Chapter 4: Chemical Thermodynamics
In the 1840s, a young Englishman named James Prescott Joule, who worked in his family’s brewery in Manchester, set out to answer a question that sounds deceptively simple: when work seems to disappear, where does it go?
He built an apparatus in which falling weights turned a paddle wheel inside an insulated container of water, as shown in Fig. 4.0. As the weights came down, the paddles churned the water, and Joule measured the very small rise in temperature that followed, using thermometers of unusual precision for his time. He repeated the experiment patiently — with different liquids, different weights, different heights of fall. The result was always the same. A fixed amount of mechanical work always produced a fixed amount of heat.
Most scientists then pictured heat as a weightless fluid that flowed from hot bodies into cold ones. Joule’s measurements pointed somewhere else. Heat and work were not two different substances at all; they were two ways of transferring the same thing — energy. Nothing was destroyed when the weights fell. Energy had simply changed its form.
That single conclusion grew into the first law of thermodynamics, and from it chemists learnt to measure the heat of a reaction, to predict how much work a gas can perform, and eventually to decide whether a reaction will happen on its own.
Diagram
Fig. 4.0 – Joule’s paddle-wheel experiment: falling weight and pulley on the left, insulated vessel of water on the right containing a paddle wheel on a vertical shaft, thermometer inserted through the lid; label weight, pulley, paddle wheel, insulated water container, thermometer.
Keep these questions in mind as you read. This chapter will give you everything you need to reason out each one for yourself.
4.1 Introduction
Energy Changes All Around Us
Think of some very ordinary changes. Liquid water turns into vapour when it is heated. Solid ice turns into liquid water when it is left out of the freezer. Carbon burns in air and forms carbon dioxide, \(\mathrm{CO_2}\). Not one of these changes happens quietly on its own account — each of them is accompanied by a change in energy. Something is either absorbed from the surroundings or given out to them.
Energy does not stay locked in one form. It moves from one form into another, and chemistry gives us clear examples of this. In a dry cell, the chemical energy stored in the substances inside the cell is converted into electrical energy, which is why the cell can light a bulb. In the electroplating of metals the traffic runs the other way: electrical energy drawn from a source is converted into chemical energy as metal ions are deposited on the object being plated.
Once you notice this pattern, it becomes reasonable to say something quite general about nature — energy can be transformed from one form into another. The sequence of everyday conversions is put together in FC 4.1.
Flowchart
FC 4.1 – Interconversion of energy: chemical energy (substances in a dry cell) → electrical energy → glowing bulb; branch: electrical energy → chemical energy (electroplating of a metal object); branch: potential energy of stored water → kinetic energy of falling water.
What Thermodynamics Studies
The branch of science that keeps account of all these energy changes is thermodynamics. Physical changes such as melting and boiling, and chemical changes such as combustion, both fall within its scope.
The Question Thermodynamics Will Not Answer
It is just as important to know what thermodynamics does not do. Thermodynamics gives no information whatsoever about the rates of physical or chemical processes, nor about the mechanisms underlying them. It can tell you how much energy a reaction releases, and later in this chapter it will even tell you whether a reaction can occur on its own — but it will not tell you whether that reaction takes a microsecond or a century, and it will not tell you the step-by-step path the molecules follow.
This is why chemical kinetics is studied as a separate subject. Thermodynamics answers “how much energy?” and “will it happen at all?”; kinetics answers “how fast?” and “by what route?”. Keeping these two questions apart from the very beginning will save you a great deal of confusion later, and questions that deliberately mix them up are a favourite in assertion–reason items.
4.2 Terms Used in Thermodynamics
Every subject has a small set of words that must be pinned down before anything else can be said clearly. Thermodynamics is no exception. The terms in this section will be used in every remaining page of the chapter, so it is worth learning them properly the first time.
4.2.1 System and Surroundings
Consider a gas enclosed in a cylinder fitted with a movable piston, as shown in Fig. 4.1. Suppose we decide to study how the volume of this gas changes, and how much energy the gas releases or gains, when the pressure on it is varied by placing a certain mass on the piston. In such a study, the gas is the thing we have chosen to keep our eyes on. The gas under study is called the system.
Diagram
Fig. 4.1 – System and surroundings: a vertical cylinder containing gas closed by a movable piston with a small mass resting on it; label the gas as “Gas (system)” and label the cylinder walls, piston and the room around it as “Surrounding”.
Everything else — the cylinder itself, the piston, the air of the room, the table it stands on — is not being studied, but it is very much present, and energy or matter may pass between it and the gas. All of this is called the surroundings.
Notice that the boundary between the two is drawn by us, for our own convenience. If we had decided to study the cylinder and the gas together, the cylinder would then have been part of the system. What matters is that once you have drawn the boundary, you keep it fixed for the rest of the problem.
4.2.2 Types of System
Systems are classified by what is allowed to cross the boundary — energy, matter, both, or neither. Three types of systems are shown in Fig. 4.2, and each is illustrated with the same everyday object, a cup of hot coffee.
Diagram
Fig. 4.2 – Kinds of systems: (a) open system, an uncovered cup of hot coffee with arrows for both matter (vapour) and energy leaving; (b) closed system, the same cup covered with a saucer with an arrow for energy only; (c) isolated system, the covered cup enclosed in an insulating jacket with matter and energy both blocked; label (a) Open system, (b) Closed system, (c) Isolated system.
i. Open system: Fig. 4.2(a) shows an open cup containing hot coffee placed in a room. You observe that the coffee cools down, releasing heat to the surroundings. At the same time, water vapour from the coffee passes into the surroundings. Both energy and matter have crossed the boundary. Such a system, which exchanges both energy and matter with the surroundings, is called an open system.
ii. Closed system: In Fig. 4.2(b), the cup containing hot coffee is covered with a saucer. The coffee still cools down by giving away heat to the surroundings, but the water vapour from the coffee now does not pass into the surroundings — the saucer stops it. Such a system, which exchanges energy but not matter with the surroundings, is called a closed system.
iii. Isolated system: As you see in Fig. 4.2(c), a cup containing hot coffee is covered with a saucer and also insulated from the surroundings. Now the coffee does not cool down. Moreover, there is no escape of water vapour into the surroundings. Such a system, which does not allow the exchange of either energy or matter with the surroundings, is an isolated system.
4.2.3 Properties of System
The measurable characteristics of a system — its mass, volume, pressure, temperature and so on — are its properties. These fall into two groups, and the test that separates them is simple: ask what happens to the property if you take exactly half of the system.
i. Extensive property: A property which depends on the amount of matter present in a system is called an extensive property.
Examples: mass, volume, internal energy, heat capacity, number of moles.
ii. Intensive property: A property which is independent of the amount of matter in a system is called an intensive property.
Examples: pressure, temperature, surface tension, viscosity, melting point, boiling point, specific heat.
Apply the halving test and the difference becomes obvious. Take a beaker of water at 300 K and pour out exactly half. The volume has halved and so has the mass, so both are extensive. But the temperature of the remaining water is still 300 K, and its boiling point is still what it was, so both of these are intensive.
A useful point worth carrying forward: the ratio of two extensive properties is intensive. Density is mass divided by volume, and both mass and volume are extensive; halve the system and both halve together, leaving the density unchanged. The same reasoning explains why molar quantities, such as molar volume or molar enthalpy, are intensive even though volume and enthalpy themselves are extensive.
4.2.4 State Functions
As shown in Fig. 4.1, a certain amount of gas is enclosed in a cylinder fitted with a movable piston. Suppose the pressure of the gas is 1 bar \((P_1)\), its volume is 1 dm³ \((V_1)\) and its temperature is 300 K \((T_1)\) in the beginning. This initial state of the system is fully defined by specifying the values of these properties. Such properties, which define the state of a system, are state functions.
Now suppose the pressure of the system is increased to 2 bar \((P_2)\), the volume changes to 0.5 dm³ \((V_2)\), and the temperature is maintained at 300 K \((T_1)\). This is the final state of the system, and it is different from the initial state. A change in the state functions of the system brings about a change in its state. This is shown in Fig. 4.3.
Diagram
Fig. 4.3 – Change of state: left box labelled “Initial state” containing P₁, V₁, T₁, n; right box labelled “Final state” containing P₂, V₂, T₂, n; arrow from left to right labelled “Change in variables” above and “Change of state” below.
The final state of the system in Fig. 4.3 is described by a pressure of 2 bar \((P_2)\), a volume of 0.5 dm³ \((V_2)\) and a temperature of 300 K \((T_1)\). A system continues to be in such a state for as long as its state functions are unchanged.
Here comes the point that gives state functions their importance. How the pressure of 2 bar was attained — whether by increasing it from 1 bar to 2 bar, or by decreasing it from 5 bar to 2 bar — would not matter at all. The system’s state is the same either way.
Process and Path
The term process means a physical or chemical change in a system on going from one state to another. This can be achieved by a number of paths, by some operation. A path here refers to a sequence of situations the system undergoes during the accomplishment of the change.
In other words, the process in general may not necessarily determine the change in a unique way — several different sequences of situations can carry the system from the same initial state to the same final state. Only isothermal and adiabatic reversible processes follow a unique path to bring about the change of state of the system.
4.2.5 Path Functions
Not every quantity in thermodynamics enjoys the convenience of being a state function. Some depend very much on the route taken.
This distinction matters more than it may appear. Two students can take the same gas from the same initial state to the same final state and obtain different amounts of work and different amounts of heat, simply by choosing different paths — and yet the change in pressure, volume and temperature will be identical for both. You will see this happen quite literally in Problem 4.8 of a later section, where the same expansion carried out in three different ways gives three different values of work.
4.2.6 Thermodynamic Equilibrium
Consider again a gas enclosed in a cylinder fitted with a movable piston, shown in Fig. 4.1. The gas has temperature \(T_1\), pressure \(P_1\) and volume \(V_1\). These state functions continue to be constant as long as the piston is motionless and no heat exchange takes place. This is an equilibrium state.
Now move the piston in the upward direction so that the gas expands. During the movement it passes through states for which the pressure, volume and temperature are not specified and vary continuously. The gas would then be in a non-equilibrium state — you cannot even write down a single value of pressure for it.
Stop the movement of the piston. Suppose at this stage the pressure and volume of the gas are respectively \(P_2\) and \(V_2\), and the temperature is constant at \(T_1\). The state functions are constant once again, since the piston is motionless. The gas is then in another equilibrium state.
The thermodynamics considered here is limited to equilibrium states. This is an honest limitation and worth remembering: our equations describe the system where it starts and where it settles, not the turbulent in-between.
4.2.7 Process and Its Types
A transition from one equilibrium state to another is called a process. Processes are of different types, and each type is named after the quantity that is deliberately held constant.
i. Isothermal process: It is the process in which the temperature of the system remains constant throughout the transformation.
In such a process, heat flows from the system to the surroundings and vice versa, so as to keep the temperature constant. For a given temperature the internal energy \((U)\) of the system remains constant. Thus, \(\Delta T = 0\) and \(\Delta U = 0\).
ii. Isobaric process: In an isobaric process the pressure remains constant during the transformation. In the laboratory, chemical reactions are carried out in open containers at constant atmospheric pressure, or \(\Delta P = 0\).
iii. Isochoric process: It is a process during which the volume of the system remains constant during the transformation. A chemical reaction carried out in a closed container is isochoric. For an isochoric process \(\Delta V = 0\).
iv. Adiabatic process: A process in which there is no exchange of heat between the system and the surroundings is an adiabatic process, that is \((Q = 0)\). In an adiabatic process the system is completely insulated from the surroundings.
