Chapter 12: Chemical Equilibrium
In the closing years of the eighteenth century, the French chemist Claude Louis Berthollet travelled to Egypt with Napoleon’s expedition. Along the edges of certain salt lakes he found thick white crusts of sodium carbonate — the mineral known there as natron. This puzzled him. In a laboratory, sodium carbonate solution poured into calcium chloride solution gives a precipitate of calcium carbonate, and the reaction seems to go only that way. Yet in the desert, nature appeared to be running the same reaction backwards: limestone in contact with a concentrated salt solution was producing sodium carbonate.
Berthollet drew a conclusion that changed chemistry. A reaction is not locked into one direction by its own nature. The direction it actually takes depends on how much of each substance is present. With a vast excess of salt water standing against limestone for years, the “reverse” reaction could be pushed forward until it built up visible deposits.
That single insight — that chemical change can run both ways, and that quantities decide which way — later grew into the law of mass action, worked out mathematically by Guldberg and Waage, and into Le Chatelier’s principle, which predicts how a balanced system answers back when it is disturbed. The same idea today governs how ammonia is manufactured on an industrial scale, how carbon dioxide is carried in your blood, and why an opened bottle of soda goes flat. A comparison of the two situations Berthollet was thinking about is shown in Fig. 12.0.
Diagram
Fig. 12.0 – Two panels side by side: (left) a test tube where Na₂CO₃ solution + CaCl₂ solution gives a white CaCO₃ precipitate, arrow pointing right; (right) a desert salt lake with limestone rock at the shoreline and white natron crust forming, arrow pointing left; label both arrows “forward” and “reverse” and add the caption “the same reaction, opposite directions”.
Keep these questions in mind as you read. By the end of this chapter you will be able to reason out the answer to each one for yourself.
12.1 Introduction
You have already learnt that a change may be physical or chemical. In a physical change the substance itself stays the same and only its state, shape or distribution changes. In a chemical change the substances present are actually converted into new substances with a different composition.
You have also learnt that a change may be reversible or irreversible. A reversible change can be made to go back to its starting condition; an irreversible change cannot. Water falling down a slope, smoke spreading through a room, and fragrance diffusing through the air are all changes of one kind: no new substance is formed, and none of them turns around and goes back on its own. All the changes listed above are therefore irreversible physical changes.
A chemical change can be written down as a chemical reaction once we know the exact chemical composition of the reactants and the products. That is what a balanced chemical equation records — which substances are consumed, which are formed, and in what mole ratio.
In this chapter we are going to look closely at a particular class of chemical changes: reversible chemical reactions. These are reactions that can travel in both directions, and studying them will lead us to one of the most useful ideas in chemistry — chemical equilibrium.
12.1.1 Reversible reaction
A colour change you can reverse
A simple activity makes the idea visible. Dissolve 4 g of cobalt chloride in 40 ml of water. A reddish pink solution is formed. Now add 60 ml of concentrated HCl to it; the solution turns violet.
Take 5 ml of this violet solution in a test tube and place the test tube in a beaker containing an ice–water mixture. The colour of the solution becomes pink. Place the same test tube in a beaker containing water at 90 °C. The colour of the solution turns blue. The two test tubes are drawn side by side in Fig. 12.1.1.
Diagram
Fig. 12.1.1 – Two test tubes of cobalt chloride solution: left tube standing in a beaker of ice–water showing pink solution labelled Co(H₂O)₆²⁺, right tube standing in a beaker of hot water at 90 °C showing blue solution labelled CoCl₄²⁻; between them two horizontal arrows, upper arrow pointing right labelled “Heat” and lower arrow pointing left labelled “Cool”.
Cooling brings the pink colour back; heating brings the blue colour back. The colour can be driven either way, again and again, simply by changing the temperature. The change in colour of the solution is caused by a chemical reaction which reverses its direction with change of temperature.
\[\mathrm{Co(H_2O)_6^{2+}}\text{(aq)} + 4\,\mathrm{Cl^{-}}\text{(aq)} \underset{\text{Cool}}{\overset{\text{Heat}}{\rightleftharpoons}} \mathrm{CoCl_4^{2-}}\text{(aq)} + 6\,\mathrm{H_2O}\text{(l)}\]
Here the pink colour belongs to \(\mathrm{Co(H_2O)_6^{2+}}\) and the blue colour belongs to \(\mathrm{CoCl_4^{2-}}\). Two different cobalt species exist in the same solution, and the temperature decides which of them dominates.
The reaction in the above activity is an example of a reversible reaction.
Reactions that travel only one way
There are many chemical reactions which appear to proceed in a single direction. For example:
\[\mathrm{C}\text{(s)} + \mathrm{O_2}\text{(g)} \xrightarrow{\ \Delta\ } \mathrm{CO_2}\text{(g)}\]
\[2\,\mathrm{KClO_3}\text{(s)} \xrightarrow{\ \Delta\ } 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}\]
(The symbol \(\Delta\) over the arrow simply means that the reaction is carried out on heating.) These are called irreversible reactions. They proceed only in a single direction until one of the reactants is exhausted. Their direction is indicated by an arrow (\(\rightarrow\)) pointing towards the products in the chemical equation.
On the contrary, reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.
A reversible reaction is denoted by drawing two arrows, such that one arrow points in the forward direction and the other in the reverse direction (\(\rightleftharpoons\)). These two arrows are also referred to as a double arrow. Seeing a double arrow in an equation should immediately tell you that neither substance in that equation is ever fully used up.
For example:
\[\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\]
\[\mathrm{CH_3COOH}\text{(aq)} + \mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{CH_3COO^{-}}\text{(aq)} + \mathrm{H_3O^{+}}\]
The second example is worth a moment’s thought. Acetic acid does not hand over all its protons to water; some molecules ionise, some ions recombine, and both processes keep running side by side. That is exactly why acetic acid behaves as a weak acid.
Open, closed and isolated systems
Consider an example of decomposition of calcium carbonate. Calcium carbonate, when heated strongly, decomposes to form calcium oxide and carbon dioxide. Let us consider what happens if this reaction is carried out in a closed container or system, and what happens in an open container or system. Before that, we need to be clear about what these words mean, because in chemistry a “system” simply means the part of the universe we have chosen to study, and everything else is the surroundings.
Reactions are chemically represented as follows.
General representation:
\[\mathrm{CaCO_3}\text{(s)} \xrightarrow{\ \text{heat}\ } \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\]
This represents the irreversible reaction in an open container. On the other hand, the reversible reaction in a closed container is represented as
\[\mathrm{CaCO_3}\text{(s)} \underset{\text{backward}}{\overset{\text{forward}}{\rightleftharpoons}} \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\]
The two routes are compared in FC 12.1.
Flowchart
FC 12.1 – Heating calcium carbonate: CaCO₃(s) + strong heating → branch point “is the vessel closed or open?”; closed vessel branch → CO₂ is retained → CaO + CO₂ recombine → CaCO₃, CaO and CO₂ all present together, amounts constant → equilibrium; open vessel branch → CO₂ escapes to the surroundings → no recombination possible → CaCO₃ completely decomposed → CaCO₃ cannot be obtained back.
What we observe in a closed vessel
If we perform the above reaction at high temperature in a closed system, we find that after a certain time we have some calcium carbonate present along with calcium oxide and carbon dioxide. The reaction has clearly not gone to completion, even though there was nothing to stop it.
If we continue the experiment over a longer period of time at the same temperature, we find that the concentrations of calcium carbonate, calcium oxide and carbon dioxide are unchanged. The reaction thus appears to have stopped, and we say the system has attained the equilibrium.
Actually, the reaction does not stop, but proceeds in both the directions with equal rates. In other words, calcium carbonate decomposes to give calcium oxide and carbon dioxide at a particular rate. Exactly at the same rate the calcium oxide and carbon dioxide recombine and form calcium carbonate.
Such reactions which do not go to completion and occur in both the directions simultaneously are reversible reactions. A reversible reaction may be represented in general terms as:
\[\underset{\text{reactants}}{\mathrm{A} + \mathrm{B}} \underset{\text{backward}}{\overset{\text{forward}}{\rightleftharpoons}} \underset{\text{products}}{\mathrm{C} + \mathrm{D}}\]
The double arrow indicates that the reaction is reversible.
What happens in an open vessel
Consider the reaction of decomposition of calcium carbonate occurring in an open system or container. It is seen that during decomposition of calcium carbonate in an open vessel, carbon dioxide can escape away. Hence calcium carbonate cannot be obtained back.
The reason is worth stating plainly: the backward reaction needs both of its reactants. Once the carbon dioxide has drifted out into the surroundings, calcium oxide has nothing left to combine with, so the reverse change simply cannot occur. Such a reaction is an irreversible reaction, which occurs only in one direction, namely, from reactants to products.
Notice how much depends on conditions rather than on the identity of the substances. The very same chemical change — limestone breaking up — is irreversible in one vessel and reversible in another. Examiners like this point, and it appears regularly as an assertion–reason question or as a “differentiate between reversible and irreversible reactions” question in board papers.
12.2 Equilibrium in Physical Processes
Equilibrium is not something that belongs to chemical reactions alone. A purely physical change can also settle into a state in which two opposing processes run at exactly the same rate, so that nothing appears to happen. These physical cases are the easiest place to see how equilibrium works, so we begin with them.
(a) Liquid – Vapour Equilibrium
Let us now look at a reversible physical process of evaporation of liquid water into water vapour in a closed vessel. The two stages of the experiment are shown in Fig. 12.1.
Diagram
Fig. 12.1 – Two identical stoppered conical flasks partly filled with water: the “Initially” flask shows only a few scattered vapour molecules above the water surface, the “Later” flask shows a dense, uniform population of vapour molecules; label the liquid “Water”, the space above it “Vapour”, and mark the two flasks “Initially” and “Later”.
Initially there is practically no vapour in the vessel. When a liquid evaporates in a closed container, the liquid molecules escape from the liquid surface into the vapour phase, building up vapour pressure. They also condense back into the liquid state because the container is closed — nothing can leave the flask.
In the beginning the rate of evaporation is high and the rate of condensation is low. That is simply because there are hardly any vapour molecules present yet to condense. But with time, as more and more vapour is formed, the rate of evaporation goes down and the rate of condensation increases.
Eventually the two rates become equal. This gives rise to a constant vapour pressure. This state is known as an ‘equilibrium state’.
