adsorbs moisture and nourishing materials due to its colloidal nature — which is a direct link back to the first half of this chapter, since a colloidal soil offers an enormous surface on which water and nutrient ions can be held instead of being washed away.

Fog, mist and rain. Mist is caused by small droplets of water dispersed in air. Fog is formed whenever there is a temperature difference between the ground and the air. A large portion of air containing dust particles gets cooled below its dew point; the moisture from the air condenses on the surface of these particles, which form fine droplets. These droplets are colloid particles and float in air as fog or mist.

The dew point is the temperature at which air becomes saturated with water vapour, so that any further cooling forces the vapour to condense. Note that the dust particle is not incidental here — it provides the surface on which condensation begins, exactly as the cold bottle did in Section 11.1.

11.9.2 Classification of Colloids

Colloids are classified in three different ways: by the physical states of the two phases, by the affinity between them, and by the size of the molecules involved.

Classification Based on Physical State

Table 11.4 illustrates the types of colloids in accordance with the physical states of the dispersed phase and the dispersion medium.

Table 11.4: Types of colloids based on physical state

Dispersed phase Dispersion medium Type of colloid Examples
solid solid solid sol coloured glasses, gem stones, porcelain, paper
solid liquid sols and gels paints, cell fluids, gelatin, muddy water, starch solution
solid gas aerosol smoke, dust
liquid solid gel cheese, butter, jellies
liquid liquid emulsion milk, hair cream
liquid gas aerosol fog, mist, cloud, hair sprays, insecticide sprays
gas solid solid sol pumice stone, foam rubber, plaster
gas liquid foam froth, whipped cream, soap lather

Two habits will keep this table straight. Always name the dispersed phase first and the dispersion medium second, and always ask which component is present in the larger proportion — that one is the medium. Butter and milk both contain fat and water, but in butter the water is dispersed in fat, and in milk the fat is dispersed in water.

For each of the following colloids, identify the dispersed phase and the dispersion medium, and name the type of colloid: (a) smoke, (b) cheese, (c) whipped cream, (d) hair cream.

Solution.

  1. Smoke. Fine solid particles of carbon and dust are dispersed in air. Dispersed phase: solid; dispersion medium: gas. Type: aerosol.

  2. Cheese. Droplets of liquid are held within a solid framework. Dispersed phase: liquid; dispersion medium: solid. Type: gel.

  3. Whipped cream. Air has been beaten into a liquid. Dispersed phase: gas; dispersion medium: liquid. Type: foam.

  4. Hair cream. One liquid is dispersed in another. Dispersed phase: liquid; dispersion medium: liquid. Type: emulsion.

The method is always the same: decide the physical state of the material that is broken up and spread about, then the state of the material it is spread through, and read the name off Table 11.4.

(question only — no solution)

Identify the dispersed phase, the dispersion medium and the type of colloid in each of the following: (a) fog, (b) pumice stone, (c) coloured glass, (d) muddy water, (e) soap lather.

Classification Based on Interaction or Affinity of Phases

On the basis of interaction or affinity of phases, a colloidal solution is classified as lyophilic and lyophobic. If water is the dispersion medium, the terms hydrophilic and hydrophobic are used.

Lyophilic colloids. Lyo means liquid and philic means loving. A colloidal solution in which the particles of the dispersed phase have a great affinity for the dispersion medium is a lyophilic colloid.

If a lyophilic sol is evaporated, the dispersed phase separates. But if it is remixed with the medium, the sol can be formed again. That is why such sols are called reversible sols. They are stable and difficult to coagulate.

NoteDefinition

Lyophilic colloids are colloids in which the particles of the dispersed phase have a strong affinity for the dispersion medium; they are formed by direct mixing, are reversible, and are self-stabilized.

Lyophobic colloids. A colloidal solution in which the particles of the dispersed phase have no affinity for the dispersion medium is a lyophobic colloid. Phobic means fearing, hence liquid-hating.

The common examples are Ag, Au, hydroxides like \(\mathrm{Al(OH)_3}\) and \(\mathrm{Fe(OH)_3}\), and metal sulfides. Once precipitated or coagulated, they have little or no tendency to revert back to the colloidal state.

NoteDefinition

Lyophobic colloids are colloids in which the particles of the dispersed phase have no affinity for the dispersion medium; they must be prepared by special methods, are irreversible, and need traces of a stabilizer to survive.

The vocabulary of this classification is built entirely from Greek. Lyo is from a root meaning to loosen or dissolve, philos means loving and phobos means fearing, so a lyophilic particle is one that “loves the solvent” and a lyophobic one “fears” it. When the solvent is water in particular, hydor takes the place of lyo, giving hydrophilic and hydrophobic — the same word you met in the froth flotation process, where pine oil made an ore surface water-fearing.

Table 11.5 shows a comparison of lyophilic and lyophobic colloids.

Table 11.5: Distinction between Lyophilic and Lyophobic colloids

Lyophilic colloids Lyophobic colloids
1. Formed easily by direct mixing. Formed only by special methods.
2. Reversible. Irreversible.
3. The particles are not easily visible even under an ultramicroscope. The particles are easily detected under an ultramicroscope.
4. These are self stabilized. These are unstable and hence require traces of stabilizers.
5. Addition of a large amount of electrolytes causes precipitation/coagulation. Addition of a small amount of electrolytes causes precipitation/coagulation.
6. Viscosity of dispersed phase much higher than that of the dispersion medium. Viscosity of dispersed phase is nearly the same as the dispersion medium.
7. Surface tension of dispersed phase is lower than dispersion medium. Surface tension of dispersed phase is nearly the same as the dispersion medium.

Row 5 is the one examiners return to most often, so make sure the reason is clear. A lyophilic particle carries a coat of the dispersion medium around it, and that coat must be stripped away before the particles can come together — which takes a great deal of electrolyte. A lyophobic particle has no such coat and is held apart only by its charge, so a small amount of electrolyte is enough to neutralise that charge and bring it down.

Classification Based on Molecular Size

Colloids are classified into three types in accordance with the size of their molecules.

Multimolecular colloids. The individual particles consist of an aggregate of atoms or small molecules with size less than \(10^3\) pm. For example, gold sol consists of particles of various sizes having several gold atoms. A colloidal solution in which the particles are held together with van der Waals forces of attraction is called a multimolecular colloid. For example, \(\mathrm{S_8}\) sulfur molecules.

Macromolecular colloids. The molecules of the dispersed phase are sufficiently large (macro) to be of colloidal dimensions. Examples are starch, cellulose, proteins, polythene, nylon and plastics.

The distinction between the first two types is worth stating plainly. In a multimolecular colloid, the particle is large but each molecule in it is small — many small units have clustered together. In a macromolecular colloid, a single molecule is itself already of colloidal size, so no clustering is needed at all.

Associated colloids or micelles. These substances behave as normal electrolytes at low concentration and become associated at higher concentration, forming a colloidal solution. The associated particles are called micelles. For example, soap and detergent.

A soap molecule has a long hydrophobic hydrocarbon chain, called the tail, attached to a hydrophilic ionic group (carboxylate), called the head. In water the soap molecules arrange themselves to form spherical particles that are called micelles (Fig. 11.5). In each micelle the hydrophobic tails of the soap molecules point to the centre and the hydrophilic heads lie on the surface of the sphere. As a result of this, soap dispersion in water is stable.

Diagram

Fig. 11.5 – Soap micelle in water: a spherical arrangement of soap molecules with the ionic carboxylate heads on the outside of the sphere facing the surrounding water molecules, and the long hydrocarbon tails pointing inward to the centre. Label “hydrophilic head”, “hydrophobic tail”, “H₂O” at several points around the sphere.

The arrangement in Fig. 11.5 solves a problem that neither end of the molecule could solve alone. The ionic head is at home in water and the hydrocarbon tail is not, so the molecules turn every tail inward, away from the water, and present only heads to the outside. Each micelle is then a particle of colloidal size whose surface is entirely water-friendly, which is why the dispersion is stable — and, as Section 11.9.8 notes, why soap can lift grease at all.

Multimolecular vs Macromolecular vs Associated colloids | Feature | Multimolecular | Macromolecular | Associated (micelles) | |—|—|—|—| | What the particle is | Aggregate of many atoms or small molecules | One very large single molecule | Aggregate formed only above a certain concentration | | Force holding it together | van der Waals forces | Not applicable; the molecule is already large | Attraction between hydrocarbon tails, heads facing the medium | | Behaviour at low concentration | Still colloidal | Still colloidal | Behaves as a normal electrolyte | | Examples | Gold sol, \(\mathrm{S_8}\) sulfur | Starch, cellulose, proteins, polythene, nylon | Soaps, detergents |

11.9.3 Preparation of Colloids

A few important methods for the preparation of colloids are as follows. Notice that the three routes approach the same target from opposite directions: the first two build colloidal particles up from something smaller, while the third breaks a precipitate down from something larger. The routes are collected in FC 11.5.