Because the heat has nowhere to go, it stays and shows itself as a temperature change. For an exothermic process the heat is released, which raises the temperature of the system. If the process is endothermic the temperature falls. This results in either an increase or a decrease of internal energy.
Reversible Process
v. Reversible process: Consider a gas enclosed in a cylinder fitted with a movable piston. Let the external pressure \(P_{ext}\) on the outer surface of the piston be set equal to the pressure \(P\) of the gas. Neither expansion nor compression of the gas occurs. A system is then said to be in mechanical equilibrium with the surroundings.
Now consider that \(P_{ext}\) is reduced by an infinitesimal amount. Now that \(P_{ext}\) is infinitesimally smaller than \(P\), the piston moves out slowly, allowing the gas to expand.
If instead \(P_{ext}\) is slightly increased so that it becomes infinitesimally greater than \(P\), the piston moves inward with a compression of the gas.
The word infinitesimal is doing serious work in both these sentences. It means a change so small that the system is never pushed far from balance — at every instant the inside and the outside pressures are as good as equal.
For the system in mechanical equilibrium with its surroundings, an infinitesimal change may cause the process to occur in the reverse direction. The process is then said to be thermodynamically reversible.
Features of reversible process
The driving and opposing forces differ by an infinitesimal amount.
The process can be reversed by an infinitesimal change in conditions.
A reversible process proceeds infinitely slowly and takes place in an infinite number of steps.
At the end of every step of the process, the system attains mechanical equilibrium with the surroundings.
Read feature (iii) once more, because it carries an important consequence. A process that needs an infinite number of steps would take an infinite time to complete, which means no real process is ever truly reversible. A reversible process is an idealisation — a limiting case that real processes can approach but never reach. Its value is that it sets the boundary of what is possible, and in the next section you will find that this boundary is exactly where the work done by a system becomes maximum.
4.3 Nature of Heat and Work
A system can exchange energy with its surroundings in exactly two ways — as work and as heat. Both are studied in this section, and by the end of it you will be able to attach the correct sign to each.
4.3.1 Nature of Work (W)
In mechanics, work is defined as the energy by which a body is displaced through a distance \(d\) with the application of a force. Thus,
\[W = f \times d\]
In thermodynamics the type of work involved is pressure–volume work, or \(PV\) work. That is, work is done when the system (a gas) expands or contracts against the external opposing force.
Why the Product PV Is a Work Term
At first sight, pressure multiplied by volume does not look like energy at all. A short piece of reasoning shows that it is. It may be realized that the product of pressure and volume is equal to work. Pressure is defined as force per unit area. If \(d\) is the distance, the area \(A = d^2\) and the volume \(V = d^3\). Then
\[PV = \frac{f}{A} \times V = \frac{f}{d^2} \times d^3 = f\,d = W\]
So \(PV\) carries the units of energy and behaves as a work term. This small result is the reason the whole of thermodynamics can be written using pressure and volume, which are easy to measure, rather than using force and displacement, which for a swarm of gas molecules would be impossible to track.
Now let us explore \(PV\) work with two chemical reactions carried out in a cylinder equipped with a frictionless movable piston, with a certain mass resting on its outer surface. The mass is important — it is the object in the surroundings that gets lifted or lowered, and watching it is how we know work has been done.
i. Decomposition of H₂O₂
Consider
\[2\,\mathrm{H_2O_2}\text{(l)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}\]
Look carefully at what changes here. The reactant is a liquid and one of the products is a gas, so a substance of very small volume has turned into a substance of very large volume. The gas produced in the above reaction pushes the piston upwards, so that the mass in the surroundings is raised, as shown in Fig. 4.4.
Diagram
Fig. 4.4 – Decomposition of H₂O₂: two cylinders side by side, each with a movable piston carrying a small mass; (a) “Before reaction” showing liquid H₂O₂ with the piston low, (b) “After reaction” showing O₂(g) collected below a raised piston with the mass lifted higher; upward arrow between them.
In lifting the mass, the system loses energy to the surroundings, or it performs work on the surroundings. With no heat being transferred, a loss of energy by the system is equal to the work done by the system on the surroundings. This is \(PV\) expansion.
ii. Reaction between NH₃ Gas and HCl Gas
Now, consider
\[\mathrm{NH_3}\text{(g)} + \mathrm{HCl}\text{(g)} \rightarrow \mathrm{NH_4Cl}\text{(s)}\]
Here the accounting runs the other way. Two gases are being converted into a solid. As the reaction progresses the gases are consumed, resulting in a decrease of volume. The piston moves down. A decrease in the height of the mass is shown in Fig. 4.5.
Diagram
Fig. 4.5 – Reaction between NH₃(g) and HCl(g): two cylinders side by side with pistons carrying a mass; (c) “before reaction” showing NH₃(g) + HCl(g) filling the cylinder with the piston high, (d) “after reaction” showing white solid NH₄Cl at the bottom with the piston and mass lowered; downward arrow between them.
In the process the surroundings lose energy to the system and perform work on the system. If no heat transfer occurs, the work done by the surroundings is equal to the gain in energy by the system. This is \(PV\) work.
Thus, work refers to a way by which a system exchanges energy with the surroundings.
4.3.2 Nature of Heat (Q)
Like work, heat is a form of energy by which the system exchanges energy with its surroundings. When the system and its surroundings are at different temperatures, heat either flows into the system or is let out of it.
Notice the condition attached to that statement — a difference in temperature. Heat is not something a body stores; it is something that moves, and it moves only when there is a temperature difference to drive it. Once the system and surroundings reach the same temperature, the flow of heat stops even though both still possess energy.
4.3.3 Sign Conventions of W and Q
Since energy can travel in either direction, we need a rule that tells us when to write a plus sign and when to write a minus sign. Once fixed, this rule is never changed. The energy changes for the system are considered hereafter — that is, we always take the point of view of the system, never of the surroundings.
The energy entering the system from the surroundings has a positive value, while the energy leaving the system and flowing into the surroundings is negative. This is shown in Fig. 4.6.
Diagram
Fig. 4.6 – Sign conventions: central box labelled SYSTEM; arrow entering from the left labelled “On the system” marked +W and arrow leaving to the left labelled “By the system” marked –W, both from the side marked “surroundings”; arrow entering from the right labelled “Heat absorbed” marked +Q and arrow leaving to the right labelled “Heat released” marked –Q, from the side marked “surrounding”.
\(+Q\) : Heat is absorbed by the system from the surroundings.
\(-Q\) : Heat is released by the system to the surroundings.
\(+W\) : Work is done on the system by the surroundings.
\(-W\) : Work is done by the system on the surroundings.
Note that \(W\) and \(Q\) are path functions.
The whole rule reduces to a single question, and it is worth turning that question into a habit before you attempt any numerical. Ask only: is the energy coming in or going out of the system? The decision is set out in FC 4.2.
Flowchart
FC 4.2 – Fixing the sign of Q and W: identify the system → ask “is energy crossing the boundary as heat or as work?” → if heat: energy enters the system → +Q; energy leaves the system → –Q → if work: surroundings do work on the system (gas compressed) → +W; system does work on the surroundings (gas expands) → –W.
4.4 Expression for Pressure–Volume (PV) Work
In the previous section you saw that a gas does work when it pushes a piston out, and that work is done on it when the piston is pushed in. We now turn that qualitative picture into an equation that can actually be used in a calculation.
Consider a certain amount of gas at constant pressure \(P\) enclosed in a cylinder fitted with a frictionless, rigid movable piston of area \(A\). This is shown in Fig. 4.7. Let the volume of the gas be \(V_1\) at temperature \(T\).
Diagram
Fig. 4.7 – Pressure–volume work: a vertical cylinder containing gas, the gas pressure P shown by an upward arrow at the bottom; the piston shown in two positions, the lower dashed position enclosing volume V₁ and the upper position enclosing volume V₂; the distance moved by the piston marked d with a small vertical arrow.
Building the Expression Step by Step
On expansion, the force exerted by the gas is equal to the area of the piston multiplied by the pressure with which the gas pushes against the piston. This pressure is equal in magnitude and opposite in sign to the external atmospheric pressure that opposes the movement, and has its value \(-P_{ext}\). Thus,
\[f = -P_{ext} \times A \qquad \text{.......(4.1)}\]
where \(P_{ext}\) is the external atmospheric pressure.
The minus sign is not a decoration, and it is worth pausing on. The gas pushes upward, the atmosphere pushes downward, and the two are in opposite directions. Writing the force in terms of \(P_{ext}\) with a negative sign is what will eventually make the work come out negative when the gas expands — exactly as the sign convention of Section 4.3.3 demands.
If the piston moves out a distance \(d\), then the amount of work done is equal to the force multiplied by the distance.
\[W = f \times d \qquad \text{.......(4.2)}\]
Substitution from Eq. (4.1) gives
\[W = -P_{ext} \times A \times d \qquad \text{.......(4.3)}\]
The product of the area of the piston and the distance it moves is the volume change \((\Delta V)\) in the system.
\[\Delta V = A \times d \qquad \text{.......(4.4)}\]
Combining equations (4.3) and (4.4) we write
\[W = -P_{ext}\,\Delta V \qquad \text{.......(4.5)}\]
\[W = -P_{ext}\,(V_2 - V_1)\]
where \(V_2\) is the final volume of the gas.
Reading the Signs Off the Equation
When the gas expands, work is done by the system on the surroundings. Since \(V_2 > V_1\), the quantity \(\Delta V\) is positive, and therefore \(W\) is negative. When the gas is compressed, work is done on the system by the surroundings. In this case \(V_2 < V_1\), so \(\Delta V\) is negative, and \(-P_{ext}\,\Delta V\) or \(W\) is positive.
Eq. (4.5) shows that the external pressure determines the work during expansion (or compression) of the gas. Read that sentence once more, because students often expect the pressure of the gas itself to appear in the formula. It does not. What matters is what the gas is pushing against.
This leads to an important consequence. A volume change does no work unless the system is linked to the surroundings by external pressure. If there is nothing outside to push against, the gas can change its volume as much as it likes and still transfer no energy as work.
4.4.1 Free Expansion
A free expansion means expansion against zero opposing force. Such an expansion occurs in a vacuum.
The work done by a system during such an expansion is given by Eq. (4.5), \(W = -P_{ext}\,\Delta V\). When the gas expands in a vacuum, there is no opposing force, that is \(P_{ext} = 0\), and hence \(W = 0\). In other words, no work is done when the gas expands freely in a vacuum.