In this state, the number of molecules leaving the liquid surface equals the number of molecules returning to the liquid from the vapour state. Across the interface, there is a lot of activity between the liquid and the vapour. This state, when the rate of evaporation is equal to the rate of condensation, is called equilibrium state. It may be represented as:
\[\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_2O}\text{(vapour)}\]
Saturated vapour pressure and boiling point
At equilibrium, the pressure exerted by the gaseous water molecules at a given temperature remains constant, known as the equilibrium vapour pressure of water (or saturated vapour pressure of water, or aqueous tension).
The saturated vapour pressure increases with increase of temperature. This makes sense: heating gives the molecules more kinetic energy, so more of them can break free of the liquid surface, and a denser vapour exerts a greater pressure.
In the case of water, the saturated vapour pressure is 1.013 bar (1 atm) at 100 °C. Therefore, water boils at 100 °C when exposed to 1 atm pressure. A liquid boils when its vapour can push back the atmosphere pressing down on it, and that happens only when the two pressures become equal.
For any pure liquid at 1 atm pressure, the temperature at which its saturated vapour pressure equals the atmospheric pressure is called the normal boiling point of that liquid.
The boiling point of water is 100 °C at 1.013 bar pressure, whereas the boiling point of ethyl alcohol is 78 °C. Ethyl alcohol reaches the same vapour pressure at a much lower temperature because its molecules are held together less tightly than water molecules.
Diagram
Fig. 12.2.1 – Graph with time on the x-axis and rate on the y-axis: one curve starting high and falling to a steady value (rate of evaporation), a second curve starting at zero and rising to the same steady value (rate of condensation); mark the point where they meet with a dashed vertical line labelled “equilibrium attained (rates are equal)”.
(b) Solid – Liquid Equilibrium
Consider a mixture of ice and water in a perfectly insulated thermos flask at 273 K. This is an isolated system, since neither matter nor heat can pass in or out. It is an example of solid–liquid equilibrium.
Ice and water are at constant temperature. They remain in what is called solid–liquid equilibrium. Here too nothing has actually stopped: ice keeps melting and water keeps freezing, but at exactly matching rates, so the mass of ice and the mass of water both stay the same.
\[\mathrm{H_2O}\text{(s)} \rightleftharpoons \mathrm{H_2O}\text{(l)}\]
Notice that this equilibrium exists only at one temperature for a given pressure, namely the melting point, 273 K at 1 atm. Above it the ice would all melt; below it the water would all freeze.
(c) Solid – Vapour Equilibrium
A second activity makes this kind of equilibrium visible. Place some iodine crystals in a closed vessel and observe the change in colour intensity in it. After some time the vessel gets filled up with violet coloured vapour. The intensity of the violet colour becomes stable after a certain time.
What do you see in the flask (see Fig. 12.2)?
Diagram
Fig. 12.2 – A stoppered conical flask containing dark grey-black iodine crystals at the bottom and dense violet vapour filling the space above them; caption “Solid iodine in equilibrium with its vapour”.
We see both, that is, solid iodine and iodine vapour in the closed vessel. It means solid iodine sublimes to give iodine vapour and the iodine vapour condenses to form solid iodine. Sublimation is the direct change of a solid into its vapour without passing through the liquid state.
The stable intensity of the colour indicates a state of equilibrium between solid and vapour iodine. The reason the colour is such a good indicator is that the depth of the violet colour depends on how much iodine vapour is present; once the colour stops deepening, the amount of vapour has become constant. We can write the same as follows:
\[\mathrm{I_2}\text{(s)} \underset{\text{condensation}}{\overset{\text{sublimation}}{\rightleftharpoons}} \mathrm{I_2}\text{(g)}\]
Other examples showing this kind of equilibrium are:
\[1.\quad \text{Camphor}\ \text{(s)} \rightleftharpoons \text{Camphor}\ \text{(g)}\]
\[2.\quad \text{Ammonium chloride}\ \text{(s)} \rightleftharpoons \text{Ammonium chloride}\ \text{(g)}\]
Equilibrium in a saturated solution
A third activity uses nothing more than sugar and water.
- Dissolve a given amount of sugar in the minimum amount of water at room temperature.
- Increase the temperature and dissolve more amount of sugar in the same amount of water to make a thick sugar syrup solution.
- Cool the syrup to the room temperature.
Note the observation: Sugar crystals separate out.
In a saturated solution there exists dynamic equilibrium between the solute molecules in the solid state and in the dissolved state. The word dynamic is important: the equilibrium is not a dead, motionless state but a busy one in which two opposite changes continue at equal rates.
\[\text{Sugar}\ \text{(aq)} \rightleftharpoons \text{Sugar}\ \text{(s)}\]
The rate of dissolution of sugar \(=\) The rate of crystallization of sugar
This also explains the observation in the activity. Hot water can hold more dissolved sugar than cold water can, so on cooling the solution holds more sugar than the new equilibrium allows. Crystallization then runs faster than dissolution until the excess sugar has come out as crystals and the two rates match once more.
The concentration of solute in a saturated solution depends on temperature. That is why the phrase “at the given temperature” can never be dropped from the definition.
In every one of these four cases the same three features appear: the system is closed, two opposite processes are running, and their rates have become equal. Board and entrance papers often test this as a match-the-following item, or as an assertion–reason question of the type “Assertion: at equilibrium the process stops. Reason: the concentrations no longer change.” Being clear that the processes continue is what earns the mark.
12.3 Equilibrium in Chemical Process
If a reaction takes place in a closed system so that the products and reactants cannot escape, we often find that the reaction does not give a 100 % yield of products. Instead, some reactants remain after the concentrations stop changing.
This is the chemical counterpart of what we have just seen in the physical processes. A reaction that seems to have died away has in fact only reached a balance point.
When there is no further change in concentration of reactant and product, we say that the reaction has attained equilibrium, with the rates of forward and reverse reactions being equal.
Chemical equilibrium at a given temperature is characterized by constancy of measurable properties such as pressure, concentration, density etc. This is very useful in the laboratory, because it tells us what to watch: once a measurable property such as colour intensity, pressure or density has settled to a steady value at constant temperature, equilibrium has been reached.
Chemical equilibrium can be approached from either side. This is a surprising and important statement, and the two activities that follow are designed to demonstrate it.
Observe and discuss: the dinitrogen tetroxide flask
I. Colourless \(\mathrm{N_2O_4}\) taken in a closed flask is converted to \(\mathrm{NO_2}\) (a reddish brown gas). The two flasks are compared in Fig. 12.3.
\[\underset{\text{colourless}}{\mathrm{N_2O_4}\text{(g)}} \rightleftharpoons \underset{\text{reddish brown}}{2\,\mathrm{NO_2}\text{(g)}}\]
Diagram
Fig. 12.3 – Two identical round-bottomed flasks with stoppers side by side: the left flask colourless and labelled N₂O₄, the right flask filled with orange-brown gas and labelled NO₂; a thick red arrow pointing from left flask to right flask and a thick blue arrow pointing from right flask back to left flask; caption “N₂O₄ conversion to NO₂”.
The colour becomes light brown, indicating the presence of \(\mathrm{NO_2}\) in the mixture. Note carefully that the colour becomes light brown and not deep brown — the conversion has clearly not gone all the way.
The formation of \(\mathrm{NO_2}\) from \(\mathrm{N_2O_4}\) is reversible. In such a reaction the reactants react to form the products, and the products react to give back the reactants.
As soon as the forward reaction produces any \(\mathrm{NO_2}\), the reverse reaction begins and \(\mathrm{NO_2}\) starts combining back to form \(\mathrm{N_2O_4}\). At equilibrium, the concentrations of \(\mathrm{N_2O_4}\) and \(\mathrm{NO_2}\) remain unchanged and do not vary with time, because the rate of formation of \(\mathrm{N_2O_4}\) is equal to the rate of formation of \(\mathrm{NO_2}\), as shown in Fig. 12.4 (b).
The way the concentrations settle down is drawn in Fig. 12.4 (a), and the way the two rates approach each other is drawn in Fig. 12.4 (b).
Diagram
Fig. 12.4 (a) – Graph of concentration (y-axis) against time (x-axis) for the conversion of N₂O₄ to NO₂: one curve rising from zero and levelling off, labelled NO₂; a second curve starting high and falling to a steady value, labelled N₂O₄; a dashed vertical line where both curves become horizontal, labelled “Equilibrium achieved”.
Diagram
Fig. 12.4 (b) – Graph of rate (y-axis) against time (x-axis): upper curve starting high and falling to a constant value, labelled forward rate kf[N₂O₄]; lower curve starting at zero and rising to the same constant value, labelled reverse rate kr[NO₂]²; dashed vertical line where the curves meet, labelled “Equilibrium achieved (rates are equal)”.
How the rates change on the way to equilibrium
Reading these two graphs together gives the whole story of how a chemical equilibrium is set up. For any reversible reaction in a closed system, the rate of the forward reaction is high in the beginning, while the rate of the reverse reaction is very slow — at the start there is nothing on the product side to react.
As the forward reaction proceeds and products accumulate, the rate of the forward reaction slows down (its reactants are being used up) while the rate of the reverse reaction increases (its reactants are being built up). Finally the rates become equal and equilibrium is established.
Observe and discuss: approaching the same equilibrium from both sides
II. Consider the following dissociation reaction.
\[\underset{\text{(Colourless gas)}}{2\,\mathrm{HI}\text{(g)}} \rightleftharpoons \mathrm{H_2}\text{(g)} + \underset{\text{(Violet coloured gas)}}{\mathrm{I_2}\text{(g)}}\]
The reaction is carried out in a closed vessel starting with hydrogen iodide. The following observations are noted:
- At first, there is an increase in the intensity of violet colour.
- After a certain time the increase in the intensity of violet colour stops.
- When the contents in the closed vessel are analysed at this stage, it is observed that the reaction mixture contains hydrogen iodide, hydrogen and iodine with their concentrations remaining constant over time.
The violet colour is a convenient tracker here because only iodine vapour is coloured; hydrogen iodide and hydrogen are both colourless. A deepening violet colour therefore means iodine is being produced.
The rate of decomposition of HI becomes equal to the rate of combination of \(\mathrm{H_2}\) and \(\mathrm{I_2}\). At equilibrium, no net change is observed and both reactions continue to occur.
Let us now see what happens if we start with hydrogen and iodine vapour in a closed container at a certain temperature.
\[\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\]
- At first, the violet colour of \(\mathrm{I_2}\) is deep.
- The intensity of violet colour starts decreasing.
- After a certain time the decrease in the intensity of colour stops.
- Analysis of the reaction mixture at this stage shows that it contains HI, \(\mathrm{H_2}\) and \(\mathrm{I_2}\) with their concentrations remaining constant over time. In other words, the same equilibrium can be attained by starting from any side.