Flowchart

FC 11.5 – Preparation of colloids, three routes: (1) chemical methods → double decomposition, oxidation, reduction or hydrolysis → water-insoluble product molecules aggregate → sol; (2) Bredig’s arc method → electric arc struck between metal electrodes immersed in the dispersion medium → metal vaporises → vapour condenses in the medium → metal sol; (3) peptization → precipitate + dispersion medium + small amount of electrolyte, shaken → precipitate adsorbs one ion of the electrolyte → charge develops → precipitate breaks up into colloidal particles → sol. Mark routes 1 and 2 “building up from smaller units” and route 3 “breaking down from a precipitate”.

Chemical Methods

Colloids may be prepared by double decomposition, oxidation, reduction or hydrolysis. Molecules of the water-insoluble products of these reactions aggregate together and form sols.

\[\mathrm{SO_2} + 2\,\mathrm{H_2S} \xrightarrow{\ \text{Oxidation}\ } 3\,\mathrm{S} \downarrow + 2\,\mathrm{H_2O}\]

\[2\,\mathrm{AuCl_3} + 3\,\mathrm{HCHO} + 3\,\mathrm{H_2O} \xrightarrow{\ \text{Reduction}\ } 2\,\mathrm{Au} \downarrow + 3\,\mathrm{HCOOH} + 6\,\mathrm{HCl}\]

\[\mathrm{FeCl_3} + 3\,\mathrm{H_2O} \xrightarrow{\ \text{Hydrolysis}\ } \mathrm{Fe(OH)_3} \downarrow + 3\,\mathrm{HCl}\]

The downward arrow marks a product that is insoluble in water. In each case that insoluble product does not grow into a visible precipitate straight away; while its particles are still in the colloidal size range, a sol has been formed. In the second reaction, methanal (\(\mathrm{HCHO}\)) is the reducing agent and is itself oxidised to methanoic acid (\(\mathrm{HCOOH}\)) while gold(III) is reduced to metallic gold.

Electrical Disintegration by Bredig’s Arc Method

This process involves vaporization as well as condensation. Colloidal sols of metals such as gold, silver and platinum can be prepared by this method.

In this method, an electric arc is struck between electrodes of the metal immersed in the dispersion medium, as shown in Fig. 11.6. The intense heat produced vapourises the metal, which then condenses to form particles of colloidal sol.

Diagram

Fig. 11.6 – Bredig’s arc method: a vessel of water surrounded by a bath of ice, with two metal electrodes dipped into the water from above and connected to a supply, an electric arc struck between their tips below the water surface. Label “metal electrodes”, “arc”, “water”, “ice”.

Two details of the set-up in Fig. 11.6 matter. The vessel is kept in an ice bath, since the arc is fiercely hot and the surrounding liquid must stay cool enough to condense the metal vapour into very small particles rather than let them grow. And the arc is struck under the liquid, so the vapour has nowhere to escape to and condenses within the medium itself.

Peptization

During peptization a precipitate is converted into a colloidal sol by shaking with the dispersion medium in the presence of a small amount of an electrolyte. The electrolyte used is known as the peptizing agent.

During the process, the precipitate adsorbs one of the ions of the electrolyte on its surface. A positive or negative charge is developed on the precipitate, which finally breaks up into small particles of colloidal size.

NoteDefinition

Peptization is the process of converting a precipitate into a colloidal sol by shaking it with the dispersion medium in the presence of a small amount of an electrolyte, called the peptizing agent.

The mechanism is worth following, because it explains why colloidal particles stay apart at all. Adsorption of one kind of ion gives every particle the same sign of charge, and particles carrying the same charge repel one another. That repulsion first tears the loose precipitate apart and then keeps the fragments from re-uniting. Peptization is thus adsorption from Section 11.2 doing the work of dispersion.

11.9.4 Purification of Colloidal Solution

A colloidal solution generally contains an excessive amount of electrolytes and some other soluble impurities. A small quantity of an electrolyte is necessary for the stability of a colloidal solution. A large quantity of electrolyte may result in coagulation. It is also necessary to reduce soluble impurities.

The process used for reducing the amount of impurities to a requisite minimum is known as purification of the colloidal solution.

NoteDefinition

Purification of a colloidal solution is the process of reducing the amount of electrolytes and other soluble impurities in the sol to the requisite minimum.

Q. A little electrolyte keeps a sol stable, but too much destroys it. How can the same substance do both?

A. Because a small quantity supplies the ions that are adsorbed on the particles and give them their like charges, and it is that mutual repulsion which keeps the particles apart. A large quantity supplies so many ions of the opposite sign that the charge on the particles is neutralised instead, the repulsion disappears, and the particles come together and settle. The same substance stabilises in traces and coagulates in excess.

Purification of a colloidal solution can be carried out using dialysis by the following method.

Dialysis is a process of removing a dissolved substance from a colloidal solution by diffusion through a suitable membrane.

NoteDefinition

Dialysis is the process of removing a dissolved substance from a colloidal solution by allowing it to diffuse through a suitable membrane, while the colloidal particles are retained.

The apparatus used is a dialyser, drawn in Fig. 11.7. A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing. The molecules and ions diffuse through the membrane into the outer water, and pure colloidal solution is left behind.

Diagram

Fig. 11.7 – Dialysis: a bag of dialysing membrane holding the colloidal solution, suspended inside a larger vessel; fresh water entering the vessel at one side and leaving with the dissolved electrolyte at the other; small ions shown crossing the membrane outward while larger dispersed-phase particles remain inside. Label “dialysing membrane”, “dispersed phase”, “dissolved electrolyte”, “water in”, “water + electrolyte out”.

Why the method works comes down to size, and it is the very property that defines a colloid. The membrane has pores large enough for ions and small molecules to pass but too small for colloidal particles, so the impurities leave while the sol stays. The water is kept flowing for a reason as well: diffusion continues only while the outside is less concentrated than the inside, so if the water were left standing, the process would slow and stop.

This is the chemistry behind a dialysis machine. In a patient whose kidneys have failed, waste products such as urea build up in the blood because they are no longer being removed. In haemodialysis, blood is passed on one side of a membrane while a carefully prepared solution flows on the other. Small waste molecules and excess ions diffuse across, exactly as they do in Fig. 11.7, while the blood’s proteins and cells — colloidal and larger — are far too big to cross and remain in the circulation. The dialysing fluid is kept flowing to maintain the concentration difference, and its composition is adjusted so that substances the body needs are not stripped out along with the wastes. A laboratory purification method and a life-support therapy rest on the same single fact about particle size.

11.9.5 Properties of Colloidal Dispersions

Various properties exhibited by colloidal dispersions are described below. They fall into groups — general, optical, colour, kinetic and electrical — and each group turns out to be a consequence of the one fact we began with: the particles are of an in-between size.

General Properties
  1. A colloidal system is heterogeneous and consists of two phases, the dispersed phase and the dispersion medium.

  2. The dispersed phase particles pass slowly through parchment paper or animal membrane, but readily pass through ordinary filter paper.

  3. The particles usually are not detectable by a powerful microscope.

The second of these is the practical basis of dialysis, and it also warns us that filtration is useless for separating a colloid. The pores of ordinary filter paper are enormous compared with a colloidal particle, so the sol runs straight through; only a membrane with pores of nearly molecular size can hold the particles back.

Optical Property: Tyndall Effect

Tyndall observed that when light passes through a true solution, the path of light through it cannot be detected.

However, if the light passes through a colloidal dispersion, the particles scatter some light in all directions, and the path of the light through the colloidal dispersion becomes visible to an observer standing at right angles to its path. The two cases are contrasted in Fig. 11.8.

Diagram

Fig. 11.8 – Tyndall effect: a light source with a slit producing a narrow beam, directed in turn through two vessels — one holding a true solution, where the beam is not visible from the side, and one holding a colloidal dispersion, where a bright cone of scattered light is seen. An observer’s eye is drawn at right angles to the beam. Label “light source”, “beam”, “light beam not visible”, “Tyndall cone; light beam is visible”.