4.4.2 Units of Energy and Work
Since work is a form of energy, its SI unit is the joule. It is useful to see how the joule is built out of base units, because that is what allows us to convert between the units chemists actually use.
\[1\ \text{J} = 1\ \text{kg m}^2\text{ s}^{-2} = 1\ \text{Pa m}^3\]
\[1\ \text{Pa} = 1\ \text{kg m}^{-1}\text{ s}^{-2}\]
From Eq. (4.5), \(W = -P_{ext}\,\Delta V\), if the pressure is expressed in bar and \(\Delta V\) in dm³, then the work has the units of bar dm³. This is a perfectly good unit of energy, but it needs converting before an answer can be quoted in joules.
\[1\ \text{bar} = 10^5\ \text{Pa} = 10^5\ \text{kg m}^{-1}\text{ s}^{-2}\]
\[1\ \text{dm}^3\ \text{bar} = \text{dm}^3 \times 10^5\ \text{kg m}^{-1}\text{ s}^{-2}\]
\[= \text{m}^3 \times 10^{-3} \times 10^5\ \text{kg m}^{-1}\text{ s}^{-2}\]
\[= 100\ \text{kg m}^2\text{ s}^{-2} = 100\ \text{J}\]
The step in the middle simply uses \(1\ \text{dm}^3 = 10^{-3}\ \text{m}^3\). The conversion factor that emerges — 1 dm³ bar = 100 J — will be needed in almost every numerical in this chapter, so it is worth committing to memory.
4.5 Concept of Maximum Work
Eq. (4.5) shows that the amount of work performed by a system is governed by the opposing force \((P_{ext})\). The larger the opposing force, the more work is done by the system to overcome it.
This is worth thinking about slowly, because it runs against everyday intuition. We usually feel that a task is easier when there is less resistance — and it is. But “easier” here means less work, and less work means less energy transferred. A gas that meets no resistance transfers nothing at all.
How the Opposing Force Controls the Work
If the opposing force is zero, no work is involved — this is precisely the free expansion into a vacuum you met in Section 4.4.1. With an increase of the opposing force from zero, more work will be needed by the system. When the opposing force reaches its maximum, the system performs maximum work. With the opposing force being greatest, more effort would be needed to overcome it.
But the opposing force cannot be raised without limit. Thus, when the opposing force \((P_{ext})\) becomes greater than the driving force \((P)\), the process gets reversed — the piston would move inward instead of outward, and the gas would be compressed rather than expanded. Since the opposing force cannot be greater than the driving force, it should be the maximum, that is, as close to the driving force as it is possible to get without exceeding it.
If the pressure \(P\) of the gas differs from \(P_{ext}\) by a quantity \(\Delta P\), then \(P - P_{ext} = \Delta P\) and \(P_{ext} = P - \Delta P\). Eq. (4.5) then becomes
\[W = -(P - \Delta P)\,\Delta V\]
The work \((W)\) would be maximum when \(\Delta P\) is smallest. This means the opposing force \((P_{ext})\) must be infinitesimally smaller than the driving force \((P)\) for the work to be maximum. This is required for the process to be reversible. The maximum work is obtained from the change which is thermodynamically reversible.
4.5.1 Expression for the Maximum Work
Consider \(n\) moles of an ideal gas enclosed in a cylinder fitted with a frictionless movable rigid piston. It expands isothermally and reversibly from the initial volume \(V_1\) to the final volume \(V_2\) at temperature \(T\). The expansion takes place in a number of steps, illustrated in Fig. 4.8.
Diagram
Fig. 4.8 – Reversible expansion: four cylinders in a row labelled Step 1, Step 2, Step 3, Continued; each contains gas with a piston carrying a pile of small masses; from step to step one small mass is removed and the piston rises by a small amount marked dv; the first cylinder marked V₁.
During each step, the external pressure \(P_{ext}\) is made infinitesimally smaller than the pressure \(P\) of the gas, with a gradual removal of masses from the piston. The gas expands slowly and its pressure \(P\) would decrease. The expansion continues until the pressure of the gas falls to \(P_{ext}\). Beyond this, no further expansion occurs and the system attains mechanical equilibrium with its surroundings. The volume of the gas is increased by an infinitesimal quantity dv in each single step.
The process is repeated in such a way that every time \(P_{ext}\) is lowered infinitesimally, the gas undergoes a series of infinitesimal increments in volume until the volume \(V_2\) is attained.
Work Done in One Infinitesimal Step
When the volume of a gas increases by an infinitesimal amount \(dV\) in a single step, the small quantity of work done is
\[dW = -P_{ext}\,dV \qquad \text{.......(4.6)}\]
As the expansion is reversible, \(P\) is greater by a very small quantity \(dP\) than \(P_{ext}\). Thus,
\[P - P_{ext} = dP \quad \text{or} \quad P_{ext} = P - dP \qquad \text{.......(4.7)}\]
Combining equations (4.6) and (4.7),
\[dW = -(P - dP)\,dV = -P\,dV + dP\,dV\]
Neglecting the product \(dP\,dV\), which is very small, we get
\[dW = -P\,dV \qquad \text{....... (4.8)}\]
The step just taken deserves a word. Both \(dP\) and \(dV\) are already infinitesimally small; their product is a small quantity multiplied by another small quantity, and is therefore smaller still — small enough to be discarded without affecting the result. Notice also what Eq. (4.8) has achieved: the external pressure has disappeared and the pressure of the gas itself has taken its place. This substitution is legitimate only because the process is reversible, and it is what makes the integration possible.
Adding Up All the Steps
The total amount of work done during the entire expansion from volume \(V_1\) to \(V_2\) would be the sum of the infinitesimal contributions of all the steps. Since these steps are infinite in number and each is vanishingly small, ordinary addition will not do — the total work is obtained by integration of Eq. (4.8) between the limits of the initial and final states. This is the maximum work, the expansion being reversible. Thus,
\[\int_{\text{initial}}^{\text{final}} dW = -\int_{V_1}^{V_2} P\,dV\]
Hence
\[W_{max} = -\int_{V_1}^{V_2} P\,dV \qquad \text{....... (4.9)}\]
The pressure \(P\) inside the integral is not a constant — it falls steadily as the gas expands — so it cannot simply be taken outside. It must first be expressed in terms of \(V\), and the ideal gas law does exactly that.
Using the ideal gas law
\[PV = nRT\]
so that \(P = nRT/V\), and therefore
\[W_{max} = -\int_{V_1}^{V_2} nRT\,\frac{dV}{V}\]
\[= -nRT \int_{V_1}^{V_2} \frac{dV}{V} \quad \text{because } T \text{ is constant.}\]
\[= -nRT\,\ln(V)\Big|_{V_1}^{V_2}\]
\[= -nRT\,(\ln V_2 - \ln V_1)\]
\[= -nRT\,\ln \frac{V_2}{V_1}\]
\[= -2.303\,nRT\,\log_{10}\frac{V_2}{V_1} \qquad \text{....... (4.10)}\]
Two points in that chain of steps are worth pausing over. First, the natural logarithm appears because the integral of \(dV/V\) is \(\ln V\) — that is a standard result of calculus, and it is the reason logarithms turn up in this equation and in several others later in the chapter. Second, the last line converts from the natural logarithm to the base-ten logarithm using \(\ln x = 2.303 \log_{10} x\), purely because log tables and calculators are set up for base ten.
Here \(n\) is the number of moles, \(R\) is the gas constant \(= 8.314\ \text{J K}^{-1}\text{mol}^{-1}\), \(T\) is the absolute temperature in kelvin, and \(V_1\) and \(V_2\) are the initial and final volumes.
The Same Expression in Terms of Pressure
Volumes are not always the quantities given in a problem; often the initial and final pressures are known instead. Boyle’s law supplies the bridge. At constant temperature, \(P_1V_1 = P_2V_2\) or
\[\frac{V_2}{V_1} = \frac{P_1}{P_2}\]
Replacing \(V_2/V_1\) in Eq. (4.10) by \(P_1/P_2\), we have
\[W_{max} = -2.303\,nRT\,\log \frac{P_1}{P_2} \qquad \text{.......(4.11)}\]
Note carefully that the ratio is inverted — volumes appear as final over initial, but pressures appear as initial over final. Reversing this by mistake is one of the commonest errors in board and entrance examinations, and it flips the sign of the answer.
4.6 Internal Energy (U)
Every substance is associated with a definite amount of energy. This energy stored in a substance is internal energy, denoted by \(U\).
The internal energy of a system is made up of the kinetic and potential energies of the individual particles of the system. The kinetic part comes from the constant motion of the particles — their movement from place to place, their rotation, and the vibration of atoms within each molecule. The potential part comes from the forces acting between and within the particles — the attractions between neighbouring molecules and the energy locked in chemical bonds.
Why Only the Change in U Can Be Measured
Adding up the kinetic and potential energies of every particle in even a drop of water would be an impossible task, and in practice no one attempts it. Fortunately, chemistry never needs the total. What a chemist wants to know is how much the energy changed during a process, and that is written as
\[\Delta U = U_2 - U_1\]
where \(U_1\) and \(U_2\) are the internal energies of the initial and final states, respectively. \(U\) is a state function and an extensive property.
Both of those descriptions carry consequences you should be able to use. Because \(U\) is a state function, \(\Delta U\) depends only on where the system started and where it finished, and not at all on the route taken between them — a welcome contrast to \(W\) and \(Q\), which you saw in Section 4.5 take three different values along three different paths. Because \(U\) is extensive, doubling the quantity of substance doubles its internal energy.
Keeping Track of Energy Entering and Leaving
A transfer of energy, as heat or as work, from the system would change its internal energy. To know \(\Delta U\), the energy supplied to or removed from the system needs to be monitored. There are only two possibilities, and both follow the sign convention already fixed in Section 4.3.3.
The energy transferred to the system by heating it or by performing work on it is added to the system.
The energy transferred from the system by cooling it or by performing work on the surroundings is removed from the system.
The bookkeeping is set out in FC 4.3.
Flowchart
FC 4.3 – Determining ΔU: identify each energy transfer → ask “does it enter or leave the system?” → enters (heat supplied to the system, or work done on the system) → add to internal energy, contributes a positive term → leaves (heat released by the system, or work done by the system on the surroundings) → remove from internal energy, contributes a negative term → sum all contributions → ΔU.
The following examples illustrate how to determine \(\Delta U\).
30 kJ of heat is supplied to the system. It would be added to the internal energy of the system and \(\Delta U = +30\) kJ.
If 20 kJ of work is done on the system, it is added to the internal energy of the system. Consequently, \(\Delta U = +20\) kJ.
Suppose a system releases 10 kJ of heat and performs 15 kJ of work on the surroundings. These quantities are removed from the internal energy of the system and \(\Delta U = -25\) kJ.
Notice in the third example that heat and work are simply added together with their proper signs. Neither is given any special standing. This is a hint of the law that the next section states formally.
4.7 First Law of Thermodynamics
The first law of thermodynamics is simply the conservation of energy. You have met this idea before in physics; here it is put in a form a chemist can use.
According to this law, the total energy of a system and its surroundings remains constant when the system changes from an initial state to a final state. Energy may move across the boundary, and it may change its form on the way, but the books always balance.
The law is stated in different ways, as follows.
Energy of the universe remains constant.
The total internal energy of an isolated system is constant.
Energy is neither created nor destroyed and can only be converted from one form to another.
All the above statements are equivalent. Statement (ii) follows from statement (i) once you recall from Section 4.2.2 that an isolated system exchanges neither energy nor matter with its surroundings — with nothing crossing the boundary, its internal energy cannot change.
4.7.1 Formulation of First Law of Thermodynamics
A system exchanges energy with its surroundings either by the transfer of heat or by doing work — the two routes established in Section 4.3. Energy supplied to the system increases its internal energy. On the other hand, removal of heat or work from the system decreases its internal energy.