12.4 Law of Mass Action and Equilibrium Constant
So far our description of equilibrium has been qualitative — we have said that two rates become equal and that the concentrations then stop changing. The next step is to put a number on that state.
The chemical equilibrium is mathematically described in terms of what is called the equilibrium constant (\(K_C\)). We noted earlier that an equilibrium state is attained when the rate of the forward and reverse reactions become equal (refer Fig. 12.4). To build the equilibrium constant, therefore, we must first be able to write down the rate of a reaction.
12.4.1 Rate of chemical reaction
As any reaction proceeds, the concentration of the reactants decreases and the concentration of the products increases. Nothing is created or destroyed in the process; reactants are simply being converted into products, so one quantity falls exactly as the other rises.
The rate of reaction thus can be determined by measuring the extent to which the concentration of a reactant decreases in the given time interval, or the extent to which the concentration of a product increases in the given time interval. Mathematically, the rate of reaction is expressed as:
\[\text{Rate} = -\ \frac{d[\text{Reactant}]}{dT} = \frac{d[\text{Product}]}{dT}\]
where \(d[\text{reactant}]\) and \(d[\text{product}]\) are the small change in concentration during the small time interval \(dT\).
The minus sign in the first term needs a word of explanation, because students lose marks over it. The concentration of a reactant is falling, so \(d[\text{Reactant}]\) is itself a negative quantity; putting a minus sign in front of it makes the rate come out positive. A rate is always reported as a positive number, whether we track it through a reactant or through a product.
12.4.2 Law of mass action
The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentration terms of all the reactants.
The reasoning behind it is physical. A reaction can only occur when reactant particles meet; the more crowded the reactant particles are, the more often they meet, and the faster the reaction goes.
In case the balanced chemical equation shows more molecules of reactants, the concentration is raised to a power equal to the number of molecules of that reactant. If two molecules of a reactant must come together, then doubling its concentration doubles the chance of finding the first molecule and doubles the chance of finding the second, so the effect appears as a square.
A rate equation can be written for a reaction by applying the law of mass action as follows. Consider a reaction
\[\mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C}\]
Here A and B are the reactants and C is the product. The concentrations of chemical species are expressed in \(\text{mol L}^{-1}\) and denoted by putting the formula in square brackets. So \([\mathrm{A}]\) is read as “the concentration of A in moles per litre”.
By applying the law of mass action to this reaction we write a proportionality expression as:
\[\text{Rate} \ \alpha\ [\mathrm{A}][\mathrm{B}]\]
This proportionality expression is transformed into an equation by introducing a proportionality constant, \(k\), as follows:
\[\text{Rate} = k\,[\mathrm{A}][\mathrm{B}] \qquad \ldots (12.1)\]
The Eq. (12.1) is called the rate equation and the proportionality constant, \(k\), is called the rate constant of the reaction. Note the difference in what the two symbols tell us: the rate changes continuously as the reaction proceeds and the concentrations change, whereas the rate constant \(k\) stays fixed for that reaction at a fixed temperature.
12.4.3 Equilibrium constant
Consider a hypothetical reversible reaction
\[\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\]
As we have noted earlier, two reactions, namely forward and reverse reactions, occur simultaneously in a reversible chemical reaction. We, therefore, write rate equations for the forward and reverse reactions.
For the forward reaction the reactants are A and B:
\[\text{Rate}_{\text{forward}}\ \alpha\ [\mathrm{A}][\mathrm{B}]\]
\[\therefore\ \text{Rate}_{\text{forward}} = k_f\,[\mathrm{A}][\mathrm{B}] \qquad \ldots (12.2)\]
For the reverse reaction, C and D are the reactants:
\[\text{Rate}_{\text{reverse}}\ \alpha\ [\mathrm{C}][\mathrm{D}]\]
\[\therefore\ \text{Rate}_{\text{reverse}} = k_r\,[\mathrm{C}][\mathrm{D}] \qquad \ldots (12.3)\]
Here \(k_f\) is the rate constant of the forward reaction and \(k_r\) is the rate constant of the reverse reaction.
At equilibrium, the rates of forward and reverse reactions are equal. Thus,
\[\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}\]
\[\therefore\ k_f\,[\mathrm{A}][\mathrm{B}] = k_r\,[\mathrm{C}][\mathrm{D}]\]
Rearranging so that the two rate constants stand together on one side:
\[\therefore\ \frac{k_f}{k_r} = K_C = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \qquad \ldots (12.4)\]
\(K_C\) is called the equilibrium constant. Notice what has just been proved: since \(k_f\) and \(k_r\) are both constants at a fixed temperature, their ratio must also be a constant — so the concentration ratio on the right-hand side is forced to have a fixed value at that temperature, no matter what amounts we started with.
Equilibrium is dynamic, not dead
At equilibrium the ratio of the product multiplicative term denoting the ratio of concentration of products to that of the reactants is unchanged and equals \(K_C\). The value of \(K_C\) depends upon the temperature.
It is interesting to note that though the concentration ratio remains unchanged, both the forward as well as reverse reactions do proceed at equilibrium, but at the same rate. Therefore the chemical equilibrium is a dynamic equilibrium.
How the balanced equation fixes the expression
The equilibrium constant depends on the form of the balanced chemical equation. Consider the general reversible reaction
\[a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}\]
The equilibrium constant
\[K_C = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \qquad \ldots (12.5)\]
Products always go in the numerator, reactants in the denominator, and every coefficient from the balanced equation becomes an exponent.
If the equilibrium is written as
\[c\mathrm{C} + d\mathrm{D} \rightleftharpoons a\mathrm{A} + b\mathrm{B},\]
\[K'_C = \frac{[\mathrm{A}]^a[\mathrm{B}]^b}{[\mathrm{C}]^c[\mathrm{D}]^d} = \frac{1}{K_C}\]
Thus the equilibrium constant of the reverse chemical reaction, \(K'_C\), is the reciprocal of the equilibrium constant \(K_C\) of the forward reaction. This is why a numerical value of an equilibrium constant is meaningless unless the balanced equation it belongs to is written alongside it.
Let us consider the equilibrium that occurs in the Haber process for the synthesis of ammonia:
\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}\]
\[K_C = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}\]
The power 2 on \([\mathrm{NH_3}]\) is the coefficient of \(\mathrm{NH_3}\) in the balanced equation, and the power 3 on \([\mathrm{H_2}]\) is the coefficient of \(\mathrm{H_2}\). Nitrogen has a coefficient of 1, so its concentration carries the power 1, which we do not bother to write.
And for the equilibrium reaction written as
\[2\,\mathrm{NH_3}\text{(g)} \rightleftharpoons \mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)},\]
\[K'_C = \frac{[\mathrm{N_2}][\mathrm{H_2}]^3}{[\mathrm{NH_3}]^2}\]
12.4.3a Equilibrium constant with respect to partial pressure (\(K_P\))
For reactions involving gases, it is convenient to express the equilibrium constant in terms of partial pressure. The practical reason is straightforward: the pressure of a gas in a vessel can be read off a gauge, whereas its concentration would have to be worked out indirectly.
\(\therefore\) For the reaction,
\[a\mathrm{A}\text{(g)} + b\mathrm{B}\text{(g)} \rightleftharpoons c\mathrm{C}\text{(g)} + d\mathrm{D}\text{(g)},\]
the equilibrium constant can be expressed using partial pressures (\(K_P\)) as given by
\[K_P = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \qquad \ldots (12.6)\]
where \(P_A\), \(P_B\), \(P_C\) and \(P_D\) are equilibrium partial pressures of A, B, C and D, respectively. The partial pressure of a gas in a mixture is the pressure that gas alone would exert if it occupied the whole vessel at the same temperature.
That equation is the bridge we shall need in a moment:
\[PV = nRT\]
where \(P\) is pressure in pascal, \(n\) is the number of moles of gas, \(V\) is volume in \(\text{dm}^3\) or L, \(R\) is the molar gas constant in \(\text{Pa m}^3\ \text{K}^{-1}\text{mol}^{-1}\), and \(T\) is absolute temperature in kelvin.
12.4.4 Relationship between partial pressure and concentration
For a mixture of ideal gases, the partial pressure of each component is directly proportional to its concentration at constant temperature.
For component A,
\[P_A V = n_A RT\]
\(\dfrac{n_A}{V}\) is the molar concentration of A in \(\text{mol dm}^{-3}\).
Dividing both sides by \(V\) isolates that ratio:
\[P_A = \frac{n_A}{V} \times RT, \qquad \text{where}\ \frac{n_A}{V} = [\mathrm{A}]\]
\[\therefore\ P_A = [\mathrm{A}]RT\]
Similarly, for components B, C and D the partial pressures are
\[P_B = [\mathrm{B}]RT\] \[P_C = [\mathrm{C}]RT\] \[P_D = [\mathrm{D}]RT\]
12.4.5 Relationship between \(K_P\) and \(K_C\)
Consider a general reversible reaction:
\[a\mathrm{A}\text{(g)} + b\mathrm{B}\text{(g)} \rightleftharpoons c\mathrm{C}\text{(g)} + d\mathrm{D}\text{(g)}\]
Now substituting the expressions for \(P_A\), \(P_B\), \(P_C\), \(P_D\) in Eq. (12.6):
\[K_P = \frac{[P_C]^c [P_D]^d}{[P_A]^a [P_B]^b} = \frac{[\mathrm{C}]^c (RT)^c [\mathrm{D}]^d (RT)^d}{[\mathrm{A}]^a (RT)^a [\mathrm{B}]^b (RT)^b}\]
Collecting the \(RT\) factors in the numerator and in the denominator:
\[K_P = \frac{[\mathrm{C}]^c [\mathrm{D}]^d (RT)^{c+d}}{[\mathrm{A}]^a [\mathrm{B}]^b (RT)^{a+b}}\]
Since dividing powers of the same quantity means subtracting the indices:
\[\therefore\ K_P = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b} \times (RT)^{(c+d)-(a+b)}\]
\[\therefore\ K_P = \frac{[\mathrm{C}]^c [\mathrm{D}]^d}{[\mathrm{A}]^a [\mathrm{B}]^b} \times (RT)^{\Delta n}\]
But
\[K_C = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \qquad \text{from equation (12.5)}\]
\[\therefore\ \boldsymbol{K_P = K_C\,(RT)^{\Delta n}} \qquad \ldots (12.7)\]
where \(\Delta n\) = (number of moles of gaseous products) \(-\) (number of moles of gaseous reactants) in the balanced chemical equation.
\[R = 8.314\ \text{Pa m}^3\ \text{K}^{-1}\text{mol}^{-1}\]
While calculating the value of \(K_P\), pressure should be expressed in bar, because standard pressure is 1 bar.
\[[\ 1\ \text{pascal (Pa)} = 1\ \text{Nm}^{-2}\ \text{and}\ 1\ \text{bar} = 10^5\ \text{Pa}\ ]\]
Relations of this kind are heavily used in entrance papers, usually in one of two forms: you are given a balanced equation and asked to state the relation between \(K_P\) and \(K_C\), or you are given one constant and asked to compute the other. In both cases the single decisive step is getting \(\Delta n\) right, including its sign.