The phenomenon of scattering of light by colloidal particles, and making the path of light visible through the dispersion, is referred to as the Tyndall effect.

NoteDefinition

The Tyndall effect is the scattering of light by colloidal particles, by which the path of a beam of light passing through the dispersion becomes visible when viewed at right angles to the beam.

The bright cone of the light is called the Tyndall cone. The Tyndall effect is observed only when the following conditions are satisfied.

  1. The diameter of the dispersed particles is not much smaller than the wavelength of light used.

  2. The refractive indices of the dispersed phase and the dispersion medium differ largely.

Both conditions have a clear physical meaning. A particle very much smaller than the wavelength of light barely disturbs the wave at all, which is why a true solution shows no cone; and if the two phases bend light to nearly the same extent, a light ray crossing the boundary hardly changes direction, so little is scattered even when the particles are of the right size.

Importance of the Tyndall effect:

  1. It is useful in determining the number of particles in a colloidal system and the particle size therein.

  2. It is used to distinguish between a colloidal dispersion and a true solution.

Q3. Milk looks white, and fog makes the beam of a headlight visible in the air, but a clear salt solution does neither. What must be different about the particles inside them?

A. Their size. In a salt solution the particles are individual ions, 0.1 to 2 nm across — very much smaller than the wavelength of visible light — so they scatter almost nothing and the liquid stays transparent with no visible beam. In milk and in fog the dispersed particles are 2 to 500 nm across, comparable with the wavelength of light, and their refractive index differs sharply from that of the surrounding medium. Such particles scatter light in all directions. Seen from the side, that scattered light makes the beam visible as a Tyndall cone in fog; seen in bulk, scattering from countless fat droplets sends light back at us from every angle and milk looks white. Nothing has been added to milk to colour it — the whiteness is scattered light, and the test for a colloid is exactly this.

John Tyndall was a nineteenth-century British physicist with an interest in why the sky is blue. He passed beams of light through tubes of air containing fine suspended particles and watched the beam light up along its path, and he found that the scattered light was richer in blue than the light that came through. The scattering of shorter wavelengths more strongly than longer ones is the reason the daytime sky is blue while the setting sun, seen through a long path of atmosphere from which the blue has been scattered away, appears red. His experiments also settled a practical dispute of the time: air that scatters no light at all is air free of floating particles, which gave a way of testing whether air had truly been sterilised.

Colour

The colour of a colloidal solution depends on the wavelength of light scattered by the dispersed particles.

The colour of a colloidal dispersion also changes with the manner in which the observer receives the light. For example, a mixture of a few drops of milk and a large amount of water appears blue when viewed by the scattered light and red when viewed by the transmitted light.

It also depends on the size of the colloidal particles. For example, the finest gold sol is red in colour, whereas with an increase in size it appears purple.

This last observation deserves a moment’s thought, because it shows how sensitive a colloid is to particle size. The chemical substance in both sols is nothing but gold; no reaction has taken place and no new compound has formed. Only the size of the aggregates has changed, and with it the wavelengths scattered — which is why a colour change can be used as evidence that colloidal particles have grown.

Kinetic Property: Brownian Motion

The colloidal or microscopic particles undergo ceaseless random zig-zag motion in all directions in a fluid. This motion of the dispersed phase particles is called Brownian motion, and it is illustrated in Fig. 11.9.

Diagram

Fig. 11.9 – Brownian motion: a cube of fluid containing several colloidal particles, each drawn with a jagged zig-zag arrow trail showing its random path in all directions, with no two paths alike. Label “colloidal particle”, “random zig-zag path”.

NoteDefinition

Brownian motion is the ceaseless, random zig-zag motion of colloidal particles in all directions in a fluid.

The British botanist Robert Brown observed such motion of pollen grains under a microscope. The random motion was explained by Albert Einstein in 1905.

Cause of Brownian motion:

  1. Constant collision of the particles of the dispersed phase with the fast-moving molecules of the dispersion medium (fluid).

  2. Due to this, the dispersed phase particles acquire kinetic energy from the molecules of the dispersion medium. This kinetic energy brings forth Brownian motion.

The collisions come from every side, but at any instant they do not balance exactly, and a particle small enough feels the difference and is pushed. Two consequences follow, both worth remembering. Brownian motion is stronger for smaller particles and at higher temperature, since the medium’s molecules then move faster. And because the particles are kept perpetually in motion, they never get the chance to settle — which is why a colloid, unlike a suspension, does not separate on standing.

Robert Brown was studying pollen grains in water under his microscope in the 1820s when he noticed that fine particles released from them jittered about endlessly and would not come to rest. He tested carefully whether the motion was a sign of life, using particles of ground-up mineral matter as well, and found the same restless movement — so the cause was not biological. The explanation waited nearly eighty years. In 1905 Albert Einstein showed that the jitter was exactly what one should expect if the surrounding fluid were made of molecules in rapid random motion, and he worked out how far such a particle should wander in a given time. When experiments confirmed his prediction, the debate over whether molecules were real objects or a convenient bookkeeping device was effectively over. A botanist’s puzzle about pollen had become the evidence for the existence of molecules.

Electrical Properties

Charge on colloidal particles. Colloidal particles carry an electric charge. The nature of this charge is the same on all particles for a given colloidal solution, which can be either positive or negative. Some common sols with the nature of the charge on the particles are listed in Table 11.6.

Table 11.6: Charge on dispersed particles

Positively charged sols Negatively charged sols
1. Hydrated metallic oxides \(\mathrm{Al_2O_3 \cdot xH_2O}\), \(\mathrm{CrO_3 \cdot xH_2O}\), \(\mathrm{Fe_2O_3 \cdot xH_2O}\) Metals, Cu, Ag, Au sols; metallic sulphides \(\mathrm{As_2S_3}\), \(\mathrm{Sb_2S_3}\), \(\mathrm{CdS}\)
2. Basic dye stuff, methylene blue sols Acid dye stuff, eosin, congo red sol
3. Haemoglobin (blood) Sols of starch, gum
4. Oxides: \(\mathrm{TiO_2}\) sol Gelatin, clay, gum sols

That every particle in a given sol carries the same sign of charge is what keeps the sol alive. Like charges repel, so the particles are held apart and cannot combine into larger ones — the same argument met in peptization.

Electrophoresis. The electrophoresis set-up is shown in Fig. 11.10. The figure shows a U tube set-up in which two platinum electrodes are dipped in a colloidal solution. When an electric potential is applied across the two electrodes, colloidal particles move towards one or the other electrode.

Diagram

Fig. 11.10 – Electrophoresis: a U tube filled with colloidal solution, a platinum electrode dipped in each arm and connected to a supply, one arm marked anode and the other cathode, with a reservoir above and a stop cock at the base. Charged colloidal particles are shown migrating towards one electrode and collecting on it. Label “anode”, “cathode”, “colloidal solution”, “deposited particles”, “reservoir”, “stop cock”.

The movement of colloidal particles under an applied electric potential is called electrophoresis.

NoteDefinition

Electrophoresis is the movement of colloidal particles towards one of the electrodes under an applied electric potential.

Positively charged particles move towards the cathode, while negatively charged particles migrate to the anode and get deposited on the respective electrode. The rule is the ordinary one of electrostatics — a particle is drawn to the electrode of opposite sign — and it is worth fixing in memory that the cathode is the negative electrode in such a cell, which is why positive particles travel to it.

Electroosmosis. The movement of dispersed particles can be prevented by suitable means, such as the use of a membrane. Then it is observed that the dispersion medium begins to move in an electric field. This is termed electroosmosis.

NoteDefinition

Electroosmosis is the movement of the dispersion medium in an electric field when the movement of the dispersed particles is prevented, for example by a membrane.

The pairing of the two effects is neat and often asked about: in electrophoresis the particles move through a stationary medium, while in electroosmosis the particles are held and the medium moves instead.

Applications of electrophoresis:

  1. On the basis of the direction of movement of the colloidal particles under the influence of an electric field, it is possible to know the sign of the charge on the particles.

  2. It is also used to measure the rate of migration of sol particles.

  3. A mixture of colloidal particles can be separated by electrophoresis, since different colloidal particles in a mixture migrate at different rates.

Q. Both electrophoresis and chromatography separate a mixture. What is the essential difference between them?

A. What does the separating. In chromatography the components are separated because they are adsorbed to different extents on a solid, so the separation depends on affinity for a surface. In electrophoresis they are separated because they carry different charges and sizes and so migrate at different rates in an electric field. One exploits adsorption, the other exploits charge.