Suppose \(Q\) is the heat supplied to the system and \(W\) is the work done on the system by the surroundings. Both are energy entering the system, so both carry a positive sign, and the internal energy of the system would increase.
The increase in internal energy of the system is equal to the sum of the quantity of heat supplied to the system and the amount of work done on the system, or
\[\Delta U = Q + W \qquad \text{.......(4.12)}\]
where \(\Delta U\) is the increase in internal energy of the system. Eq. (4.12) is the first law of thermodynamics.
For infinitesimal changes,
\[dU = dQ + dW \qquad \text{.......(4.13)}\]
4.7.2 First Law of Thermodynamics for Various Processes
Eq. (4.12) is completely general. Its usefulness comes from what happens to it when one of the four processes of Section 4.2.7 is imposed — in each case a term drops out, and a simpler and more powerful relation is left behind.
i. Isothermal process: The temperature is constant in such a process, so the internal energy is constant. Hence, \(\Delta U = 0\).
For an isothermal process
\[0 = Q + W \quad \text{or} \quad W = -Q \qquad \text{.......(4.14)}\]
The above equation implies that the heat absorbed by the system is entirely used for doing work on the surroundings. When work is done on the system by the surroundings, it results in the release of heat. Nothing is stored; whatever comes in through one door leaves through the other.
ii. Adiabatic process: In an adiabatic process, there is no exchange of heat between the system and its surroundings, that is, \(Q = 0\). Then
\[-\Delta U = -W \qquad \text{....... (4.15)}\]
Thus an increase in internal energy of the system is the work done on it. If the work is done by the system on the surroundings at the expense of its internal energy, the internal energy accompanying the adiabatic process would decrease.
This is precisely why an adiabatic compression raises the temperature of a gas — the work done on it has nowhere to go except into its internal energy, and there is no escape route for heat.
iii. Isochoric process: Substitution of \(W = -P_{ext}\,\Delta V\) into Eq. (4.12) gives
\[\Delta U = Q - P_{ext}\,\Delta V \qquad \text{....... (4.16)}\]
As the reaction is carried out in a closed container, the volume of the system is constant, or \(\Delta V = 0\), and the work term vanishes altogether:
\[\Delta U = Q_v \qquad \text{....... (4.17)}\]
Equation (4.17) shows that a change in internal energy of the system is due to heat transfer at constant volume. The subscript ‘V’ indicates that heat is transferred at constant volume. Further, \(U\) being a state function, \(Q_v\) is also a state function.
That last conclusion deserves attention. Heat in general is a path function, but heat measured under the specific restriction of constant volume becomes equal to \(\Delta U\), and therefore inherits its state-function character.
iv. Isobaric process: Usually chemical reactions are carried out in open containers under constant atmospheric pressure. In such reactions, \(\Delta V \neq 0\) — the gases produced or consumed are free to push the atmosphere back or let it in.
Replacing \(Q\) by \(Q_p\) and \(\Delta U\) by \(Q_p - P_{ext}\,\Delta V\) in equation (4.16) gives
\[Q_p = \Delta U + P_{ext}\,\Delta V \qquad \text{....... (4.18)}\]
Compare Eq. (4.18) with Eq. (4.17) and you can see exactly where the two heats differ. At constant volume the whole of the heat goes into internal energy; at constant pressure part of it is spent on pushing back the surroundings, and only the remainder appears as \(\Delta U\).
The reactions carried out in open containers under constant atmospheric pressure are so common in chemistry that a special symbol \(\Delta H\), the enthalpy change, is given to indicate heat changes occurring at constant pressure. This new quantity is developed fully in the next section.
4.8 Enthalpy (H)
At the end of the previous section a new quantity was promised, one designed for the constant-pressure conditions under which most chemistry is actually done. That quantity is enthalpy.
Enthalpy of a system is the sum of the internal energy of a system and the energy equivalent to \(PV\) work.
\[H = U + PV \qquad \text{....... (4.19)}\]
Change in enthalpy, \(\Delta H\), is also a state function, given by
\[\Delta H = H_2 - H_1 \qquad \text{....... (4.20)}\]
where \(H_1\) and \(H_2\) are the enthalpies of the initial and final states, respectively. Enthalpy is a state function because it is built entirely out of state functions — \(U\), \(P\) and \(V\) are each independent of the path, so any combination of them must be too.
Showing That ΔH Is the Heat at Constant Pressure
From Eq. (4.19), we write
\[H_1 = U_1 + P_1V_1 \quad \text{and} \quad H_2 = U_2 + P_2V_2\]
With these,
\[\Delta H = (U_2 + P_2V_2) - (U_1 + P_1V_1)\]
\[= (U_2 - U_1) + (P_2V_2 - P_1V_1)\]
\[= \Delta U + \Delta(PV) \qquad \text{.......(4.21)}\]
For constant pressure, \(P_1 = P_2 = P\) and
\[\Delta H = \Delta U + P\,\Delta V \qquad \text{....... (4.22)}\]
If the pressure inside and outside is the same, or \(P_{ext} = P\), Eq. (4.18) gives
\[Q_p = \Delta U + P\,\Delta V \qquad \text{....... (4.23)}\]
The right-hand sides of Eqs. (4.22) and (4.23) are now identical, so the left-hand sides must be equal as well. From equations (4.22) and (4.23),
\[\Delta H = Q_p \qquad \text{.......(4.24)}\]
Thus, the change in enthalpy of a system is equal to the heat transferred from it at constant pressure. \(H\) and \(Q_p\) are state functions.
This is the result that makes enthalpy worth defining at all. A chemist working with an open flask on a bench cannot easily measure \(\Delta U\), but can measure the heat given out or taken in — and that heat, at constant pressure, is \(\Delta H\).
4.8.1 Relationship between ΔH and ΔU for Chemical Reactions
At constant pressure, \(\Delta H\) and \(\Delta U\) are related as
\[\Delta H = \Delta U + P\,\Delta V\]
Whether the two are practically the same or noticeably different depends entirely on how large \(\Delta V\) is, and that in turn depends on the physical states of the substances taking part.
i. For reactions involving solids and liquids, \(\Delta V\) is usually very small — solids or liquids do not show a volume change with a change of pressure — and therefore
\[\Delta H = \Delta U\]
ii. For reactions involving gases, \(\Delta V\) cannot be neglected, and
\[\Delta H = \Delta U + P\,\Delta V\]
\[= \Delta U + P(V_2 - V_1)\]
\[\Delta H = \Delta U + PV_2 - PV_1 \qquad \text{......(4.25)}\]
where \(V_1\) is the volume of the gas phase reactants and \(V_2\) that of the gaseous products.
Volumes of gases are inconvenient to work with, since they change with temperature and pressure. The number of moles is a far more convenient measure, and the ideal gas equation lets us change over.
We assume the reactant and product behave ideally. Applying the ideal gas equation \(PV = nRT\), when \(n_1\) moles of gaseous reactants produce \(n_2\) moles of gaseous products, the ideal gas equation gives
\[PV_1 = n_1RT \quad \text{and} \quad PV_2 = n_2RT \qquad \text{.......(4.26)}\]
Substitution of Eq. (4.26) into Eq. (4.25) yields
\[\Delta H = \Delta U + n_2RT - n_1RT\]
\[= \Delta U + (n_2 - n_1)\,RT\]
\[= \Delta U + \Delta n_g RT \qquad \text{....... (4.27)}\]
where \(\Delta n_g\) is the difference between the number of moles of gaseous products and those of gaseous reactants.
Take care with two points when using Eq. (4.27). Only gaseous substances are counted in \(\Delta n_g\); solids and liquids are ignored, since their volumes are negligible. And \(R\) is normally quoted as \(8.314\ \text{J K}^{-1}\text{mol}^{-1}\) while \(\Delta H\) is usually in kJ, so one of them must be converted before the two are added.
4.8.2 Work Done in Chemical Reaction
The result just used in Problem 4.12 can be derived in general. The work done by a system at constant temperature and pressure is given by \(W = -P_{ext}\,\Delta V\). Assuming \(P_{ext} = P\),
\[W = -P\,\Delta V\]
\[= -P\,(V_2 - V_1)\]
\[= -PV_2 + PV_1\]
If the gases were ideal, using Eq. (4.26),
\[PV_1 = n_1RT \quad \text{and} \quad PV_2 = n_2RT\]
At constant temperature and pressure,
\[W = -n_2RT + n_1RT\]
\[= -(n_2 - n_1)\,RT\]
\[= -\Delta n_g RT \qquad \text{.......(4.28)}\]
The above equation gives the work done by the system in chemical reactions. Its great convenience is that no volume needs to be measured at all — a glance at the balanced equation gives \(\Delta n_g\), and the work follows.
The sign of \(W\) depends on \(\Delta V\). We consider the following cases.
If \(n_2 > n_1\), \(\Delta n_g\) is positive and \(W < 0\), or work is done by the system.
If \(n_1 > n_2\), \(\Delta n_g\) is negative and \(W > 0\), or work is done on the system.
If \(n_1 = n_2\), \(\Delta n_g = 0\) and \(W = 0\), or no \(PV\) work is done when the number of moles of reactants and products are equal.
The decision is summarised in FC 4.4.
Flowchart
FC 4.4 – Sign of W in a chemical reaction: balanced equation → count moles of gaseous products n₂ and moles of gaseous reactants n₁ (ignore solids and liquids) → compute Δn_g = n₂ – n₁ → branch: Δn_g positive (gas produced, volume rises) → W negative → work done by the system; Δn_g negative (gas consumed, volume falls) → W positive → work done on the system; Δn_g = 0 → W = 0 → no PV work.
4.9 Enthalpies of Physical Transformations
Enthalpy changes do not belong to chemical reactions alone. Melting, boiling, dissolving and even the removal of a single electron from an atom are all accompanied by definite enthalpy changes, and each has been given its own name and symbol. This section collects them.
4.9.1 Enthalpy of Phase Transition
In a phase transition, one phase of a substance is converted into another at constant temperature and pressure, without a change in chemical composition. Ice becoming water is a phase transition; the substance is \(\mathrm{H_2O}\) before and after, and only the arrangement of its molecules has altered.
i. Enthalpy of Fusion (ΔfusH)
The enthalpy change that occurs when one mole of a solid is converted into liquid without a change in temperature at constant pressure is the enthalpy of fusion. For example,
\[\mathrm{H_2O}\text{(s)} \rightarrow \mathrm{H_2O}\text{(l)}\]
\[\Delta_{fus}H = +6.01\ \text{kJ mol}^{-1}\ \text{at }0\ ^\circ\text{C}\]
When 1 mole of solid ice melts at 0 °C and 1 atm pressure, the change in enthalpy is 6.01 kJ. The same amount of heat is absorbed by the ice during the melting. The energy is spent in loosening the ordered arrangement of molecules in the crystal, not in raising the temperature — which is exactly why an ice–water mixture stays at 0 °C while it melts.
A reverse of fusion is the freezing of a solid.
\[\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_2O}\text{(s)}, \qquad \Delta_{freez}H = -6.01\ \text{kJ mol}^{-1}\ \text{at }0\ ^\circ\text{C}\]
Thus, when one mole of liquid water freezes at 0 °C, heat is evolved. Note that the magnitude is identical and only the sign has flipped. That is a direct consequence of enthalpy being a state function: reversing the direction of a change simply reverses the sign of \(\Delta H\).
ii. Enthalpy of Vaporization (ΔvapH)
It is the enthalpy change accompanying the vaporization of one mole of liquid without changing its temperature at constant pressure.