12.5 Homogeneous and Heterogeneous Equilibria
12.5.1 Homogeneous reactions and Heterogeneous reactions
In a homogeneous reaction all the reactants are in the same phase. A phase is any part of a system that is physically uniform throughout — a single gas mixture is one phase, a solution is one phase, and each separate solid is a phase of its own.
For example, the following is a homogeneous gas phase reaction:
\[2\,\mathrm{HI}\text{(g)} \rightleftharpoons \mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)}\]
Every species here carries the label (g), so there is only one phase and no boundary anywhere in the vessel.
A reversible reaction involving reactants and products those are in different phases is called a heterogeneous reaction.
\[\mathrm{NH_3}\text{(g)} + \mathrm{HCl}\text{(g)} \rightleftharpoons \mathrm{NH_4Cl}\text{(s)}\]
Here two gases combine to form a solid, so the system contains a gas phase and a solid phase with a visible boundary between them.
Equilibria involving homogeneous or heterogeneous reactions are called homogeneous and heterogeneous equilibria respectively.
12.5.2 Equilibrium constant for heterogeneous equilibria
As stated earlier, equilibrium having more than one phase is called heterogeneous equilibrium. For example, if ethanol is placed in a conical flask, liquid–vapour equilibrium is established, as shown in Fig. 12.5.1.
Diagram
Fig. 12.5.1 – A stoppered conical flask partly filled with ethanol: the liquid at the bottom labelled C₂H₅OH(l), the space above filled with scattered vapour molecules labelled C₂H₅OH(g), a clear horizontal line marking the phase boundary, and a pair of small arrows across the boundary labelled “evaporation” (up) and “condensation” (down).
\[\mathrm{C_2H_5OH}\text{(l)} \rightleftharpoons \mathrm{C_2H_5OH}\text{(g)}\]
For a given temperature
\[K_C = \frac{[\mathrm{C_2H_5OH}\text{(g)}]}{[\mathrm{C_2H_5OH}\text{(l)}]}\]
At any given temperature the density of a liquid is constant irrespective of the amount of liquid, and, therefore, the term in the denominator is constant.
\[[\mathrm{C_2H_5OH}\text{(l)}] = \text{constant}\]
This point deserves a moment’s thought, since it is the heart of the matter. The concentration of a pure liquid or a pure solid is fixed by its density and molar mass alone — it is the number of moles packed into each cubic decimetre of the substance itself. Pouring in twice as much ethanol gives twice the moles in twice the volume, so the concentration is unchanged. A quantity that never changes has no business sitting in an expression meant to track change, so it is absorbed into the constant.
Therefore, the modified equilibrium constant for evaporation of ethanol will be
\[K'_C = K_C \times \text{constant} = [\mathrm{C_2H_5OH}\text{(g)}]\]
For the equilibrium of sublimation of iodine
\[\mathrm{I_2}\text{(s)} \rightleftharpoons \mathrm{I_2}\text{(g)}\]
the modified equilibrium constant \(K'_C = [\mathrm{I_2}\text{(g)}]\).
Concentrations of pure solids do not change and thus expressions for equilibria including solids are simplified.
The steps for writing such an expression are collected in FC 12.2.
Flowchart
FC 12.2 – Writing K_C for an equilibrium: balanced equation → list every species with its state → branch on state: (g) or (aq) → include the concentration term, raised to its coefficient; pure (s) or pure (l) → omit the term → assemble the expression with product terms in the numerator and reactant terms in the denominator → find Δn from the terms actually written → unit of K_C = (mol dm⁻³)^Δn.
12.5.3 Units of equilibrium constant
The unit of equilibrium constant depends upon the expression of \(K_C\), which is different for different equilibria. Therefore, the unit of \(K_C\) is also different. There is no single unit for “the equilibrium constant”; you must work it out from the expression each time.
To calculate units of equilibrium constant
| Equilibrium Reaction (I) | Equilibrium Reaction (II) |
|---|---|
| \(\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\) | \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}\) |
| \(K_C = \dfrac{[\mathrm{HI}\text{(g)}]^2}{[\mathrm{H_2}\text{(g)}][\mathrm{I_2}\text{(g)}]}\ \ \ldots\text{(I)}\) | \(K_C = \dfrac{[\mathrm{NH_3}\text{(g)}]^2}{[\mathrm{N_2}\text{(g)}][\mathrm{H_2}\text{(g)}]^3}\ \ \ldots\text{(II)}\) |
| Unit of \(K_C = \dfrac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}][\text{mol dm}^{-3}]}\) | Unit of \(K_C = \dfrac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}][\text{mol dm}^{-3}]^3}\) |
| \(= \dfrac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}]^2}\) | \(= \dfrac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}]^4}\) |
| As all the units cancel out, for (I), \(K_C\) has no units | \(= [\text{mol dm}^{-3}]^{-2}\) |
| \(= \text{mol}^{-2}\ \text{dm}^{6}\) |
Hint: To calculate the units of \(K_C\), find out the difference between the number of moles in the numerator and the number of moles in the denominator in the expression for equilibrium constant.
The unit of \(K_C\) is \((\text{mol dm}^{-3})^{\Delta n}\).
Read that “difference in number of moles” carefully: it is counted from the terms that actually appear in the expression, so omitted solids and pure liquids are not counted.
With reference to expression (II) above, the difference between number of moles
\[\Delta n = 2 - (1+3)\] \[\therefore\ \Delta n = 2 - 4\] \[\therefore\ \Delta n = -2\]
\[\therefore\ \text{unit of } K_C = (\text{mol dm}^{-3})^{\Delta n}\] \[\therefore\ \text{unit of } K_C = (\text{mol dm}^{-3})^{-2}\] \[\therefore\ \text{unit of } K_C = \text{mol}^{-2}\text{dm}^{6}\]
Similarly with reference to expression (I),
\[\therefore\ \text{unit of } K_C = (\text{mol dm}^{-3})^{\Delta n} = (\text{mol dm}^{-3})^{0}\] \[\therefore\ K_C \text{ has no unit.}\]
Anything raised to the power zero equals 1, which is why the units vanish completely whenever the moles balance on the two sides.
Questions of exactly this shape — “write the \(K_C\) expression and give its unit” — are standard two-mark board questions, and the two marks are usually lost in the same two places: forgetting to omit the pure solid, and forgetting to raise a concentration to its coefficient.
12.6 Characteristics of Equilibrium Constant
Now that we can write and evaluate an equilibrium constant, it is worth collecting together the properties that make it so useful. Four characteristics matter.
The value of equilibrium constant is independent of initial concentrations of either the reactants or products. However much of each substance we start with, the reaction adjusts itself until the concentration ratio reaches the same fixed value.
Equilibrium constant is temperature dependent. Hence \(K_C\), \(K_P\) change with change in temperature. This follows directly from \(K_C = k_f / k_r\): rate constants are governed by temperature, so their ratio must be too. It is also why no value of \(K_C\) should ever be quoted without the temperature beside it.
Equilibrium constant has a characteristic value for a particular reversible reaction represented by a balanced equation at the given temperature. The phrase “represented by a balanced equation” carries real weight — writing the same chemistry the other way round, or with doubled coefficients, gives a different number.
Higher value of \(K_C\) or \(K_P\) means more product is formed and the equilibrium point is more towards right hand side and vice versa. Since the products sit in the numerator, a large constant can only mean that the numerator is large compared with the denominator.
These four characteristics are a favourite source of assertion–reason and true-or-false questions in board and entrance papers. Two traps recur: assuming that a catalyst or a change in starting amounts alters \(K_C\), and quoting a value of \(K_C\) without stating the balanced equation and temperature it belongs to.
12.7 Application of Equilibrium Constant
Some applications of equilibrium constant are discussed below. Each of them answers a practical question: which way will a mixture move, how far will a reaction go, what will the final mixture contain, and how does all this connect with the speed of the reaction.
12.7.1 Prediction of the direction of the reaction
For the reversible reaction,
\[a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D},\]
the equilibrium constant is
\[K_C = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b},\]
where all concentrations are equilibrium concentrations.
But suppose we simply mix some reactants and products together in a vessel, in any amounts we please. That mixture is almost certainly not at equilibrium, yet nothing stops us from substituting its concentrations into the same expression. The ratio is called reaction quotient, \(Q_C\), when the concentrations are not necessarily equilibrium concentrations.
\[Q_C = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\]
The reaction quotient has the same form as that of the equilibrium constant, but involves concentrations that are not necessarily equilibrium concentrations. Comparison of \(Q_C\) and \(K_C\) is very useful to decide whether the forward or the reverse reaction should occur to establish the equilibrium. It is shown diagrammatically in Fig. 12.7.1.
Diagram
Fig. 12.7.1 – Three pairs of vertical bars comparing Q_C with K_C: left pair with a short Q_C bar beside a tall K_C bar and a broad arrow labelled “movement toward equilibrium” pointing right, captioned “Reactants → Products”; middle pair with Q_C and K_C bars of equal height, captioned “Reactants and products are at equilibrium”; right pair with a tall Q_C bar beside a short K_C bar and a broad arrow pointing left, captioned “Reactants ← Products”.
If \(Q_C < K_C\), the reaction will proceed from left to right, in the forward direction, generating more product to attain the equilibrium.
If \(Q_C = K_C\), the reaction is at equilibrium and hence no net reaction occurs.
If \(Q_C > K_C\), the reaction will proceed from right to left, requiring more reactants to attain equilibrium.
The logic behind all three lines is worth stating once. \(Q_C\) measures how product-rich the mixture is at this moment, while \(K_C\) states how product-rich it must finally be. If the mixture is not product-rich enough, more product has to be made; if it is too product-rich, some product must be converted back.
Note: The prediction of the direction of the reaction on the basis of \(Q_C\) and \(K_C\) values makes no comment on the time required for attaining the equilibrium.