Coagulation

The precipitation of colloids by removal of the charge associated with colloidal particles is called coagulation.

NoteDefinition

Coagulation is the precipitation of a colloid brought about by removing the charge on the colloidal particles, so that the particles come together and settle out.

The charge on the colloidal particles is due to the preferential adsorption of ions on their surface. For precipitation of lyophobic colloids, removal of charge is required.

The situation with lyophilic colloids is a little different. The lyophilic particles first attract the molecules of the dispersion medium and form a layer of medium surrounding the particles, which may adsorb the ions. Hence for the precipitation of lyophilic colloids, removal of the charge on the layer of medium is necessary.

In short, a lyophobic sol has one defence and a lyophilic sol has two. Strip the charge from a lyophobic particle and nothing stands between it and its neighbours. A lyophilic particle still has its coat of dispersion medium, which must also be removed — and that is why, as Table 11.5 records, it takes a large amount of electrolyte to bring a lyophilic sol down.

11.9.6 Methods to Effect Coagulation

Coagulation of the lyophobic sols can be carried out in the following ways, which are gathered in FC 11.6.

Flowchart

FC 11.6 – Methods to effect coagulation of a lyophobic sol, five branches from a single node: (1) electrophoresis → particles move to the oppositely charged electrode → get discharged → precipitate; (2) mixing two oppositely charged sols in almost equal proportions → charges neutralised → mutual coagulation; (3) boiling → adsorbed layer disturbed by increased collisions → charge reduced → precipitate settles; (4) persistent dialysis → traces of electrolyte removed → sol becomes unstable → precipitate; (5) addition of excess electrolyte → charge neutralised → precipitate. All five branches converge on “coagulated precipitate”.

  1. By electrophoresis. The colloidal particles move towards oppositely charged electrodes, get discharged and precipitate.

  2. By mixing two oppositely charged sols. Oppositely charged sols, when mixed in almost equal proportions, neutralize their charges and get precipitated. Example: mixing of hydrated ferric oxide (positive sol) and arsenious sulfide (negative sol) brings them into the precipitated forms. This type of coagulation is called mutual coagulation.

  3. By boiling. When a sol is boiled, the adsorbed layer is disturbed as a result of increased collisions with molecules in the dispersion medium. This reduces the charge on the particles, and they subsequently settle down as a precipitate.

  4. By persistent dialysis. On prolonged dialysis, traces of the electrolyte present in the sol are removed almost completely. The colloids then become unstable and finally precipitate.

  5. By addition of electrolytes. When an excess of an electrolyte is added, the colloidal particles are precipitated.

Notice that the fourth method looks like the opposite of the fifth, yet both destroy the sol. Removing the electrolyte altogether takes away the ions whose adsorption gave the particles their charge; adding far too much of it neutralises that charge with ions of the opposite sign. A sol needs a trace of electrolyte and cannot survive an excess — the point made in the sidebar of Section 11.9.4.

Hardy–Schulze rule. Generally, the greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This is known as the Hardy–Schulze rule.

NotePrinciple / Law

Hardy–Schulze rule: the greater the valence of the flocculating ion added to a sol, the greater is its power to cause precipitation of that sol. The flocculating ion is the ion whose charge is opposite in sign to the charge on the colloidal particles.

In the coagulation of a negative sol, the flocculating power follows the order

\[\mathrm{Al^{3+}} > \mathrm{Ba^{2+}} > \mathrm{Na^{+}}\]

Similarly, coagulation of a positive sol reveals

\[[\mathrm{Fe(CN)_6}]^{4-} > \mathrm{PO_4^{3-}} > \mathrm{SO_4^{2-}} > \mathrm{Cl^{-}}\]

The logic behind the rule is worth stating, because it makes the two orders impossible to confuse. A negatively charged sol is brought down by cations, so we compare cations and the most highly charged one wins. A positively charged sol is brought down by anions, so we compare anions. In each case a single ion of higher charge neutralises more of the particle’s charge, so far less of it is needed.

flocculating power along a series of ions (For a negative sol, along the series \(\mathrm{Na^{+}} \rightarrow \mathrm{Ba^{2+}} \rightarrow \mathrm{Al^{3+}}\) the flocculating power increases, because the charge on the flocculating cation rises from +1 to +3 and a more highly charged ion neutralises the particle’s negative charge more effectively. For a positive sol, along the series \(\mathrm{Cl^{-}} \rightarrow \mathrm{SO_4^{2-}} \rightarrow \mathrm{PO_4^{3-}} \rightarrow [\mathrm{Fe(CN)_6}]^{4-}\) the flocculating power increases for the same reason, the anionic charge rising from −1 to −4. Rule of thumb: identify the sign of the sol, look at the oppositely charged ion, and rank by the size of its charge.)

Q4. A pinch of alum is enough to clear a bucket of muddy water, but a gelatin solution needs a large amount of salt before anything separates out. Why does one need so little and the other so much?

A. Because the two sols are protected in different ways and by different amounts. Muddy water is a lyophobic clay sol, negatively charged, and its only defence is the mutual repulsion of its like-charged particles. Alum supplies \(\mathrm{Al^{3+}}\) ions, and by the Hardy–Schulze rule a triply charged cation is a very powerful flocculating ion for a negative sol — so a small quantity neutralises the charge, the repulsion vanishes and the clay settles. Gelatin is a lyophilic sol. Its particles are not merely charged; each one carries a coat of water molecules held by strong affinity for the medium. Neutralising the charge alone achieves nothing, because the water coat still keeps the particles apart, and it takes a large amount of electrolyte to strip that coat away as well. One sol has a single line of defence and the other has two.

alum as a coagulating agent (What it does: dissolves to furnish \(\mathrm{Al^{3+}}\) ions, which act as a powerful flocculating ion for negatively charged sols such as clay and other colloidal impurities in water → Conditions it needs: only a small quantity, stirred into the water and then allowed to stand so the coagulated matter can settle; the sol must be a negative one, since a positive sol needs a highly charged anion instead → Typical use: clarifying muddy water and making natural water potable, as described in Section 11.9.8.)

Arrange \(\mathrm{NaCl}\), \(\mathrm{BaCl_2}\) and \(\mathrm{AlCl_3}\) in increasing order of their coagulating power for an arsenious sulfide sol. Name the flocculating ion in each case and give the reason for the order.

Solution.

Step 1: find the sign of the charge on the sol. From Table 11.6, \(\mathrm{As_2S_3}\) is a negatively charged sol.

Step 2: identify the flocculating ion. The flocculating ion is the one of opposite sign to the sol, so for a negative sol it is the cation. The flocculating ions are therefore \(\mathrm{Na^{+}}\) from \(\mathrm{NaCl}\), \(\mathrm{Ba^{2+}}\) from \(\mathrm{BaCl_2}\) and \(\mathrm{Al^{3+}}\) from \(\mathrm{AlCl_3}\).

Step 3: apply the Hardy–Schulze rule. Coagulating power increases with the valence of the flocculating ion, so

\[\mathrm{NaCl} < \mathrm{BaCl_2} < \mathrm{AlCl_3}\]

Reason. The charges on the flocculating cations are +1, +2 and +3 respectively. A cation of higher charge neutralises the negative charge on the sol particles more effectively, so a smaller quantity of \(\mathrm{AlCl_3}\) is needed than of \(\mathrm{BaCl_2}\), and less of \(\mathrm{BaCl_2}\) than of \(\mathrm{NaCl}\). The chloride ion is common to all three and plays no part, since it carries the same sign as the sol.

(question only — no solution)

A sol of \(\mathrm{Fe(OH)_3}\) is to be coagulated. Three electrolytes are available: \(\mathrm{NaCl}\), \(\mathrm{Na_2SO_4}\) and \(\mathrm{Na_3PO_4}\). State the sign of the charge on the sol, identify the flocculating ion in each electrolyte, and arrange the three electrolytes in decreasing order of coagulating power, giving your reason.

Coagulation shapes coastlines. A river carries an enormous load of negatively charged colloidal clay particles, held apart by their like charges over hundreds of kilometres of fresh water. Where the river meets the sea, that water suddenly mixes with a strong electrolyte solution containing \(\mathrm{Na^{+}}\), \(\mathrm{Mg^{2+}}\) and \(\mathrm{Ca^{2+}}\) ions. The charge on the clay particles is neutralised, the repulsion that kept them suspended disappears, and the clay coagulates and settles. Repeated over centuries, this deposition builds the fertile deltas at river mouths. The same chemistry is put to work deliberately in a water treatment plant, and it is also why a sample of muddy river water in the laboratory clears far faster if a little electrolyte is stirred in than if it is simply left to stand.