For example,
\[\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_2O}\text{(g)} \qquad \Delta_{vap}H = +40\ \text{kJ mol}^{-1}\ \text{at }100\ ^\circ\text{C}\]
\[\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_2O}\text{(g)} \qquad \Delta_{vap}H = +44\ \text{kJ mol}^{-1}\ \text{at }25\ ^\circ\text{C}\]
Thus, when one mole of water is vaporised at 1 atm pressure, the enthalpy change is +40 kJ at 100 °C and +44 kJ at 25 °C. The two values differ, which is why the temperature must always be quoted alongside an enthalpy of vaporization.
On the other hand, the condensation of vapour is accompanied by a release of heat.
\[\mathrm{H_2O}\text{(g)} \rightarrow \mathrm{H_2O}\text{(l)}, \qquad \Delta_{con}H = -40.7\ \text{kJ mol}^{-1}\ \text{at }100\ ^\circ\text{C}\]
iii. Enthalpy of Sublimation (ΔsubH)
It is the enthalpy change for the conversion of one mole of solid directly into vapour at constant temperature and pressure.
Consider
\[\mathrm{H_2O}\text{(s)} \rightarrow \mathrm{H_2O}\text{(g)}, \qquad \Delta_{sub}H = 51.08\ \text{kJ mol}^{-1},\ \text{at }0\ ^\circ\text{C}\]
The conversion of solid to vapour occurs in one or two steps — first the melting of solid into liquid, and second its vaporization. The enthalpy change is the same either way, since enthalpy is a state function. At 0 °C,
\[\mathrm{H_2O}\text{(s)} \rightarrow \mathrm{H_2O}\text{(l)} \qquad \Delta_{fus}H = 6.01\ \text{kJ mol}^{-1}\]
\[\mathrm{H_2O}\text{(l)} \rightarrow \mathrm{H_2O}\text{(g)} \qquad \Delta_{vap}H = 45.07\ \text{kJ mol}^{-1}\]
Adding the two equations, the liquid water cancels from both sides and what remains is
\[\mathrm{H_2O}\text{(s)} \rightarrow \mathrm{H_2O}\text{(g)} \qquad \Delta_{sub}H = 51.08\ \text{kJ mol}^{-1}\]
It follows that
\[\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H\]
The relationship is drawn as an energy scheme in Fig. 4.9. You are, in effect, meeting Hess’s law here for the first time, several sections before it is formally stated — the enthalpy change is the same whether the solid takes the direct route to vapour or the two-step route through the liquid.
Diagram
Fig. 4.9 – Representing Δfus H, Δvap H and Δsub H: three horizontal levels stacked vertically, labelled Solid at the bottom, Liquid in the middle and Gas at the top; upward arrow from Solid to Liquid labelled Fusion, Δfus H; upward arrow from Liquid to Gas labelled Vaporization, Δvap H; a longer upward arrow at the right running directly from Solid to Gas labelled Sublimation, Δsub H.
4.9.2 Enthalpy for the Atomic / Molecular Change
The changes considered so far involved whole molecules moving apart. We now look at enthalpy changes that accompany changes to individual atoms and molecules — losing an electron, gaining one, or being pulled apart into atoms.
i. Enthalpy of Ionization (ΔionH)
It is the enthalpy change accompanying the removal of an electron from one mole of gaseous atoms. For example,
\[\mathrm{Na}\text{(g)} \rightarrow \mathrm{Na^{+}}\text{(g)} + \mathrm{e^{-}} \qquad \Delta_{ion}H = 494\ \text{kJ mol}^{-1}\]
The equation signifies that when one mole of gas-phase atomic sodium is ionized to gas-phase \(\mathrm{Na^{+}}\) ions, the enthalpy change is 494 kJ. The same amount of heat would be required to ionize one mole of Na atoms. The value is positive because an electron is held by the attraction of the nucleus, and energy must be supplied to pull it away.
The phrase “gaseous atoms” in the definition is not a technicality to be skipped over. It is stated so that the atoms are far apart and independent of one another, and the measured energy refers to a single isolated atom losing an electron, uncomplicated by neighbours.
The electron gain enthalpy, on the other hand, gives the enthalpy change when one mole of gas-phase atoms of an element accept an electron to form a gaseous anion.
For example,
\[\mathrm{Cl}\text{(g)} + \mathrm{e^{-}} \rightarrow \mathrm{Cl^{-}}\text{(g)} \qquad \Delta_{eg}H = -349\ \text{kJ mol}^{-1}\]
The electron gain enthalpy of Cl is \(-349\ \text{kJ mol}^{-1}\). The negative sign shows that energy is released — a chlorine atom, one electron short of a stable octet, attracts an incoming electron and settles into a lower energy state.
ii. Enthalpy of Atomization (ΔatomH)
The enthalpy change accompanying the dissociation of one mole of a gaseous substance into atoms is the enthalpy of atomization.
Consider,
\[\mathrm{Cl_2}\text{(g)} \rightarrow \mathrm{Cl}\text{(g)} + \mathrm{Cl}\text{(g)}; \qquad \Delta_{atom}H = 242\ \text{kJ mol}^{-1}\]
\[\mathrm{CH_4}\text{(g)} \rightarrow \mathrm{C}\text{(g)} + 4\,\mathrm{H}\text{(g)}; \qquad \Delta_{atom}H = 1660\ \text{kJ mol}^{-1}\]
In the first case only one bond is broken, and in the second, four. That is why the two values are so different in size, and it is a useful clue to the idea of bond enthalpy, which appears later in Section 4.10.8.
iii. Enthalpy of Solution (ΔsolnH)
Enthalpy of solution is the enthalpy change in a process when one mole of a substance is dissolved in a specified amount of solvent.
\[\mathrm{NaCl}\text{(s)} + \text{aq} \rightleftharpoons \mathrm{NaCl}\text{(aq)} \qquad \Delta_{soln}H = 4\ \text{kJ mol}^{-1}\]
The words “specified amount” matter, because dissolving one mole of a salt in a little water is not quite the same as dissolving it in a great deal. To remove this ambiguity a limiting value is defined: the enthalpy of solution at infinite dilution is the enthalpy change when one mole of substance is dissolved in an infinite amount of solvent, so that adding any more solvent produces no further heat change.
The two steps of dissolution. An ionic compound dissolves in water in two steps, and each has its own enthalpy change. The sequence is set out in FC 4.5.
- The ions are separated from the molecule.
\[\mathrm{MX}\text{(s)} \rightarrow \mathrm{M^{+}}\text{(g)} + \mathrm{X^{-}}\text{(g)}\]
The enthalpy change for this step is the crystal lattice enthalpy, \(\Delta_L H\), which is always positive. Pulling oppositely charged ions apart against their strong electrostatic attraction always costs energy.
- The ions are hydrated, with water molecules surrounding them.
\[\mathrm{M^{+}}\text{(g)} + x\,\mathrm{H_2O} \rightarrow [\mathrm{M(H_2O)_x}]^{+}\]
\[\mathrm{X^{-}}\text{(g)} + y\,\mathrm{H_2O} \rightarrow [\mathrm{X(H_2O)_y}]^{-}\]
The enthalpy change for this step is always negative and is called the enthalpy of hydration, \(\Delta_{hyd}H\). New attractions are being formed between the ions and the polar water molecules, and forming attractions always releases energy.
Flowchart
FC 4.5 – Dissolution of an ionic compound MX in water: solid MX(s) → step 1, lattice broken, ions separated into M⁺(g) + X⁻(g); enthalpy change = lattice enthalpy Δ_L H, always positive → step 2, gaseous ions surrounded by water molecules giving [M(H₂O)x]⁺ and [X(H₂O)y]⁻; enthalpy change = hydration enthalpy Δ_hyd H, always negative → overall Δ_soln H = Δ_L H + Δ_hyd H; branch: if Δ_L H greater in magnitude → dissolution endothermic, solution cools; if Δ_hyd H greater in magnitude → dissolution exothermic, solution warms.
The enthalpy of solution is the sum of the crystal lattice enthalpy and the enthalpy of hydration.
\[\Delta_{soln}H = \Delta_L H + \Delta_{hyd}H\]
For NaCl, \(\Delta_L H = +790\ \text{kJ/mol}\) and
\[\Delta_{hyd}H = -786\ \text{kJ/mol}\]
The enthalpy of solution of NaCl is
\[\Delta_{soln}H\ (\mathrm{NaCl}) = (+790 - 786)\ \text{kJ/mol}\]
\[= +4\ \text{kJ/mol}\]
Look at how the answer arises. Two very large numbers, nearly 800 kJ each, have almost completely cancelled, leaving a small remainder of 4 kJ. This is why common salt dissolves in water with only a barely noticeable cooling — and why small differences in either quantity decide whether a salt dissolves with warming or with cooling.
4.10 Thermochemistry
Thermochemistry deals with enthalpy changes in chemical reactions. Having built the quantity \(\Delta H\) in Section 4.8, we can now put it to work on reactions — measuring it, tabulating it, and, most usefully of all, calculating it for reactions that have never been carried out in a laboratory.
4.10.1 Enthalpy of Chemical Reaction (ΔrH)
Consider a general reaction,
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\]
The enthalpy change for the reaction is
\[\Delta_r H = (cH_C + dH_D) - (aH_A + bH_B)\]
where \(H_A\), \(H_B\), \(H_C\) and \(H_D\) are the molar enthalpies of A, B, C and D, respectively. Each molar enthalpy is multiplied by the coefficient of that substance in the balanced equation, because that coefficient tells us how many moles are involved. We write
\[\Delta_r H = \sum H_{\text{products}} - \sum H_{\text{reactants}} \qquad \text{....... (4.29)}\]
Two conditions in that definition are exam-critical and are often overlooked. The physical state must be definite — the value changes if water is produced as liquid rather than as vapour. And the amounts are fixed by the balanced equation, so a \(\Delta_r H\) quoted for one equation cannot be transferred to the same reaction written with different coefficients.
4.10.2 Exothermic and Endothermic Reactions
The enthalpy of a reaction can be positive or negative, depending on \(\sum H_{\text{products}}\) and \(\sum H_{\text{reactants}}\). Everything follows from which of the two sums is larger.
Thus if \(\sum H_{\text{products}} > \sum H_{\text{reactants}}\), \(\Delta_r H\) is positive, and this signifies that the reaction is endothermic. The products hold more enthalpy than the reactants did, and the extra had to be drawn in from the surroundings as heat.
On the other hand, if \(\sum H_{\text{products}} < \sum H_{\text{reactants}}\), \(\Delta_r H\) is negative, which means that heat is released and the reaction is exothermic.
For example,
\[\mathrm{N_2}\text{(g)} + 2\,\mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{NO_2}\text{(g)}, \qquad \Delta_r H = 66.4\ \text{kJ (endothermic)}\]
\[2\,\mathrm{KClO_3}\text{(s)} \rightarrow 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}, \qquad \Delta_r H = -78\ \text{kJ (exothermic)}\]
The two situations are drawn as enthalpy-level diagrams in Fig. 4.10, and reading such a diagram correctly is a skill worth practising.