12.7.2 To know the extent of reaction
Let us recall an expression for equilibrium constant \(K_C\) (see Fig. 12.5). It indicates that the magnitude of \(K_C\) is
- directly proportional to the equilibrium concentrations of the product.
- inversely proportional to the equilibrium concentrations of the reactants.
So the size of the number alone tells us how far the reaction travels before it halts.
Diagram
Fig. 12.5 – A horizontal arrow scale, coloured blue at the left end grading to red at the right end, marked with the values 10⁻³, 1 and 10³ along it; label the extreme left “Very small” with “Reaction proceeds hardly at all” beneath, the middle “Appreciable concentration of both reactants and products are present at equilibrium”, and the extreme right “Very large” with “Reaction proceeds nearly to completion” beneath; heading on the figure: K_C.
The larger the value of the equilibrium constant \(K_C\), the farther the reaction proceeds to the right before reaching the equilibrium state. As a rough working guide, values above about \(10^3\) mean the products dominate, values below about \(10^{-3}\) mean the reactants dominate, and values in between mean that appreciable amounts of both are present at equilibrium.
Consider, for example, the following two reversible reactions.
| \(2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightleftharpoons 2\,\mathrm{H_2O}\text{(g)}\) | \(2\,\mathrm{H_2O}\text{(g)} \rightleftharpoons 2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)}\) |
|---|---|
| Equilibrium constant values | \(K_C = \dfrac{[\mathrm{H_2}\text{(g)}]^2[\mathrm{O_2}\text{(g)}]}{[\mathrm{H_2O}\text{(g)}]^2}\) |
| \(K_C = \dfrac{[\mathrm{H_2O}\text{(g)}]^2}{[\mathrm{H_2}\text{(g)}]^2[\mathrm{O_2}\text{(g)}]}\) | \(K_C = \dfrac{1}{K_C} = \dfrac{1}{2.4 \times 10^{47}}\) |
| \(K_C = 2.4 \times 10^{47}\) At 500 K | \(K_C = 4.1 \times 10^{-48} = 0.41 \times 10^{-47}\) at 500 K |
| 1. Value of \(K_C\) is very high (\(K_C > 10^3\)). | 1. Value of \(K_C\) is very low (\(K_C < 10^{-3}\)). |
| 2. At equilibrium there is a high proportion of products compared to reactants. | 2. At equilibrium, only a small fraction of the reactants are converted into products. |
| 3. Forward reaction is favoured. | 3. Reverse reaction is favoured. |
| 4. Reaction is in favour of products and nearly goes to completion. | 4. Reaction is in favour of reactants. |
| \(K_C >>> 1\). Reaction proceeds almost totally towards products. | \(K_C <<< 1\). Reaction hardly proceeds towards the products. |
The two columns are of course the same chemistry read in opposite directions, and the two constants are reciprocals of each other, exactly as section 12.4.3 predicted. Hydrogen and oxygen combine almost completely to give water vapour; water vapour, left to itself at 500 K, hardly decomposes at all.
12.7.3 To calculate equilibrium concentrations
The equilibrium constant can be used to calculate the composition of an equilibrium mixture. This is the most quantitative of its applications, and the method is always the same: tabulate the initial amounts, express the equilibrium amounts in terms of one unknown, and substitute into the \(K_C\) expression.
Consider an equilibrium reaction,
\[\underset{\text{(ethanoic acid)}}{\mathrm{CH_3COOH}\text{(aq)}} + \underset{\text{(ethanol)}}{\mathrm{C_2H_5OH}\text{(aq)}} \rightleftharpoons \underset{\text{(ethyl ethanoate)}}{\mathrm{CH_3COOC_2H_5}\text{(aq)}} + \mathrm{H_2O}\text{(aq)}\]
The equilibrium constant is 4.0 at a certain temperature.
If we start with 2.0 mol of ethanoic acid and 2.0 mol of ethanol in ‘V’ litres, let us find out the composition of the equilibrium mixture. Consider \(x\) mol of ethyl ethanoate at equilibrium.
Since the balanced equation has a 1 : 1 : 1 : 1 mole ratio, forming \(x\) mol of ester must consume \(x\) mol of acid and \(x\) mol of alcohol, and must also form \(x\) mol of water.
| \(\mathrm{CH_3COOH}\) | \(\mathrm{C_2H_5OH}\) | \(\mathrm{CH_3COOC_2H_5}\) | \(\mathrm{H_2O}\) | |
|---|---|---|---|---|
| Initial (No. of moles) | 2.0 | 2.0 | 0 | 0 |
| At equilibrium (No. of moles) | \((2.0 - x)\) | \((2.0 - x)\) | \(x\) | \(x\) |
| Equilibrium concentrations | \(\dfrac{(2.0-x)}{V}\) | \(\dfrac{(2.0-x)}{V}\) | \(\dfrac{x}{V}\) | \(\dfrac{x}{V}\) |
Equilibrium constant
\[K_C = \frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}\]
\[\therefore\ 4.0 = \frac{\dfrac{x}{V} \times \dfrac{x}{V}}{\dfrac{(2.0-x)}{V} \times \dfrac{(2.0-x)}{V}}\]
substituting the value of equilibrium concentration. Because the same volume \(V\) appears twice in the numerator and twice in the denominator, it cancels completely — which is why this particular problem can be solved without knowing \(V\) at all.
\[\therefore\ 4.0 = \frac{x^2}{(2.0-x)^2}\]
Both sides are perfect squares, so we may take the square root of each side:
\[\therefore\ \sqrt{4} = \frac{x}{2.0-x}\]
\[\therefore\ 2 = \frac{x}{2.0-x}\]
\[\therefore\ 4 - 2x = x\]
\[\therefore\ x = \frac{4}{3}\]
\[\therefore\ x = 1.33\]
\[\therefore\ (2.0 - x) = 0.67\]
Therefore, equilibrium concentrations are 0.67 mol of ethanoic acid, 0.67 mol of ethanol and 1.33 mol of ethyl ethanoate in \(V\) litres.
Notice that the reaction has stopped well short of completion: a third of the acid is still unreacted even though nothing prevents it from reacting. That is what a moderate value of \(K_C\) looks like in practice.
12.7.4 Link between chemical equilibrium and chemical kinetics
We have deduced in section 12.4.3
\[K_C = \frac{k_f}{k_r} \qquad \ldots (12.4)\]
where \(k_f\) and \(k_r\) are velocity or rate constants of the forward and reverse reactions respectively. This equation can be used to determine the composition of the reaction mixture. In other words, a single number links how fast the two opposing reactions are with how far the overall reaction goes.
| \(k_f > k_r\) | \(k_f \approx k_r\) |
|---|---|
| \(\therefore\ K_C\) is very large. | \(k_f\) and \(k_r\) have comparable values, \(\therefore\ K_C\) is nearly one. |
| \(\therefore\ \) Reaction goes almost to completion. | Reaction never goes to completion. |
| \(\therefore\ \) If \(k_f\) is much larger than \(k_r\), the reaction may be irreversible (the reverse reaction is too slow to be detected). | Comparable concentrations of reactants and products are present at equilibrium. |
This table also settles a question raised right at the start of the chapter. A reaction we call “irreversible” is not one whose reverse is forbidden; it is one whose reverse is so slow, compared with the forward reaction, that it cannot be detected at all.
12.8 Le Chatelier’s Principle and Factors Altering the Composition at Equilibrium
We have learnt that a reaction attains a state of equilibrium under a certain set of conditions (temperature, pressure, concentration and catalyst). The obvious next question is what happens when we deliberately change one of those conditions.
In general, if we add more reactant, the system will react to remove it. If we remove a product, the system will react to replenish it. Under these changed conditions, a new equilibrium will be established with a different composition from the earlier equilibrium mixture.
The principal goal of chemical synthesis is to achieve maximum conversion of reactants to products with minimum expenditure of energy. To achieve this goal, the reaction conditions must be adjusted. This is not merely an academic exercise — for an industrial plant it is the difference between profit and loss.
The qualitative effect of various factors on the composition of the equilibrium mixture are described through the Le Chatelier’s principle.
The idea of stress
If a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction which relieves the stress. Stress means any change in concentration, pressure, volume or temperature which disturbs the original equilibrium.
The direction that the reaction takes is the one that reduces the stress. For example, if the concentration of the reactants is increased, reaction goes in a direction that tends to decrease the concentration, that is, in the direction of the forward reaction.
Le Chatelier’s Principle
Notice how modest the claim is. The system does not undo the change — it only reduces it. Adding hydrogen to an ammonia mixture leaves more hydrogen present than before; some of the added hydrogen is consumed, not all of it.
One more point of language, because a great deal of confusion in examinations comes from it. The equilibrium refers to the relative amounts of reactants and products, and thus a shift in equilibrium in a particular direction will imply that the reaction in that direction will be favoured. “Shifts to the right” therefore means “more product at the new equilibrium”.
12.8.1 Factors affecting equilibrium
(a) Change of concentration
Consider the reversible reaction representing production of ammonia (\(\mathrm{NH_3}\)):
\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)} + \text{Heat}\]
The reaction proceeds with a decrease in the number of moles (\(\Delta n = -2\)) and the forward reaction is exothermic. Iron (containing a small quantity of molybdenum) is the catalyst. Both of these facts will matter when we come to pressure and temperature.
At equilibrium
\[Q_C = K_C = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} \qquad \ldots \text{(I)}\]
Let us follow what happens in four stages when extra hydrogen is pumped in. The sequence is set out in FC 12.3.
Stage I: system at equilibrium.
Stage II: Equilibrium is disturbed by making \(Q_C < K_C\) by adding \(\mathrm{H_2}\). Look at expression (I) to see why the inequality goes this way: \([\mathrm{H_2}]\) sits in the denominator, so increasing it makes the whole ratio smaller. Due to addition of extra hydrogen, the system is no longer at equilibrium. The system regains its equilibrium in stage III.
Stage III: Added hydrogen is to be used up and converted to more \(\mathrm{NH_3}\). Hence the forward reaction is favoured. This is exactly what section 12.7.1 predicted for \(Q_C < K_C\).
Stage IV: New state of equilibrium is established.
Flowchart
FC 12.3 – Response of the ammonia equilibrium to added hydrogen: Stage I, system at equilibrium (Q_C = K_C) → Stage II, extra H₂ added, denominator of the ratio increases, so Q_C < K_C, system no longer at equilibrium → Stage III, forward reaction favoured, added H₂ consumed along with N₂ to form more NH₃, Q_C rises → Stage IV, Q_C = K_C again, new equilibrium with a higher yield of NH₃; note beside Stage IV: K_C itself is unchanged throughout.