11.9.7 Emulsions

A colloidal system in which one liquid is dispersed in another immiscible liquid is called an emulsion. These are liquid–liquid colloidal systems in which both liquids are completely or partially immiscible.

NoteDefinition

An emulsion is a colloidal system in which one liquid is dispersed as fine droplets in another liquid with which it is immiscible.

Types of emulsions: there are two types of emulsions, shown in Fig. 11.11.

Diagram

Fig. 11.11 – Emulsions: two beakers side by side in section. The left beaker shows small dark oil droplets dispersed throughout a pale water medium, labelled “oil in water”; the right beaker shows small pale water droplets dispersed throughout a darker oil medium, labelled “water in oil”.

Emulsion of oil in water (o/w type). An emulsion in which the dispersed phase is oil and the dispersion medium is water is called an emulsion of oil in water. For example: milk, vanishing cream, paint, etc. Milk consists of particles of fat dispersed in water.

Emulsion of water in oil (w/o type). An emulsion in which the dispersed phase is water and the dispersion medium is oil is called an emulsion of water in oil. For example, cod liver oil consists of particles of water dispersed in oil. Some other examples of this type include butter and cream.

Milk and butter make the pair easy to remember, since both contain the same two substances in reversed roles. Whichever liquid is present as droplets is the dispersed phase, and whichever forms the continuous liquid around them is the dispersion medium.

Table 11.7 describes the difference between the two types of emulsion.

Table 11.7: Distinction between oil in water and water in oil emulsions

Oil in water Water in oil
1. Oil is the dispersed phase and water is the dispersion medium. Water is the dispersed phase and oil is the dispersion medium.
2. If water is added, it will be miscible with the emulsion. If oil is added, it will be miscible with the emulsion.
3. An addition of a small amount of an electrolyte makes the emulsion conducting. Addition of a small amount of an electrolyte has no effect on conducting power.
4. Water is the continuous phase. Oil is the continuous phase.
5. Basic metal sulfates and water soluble alkali metal soaps are used as emulsifiers. Water insoluble soaps, such as those of Zn, Al, Fe and alkaline earth metals, are used as emulsifiers.

Rows 2 and 3 of Table 11.7 are really two laboratory tests for deciding which type you have. Extra dispersion medium mixes in freely while extra dispersed phase does not, so the liquid that dilutes the emulsion is the medium. Similarly, an electrolyte can only conduct if it dissolves in a continuous phase of water, so an emulsion made conducting by a pinch of salt must be of the oil-in-water type.

Properties of emulsion:

  1. An emulsion can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid, when mixed, forms a separate layer.

  2. The droplets in emulsions are often negatively charged and can be precipitated by electrolytes.

  3. Emulsions show Brownian movement and the Tyndall effect.

  4. The two liquids in emulsions can be separated by heating, freezing or centrifuging, etc.

The third property is a reminder that an emulsion is a colloid in the full sense, obeying everything established in Section 11.9.5. The fourth is why cream separates when milk is spun in a centrifuge, and why an emulsified sauce breaks if it is overheated.

(question only — no solution)

Butter and milk both consist of fat and water. Classify each as an oil-in-water or a water-in-oil emulsion, and describe one test based on Table 11.7 by which you could confirm your classification in the laboratory.

11.9.8 Applications of Colloids

Colloids find applications in industry and in daily life. The following are some examples.

Electrical precipitation of smoke. Smoke is a colloidal solution of solid particles of carbon, arsenic compounds, dust, etc. in air. When smoke is allowed to pass through a chamber containing plates having charge, the smoke particles lose their charge and get precipitated. The particles settle down on the floor of the chamber. The precipitator is called a Cottrell precipitator, and its arrangement is shown in Fig. 11.27.

Diagram

Fig. 11.27 – Cottrell precipitator: a chimney-like chamber with charged plates or electrodes mounted inside, smoke entering at the bottom, particles losing their charge at the plates and falling to the floor of the chamber, and clean gas leaving at the top. Label “smoke in”, “charged plates”, “precipitated particles”, “clean gas out”.

Note that this is coagulation by discharging the particles, exactly as in Section 11.9.6, carried out on an industrial scale and in a gaseous medium rather than a liquid one.

Purification of drinking water. Water obtained from natural sources contains colloidal impurities. By the addition of alum to such water, colloidal impurities get coagulated and settle down. This makes the water potable.

Medicines. Usually medicines are colloidal in nature. Colloidal medicines are more effective owing to the large surface area to volume ratio of a colloidal particle, and easy assimilation. Argyrol is a silver sol used as an eye lotion. Milk of magnesia, an emulsion, is used in stomach disorders.

Rubber industry. Rubber is obtained by coagulation of latex.

Cleansing action of soaps and detergents. The micelles of Fig. 11.5 take the grease into their hydrocarbon interiors while presenting water-friendly heads to the surrounding water, so the grease can be carried away with the wash water.

Photographic plates and films, and industrial products like paints, inks, synthetic plastics, rubber, graphite lubricants and cement, are colloids.

A spherical particle has radius \(r\). Show that its surface area to volume ratio is \(3/r\), and calculate the ratio for (a) a sphere of radius 1.0 cm and (b) a colloidal particle of radius 100 nm. Comment on why colloidal medicines act more effectively than the same substance in larger lumps.

Solution.

Step 1: obtain the ratio. For a sphere of radius \(r\), the surface area is \(A = 4\pi r^2\) and the volume is \(V = \dfrac{4}{3}\pi r^3\). Therefore

\[\frac{A}{V} = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}\]

Everything else cancels, so the ratio depends only on the radius — and it grows as the radius shrinks.

Step 2: the large sphere. With \(r = 1.0\ \text{cm}\),

\[\frac{A}{V} = \frac{3}{1.0\ \text{cm}} = 3.0\ \text{cm}^{-1}\]

Step 3: the colloidal particle. First convert the radius to centimetres: \(100\ \text{nm} = 100 \times 10^{-9}\ \text{m} = 1.0\times10^{-7}\ \text{m} = 1.0\times10^{-5}\ \text{cm}\). Then

\[\frac{A}{V} = \frac{3}{1.0\times10^{-5}\ \text{cm}} = 3.0\times10^{5}\ \text{cm}^{-1}\]

Step 4: compare. The colloidal particle has a surface area to volume ratio \(10^{5}\) times larger than the centimetre-sized sphere.

Comment. A medicine acts where its surface meets the body’s fluids. For the same total mass of substance, a colloidal preparation offers a hundred thousand times more surface for dissolution and absorption than a coarse lump would, so it is assimilated far more readily. This is the same argument used in Section 11.4 for finely divided adsorbents, applied now to a therapeutic dose.

Activity: Calculate the surface area to volume ratio of a spherical particle. See how the ratio increases with the reduction of the radius of the particle. Plot the ratio against the radius.

(question only — no solution)

Calculate the surface area to volume ratio of a spherical colloidal particle of radius 50 nm, and state how many times greater it is than that of a spherical particle of radius 5 μm made of the same material.

Collect information about surface chemistry, about adsorption, and about Brownian motion.

11.10 Summary

  1. Particles at the surface of a solid or liquid have unbalanced forces, because they have neighbours on some sides only. This unsaturation makes surfaces attract gases and dissolved substances, which then accumulate on the surface. This accumulation is called adsorption.

  2. The substance that provides the surface is the adsorbent; the substance that gathers on it is the adsorbate. Adsorption is caused by London dispersion or van der Waals forces, which are short range and additive.

  3. Surface tension is the energy required to increase the surface of a liquid by unit area. A system tends towards minimum surface tension, that is, towards lower free energy, and adsorption lowers the free energy of a surface.

  4. Adsorption is a surface phenomenon; absorption is a bulk phenomenon in which the substance is distributed uniformly through the material. When both occur together, as with a chalk dipped in ink, the process is called sorption. Removal of an adsorbed substance from a surface is desorption.

  5. Physisorption is adsorption by weak van der Waals forces. It is not specific, has a low heat of adsorption in the range 20 – 40 kJ mol\(^{-1}\), is favoured at low temperature, reaches equilibrium rapidly, is reversible, and can form a multimolecular layer at high pressure.

  6. Chemisorption is adsorption by chemical bonds, covalent or ionic. It is highly specific, has a higher heat of adsorption in the range 40 – 200 kJ mol\(^{-1}\), needs a large activation energy and so is called activated adsorption, forms only a monomolecular layer, and is irreversible. It increases with temperature at first and then decreases as the bonds break. The adsorption of hydrogen on glass is endothermic, because the hydrogen must first dissociate.