Diagram
Fig. 4.10 – Enthalpy level diagrams: left panel labelled “Exothermic reaction” with a higher horizontal level marked Reactants and a lower level marked Products, downward arrow between them labelled ΔrH negative, “heat released”; right panel labelled “Endothermic reaction” with a lower level marked Reactants and a higher level marked Products, upward arrow labelled ΔrH positive, “heat absorbed”; enthalpy H marked on the vertical axis of both panels.
4.10.3 Standard Enthalpy of Reaction (ΔrH°)
To compare the enthalpy changes of different reactions, they have to be reported under a similar set of conditions. An enthalpy change measured on a cold morning in a hill station and one measured in a coastal laboratory in May would not otherwise be comparable, so chemists have agreed on a reference condition.
Thermodynamic standard state: The standard state of a substance is the form in which the substance is most stable at a pressure of 1 bar and at a temperature of 298 K. If the reaction involves species in solution, its standard state refers to 1 M concentration.
Standard states of certain elements and compounds are \(\mathrm{H_2}\text{(g)}\), \(\mathrm{Hg}\text{(l)}\), \(\mathrm{Na}\text{(s)}\), \(\mathrm{C}\)(graphite), \(\mathrm{C_2H_5OH}\text{(l)}\), \(\mathrm{H_2O}\text{(l)}\), \(\mathrm{CaCO_3}\text{(s)}\) and \(\mathrm{CO_2}\text{(g)}\), all referring to 1 bar and 25 °C.
Notice that carbon is listed as graphite and not as diamond. The standard state is always the most stable form under those conditions, and for carbon at 1 bar and 298 K that form is graphite.
4.10.4 Thermochemical Equation
A thermochemical equation carries more information than an ordinary balanced equation — it tells you not only what reacts with what, but in what physical state, in what amount, and with what accompanying enthalpy change.
Here follow the guidelines for writing thermochemical equations.
Consider the balanced equation for the reactants and products.
The value and appropriate sign of the enthalpy change is given on the right hand side. This value is \(\Delta_r H^\circ\).
The physical states of reactants and products are specified by the letters s (solid), l (liquid), g (gas) and aq (aqueous). The \(\Delta_r H^\circ\) value refers to the physical states of the substances that appear in the equation.
The given value of \(\Delta_r H^\circ\) assumes that the reaction occurs in a given direction. \(\Delta_r H^\circ\) for the reverse reaction is equal in magnitude and opposite in sign to that of the forward reaction. An exothermic reaction on reversal becomes endothermic, and vice versa.
When the coefficients indicating the number of moles of all substances in a thermochemical equation are multiplied or divided by a certain numerical factor, the corresponding \(\Delta_r H^\circ\) needs to be multiplied or divided by the same factor.
Guidelines (iv) and (v) both follow from a single fact you already know — enthalpy is a state function and an extensive property. Reversing a change reverses the sign of \(\Delta H\); doubling the amount of substance doubles the enthalpy change.
Example of thermochemical equation
\[\mathrm{CH_4}\text{(g)} + 2\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)}, \qquad \Delta_r H^\circ = -890\ \text{kJ}\]
The equation signifies that when 1 mole of gaseous \(\mathrm{CH_4}\) and 2 moles of \(\mathrm{O_2}\) in their standard states produce 1 mole of \(\mathrm{CO_2}\) gas and 2 moles of liquid water, also in their standard states, the enthalpy change would be \(-890\) kJ.
\[\mathrm{C_2H_2}\text{(g)} + \tfrac{5}{2}\,\mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{CO_2}\text{(g)} + \mathrm{H_2O}\text{(l)}, \qquad \Delta_r H^\circ = -1300\ \text{kJ}\]
write the thermochemical equations when i. the coefficients of the substances are multiplied by 2, and ii. the equation is reversed.
4.10.5 Standard Enthalpy of Formation (ΔfH°)
Consider
\[\mathrm{H_2}\text{(g)} + \tfrac{1}{2}\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{H_2O}\text{(l)}, \qquad \Delta_r H^\circ = -286\ \text{kJ}\]
For the reaction where one mole of liquid water in its standard state is formed from \(\mathrm{H_2}\) and \(\mathrm{O_2}\) gases in their standard states, the enthalpy change for the reaction would be the standard enthalpy of formation of water. \(\Delta_f H\) of water is \(-286\ \text{kJ mol}^{-1}\).
Every word of that definition is doing work. Exactly one mole of the compound must be formed, which is why the fractional coefficient \(\tfrac{1}{2}\) appears in front of \(\mathrm{O_2}\) above rather than writing the equation with whole numbers. The starting materials must be elements, not other compounds. And everything must be in its standard state.
The formation of one mole of \(\mathrm{CH_4}\) in its standard state from the elements carbon and hydrogen in their standard states is represented by
\[\mathrm{C}\text{(graphite)} + 2\,\mathrm{H_2}\text{(g)} \rightarrow \mathrm{CH_4}\text{(g)}, \qquad \Delta_r H^\circ = -74.8\ \text{kJ}\]
or \(\Delta_f H^\circ (\mathrm{CH_4}) = -74.8\ \text{kJ mol}^{-1}\).
A fair question arises here, and it is worth confronting directly. The reaction to form a substance from its constituent elements is often hypothetical. It is not possible to simply combine carbon and hydrogen in the laboratory to prepare methane — the reaction as written does not run. The enthalpy of reaction for the formation of \(\mathrm{CH_4}\) is instead obtained indirectly, by knowing the standard enthalpy changes of reactions that can be carried out and combining them. The value \(-74.8\ \text{kJ mol}^{-1}\) therefore corresponds to a hypothetical reaction, and it is no less useful for that. How such an indirect calculation is performed is the subject of Section 4.10.9.
4.10.6 Standard Enthalpy of Reaction from Standard Enthalpies of Formation
The standard enthalpies of formation of compounds are used to determine the standard enthalpies of reactions. This is one of the most heavily used calculations in the whole chapter, and it rests on two simple statements.
Calculations of \(\Delta_r H^\circ\) from \(\Delta_f H^\circ\) of compounds are based on the following.
- The standard enthalpy of formation of an element is zero.
\[\Delta_f H^\circ(\mathrm{H_2}) = \Delta_f H^\circ(\mathrm{Cl_2}) = \Delta_f H^\circ(\mathrm{C}) = 0\]
This is not a measured fact but a sensible convention. Forming an element from itself involves no change at all, so the enthalpy change must be zero. It gives us a common baseline from which every compound’s enthalpy can be counted.
- The standard enthalpy of formation of a compound is equal to its standard enthalpy.
\[\Delta_f H^\circ(\text{compound}) = H^\circ(\text{compound})\]
Consider the reaction
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\]
The standard enthalpy of the reaction is given by
\[\Delta_r H^\circ = (cH^\circ_C + dH^\circ_D) - (aH^\circ_A + bH^\circ_B)\]
\[= [c\,\Delta_f H^\circ(\mathrm{C}) + d\,\Delta_f H^\circ(\mathrm{D})] - [a\,\Delta_f H^\circ(\mathrm{A}) + b\,\Delta_f H^\circ(\mathrm{B})]\]
\[= \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants}) \qquad \text{....... (4.30)}\]
4.10.7 Standard Enthalpy of Combustion (ΔcH°)
Consider the reaction
\[\mathrm{C_2H_2}\text{(g)} + \tfrac{5}{2}\,\mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{CO_2}\text{(g)} + \mathrm{H_2O}\text{(l)}, \qquad \Delta_r H^\circ = -1300\ \text{kJ}\]
In the above reaction, the standard enthalpy change of the oxidation reaction, \(-1300\) kJ, is the standard enthalpy of combustion of \(\mathrm{C_2H_2}\text{(g)}\).
The word “completely” is important. Combustion must be carried through to the fully oxidised products — carbon to \(\mathrm{CO_2}\) and hydrogen to \(\mathrm{H_2O}\). If carbon monoxide were formed instead, the oxidation would be incomplete and the enthalpy change would be different. Enthalpies of combustion are always negative, since burning always releases heat.
4.10.8 Bond Enthalpy
So far the enthalpy of a reaction has been built from tabulated values for whole substances. There is another route, which looks instead at what actually happens during a reaction — bonds break and bonds form.
Consider the reaction
\[\mathrm{H_2}\text{(g)} \rightarrow \mathrm{H}\text{(g)} + \mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 436.4\ \text{kJ}\]
It shows that the H–H bond in one mole of \(\mathrm{H_2}\text{(g)}\) is decomposed, producing gaseous H atoms. The enthalpy change of the reaction, 436.4 kJ, is the bond enthalpy of the H–H bond.
Bond enthalpies are always positive, because breaking a bond always requires energy to be supplied. Conversely, forming a bond always releases energy — and this is the fact behind every exothermic reaction.
HCl molecule dissociates as
\[\mathrm{HCl}\text{(g)} \rightarrow \mathrm{H}\text{(g)} + \mathrm{Cl}\text{(g)}, \qquad \Delta_r H^\circ = 431.9\ \text{kJ}\]
\[\Delta H^\circ(\text{H–Cl bond}) = 431.9\ \text{kJ mol}^{-1}\]
Average Bond Enthalpy in Polyatomic Molecules
Each covalent bond in polyatomic molecules is associated with its own specific bond enthalpy. The thermochemical equation for the dissociation of \(\mathrm{H_2O}\) molecules is
\[\mathrm{H_2O}\text{(g)} \rightarrow 2\,\mathrm{H}\text{(g)} + \mathrm{O}\text{(g)}, \qquad \Delta_r H^\circ = 927\ \text{kJ}\]
The above equation implies that the enthalpy change for breaking of two O–H bonds in one mole of gaseous \(\mathrm{H_2O}\) molecules is 927 kJ. Although the two O–H bonds in \(\mathrm{H_2O}\) are identical, the energies needed to break the individual O–H bonds are different.
This sounds contradictory until you look at the two steps separately. The bonds in \(\mathrm{H_2O}\) are broken in successive steps as shown.
\(\mathrm{H_2O}\text{(g)} \rightarrow \mathrm{OH}\text{(g)} + \mathrm{H}\text{(g)} \qquad \Delta_r H^\circ = 499\ \text{kJ}\)
\(\mathrm{OH}\text{(g)} \rightarrow \mathrm{O}\text{(g)} + \mathrm{H}\text{(g)} \qquad \Delta_r H^\circ = 428\ \text{kJ}\)
Adding the two steps:
\[\mathrm{H_2O}\text{(g)} \rightarrow 2\,\mathrm{H}\text{(g)} + \mathrm{O}\text{(g)} \qquad \Delta_r H^\circ = 927\ \text{kJ}\]
The reason the two steps differ is that the second bond is being broken in a different species altogether — an OH radical, not a water molecule — and its surroundings within the molecule are no longer the same. The total enthalpy change, 927 kJ, is therefore not twice as large as the enthalpy of the first O–H bond broken. What, then, is the enthalpy of the O–H bond in the \(\mathrm{H_2O}\) molecule?
For polyatomic molecules, the average bond enthalpy of a particular bond would be considered. Thus, the average bond enthalpy of the O–H bond \(= 927/2 = 463.5\) kJ, or \(\Delta H^\circ(\text{O–H}) = 463.5\ \text{kJ mol}^{-1}\).