According to Le Chatelier’s principle, the effect of addition of \(\mathrm{H_2}\) (or \(\mathrm{N_2}\) or both) is reduced by shifting the equilibrium from left to right so that the added \(\mathrm{N_2}\) or \(\mathrm{H_2}\) is consumed.
The forward reaction occurs to a larger extent than the reverse reaction until the new equilibrium is established. This results in an increased yield of \(\mathrm{NH_3}\).
(b) Effect of Pressure
You know that the ideal gas equation is
\[PV = nRT\]
On rearranging we get,
\[P = RT\,\frac{n}{V}\]
\[\therefore\ P \propto \frac{n}{V},\]
where the ratio \(\dfrac{n}{V}\) is an expression for the concentration of the gas in \(\text{mol dm}^{-3}\).
This proportionality is the key to the whole discussion: for gases, changing the pressure is changing the concentrations. Squeezing a gas mixture into a smaller volume raises the concentration of every gas in it.
1. An equilibrium mixture of dinitrogen tetroxide (colourless gas) and nitrogen dioxide (brown gas) is set up in a sealed flask at a particular temperature (refer to Fig. 12.3).
\[\mathrm{N_2O_4}\text{(g)} \rightleftharpoons 2\,\mathrm{NO_2}\text{(g)}\]
The effect of change of pressure on this gaseous equilibrium can be followed by observing the change in its colour intensity, as in Fig. 12.8.1. See Table 12.1.
Diagram
Fig. 12.8.1 – A gas syringe containing the brown N₂O₄/NO₂ mixture shown in three positions: plunger at rest (medium brown), plunger pushed in to halve the volume (pale, almost colourless gas), plunger pulled out to double the volume (deep brown); arrows labelled “increase pressure” and “decrease pressure”, with a note that the initial darkening on compression is momentary.
Table 12.1: Change in colour intensity
| Change in pressure | Change in colour intensity | Shift in equilibrium position |
|---|---|---|
| 1. Decrease in pressure | The colour deepens | To the right, the side with more molecules |
| 2. Increase in pressure | The colour lightens to almost colourless | To the left, the side with fewer molecules |
In this reversible reaction, the forward reaction takes place with an increase in the number of molecules, whereas the reverse reaction takes place with a decrease in the number of molecules. One molecule of \(\mathrm{N_2O_4}\) becomes two molecules of \(\mathrm{NO_2}\), so the forward direction is the crowded direction.
That is why raising the pressure drives the system leftwards: forming \(\mathrm{N_2O_4}\) reduces the total number of gas molecules, and fewer molecules exert less pressure, which relieves the applied stress.
2. Consider the reaction,
\[\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\]
As there is the same number of molecules of gas on both sides, change of pressure has no effect on the equilibrium. Here, there is no change in the value of \(K_C\) during any change in pressure of the equilibrium reaction mixture.
(c) Effect of Temperature
Before dealing with temperature, one fact about reversible reactions must be stated clearly, because everything that follows depends on it.
So if the forward reaction gives out a certain quantity of heat, the reverse reaction takes in exactly the same quantity. Heat can therefore be treated almost like a reactant or a product: supplying heat pushes the system in whichever direction absorbs it.
Consider the equilibrium reaction,
\[\mathrm{PCl_5}\text{(g)} \rightleftharpoons \mathrm{PCl_3}\text{(g)} + \mathrm{Cl_2}\text{(g)} + 92.5\ \text{kJ}\]
The forward reaction is exothermic. The reverse reaction is endothermic. An endothermic reaction consumes heat. Therefore, an increase in temperature results in shifting the equilibrium in the reverse direction to use up the added heat (heat energy converted to chemical energy).
Now let us study the effect of change of temperature on the equilibrium constant.
1. Consider the equilibrium reaction:
\[\mathrm{CO}\text{(g)} + 2\,\mathrm{H_2}\text{(g)} \rightleftharpoons \mathrm{CH_3OH}\text{(g)} + \text{Heat}\] \[\Delta H = -90\ \text{kJ mol}^{-1}\]
The equilibrium constant \(K_C\) for the reaction is given in Table 12.2.
\[K_C = \frac{[\mathrm{CH_3OH}\text{(g)}]}{[\mathrm{CO}\text{(g)}][\mathrm{H_2}\text{(g)}]^2}\]
Table 12.2
| Temperature (K) | \(K_C\) |
|---|---|
| 298 | \(1.7 \times 10^{17}\) |
| 500 | \(1.1 \times 10^{11}\) |
| 1000 | \(2.1 \times 10^{6}\) |
The forward reaction is exothermic. According to Le Chatelier’s principle, an increase in temperature shifts the position of equilibrium to the left. Therefore, the concentration of \([\mathrm{CH_3OH}\text{(g)}]\) decreases and the concentration of \(\mathrm{CO}\text{(g)}\) and \(\mathrm{H_2}\text{(g)}\) increases.
\[K_C = \frac{[\mathrm{CH_3OH}\text{(g)}]}{[\mathrm{CO}\text{(g)}][\mathrm{H_2}\text{(g)}]^2}\]
Reading the expression alongside those two changes settles the matter: the numerator has become smaller and the denominator larger. Therefore, the value of \(K_C\) decreases as the temperature is increased, and the figures in Table 12.2 fall by eleven orders of magnitude between 298 K and 1000 K.
This is the one factor that alters the equilibrium constant itself, and it is worth being able to predict the direction of that change.
(d) Effect of Catalyst
The rate of a chemical reaction increases with the use of a catalyst. A catalyst is a substance that speeds up a reaction without itself being used up.
Consider an esterification reaction:
\[\underset{\text{(ethanoic acid)}}{\mathrm{CH_3COOH}\text{(l)}} + \underset{\text{(ethanol)}}{\mathrm{C_2H_5OH}\text{(l)}} \rightleftharpoons \underset{\text{(ethyl ethanoate)}}{\mathrm{CH_3COOC_2H_5}\text{(l)}} + \underset{\text{(water)}}{\mathrm{H_2O}\text{(l)}}\]
If the above reaction is carried out without a catalyst, it would take many days to reach equilibrium. However, the addition of hydrogen ions (\(\mathrm{H^+}\)) as catalyst reduces the time to only a few hours.
Note carefully what has and has not changed. The mixture obtained after a few hours with the catalyst is the same mixture that would have been obtained after many days without it. The catalyst changes the waiting time, not the destination.
The reason is drawn in Fig. 12.8.2.
Diagram
Fig. 12.8.2 – Reaction-coordinate (potential energy vs progress of reaction) diagram: reactant level on the left, product level on the right, an upper solid curve peaking at the uncatalysed activation energy and a lower dashed curve peaking at the catalysed activation energy; mark Ea(forward) and Ea(reverse) on both curves with arrows, and label ΔH between the reactant and product levels, noting that both peaks are lowered by the same amount while ΔH is unchanged.
A catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
Because both barriers are lowered equally, both rate constants are multiplied by the same factor, and their ratio \(k_f/k_r\) — that is, \(K_C\) — is untouched. A catalyst does not affect the equilibrium constant and the equilibrium composition of a reaction mixture.
A catalyst does not appear in the balanced chemical equation and in the equilibrium constant expression.
Summary of the four factors
Let us summarize the effects of all four factors on the position of equilibrium and the value of \(K_C\).
| Effect of | Position of equilibrium | Value of \(K_C\) |
|---|---|---|
| Concentration | Changes | No change |
| Pressure | Changes if reaction involves change in number of gas molecules | No change |
| Temperature | Changes | Changes |
| Catalyst | No change | No change |
This table is one of the most heavily examined pieces of the chapter. Questions usually take the form “state the effect of … on the position of equilibrium and on the value of the equilibrium constant”, and the two halves of the answer must be given separately — a shift in position does not imply a change in \(K_C\).
12.9 Industrial Application
The Haber process (Industrial preparation of ammonia)
Industrial processes can be made efficient and profitable by applying ideas of rate of reaction and equilibrium. Everything in this chapter now comes together in a single example: a reaction whose conditions were chosen, quite deliberately, by balancing what equilibrium wants against what speed and cost allow.
The Haber process is the process of synthesis of ammonia gas by reacting together hydrogen gas and nitrogen gas in a particular stoichiometric ratio by volumes and at a selected optimum temperature.
\[\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)} + \text{Heat}\]
The gases are mixed in the ratio 1 : 3 by volume because, by Avogadro’s law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules — so a volume ratio of 1 : 3 is also a mole ratio of 1 : 3, exactly as the balanced equation demands. Feeding the gases in this ratio means neither reactant is left in useless excess.
The reaction proceeds with a decrease in the number of moles (\(\Delta n = -2\)) and the forward reaction is exothermic. Iron (containing a small quantity of molybdenum) is used as catalyst.
The whole plant is laid out in Fig. 12.6, and the order of operations is summarised in FC 12.4.
Diagram
Fig. 12.6 – Flow diagram of the Haber process: box “nitrogen from the air” and box “hydrogen from natural gas” both feeding a line labelled “nitrogen and hydrogen, 1:3 by volume” into a reactor box labelled “450 °C, 200 atm, Iron catalyst”; outlet from the reactor going to a box “gases are cooled and ammonia turns to liquid” and then down to “liquid ammonia”; a side branch labelled “unreacted gases recycled” looping from the cooler back to the reactor inlet; the equation N₂ + 3H₂ ⇌ 2NH₃ written above the reactor.
Flowchart
FC 12.4 – Haber process sequence: nitrogen from air + hydrogen from natural gas → mixed 1 : 3 by volume → compressed to the optimum pressure → passed over iron (with a little molybdenum) catalyst at the optimum temperature → equilibrium mixture of N₂, H₂ and NH₃ → cooled so that ammonia liquefies and is drawn off as liquid ammonia → branch: unreacted N₂ and H₂ recycled back to the reactor inlet, forming a loop.
We studied earlier in section 12.8.1 (a) how the change in concentration of the reactants affects the yield of ammonia. Now let us consider the effect of temperature and pressure on the synthesis of ammonia.
a. Effect of temperature
The formation of ammonia is an exothermic reaction. Hence, lowering of temperature will shift the equilibrium to the right.
At low temperature, however, the rate of reaction is small and a longer time would be required to attain the equilibrium. This is the crucial tension in the whole process: equilibrium and rate pull in opposite directions, because Le Chatelier’s principle tells us only where the reaction will end up, not how long it will take to get there.