  7. Five factors govern the adsorption of a gas on a solid: the nature of the adsorbate, where a gas of higher critical temperature is adsorbed more; the nature of the adsorbent, where porous solids such as silica gel and charcoal are effective; the surface area of the adsorbent, which increases adsorption; temperature, which decreases adsorption since the process is exothermic and, by Le Chatelier’s principle, is favoured at low temperature; and pressure, which increases adsorption until the surface is nearly covered and the extent becomes independent of pressure.

  8. An adsorption isotherm is the relation between the amount adsorbed per unit mass of adsorbent and the equilibrium pressure of a gas, or the concentration of a solution, at constant temperature.

  9. The Freundlich adsorption isotherm is an empirical relation valid over a limited range of pressures:

\[\frac{x}{m} = k\,P^{1/n} \qquad (n > 1) \qquad \ldots (11.1)\]

  1. For adsorption from solution the pressure is replaced by concentration, and taking logarithms gives a straight-line form:

\[\frac{x}{m} = k\,C^{1/n} \qquad \ldots (11.2)\]

\[\log \frac{x}{m} = \log k + \frac{1}{n}\log C \qquad \ldots (11.3)\]

  1. A plot of \(\log \dfrac{x}{m}\) against \(\log C\) or \(\log P\) is a straight line whose slope is \(\dfrac{1}{n}\) and whose intercept on the y-axis is \(\log k\). The factor \(\dfrac{1}{n}\) lies between 0 and 1. When \(\dfrac{1}{n} \to 0\), \(\dfrac{x}{m}\) is constant and adsorption is independent of pressure; when \(\dfrac{1}{n} = 1\), \(\dfrac{x}{m} = kP\) and adsorption varies directly with pressure. Experimental isotherms tend to saturate at high pressure.

  2. Adsorption is applied in heterogeneous catalysis, in gas masks of activated charcoal, in the control of humidity by silica and alumina gels, in producing a high vacuum with charcoal cooled by liquid air, in adsorption indicators such as eosin and fluorescein for precipitation titrations, in separating noble gases on coconut charcoal, in the froth flotation of sulphide ores using pine oil, and in chromatographic analysis on silica or alumina gel. Surface area determination and the purification of water are further applications.

  3. A catalyst is a substance which, added to a reacting system, increases the rate of a reaction without itself undergoing any permanent chemical change. Catalysts lower the reaction temperature and hence the energy cost of a process.

  4. In homogeneous catalysis the catalyst and reactants are in the same phase, as with iodide ions in the decomposition of aqueous hydrogen peroxide, nitric oxide in the oxidation of \(\mathrm{SO_2}\) to \(\mathrm{SO_3}\) in the lead chamber process, and \(\mathrm{H^{+}}\) from sulphuric acid in the hydrolysis of sucrose to glucose and fructose. Enzyme catalysis is an important type of homogeneous catalysis.

  5. In heterogeneous catalysis the catalyst and reactants are in different phases, the catalyst usually being a solid on whose surface the reaction occurs. Examples are finely divided iron in the Haber process for ammonia, nickel, palladium or platinum in the hydrogenation of vegetable oils to vanaspati ghee, and the metal catalyst of an automobile catalytic converter.

  6. In the Haber process \(\mathrm{Al_2O_3}\) and \(\mathrm{K_2O}\) act as promoters of the iron catalyst: \(\mathrm{Al_2O_3}\) prevents fusion of the iron particles and \(\mathrm{K_2O}\) causes chemisorption of nitrogen atoms. Molybdenum is also used as a promoter. A catalytic converter is poisoned by adsorbed lead, which is why unleaded petrol must be used.

  7. Inhibitors decrease the rate of a reaction. Ethanol added to chloroform inhibits its air oxidation to the poisonous carbonyl chloride, and dilute acid or glycerol inhibits the decomposition of hydrogen peroxide.

  8. The adsorption theory of heterogeneous catalysis explains catalytic action in five steps on the catalyst surface: diffusion of reactants to the surface, adsorption of the reactant molecules, chemical reaction on the surface with formation of an intermediate, formation of products, and desorption of the products. Fresh reactant molecules then replace the products and the cycle repeats, which is why the catalyst is unchanged in mass and composition at the end.

Flowchart

FC 11.7 – Chapter overview: unbalanced forces at a surface → adsorption (adsorbent + adsorbate) → branch A: types of adsorption → physisorption / chemisorption → factors affecting adsorption → adsorption isotherm (Freundlich) → applications of adsorption; branch B: adsorption on a solid surface → heterogeneous catalysis → five-step catalytic cycle (diffusion → adsorption → reaction and intermediate → product formation → desorption → back to diffusion) → features of solid catalysts (activity, selectivity, shape selectivity); branch C: large surface of finely divided matter → colloids → classification → preparation → purification → properties → coagulation → emulsions → applications.

  1. The catalytic activity of a solid depends on the strength of chemisorption, which must be neither too weak, or the reactant is not held, nor too strong, or the adsorbed molecules are immobilized and the products cannot desorb. The d-block metals Fe, V and Cr are strongly active towards \(\mathrm{O_2}\), \(\mathrm{C_2H_2}\), \(\mathrm{C_2H_4}\), \(\mathrm{CO}\), \(\mathrm{H_2}\), \(\mathrm{CO_2}\) and \(\mathrm{N_2}\); Mn and Cu cannot adsorb \(\mathrm{N_2}\) and \(\mathrm{CO_2}\); Mg and Li adsorb \(\mathrm{O_2}\) selectively.

  2. Solid catalysts show catalytic selectivity: the same reactants give different products over different catalysts. Ethylene and dioxygen give acetaldehyde over \(\mathrm{Pd/Al_2O_3}\) but ethylene oxide over \(\mathrm{Ag/Al_2O_3}\); carbon monoxide with hydrogen gives methane and water over nickel but methanol over \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\).

  3. Zeolites are aluminosilicates with a three-dimensional silicate network in which some silicon atoms are replaced by aluminium, giving an Al–O–Si framework and a microporous structure. Because reaction depends on the size and shape of the reactant and product molecules relative to the pores and cavities, zeolites are shape selective catalysts. The zeolite ZSM-5 converts alcohols directly to gasoline by dehydration.

  4. Colloids are heterogeneous mixtures in which the dispersed phase is distributed through the dispersion medium, the medium being the component present in the largest proportion. Colloidal particles are larger than molecules but too small to be seen with the naked eye.

  5. Particle size is the essential difference between the three kinds of mixture. Solutions contain particles of diameter 0.1 to 2 nm and are transparent; colloids contain particles of diameter 2 to 500 nm and are translucent; neither separates on standing, whereas a suspension does.

  6. Everyday examples of colloids include the blue colour of the sky, blood as a dispersion of plasma proteins and antibodies in water, fertile soils in which humus acts as a protective colloid, and fog and mist formed when moisture condenses on dust particles below the dew point.

  7. Colloids are classified in three ways. By physical state, giving solid sols, sols and gels, aerosols, gels, emulsions and foams. By affinity of the phases, giving lyophilic colloids, which are reversible, self-stabilized and need a large amount of electrolyte to coagulate, and lyophobic colloids, which are irreversible, need stabilizers and coagulate with a small amount of electrolyte; when the medium is water the terms hydrophilic and hydrophobic are used. By molecular size, giving multimolecular colloids such as gold sol and \(\mathrm{S_8}\) held by van der Waals forces, macromolecular colloids such as starch, cellulose, proteins, polythene and nylon, and associated colloids or micelles such as soaps and detergents.

  8. A soap molecule has a hydrophobic hydrocarbon tail and a hydrophilic ionic carboxylate head. In water the molecules form spherical micelles with the tails pointing inward and the heads on the surface, which makes the dispersion stable and explains the cleansing action of soaps and detergents.