The same behaviour appears in methane, and on a larger scale. In the \(\mathrm{CH_4}\) molecule there are four identical C–H bonds, and yet the bond enthalpies of all 4 C–H bonds are different. The breaking of C–H bonds in \(\mathrm{CH_4}\) occurs in four steps as follows.
\[\mathrm{CH_4}\text{(g)} \rightarrow \mathrm{CH_3}\text{(g)} + \mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 427\ \text{kJ}\]
\[\mathrm{CH_3}\text{(g)} \rightarrow \mathrm{CH_2}\text{(g)} + \mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 439\ \text{kJ}\]
\[\mathrm{CH_2}\text{(g)} \rightarrow \mathrm{CH}\text{(g)} + \mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 452\ \text{kJ}\]
\[\mathrm{CH}\text{(g)} \rightarrow \mathrm{C}\text{(g)} + \mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 347\ \text{kJ}\]
Adding all four steps gives
\[\mathrm{CH_4}\text{(g)} \rightarrow \mathrm{C}\text{(g)} + 4\,\mathrm{H}\text{(g)}, \qquad \Delta_r H^\circ = 1665\ \text{kJ}\]
Average C–H bond enthalpy \(= 1665\ \text{kJ}/4 = 416\) kJ
Hence, \(\Delta_r H^\circ(\text{C–H}) = 416\ \text{kJ mol}^{-1}\).
Reaction and Bond Enthalpies
In a chemical reaction, bonds are broken and formed. The enthalpies of reactions involving substances having covalent bonds are calculated by knowing the bond enthalpies of the reactants and those in the products. The calculations assume that all the bonds of a given type are identical — an assumption you have just seen is only approximately true.
Enthalpy change of a reaction
\[\Delta_r H^\circ = \sum \Delta H^\circ (\text{reactant}) - \sum \Delta H^\circ (\text{product}) \qquad \text{....... (4.31)}\]
Take a moment over the order of the terms, because it is the reverse of Eq. (4.30) and mixing the two up is a common source of lost marks. In Eq. (4.30) we subtracted reactants from products, because we were dealing with enthalpies of substances. Here we subtract products from reactants, because we are dealing with bonds broken (a cost, in the reactants) and bonds formed (a return, in the products).
Consider the reaction
\[\mathrm{H_2}\text{(g)} + \mathrm{I_2}\text{(g)} \rightarrow 2\,\mathrm{HI}\text{(g)}\]
The enthalpy is given by
\[\Delta_r H^\circ = [\Delta H^\circ(\text{H–H}) + \Delta H^\circ(\text{I–I})] - [2\,\Delta H^\circ(\text{H–I})]\]
If the reactants and products are diatomic molecules, Eq. (4.31) gives accurate results, since the bond enthalpies are known accurately. For reactions involving polyatomic molecules, the reaction enthalpies calculated via Eq. (4.31) would be approximate and refer to average bond enthalpies.
4.10.9 Hess’s Law of Constant Heat Summation
The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy is a state function. The enthalpy change of a reaction depends only on the initial and final states, and not on the path by which the reaction occurs. This is illustrated as an energy cycle in Fig. 4.11.
Diagram
Fig. 4.11 – Hess’s law cycle: a horizontal arrow at the top from a box marked Reactants to a box marked Products labelled “direct route, ΔrH”; below it, two arrows forming a path from Reactants down to an intermediate box and then up to Products, labelled ΔrH₁ and ΔrH₂; caption note ΔrH = ΔrH₁ + ΔrH₂.
To determine the overall equation of a reaction, the reactants and products in the individual steps are added or subtracted like algebraic entities. Species appearing on opposite sides of the steps cancel, exactly as terms cancel in an algebraic sum.
Consider the synthesis of \(\mathrm{NH_3}\).
\(2\,\mathrm{H_2}\text{(g)} + \mathrm{N_2}\text{(g)} \rightarrow \mathrm{N_2H_4}\text{(g)}, \qquad \Delta_r H_1^\circ = +95.4\ \text{kJ}\)
\(\mathrm{N_2H_4}\text{(g)} + \mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}, \qquad \Delta_r H_2^\circ = -187.6\ \text{kJ}\)
Adding the two, the hydrazine formed in step (i) is consumed in step (ii) and cancels out:
\[3\,\mathrm{H_2}\text{(g)} + \mathrm{N_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}, \qquad \Delta_r H^\circ = -92.2\ \text{kJ}\]
The sum of the enthalpy changes for steps (i) and (ii) is equal to the enthalpy change for the overall reaction.
Application of Hess’s law
Hess’s law has been useful to calculate the enthalpy changes for reactions with their enthalpies being not known experimentally. This includes reactions that are too slow to measure, reactions that give side products, and reactions — like the direct formation of methane from its elements — that cannot be made to happen at all.
The working method is set out in FC 4.6.
Flowchart
FC 4.6 – Using Hess’s law: write the target equation and identify which substance must appear on which side → examine each given equation → branch: if a substance appears on the wrong side, reverse that equation and reverse the sign of its ΔrH°; if the coefficient does not match, multiply or divide the equation and its ΔrH° by the same factor → add the adjusted equations so that all intermediates cancel → the sum of the adjusted ΔrH° values is the ΔrH° of the target reaction.
\(\mathrm{C}\text{(graphite)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)}, \qquad \Delta_r H^\circ = -393.5\ \text{kJ}\)
\(\mathrm{CO}\text{(g)} + \tfrac{1}{2}\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)}, \qquad \Delta_r H^\circ = -283.0\ \text{kJ}\)
4.11 Spontaneous (Irreversible) Process
Everything so far has been about how much energy a change involves. We now come to a different and more searching question: will the change happen at all, if simply left to itself?
Spontaneous processes have a natural tendency to occur and do not require any external influence for their occurrence.
Four familiar examples make the idea concrete.
- Aqueous NaOH and HCl solutions are mixed together. NaOH immediately combines with HCl to form NaCl and water.
\[\mathrm{NaOH}\text{(aq)} + \mathrm{HCl}\text{(aq)} \rightarrow \mathrm{NaCl}\text{(aq)} + \mathrm{H_2O}\text{(l)}, \qquad \Delta_r H^\circ = -57\ \text{kJ}\]
No external force or energy is required for the reaction to occur. This is spontaneous. The process stops when the HCl or the NaOH is consumed. Note also what does not happen: when NaCl is dissolved in water, it does not react with water to produce NaOH and HCl. The change runs one way only.
Water flows from a higher level to a lower level. It is not necessary to apply an external force. It is a spontaneous process. The flow ceases when the two levels become equal, or when equilibrium is reached.
Ice melts spontaneously above 0 °C.
Hot coffee in a cup placed in a room cools down, releasing heat to the surroundings. This is spontaneous.
Key points of spontaneous process
It occurs of its own and does not require any external agency to occur.
It proceeds in one direction and cannot take place in the opposite direction unless an external stimulant is present.
The spontaneous processes can be rapid or slow — spontaneity is not concerned with the rate of the reaction.
The process continues till equilibrium is reached. The spontaneous (natural) processes tend to occur in a direction that leads to equilibrium.
Point (iii) is the one students most often get wrong, and it is worth stating plainly: spontaneous does not mean fast. Thermodynamics tells you whether a change is allowed to happen; it says nothing whatever about how long it will take. That limitation was announced right at the start of this chapter, in Section 4.1.
4.11.1 Energy and Spontaneity
A reasonable first guess is that reactions happen when they release energy. The spontaneous reaction takes place in a direction in which the energy of the system is lowered. It is accompanied by a release of energy. The reaction between NaOH and HCl is exothermic \((\Delta_r H^\circ = -57\ \text{kJ})\) and is spontaneous.
On the other hand, consider these two cases.
Ice melts spontaneously above 0 °C by absorbing heat from the surroundings. It is endothermic.
Likewise, NaCl dissolves spontaneously in water with the absorption of heat from the surroundings.
\[\mathrm{NaCl}\text{(s)} + \text{aq} \rightarrow \mathrm{Na^{+}}\text{(aq)} + \mathrm{Cl^{-}}\text{(aq)} \qquad \Delta H^\circ = +3.9\ \text{kJ mol}^{-1}\]
These are endothermic and spontaneous. It is therefore clear that exothermicity is not the sufficient criterion for deciding spontaneity. There needs to be another factor to describe spontaneity.
The guess has failed, and it has failed in an instructive way. Something other than energy must also be pushing these changes forward.
4.11.2 Entropy
To know what entropy is, consider the following processes.
In the solid state, water molecules in ice are arranged in a definite order.
When ice melts, this highly crystalline arrangement of water molecules collapses. The molecules become free in the liquid state. An ordered state thus tends to become more disordered.
When liquid water vaporises, gaseous water molecules move freely and randomly in the available space. A less disordered state becomes highly disordered, as shown in Fig. 4.12.
Diagram
Fig. 4.12 – Increasing disorder: three panels side by side, each showing circles representing water molecules; left panel with circles packed in a regular close-packed array labelled “Ice, highly ordered state”; middle panel with circles loosely and irregularly arranged labelled “H₂O(l), disordered state”; right panel with circles widely and randomly scattered labelled “H₂O(g), highly disordered state”.
During the melting of ice or the vaporisation of liquid water, the disorder or randomness increases. The disorder or randomness is measured by entropy, denoted by S. The greater the disorder of a system, the larger is its entropy. The melting of ice and vaporisation of liquid water show that disorder, and hence entropy, of a substance increases as it passes from solid to liquid to gas.
In both processes the entropy change \(\Delta S > 0\). Look at the following processes.
- Dissolution of solid \(\mathrm{I_2}\) in water:
\[\mathrm{I_2}\text{(s)} + \text{aq.} \rightarrow \mathrm{I_2}\text{(aq)} \qquad (\Delta S \text{ is positive})\]
ordered state \(\rightarrow\) disordered state
When solid iodine dissolves in water, the \(\mathrm{I_2}\) molecules move randomly. Thus the disorder, and hence the entropy of the system, increases, or \(\Delta S\) is positive for the dissolution process.
- Dissociation of \(\mathrm{H_2}\) molecule into atoms:
\[\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{H}\text{(g)} \qquad (\Delta S \text{ is positive})\]
One mole of \(\mathrm{H_2}\) gas is converted into two H atoms. A larger disorder is associated with separated H atoms than with the \(\mathrm{H_2}\) molecule. Thus, disorder and hence entropy increases, or \(\Delta S\) is positive.
\(2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}\)
\(\mathrm{CaCO_3}\text{(s)} \xrightarrow{\ \Delta\ } \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\)
Quantitative Definition of Entropy
Entropy is a measure of molecular disorder or randomness. An entropy change of a system is equal to the amount of heat transferred \((Q_{rev})\) to it in a reversible manner, divided by the temperature in kelvin \(T\) at which the transfer takes place. Thus
\[\Delta S = \frac{Q_{rev}}{T} \qquad \text{....... (4.32)}\]
The \(\Delta S\) is thus expressed in \(\text{J K}^{-1}\).
Entropy, or its change \(\Delta S\), is a state function and depends on the initial and final states of the system and not on the path connecting the two states. Notice that \(Q_{rev}\) carries the subscript “rev” — the heat must be the reversible heat, the maximum-work path of Section 4.5. This is what rescues \(\Delta S\) from inheriting the path dependence of ordinary heat.