At high temperatures, the reaction occurs rapidly with appreciable decomposition of ammonia. Hence, the optimum temperature has to be used. The optimum temperature is about 773 K.
b. Effect of pressure
The forward reaction is favoured with high pressure as it proceeds with a decrease in the number of moles. Four moles of gas on the left become two moles on the right, so shifting to the right relieves an applied pressure.
At high pressure, the catalyst becomes inefficient. Therefore an optimum pressure needs to be used. The optimum pressure is about 250 atm.
12.10 Summary
A chemical change may be irreversible, proceeding in one direction only until a reactant is exhausted and shown by a single arrow, or reversible, proceeding in both directions at once and shown by a double arrow, \(\rightleftharpoons\).
The direction from reactants to products is the forward reaction; from products to reactants it is the reverse or backward reaction.
Whether a reaction behaves reversibly often depends on the container. A closed system exchanges heat but not matter with the surroundings; an open system exchanges both; an isolated system exchanges neither. Limestone heated in a closed vessel reaches equilibrium, \(\mathrm{CaCO_3}\text{(s)} \rightleftharpoons \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\), but in an open vessel the escaping \(\mathrm{CO_2}\) makes the decomposition irreversible.
Physical processes also reach equilibrium in a closed system: liquid–vapour (\(\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H_2O}\text{(vapour)}\)), solid–liquid (ice and water at 273 K), solid–vapour (\(\mathrm{I_2}\text{(s)} \rightleftharpoons \mathrm{I_2}\text{(g)}\)), and dissolution–crystallization in a saturated solution.
The constant pressure exerted by a vapour in equilibrium with its liquid at a given temperature is the saturated vapour pressure, or aqueous tension in the case of water. It increases with temperature. The normal boiling point is the temperature at which the saturated vapour pressure equals 1 atm (1.013 bar) — 100 °C for water and 78 °C for ethyl alcohol.
A saturated solution is one in which no more solute can dissolve at the given temperature; in it the rate of dissolution equals the rate of crystallization.
Chemical equilibrium is reached when the concentrations of reactants and products stop changing because the forward and reverse rates have become equal. It is recognised by the constancy of measurable properties such as concentration, pressure and density at a given temperature, and it can be approached from either side — the same mixture is obtained whether one starts with HI or with \(\mathrm{H_2}\) and \(\mathrm{I_2}\).
Equilibrium is dynamic, not static: both reactions continue to occur, at equal rates.
The rate of a reaction is the fall in reactant concentration, or the rise in product concentration, per unit time: \[\text{Rate} = -\ \frac{d[\text{Reactant}]}{dT} = \frac{d[\text{Product}]}{dT}\]
Law of mass action: the rate of a reaction at each instant is proportional to the product of the concentrations of all the reactants, each raised to a power equal to the number of molecules of that reactant in the balanced equation. For \(\mathrm{A} + \mathrm{B} \longrightarrow \mathrm{C}\) this gives the rate equation \(\text{Rate} = k\,[\mathrm{A}][\mathrm{B}]\), where \(k\) is the rate constant.
Equating the forward and reverse rates at equilibrium for \(\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}\) gives \[\frac{k_f}{k_r} = K_C = \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]}\]
For the general reaction \(a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}\), \[K_C = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\] and the equilibrium constant of the reverse reaction is its reciprocal, \(K'_C = 1/K_C\).
For gaseous reactions the constant may be written with equilibrium partial pressures: \[K_P = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}\]
Since \(P_A = [\mathrm{A}]RT\) for each component of an ideal gas mixture, the two constants are linked by \[K_P = K_C\,(RT)^{\Delta n}\] where \(\Delta n\) = moles of gaseous products \(-\) moles of gaseous reactants. When \(\Delta n = 0\), \(K_P = K_C\). Pressures are expressed in bar, since standard pressure is 1 bar.
An equilibrium is homogeneous if all species are in the same phase, and heterogeneous if they are not. In writing \(K_C\) for a heterogeneous equilibrium, only gases (g) and dissolved species (aq) are included; pure solids and pure liquids are omitted because their concentrations are constant at a given temperature.
The unit of \(K_C\) is \((\text{mol dm}^{-3})^{\Delta n}\), counted from the terms that actually appear in the expression. When the moles balance, \(K_C\) has no unit.
Characteristics of the equilibrium constant: it is independent of the initial concentrations; it is temperature dependent; it has a characteristic value for a particular balanced equation at a given temperature; and a higher value means more product at equilibrium.
The reaction quotient \(Q_C\) has the same form as \(K_C\) but uses the concentrations present at any moment. If \(Q_C < K_C\) the net reaction is forward; if \(Q_C = K_C\) the system is at equilibrium; if \(Q_C > K_C\) the net reaction is backward. The comparison says nothing about the time needed.
The magnitude of \(K_C\) measures the extent of reaction. Values above about \(10^3\) mean the reaction goes nearly to completion; values below about \(10^{-3}\) mean it hardly proceeds; intermediate values mean appreciable amounts of both reactants and products are present.
\(K_C\) can be used to calculate the composition of an equilibrium mixture by tabulating initial moles, expressing the equilibrium moles in terms of one unknown, and substituting into the \(K_C\) expression.
Because \(K_C = k_f/k_r\), a reaction with \(k_f \gg k_r\) has a very large \(K_C\) and goes almost to completion, while \(k_f \approx k_r\) gives \(K_C\) near one and comparable amounts of reactants and products. A reaction called irreversible is one whose reverse reaction is too slow to detect.
Le Chatelier’s principle: when a system at equilibrium is subjected to a change in any of the factors determining the equilibrium conditions, the system responds so as to minimize the effect of that change. A stress is any change in concentration, pressure, volume or temperature.
Increasing the concentration of a species shifts the equilibrium in the opposite direction so as to consume it; \(K_C\) is unchanged.
Increasing the pressure shifts the equilibrium towards the side with fewer gas molecules; a reaction with equal numbers of gas molecules on both sides, such as \(\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\), is unaffected. \(K_C\) is unchanged.
Raising the temperature shifts the equilibrium in the endothermic direction, and this is the only factor that alters the value of \(K_C\). For the exothermic reaction \(\mathrm{CO}\text{(g)} + 2\,\mathrm{H_2}\text{(g)} \rightleftharpoons \mathrm{CH_3OH}\text{(g)} + \text{Heat}\), \(K_C\) falls from \(1.7 \times 10^{17}\) at 298 K to \(2.1 \times 10^{6}\) at 1000 K.
A catalyst lowers the activation energy of the forward and reverse reactions by exactly the same amount. It therefore shortens the time taken to reach equilibrium but changes neither the equilibrium composition nor \(K_C\), and it does not appear in the balanced equation or in the \(K_C\) expression.
The steps for predicting the effect of any imposed change are collected in FC 12.5.
Flowchart
FC 12.5 – Applying Le Chatelier’s principle: identify the stress → branch on its type: (a) concentration changed → equilibrium shifts away from the species added, or towards the species removed; K_C unchanged; (b) pressure increased → compare moles of gas on the two sides: fewer moles on the product side → shifts forward, fewer on the reactant side → shifts backward, equal moles (Δn = 0) → no shift; K_C unchanged; (c) temperature raised → shifts in the endothermic direction, so for an exothermic forward reaction it shifts backward and K_C decreases, for an endothermic forward reaction it shifts forward and K_C increases; (d) catalyst added → no shift, K_C unchanged, only the time to reach equilibrium is reduced → in every case a new equilibrium is established.
- In the Haber process, \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)} + \text{Heat}\), nitrogen and hydrogen are combined in the ratio 1 : 3 by volume over iron containing a little molybdenum. The forward reaction is exothermic and proceeds with \(\Delta n = -2\), so a low temperature and a high pressure favour the yield; because a low temperature makes the reaction too slow and too high a pressure makes the catalyst inefficient, optimum conditions of about 773 K and about 250 atm are used, ammonia is removed by liquefaction, and unreacted gases are recycled.
12.11 Check Your Concepts
Assertion: At equilibrium, a reaction appears to have stopped. Reason: At equilibrium the rate constants of the forward and reverse reactions both become zero. State whether the assertion and the reason are true or false, and whether the reason correctly explains the assertion.
A sealed bottle half filled with water never dries up, while an identical open bottle empties within a few days. Explain both observations in terms of the rates of evaporation and condensation.
Why can calcium carbonate be recovered when it is heated in a closed vessel but not when it is heated in an open kiln? Answer in terms of the requirements of the backward reaction.
Give a reason: pure solids and pure liquids are omitted from the equilibrium constant expression of a heterogeneous equilibrium.
For a certain reaction \(K_C = 4.0\) at a given temperature. What will be the value of the equilibrium constant if the balanced equation is (a) reversed, and (b) written with all its coefficients doubled? Justify each answer from the form of the \(K_C\) expression.
Explain why the equilibrium constant of one reaction has no unit while that of another has the unit \(\text{mol}^{-2}\text{dm}^{6}\). Under what condition is \(K_C\) unitless?
Distinguish between the reaction quotient and the equilibrium constant, and state how their comparison is used to predict the direction of a net reaction.
Why does a change of pressure have no effect on the position of the equilibrium \(\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\), while it strongly affects \(\mathrm{N_2O_4}\text{(g)} \rightleftharpoons 2\,\mathrm{NO_2}\text{(g)}\)?
Assertion: A catalyst increases the yield of product at equilibrium. Reason: A catalyst lowers the activation energy of the forward reaction. State whether each statement is true or false and explain your reasoning.
The forward reaction in the Haber process is exothermic, yet the plant is operated at about 773 K. Explain this choice of temperature, and state what would happen to the yield and to the time taken if the temperature were lowered to 400 K.
For the equilibrium \(\mathrm{CaCO_3}\text{(s)} \rightleftharpoons \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\), the forward reaction is endothermic. Predict the effect on the position of the equilibrium and on the value of \(K_C\) of (a) raising the temperature, (b) increasing the pressure, (c) adding more \(\mathrm{CaCO_3}\text{(s)}\), (d) removing \(\mathrm{CO_2}\) as it forms.
Match each system in the first list with the type of equilibrium in the second.
| Column A | Column B |
|---|---|
| a. ice and water in a thermos flask at 273 K | i. liquid–vapour |
| b. iodine crystals in a stoppered jar | ii. solid–vapour |
| c. sugar syrup standing over sugar crystals | iii. solid–liquid |
| d. ethanol in a stoppered conical flask | iv. saturated solution |
Sugar is dissolved in the minimum amount of hot water and the solution is then cooled to room temperature. Crystals separate out. Explain this observation using the idea of dynamic equilibrium in a saturated solution.