  9. Colloids are prepared by chemical methods using double decomposition, oxidation, reduction or hydrolysis:

\[\mathrm{SO_2} + 2\,\mathrm{H_2S} \xrightarrow{\ \text{Oxidation}\ } 3\,\mathrm{S} \downarrow + 2\,\mathrm{H_2O}\]

\[2\,\mathrm{AuCl_3} + 3\,\mathrm{HCHO} + 3\,\mathrm{H_2O} \xrightarrow{\ \text{Reduction}\ } 2\,\mathrm{Au} \downarrow + 3\,\mathrm{HCOOH} + 6\,\mathrm{HCl}\]

\[\mathrm{FeCl_3} + 3\,\mathrm{H_2O} \xrightarrow{\ \text{Hydrolysis}\ } \mathrm{Fe(OH)_3} \downarrow + 3\,\mathrm{HCl}\]

  1. In Bredig’s arc method an electric arc struck between metal electrodes immersed in the dispersion medium vaporises the metal, which condenses to colloidal particles; sols of gold, silver and platinum are made this way. In peptization a precipitate is shaken with the dispersion medium and a small amount of electrolyte, called the peptizing agent; the precipitate adsorbs one of the ions, acquires a charge and breaks up into colloidal particles.

  2. A sol needs a small quantity of electrolyte for stability but is coagulated by a large quantity. Purification of a colloidal solution reduces electrolytes and soluble impurities to the requisite minimum, and is carried out by dialysis, in which a dissolved substance diffuses out through a suitable membrane in a dialyser while the colloidal particles are retained.

  3. Colloidal systems are heterogeneous and consist of two phases. Their particles pass slowly through parchment paper or animal membrane but readily through ordinary filter paper, and are usually not detectable by a powerful microscope.

  4. The Tyndall effect is the scattering of light by colloidal particles, making the path of the beam visible at right angles as a Tyndall cone. It requires that the particle diameter be not much smaller than the wavelength of light used, and that the refractive indices of the two phases differ largely. It is used to find the number and size of the particles and to distinguish a colloid from a true solution.

  5. The colour of a colloid depends on the wavelength of light scattered, on whether the light is scattered or transmitted, and on particle size. Dilute milk appears blue by scattered light and red by transmitted light, and the finest gold sol is red while larger particles appear purple.

  6. Brownian motion is the ceaseless random zig-zag motion of colloidal particles, first observed by Robert Brown in pollen grains and explained by Albert Einstein in 1905. It is caused by constant collisions of the dispersed particles with fast-moving molecules of the dispersion medium, from which the particles acquire kinetic energy.

  7. Colloidal particles carry an electric charge of the same sign throughout a given sol, arising from preferential adsorption of ions. Electrophoresis is the movement of these particles under an applied electric potential, positive particles moving to the cathode and negative particles to the anode; it is used to find the sign of the charge, to measure the rate of migration, and to separate a mixture of colloidal particles. Electroosmosis is the movement of the dispersion medium in an electric field when the particles are prevented from moving.

  8. Coagulation is the precipitation of a colloid by removal of the charge on its particles. Lyophobic sols are coagulated by electrophoresis, by mixing two oppositely charged sols in nearly equal proportions, which is mutual coagulation, by boiling, by persistent dialysis and by adding an excess of electrolyte. Lyophilic sols also carry a layer of dispersion medium which must be removed as well.

  9. By the Hardy–Schulze rule, the greater the valence of the flocculating ion, the greater its power to cause precipitation. For a negative sol the order is \(\mathrm{Al^{3+}} > \mathrm{Ba^{2+}} > \mathrm{Na^{+}}\), and for a positive sol it is \([\mathrm{Fe(CN)_6}]^{4-} > \mathrm{PO_4^{3-}} > \mathrm{SO_4^{2-}} > \mathrm{Cl^{-}}\).

  10. An emulsion is a colloidal system of one liquid dispersed in another immiscible liquid. In the oil-in-water type, such as milk, vanishing cream and paint, water is the continuous phase, a small amount of electrolyte makes the emulsion conducting, and water-soluble alkali metal soaps or basic metal sulfates are the emulsifiers. In the water-in-oil type, such as cod liver oil, butter and cream, oil is the continuous phase, an electrolyte has no effect on conducting power, and water-insoluble soaps of Zn, Al, Fe and alkaline earth metals are the emulsifiers.

  11. An emulsion can be diluted with any amount of the dispersion medium, its droplets are often negatively charged and can be precipitated by electrolytes, it shows Brownian movement and the Tyndall effect, and its two liquids can be separated by heating, freezing or centrifuging.

  12. Colloids are applied in the electrical precipitation of smoke in a Cottrell precipitator, in making water potable by coagulating colloidal impurities with alum, in medicines such as the silver sol Argyrol and the emulsion milk of magnesia, in obtaining rubber by coagulation of latex, in the cleansing action of soaps and detergents, and in photographic plates and films as well as paints, inks, plastics, rubber, graphite lubricants and cement.

11.11 Check Your Concepts

  1. Why does a molecule lying at the surface of a liquid experience a net inward pull, while a molecule in the interior does not? Use your answer to explain why surfaces adsorb at all.

  2. A student writes that ink is “adsorbed” by blotting paper. Correct the statement and justify your correction using the difference between a surface phenomenon and a bulk phenomenon.

  3. Assertion: Adsorption is always accompanied by a decrease in the free energy of the surface. Reason: Adsorption uses up some of the unsatisfied forces of the surface particles. State whether each statement is true, and whether the reason correctly explains the assertion.

  4. Both physisorption and chemisorption release heat, yet only chemisorption needs an activation energy. Explain why, and state what this implies about the effect of a rise in temperature on each.

  5. The adsorption of hydrogen on glass is endothermic, although chemisorption is usually exothermic. Account for this exception.

  6. A gas is found to form a multimolecular layer on a solid at high pressure. Which type of adsorption is this, and why can the other type not form more than one layer?

  7. Arrange the gases \(\mathrm{H_2}\), \(\mathrm{N_2}\), \(\mathrm{NH_3}\) and \(\mathrm{SO_2}\) in increasing order of the extent to which they are adsorbed on a given mass of activated charcoal at constant temperature and pressure, and state the property of the gas that decides the order.

  8. Assertion: Adsorption of a gas on charcoal decreases as the temperature is raised. Reason: Adsorption is an exothermic process. Comment on both statements and name the principle that connects them.

  9. Explain why the extent of adsorption of a gas on a solid becomes independent of pressure at high pressure, and identify the value approached by \(1/n\) in the Freundlich equation in this region.

  10. Why is the Freundlich adsorption isotherm described as an empirical relation, and what limitation does this place on its use?

  11. Explain the purpose of plotting \(\log(x/m)\) against \(\log P\) instead of plotting \(x/m\) against \(P\), and state what the slope and the intercept of the resulting line give.

  12. In a precipitation titration of sodium chloride against silver nitrate, fluorescein remains pale yellow for most of the titration and turns reddish pink only at the end point. Explain this behaviour in terms of preferential adsorption.

  13. In the froth flotation process, the sulphide ore rises into the froth while silica settles at the bottom. Explain the role of pine oil, and state why no chemical change of the ore is involved.

  14. A catalyst increases the rate of a reaction but does not change the amount of product obtainable at equilibrium. Explain this using the idea of activation energy, and state what a catalyst does not alter in a reaction-coordinate diagram.

  15. Classify each of the following as homogeneous or heterogeneous catalysis, giving a reason based on phases: (a) hydrolysis of sucrose in the presence of sulphuric acid; (b) hydrogenation of vegetable oil over nickel; (c) oxidation of \(\mathrm{SO_2}\) by \(\mathrm{O_2}\) in the presence of nitric oxide; (d) removal of \(\mathrm{CO}\) from car exhaust.

  16. Distinguish between a promoter, a catalyst poison and an inhibitor, giving one example of each from this chapter.

  17. Complete the following, writing the balanced equation and naming the product in each case: (a) \(\mathrm{C_2H_4} + \mathrm{O_2}\) over \(\mathrm{Pd/Al_2O_3}\); (b) \(\mathrm{C_2H_4} + \mathrm{O_2}\) over \(\mathrm{Ag/Al_2O_3}\); (c) \(\mathrm{CO} + \mathrm{H_2}\) over Ni; (d) \(\mathrm{CO} + \mathrm{H_2}\) over \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\).

  18. A solid adsorbs a reactant gas so strongly that the adsorbed molecules cannot move on its surface. Explain, with reference to the five steps of heterogeneous catalysis, why this solid is a poor catalyst.

  19. Why are zeolites described as shape selective catalysts? Explain how their selectivity differs in origin from the selectivity shown by \(\mathrm{Pd/Al_2O_3}\) and \(\mathrm{Ag/Al_2O_3}\) towards ethylene.

  20. Match each item in column I with the correct item in column II.

Column I Column II
a. \(\mathrm{K_2O}\) in the Haber process i. inhibitor
b. lead in a catalytic converter ii. shape selective catalyst
c. ethanol added to chloroform iii. promoter
d. ZSM-5 iv. catalyst poison
  1. Both a true solution and a colloid remain unchanged on standing, yet only one of them shows a visible cone when a beam of light is passed through it. Name the effect, state the two conditions necessary for it to be observed, and explain why one system fails to show it.

  2. A colloidal sol passes readily through ordinary filter paper but only slowly through parchment paper. Explain, and name the purification technique that makes use of this difference.

  3. A small quantity of an electrolyte stabilises a sol while a large quantity coagulates it. Explain this apparent contradiction, and state why persistent dialysis also destroys a sol.

  4. Explain why a lyophilic sol needs a large amount of electrolyte to coagulate, whereas a lyophobic sol needs only a small amount.

  5. For each of the following, state the sign of the charge on the sol and name the ion that would act as the flocculating ion: (a) \(\mathrm{As_2S_3}\) sol, (b) \(\mathrm{Fe(OH)_3}\) sol, (c) gold sol, (d) haemoglobin.

  6. Arrange \(\mathrm{NaCl}\), \(\mathrm{Na_2SO_4}\) and \(\mathrm{Na_3PO_4}\) in increasing order of coagulating power for a positively charged sol, and name the rule you have applied.

  7. Assertion: Colloidal particles do not settle out on standing. Reason: The particles are kept in continuous random motion by collisions with the molecules of the dispersion medium. State whether each is true and whether the reason explains the assertion, and name the motion referred to.

  8. Distinguish between electrophoresis and electroosmosis. State one application of electrophoresis that depends on different particles migrating at different rates.

  9. Milk and butter contain the same two substances. Classify each as an oil-in-water or water-in-oil emulsion, and describe two tests that would let you tell the two types apart in the laboratory.

  10. Explain how a soap molecule is able to remove grease from cloth, referring to the structure of a micelle and to the terms hydrophobic and hydrophilic.

  11. Explain, in terms of coagulation, (a) why alum is added to muddy water to make it potable, and (b) why smoke can be removed from flue gas by passing it between charged plates.

  12. Classify each of the following as multimolecular, macromolecular or associated colloid, giving a reason: (a) gold sol, (b) starch solution, (c) soap solution above a certain concentration, (d) a sol of \(\mathrm{S_8}\) molecules.

  13. Name the type of adsorption in which van der Waals forces are present.

  14. Name the type of adsorption in which a compound is formed.

  15. Write an equation for the Freundlich adsorption isotherm.

  16. Define the terms —

    1. Inhibition
    2. Electrophoresis
    3. Catalysis.
  17. Define adsorption. Why can students read a blackboard written on with chalk?

  18. Write the characteristics of adsorption.

  19. Distinguish between lyophobic and lyophilic sols.

  20. Identify the dispersed phase and the dispersion medium in the following colloidal dispersions —

    1. milk
    2. blood
    3. printing ink
    4. fog.
  21. Write notes on —

    1. Tyndall effect
    2. Brownian motion
    3. Types of emulsion
    4. Hardy–Schulze rule.
  22. Explain electrophoresis in brief with the help of a diagram. What are its applications?

  23. Explain why a finely divided substance is more effective as an adsorbent.

  24. What is the adsorption isotherm?

  25. An aqueous solution of raw sugar, when passed over beds of animal charcoal, becomes colourless. Explain.

  26. What happens when a beam of light is passed through a colloidal sol?

  27. Mention the factors affecting adsorption of a gas on solids.

  28. Give four uses of adsorption.

  29. Explain Bredig’s arc method.

  30. Explain the term emulsions and the types of emulsions.

  31. Explain the following:

    1. A finely divided substance is more effective as an adsorbent.
    2. Freundlich adsorption isotherm, with the help of a graph.
  32. Distinguish between the following:

    1. Adsorption and absorption. Give one example.
    2. Physisorption and chemisorption. Give one example.
  33. Adsorption is a surface phenomenon. Explain.

  34. Explain how the adsorption of a gas on a solid varies with —

    1. nature of adsorbate and adsorbent
    2. surface area of adsorbent.
  35. Explain two applications of adsorption.

  36. Explain micelle formation in soap solution.

  37. Draw labelled diagrams of the following —

    1. Tyndall effect
    2. Dialysis
    3. Bredig’s arc method
    4. Soap micelle.

11.12 Practice With Numericals

  1. For the adsorption of a gas on charcoal at constant temperature, a plot of \(\log(x/m)\) against \(\log P\) is a straight line of slope 0.25 with an intercept of 1.0 on the \(\log(x/m)\) axis, \(P\) being in kPa. Calculate (a) the values of \(n\) and \(k\) and (b) the value of \(x/m\) when the equilibrium pressure is 16 kPa.

  2. At a fixed temperature, 1.0 g of an adsorbent adsorbs 0.80 g of a gas at an equilibrium pressure of 25 kPa. For this system \(1/n = 0.5\). Calculate the mass of gas adsorbed by 1.0 g of the same adsorbent at an equilibrium pressure of 100 kPa.

  3. A cube of a solid adsorbent of edge length 3.0 cm is ground into cubes of edge length 0.030 cm. Calculate (a) the number of small cubes formed, (b) the total surface area before and after grinding, and (c) the factor by which the surface area has increased. State whether the volume changes.

  4. Show that the surface area to volume ratio of a sphere of radius \(r\) is \(3/r\). Hence calculate this ratio for (a) a spherical particle of radius 2.0 mm and (b) a colloidal particle of radius 20 nm, and find how many times larger the ratio is for the colloidal particle.

  5. 30.0 cm\(^3\) of a sodium chloride solution containing a little fluorescein required 25.0 cm\(^3\) of 0.0500 M silver nitrate solution to reach the end point. Calculate (a) the molarity of the sodium chloride solution and (b) the mass of silver chloride precipitated. Molar mass of \(\mathrm{AgCl} = 143.5\ \text{g mol}^{-1}\).

  6. 1.0 g of activated charcoal adsorbs 380 cm\(^3\) of \(\mathrm{SO_2}\) measured at STP. Calculate (a) the number of moles and (b) the mass of \(\mathrm{SO_2}\) adsorbed per gram of charcoal. Molar mass of \(\mathrm{SO_2} = 64\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).

  7. A sample of vegetable oil contains 0.25 mol of molecules, each carrying two carbon–carbon double bonds. Calculate (a) the number of moles of hydrogen needed for complete hydrogenation, (b) the mass of hydrogen required and (c) the volume this hydrogen occupies at STP. Molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).

  8. 14 g of carbon monoxide is completely converted to methanol by reaction with hydrogen over a \(\mathrm{Cu/ZnO}\text{-}\mathrm{Cr_2O_3}\) catalyst. Calculate (a) the volume of hydrogen consumed at STP and (b) the mass of methanol formed. Molar masses: \(\mathrm{CO} = 28\ \text{g mol}^{-1}\), \(\mathrm{CH_3OH} = 32\ \text{g mol}^{-1}\); molar volume at STP \(= 22.4\ \text{L mol}^{-1}\).

  9. When 2.0 mol of a gas is adsorbed on the surface of a metal, 240 kJ of heat is released. Calculate the heat of adsorption per mole of the gas, and state, with reason, whether the process is physisorption or chemisorption.

  10. The particles of a certain dispersion have a diameter of 250 nm. Express this diameter in metres, and state with reason whether the dispersion is a true solution, a colloid or a suspension.

  11. A gold sol is prepared by the reduction of \(\mathrm{AuCl_3}\) using methanal, as given in the chapter. Calculate the mass of gold obtained when 0.020 mol of \(\mathrm{AuCl_3}\) is completely reduced, and the number of moles of methanoic acid formed at the same time. Molar mass of \(\mathrm{Au} = 197\ \text{g mol}^{-1}\).

11.13 Objective Questions

  1. The size of colloidal particles lies between —
    1. \(10^{-10}\) m and \(10^{-9}\) m
    2. \(10^{-9}\) m and \(10^{-6}\) m
    3. \(10^{-6}\) m and \(10^{-4}\) m
    4. \(10^{-5}\) m and \(10^{-2}\) m
  2. Gum in water is an example of —
    1. true solution
    2. suspension
    3. lyophilic sol
    4. lyophobic sol
  3. In the Haber process of production of ammonia, \(\mathrm{K_2O}\) is used as —
    1. catalyst
    2. inhibitor
    3. promotor
    4. adsorbate
  4. Fruit jam is an example of —
    1. sol
    2. gel
    3. emulsion
    4. true solution