Explain how the same equilibrium mixture can be obtained by starting either from hydrogen iodide alone or from hydrogen and iodine, and state what this observation proves about the nature of equilibrium.
Assertion: Increasing the concentration of a reactant increases the value of \(K_C\). Reason: Increasing the concentration of a reactant shifts the equilibrium in the forward direction. Comment on both statements.
Using the relation \(K_C = k_f / k_r\), explain why some reactions are described as irreversible even though a reverse reaction is in principle possible.
Give reasons for each of the following: (a) a value of \(K_C\) must always be quoted along with the temperature and the balanced equation, (b) a catalyst does not appear in the \(K_C\) expression, (c) nitrogen and hydrogen are fed to the Haber reactor in the ratio 1 : 3 by volume.
State whether each of the following equilibria is homogeneous or heterogeneous, and write the \(K_C\) expression for each: (a) \(2\,\mathrm{HI}\text{(g)} \rightleftharpoons \mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)}\), (b) \(\mathrm{NH_4Cl}\text{(s)} \rightleftharpoons \mathrm{NH_3}\text{(g)} + \mathrm{HCl}\text{(g)}\), (c) \(\mathrm{C}\text{(s)} + \mathrm{CO_2}\text{(g)} \rightleftharpoons 2\,\mathrm{CO}\text{(g)}\).
A syringe containing an equilibrium mixture of \(\mathrm{N_2O_4}\) and \(\mathrm{NO_2}\) is compressed to half its volume at constant temperature. Describe and explain the change in the colour of the gas, and state whether \(K_C\) has changed.
For an exothermic gaseous reaction, state and justify the conditions of temperature and pressure that would give the maximum equilibrium yield of product, and explain why those ideal conditions are usually not the ones used in industry.
State the Law of Mass action.
Write an expression for equilibrium constant with respect to concentration.
Derive mathematically the value of \(K_p\) for \(\mathrm{A}\text{(g)} + \mathrm{B}\text{(g)} \rightleftharpoons \mathrm{C}\text{(g)} + \mathrm{D}\text{(g)}\).
Write expressions of \(K_c\) for the following chemical reactions —
- \(2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightleftharpoons 2\,\mathrm{SO_3}\text{(g)}\)
- \(\mathrm{N_2O_4}\text{(g)} \rightleftharpoons 2\,\mathrm{NO_2}\text{(g)}\)
Mention various applications of equilibrium constant.
How does the change of pressure affect the value of equilibrium constant?
Differentiate irreversible and reversible reaction.
Write suitable conditions of concentration, temperature and pressure used during the manufacture of ammonia by the Haber process.
Relate the terms reversible reactions and dynamic equilibrium.
For the equilibrium \(\mathrm{BaSO_4}\text{(s)} \rightleftharpoons \mathrm{Ba^{2+}}\text{(aq)} + \mathrm{SO_4^{2-}}\text{(aq)}\) state the effect of — a. Addition of \(\mathrm{Ba^{2+}}\) ion b. Removal of \(\mathrm{SO_4^{2-}}\) ion c. Addition of \(\mathrm{BaSO_4}\text{(s)}\) — on the equilibrium.
Explain:
- Dynamic nature of chemical equilibrium with a suitable example.
- Relation between \(K_c\) and \(K_p\).
- State and explain Le Chatelier’s principle with reference to —
- change in temperature
- change in concentration.
- Reversible reaction
- Rate of reaction
- What is the effect of adding chloride on the position of the equilibrium \(\mathrm{AgCl}\text{(s)} \rightleftharpoons \mathrm{Ag^{+}}\text{(aq)} + \mathrm{Cl^{-}}\text{(aq)}\)?
12.12 Practice With Numericals
For the equilibrium \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NH_3}\text{(g)}\) the equilibrium concentrations at a certain temperature are \([\mathrm{N_2}] = 0.20\ \text{mol dm}^{-3}\), \([\mathrm{H_2}] = 0.10\ \text{mol dm}^{-3}\) and \([\mathrm{NH_3}] = 0.40\ \text{mol dm}^{-3}\). Calculate the value of \(K_C\) and deduce its unit.
2.0 mol of hydrogen iodide is heated in a closed vessel of volume \(2.0\ \text{dm}^3\) until equilibrium is reached, when 0.20 mol of hydrogen is found to be present. For \(2\,\mathrm{HI}\text{(g)} \rightleftharpoons \mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)}\), calculate the equilibrium concentration of each of the three gases and the value of \(K_C\).
1.0 mol of \(\mathrm{N_2O_4}\) is placed in a \(1.0\ \text{dm}^3\) closed flask and is found to be 20 % dissociated at equilibrium. For \(\mathrm{N_2O_4}\text{(g)} \rightleftharpoons 2\,\mathrm{NO_2}\text{(g)}\), calculate the equilibrium concentrations of both gases and the value of \(K_C\) with its unit.
A mixture of 3.0 mol of ethanoic acid and 3.0 mol of ethanol in \(V\) litres is allowed to reach equilibrium at a temperature at which \(K_C = 4.0\) for \(\mathrm{CH_3COOH}\text{(aq)} + \mathrm{C_2H_5OH}\text{(aq)} \rightleftharpoons \mathrm{CH_3COOC_2H_5}\text{(aq)} + \mathrm{H_2O}\text{(aq)}\). Calculate the number of moles of ester formed and the number of moles of unreacted acid at equilibrium.
For \(\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightleftharpoons 2\,\mathrm{HI}\text{(g)}\), \(K_C = 49\) at a certain temperature. Equal concentrations of hydrogen and iodine are mixed and allowed to reach equilibrium, at which the concentration of HI is \(0.70\ \text{mol dm}^{-3}\). Calculate the equilibrium concentrations of \(\mathrm{H_2}\) and \(\mathrm{I_2}\).
For \(\mathrm{N_2O_4}\text{(g)} \rightleftharpoons 2\,\mathrm{NO_2}\text{(g)}\), \(K_C = 0.53\ \text{mol dm}^{-3}\) at a certain temperature. A vessel at the same temperature contains \([\mathrm{N_2O_4}] = 0.40\ \text{mol dm}^{-3}\) and \([\mathrm{NO_2}] = 0.80\ \text{mol dm}^{-3}\). Calculate \(Q_C\) and predict the direction in which the net reaction will proceed.
2.0 mol of \(\mathrm{PCl_5}\) is heated in a closed vessel of volume \(2.0\ \text{dm}^3\). At equilibrium 0.50 mol of chlorine is present. For \(\mathrm{PCl_5}\text{(g)} \rightleftharpoons \mathrm{PCl_3}\text{(g)} + \mathrm{Cl_2}\text{(g)}\), calculate (a) the percentage of \(\mathrm{PCl_5}\) dissociated and (b) the value of \(K_C\) with its unit.
For each of the following equilibria, state the value of \(\Delta n\) and write the relation between \(K_P\) and \(K_C\): (a) \(2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightleftharpoons 2\,\mathrm{SO_3}\text{(g)}\), (b) \(\mathrm{PCl_5}\text{(g)} \rightleftharpoons \mathrm{PCl_3}\text{(g)} + \mathrm{Cl_2}\text{(g)}\), (c) \(\mathrm{N_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightleftharpoons 2\,\mathrm{NO}\text{(g)}\), (d) \(2\,\mathrm{NH_3}\text{(g)} \rightleftharpoons \mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)}\).
Deduce the unit of \(K_C\) for each of the following: (a) \(\mathrm{PCl_3}\text{(g)} + \mathrm{Cl_2}\text{(g)} \rightleftharpoons \mathrm{PCl_5}\text{(g)}\), (b) \(\mathrm{CaCO_3}\text{(s)} \rightleftharpoons \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\), (c) \(2\,\mathrm{H_2S}\text{(g)} \rightleftharpoons 2\,\mathrm{H_2}\text{(g)} + \mathrm{S_2}\text{(g)}\).
For a gaseous equilibrium in which the number of moles of gas is the same on both sides of the balanced equation, \(K_P = 3.6 \times 10^{2}\) at 500 K. State the value of \(K_C\) at this temperature and justify your answer using \(K_P = K_C\,(RT)^{\Delta n}\).
For a reversible reaction at 600 K, the equilibrium constant is \(K_C = 2.5 \times 10^{2}\) and the rate constant of the reverse reaction is \(k_r = 4.0 \times 10^{-4}\). Calculate the rate constant of the forward reaction, and state whether the equilibrium mixture is richer in reactants or in products.
Limestone is heated in a closed vessel of volume \(5.0\ \text{dm}^3\) at a high constant temperature. At equilibrium the vessel is found to contain 0.50 mol of carbon dioxide together with solid calcium carbonate and solid calcium oxide. Write the \(K_C\) expression for \(\mathrm{CaCO_3}\text{(s)} \rightleftharpoons \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\) and calculate the value of \(K_C\) with its unit.
12.13 Objective Questions
- The equilibrium \(\mathrm{H_2O}\text{(l)} \rightleftharpoons \mathrm{H^{+}}\text{(aq)} + \mathrm{OH^{-}}\text{(aq)}\) is —
- dynamic
- static
- physical
- mechanical
- For the equilibrium \(\mathrm{A} \rightleftharpoons 2\mathrm{B} + \text{Heat}\), the number of ‘A’ molecules increases if —
- volume is increased
- temperature is increased
- catalyst is added
- concentration of B is decreased
- For the equilibrium \(\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{NO}\text{(g)} \rightleftharpoons 2\,\mathrm{NOCl}\text{(g)}\) the concentration of NOCl will increase if the equilibrium is disturbed by —
- adding \(\mathrm{Cl_2}\)
- removing NO
- adding NOCl
- removal of \(\mathrm{Cl_2}\)
- The relation between \(K_c\) and \(K_p\) for the reaction \(\mathrm{A}\text{(g)} + \mathrm{B}\text{(g)} \rightleftharpoons 2\,\mathrm{C}\text{(g)} + \mathrm{D}\text{(g)}\) is —
- \(K_c = \sqrt{K_p}\)
- \(K_p = K_c^2\)
- \(K_c = \dfrac{1}{\sqrt{K_p}}\)
- \(K_p / K_c = 1\)
- When the volume of the equilibrium reaction \(\mathrm{C}\text{(g)} + \mathrm{H_2O}\text{(g)} \rightleftharpoons \mathrm{CO}\text{(g)} + \mathrm{H_2}\text{(g)}\) is increased at constant temperature the equilibrium will —
- shift from left to right
- shift from right to left
- be unaltered
- can not be predicted