Two features of Eq. (4.32) deserve explanation, since the equation is not merely a definition pulled out of the air.
When heat is added to a system, the molecular motions increase owing to the increase of their kinetic energies. This results in increased molecular disorder, and thus in the entropy of the system. \(\Delta S\) is proportional to \(Q_{rev}\) — which is why \(Q_{rev}\) appears in the numerator.
The effectiveness of the addition of heat in increasing randomness depends on temperature. If a certain amount of heat is added to a system at a higher temperature, then the disorder caused is less than that caused by adding the same amount of heat to the system at a lower temperature. Thus, \(\Delta S\) relates reciprocally to the temperature at which the heat is added — which is why \(T\) appears in the denominator.
The reasoning behind (ii) is worth carrying with you. A system that is already hot is already highly disordered, so a given quantity of heat makes proportionately little difference to it. The same heat delivered to a cold, quiet system disturbs it far more.
4.11.3 Entropy and Spontaneity (Second Law of Thermodynamics)
Look at the following examples.
The entropy increases when ice melts above 0 °C and water vaporizes at 100 °C. Both are spontaneous.
Consider the spontaneous reaction at room temperature
\[2\,\mathrm{H_2O_2}\text{(l)} \rightarrow 2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)}, \qquad \Delta S = +126\ \text{J K}^{-1}\]
Entropy increases due to the formation of \(\mathrm{O_2}\) gas.
From the above examples, it is clear that the entropy of the system increases in spontaneous processes. But now consider the reaction
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}, \qquad \Delta S = -327\ \text{J K}^{-1}\]
The entropy of the system decreases. Note that the reaction is nevertheless spontaneous — hydrogen and oxygen burn readily to form water.
So the entropy of the system is not by itself a reliable guide either, any more than the enthalpy was. The resolution lies in widening our view.
4.11.4 Second Law of Thermodynamics
For the process to be spontaneous,
\[\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0 \qquad (4.33)\]
The word total is the key. The entropy of a system may fall, provided the entropy of the surroundings rises by more.
Consider
\[2\,\mathrm{H_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O}\text{(l)}\]
\(\Delta S = -327\ \text{J K}^{-1}\), and \(\Delta H = -572\) kJ (both at 298 K).
To find \(\Delta S_{total}\), we need to know \(\Delta S_{surr}\). \(\Delta H\) for the reaction is \(-572\) kJ. When 2 moles of \(\mathrm{H_2}\) and 1 mole of \(\mathrm{O_2}\) gas combine to form 2 moles of liquid water, 572 kJ of heat is released, which is received by the surroundings at constant pressure and 298 K. The entropy change of the surroundings is
\[\Delta S_{surr} = \frac{Q_{rev}}{T} = \frac{572 \times 10^3\ \text{J}}{298\ \text{K}} = 1919\ \text{J K}^{-1}\]
\[\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}\]
\[= -327\ \text{J K}^{-1} + 1919\ \text{J K}^{-1}\]
\[= +1592\ \text{J K}^{-1}\]
\[\Delta S_{total} > 0.\]
The reaction is thus spontaneous. It follows that to decide the spontaneity of reactions, we need to consider the entropy of the system and its surroundings.
Look closely at where the large positive term came from. The reaction is strongly exothermic, and all that heat is delivered to the surroundings, stirring them into greater disorder. The modest ordering within the system, \(-327\ \text{J K}^{-1}\), is swamped by it.
The total entropy increases during a spontaneous process that finally reaches equilibrium. The equilibrium corresponds to maximum total entropy. The total entropy change, \(\Delta S_{total}\), must be zero for a process at equilibrium.
From the above,
\(\Delta S_{total} > 0\), the process is spontaneous.
\(\Delta S_{total} < 0\), the process is nonspontaneous.
\(\Delta S_{total} = 0\), the process is at equilibrium.
4.11.5 Gibbs Energy
As pointed out in the preceding section, it is necessary to determine both \(\Delta S_{sys}\) and \(\Delta S_{surr}\) for predicting the spontaneity of a reaction. This is inconvenient. We are more interested in the system, the reaction mixture, and measuring what happens to the entire surroundings is hardly practical. It is therefore convenient to consider the criterion of spontaneity in terms of the thermodynamic properties of the system alone.
This problem was solved by the American theoretician J. W. Gibbs. He introduced a new thermodynamic property called Gibbs energy, usually denoted by G.
The Gibbs energy is defined as
\[G = H - TS \qquad \text{....... (4.34)}\]
where \(H\) is the enthalpy and \(S\) the entropy of the system. Since \(H\), \(T\) and \(S\) are state functions, \(G\) is a state function. A change in Gibbs energy depends on the initial and final states of the system and not on a path connecting the two states.
The change in Gibbs energy at constant temperature and constant pressure is given by
\[\Delta G = \Delta H - T\,\Delta S \qquad \text{....... (4.35)}\]
4.11.6 Gibbs Energy and Spontaneity
The total entropy change that accompanies a process is given by
\[\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}\]
\[= \Delta S + \Delta S_{surr} \qquad \text{....... (4.36)}\]
The subscript sys that refers to the system is dropped hereafter.
Relation between ΔG and ΔStotal
According to the second law of thermodynamics, for a process to be spontaneous, \(\Delta S_{total} > 0\).
If \(\Delta H\) is the enthalpy change accompanying a reaction (the system), the enthalpy change of the surroundings is \(-\Delta H\) — whatever heat the system gives out, the surroundings take in. With this,
\[\Delta S_{surr} = -\frac{\Delta H}{T} \qquad \text{....... (4.37)}\]
Substituting the above into Eq. (4.36),
\[\Delta S_{total} = \Delta S - \frac{\Delta H}{T} \qquad \text{....... (4.38)}\]
Thus \(\Delta S_{total}\) is expressed in terms of the properties of the system only — which was exactly the goal. Multiplying throughout by \(T\) and rearranging,
\[-T\,\Delta S_{total} = \Delta H - T\,\Delta S \qquad \text{.......(4.39)}\]
Substituting in Eq. (4.35),
\[\Delta G = -T\,\Delta S_{total} \qquad \text{....... (4.40)}\]
Eq. (4.40) is the bridge between the two criteria. Since \(T\) is always positive, \(\Delta G\) and \(\Delta S_{total}\) must always carry opposite signs.
For a spontaneous reaction \(\Delta S_{total} > 0\) and hence \(\Delta G < 0\). At constant temperature and pressure, the Gibbs energy of the system decreases in a spontaneous process.
The second law leads to the conditions of spontaneity, which are summarised here.
\(\Delta S_{total} > 0\) and \(\Delta G < 0\), the process is spontaneous.
\(\Delta S_{total} < 0\) and \(\Delta G > 0\), the process is nonspontaneous.
\(\Delta S_{total} = 0\) and \(\Delta G = 0\), the process is at equilibrium.
4.11.7 Spontaneity and ΔH or ΔS
From \(\Delta G = \Delta H - T\,\Delta S\) (at constant \(T\) and \(P\)), the temperature term determines the relative contributions of \(\Delta H\) and \(\Delta S\) to \(\Delta G\). Since \(\Delta S\) is multiplied by \(T\) but \(\Delta H\) is not, raising the temperature magnifies the entropy contribution while leaving the enthalpy contribution untouched. Four combinations of signs are possible.
If \(\Delta H\) and \(\Delta S\) are both negative, then \(\Delta G\) will be negative only when \(\Delta H\) is more negative than \(T\Delta S\). This is possible at low temperatures only.
If \(\Delta H\) and \(\Delta S\) are both positive, \(\Delta G\) will be negative only if \(T\Delta S > \Delta H\). This is possible only at high temperatures.
For \(\Delta H\) negative and \(\Delta S\) positive, it follows that \(\Delta G\) is negative regardless of temperature. Such reactions are spontaneous at all temperatures.
For \(\Delta H\) positive and \(\Delta S\) negative, \(\Delta G\) is positive regardless of temperature. Such reactions are nonspontaneous at all temperatures.
Flowchart
FC 4.7 – Deciding spontaneity from the signs of ΔH and ΔS: find the sign of ΔH and the sign of ΔS → branch: ΔH negative and ΔS positive → ΔG negative at every T → spontaneous always; ΔH positive and ΔS negative → ΔG positive at every T → nonspontaneous always; ΔH negative and ΔS negative → compare |ΔH| with |TΔS| → spontaneous only below the cross-over temperature; ΔH positive and ΔS positive → compare TΔS with ΔH → spontaneous only above the cross-over temperature.
4.11.8 Temperature of Equilibrium
For equilibrium, \(\Delta G = 0\). Then \(\Delta G = \Delta H - T\Delta S\) gives
\[\therefore\ T = \frac{\Delta H}{\Delta S} \qquad \text{....... (4.41)}\]
\(T\) is the temperature at which the change over from spontaneous to nonspontaneous behaviour occurs. It applies to cases 1 and 2 of the previous section, where the sign of \(\Delta G\) depends on temperature; in cases 3 and 4 there is no such cross-over to find.
\(\Delta H\) and \(\Delta S\) are assumed to be independent of temperature in Eq. (4.41). Introducing the temperature dependence of \(\Delta H\) or \(\Delta S\) would not cause significant error for a moderate temperature range.
4.11.9 Gibbs Function and Equilibrium Constant
Gibbs energy change for a chemical reaction is given by
\[\Delta G = \Delta G^\circ + RT \ln Q \qquad \text{.......(4.42)}\]
where \(\Delta G^\circ\) is the standard Gibbs energy change, that is, the Gibbs energy change when the reactants and products in a reaction are in their standard states. \(Q\) is called the reaction quotient. \(Q\) is analogous to the equilibrium constant and involves nonequilibrium concentrations, or partial pressures in the case of a gaseous reaction.
The distinction between \(Q\) and \(K\) is worth fixing firmly. Both are written from the same expression, but \(Q\) can be evaluated at any moment during a reaction, while \(K\) has that one particular value the expression reaches when the reaction has settled at equilibrium.
Consider
\[a\mathrm{A} + b\mathrm{B} \rightarrow c\mathrm{C} + d\mathrm{D}\]
\[\Delta G = \Delta G^\circ + RT \ln Q_c\]
\[= \Delta G^\circ + RT \ln \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b} \qquad \text{......(4.43)}\]
or
\[\Delta G = \Delta G^\circ + RT \ln Q_p\]
\[= \Delta G^\circ + RT \ln \frac{P_C^{\,c} \times P_D^{\,d}}{P_A^{\,a} \times P_B^{\,b}} \qquad \text{......(4.44)}\]
When the reaction reaches equilibrium, \(\Delta G = 0\), and \(Q_c\) and \(Q_p\) become \(K_c\) and \(K_p\), respectively. Thus,
\[0 = \Delta G^\circ + RT \ln K_c \quad \text{and} \quad 0 = \Delta G^\circ + RT \ln K_p\]
or
\[\Delta G^\circ = -RT \ln K_c \quad \text{and} \quad \Delta G^\circ = -RT \ln K_p \qquad \text{......(4.45)}\]
or
\[\Delta G^\circ = -2.303\,RT \log_{10} K_c\]
and
\[\Delta G^\circ = -2.303\,RT \log_{10} K_p \qquad \text{....... (4.46)}\]
Eq. (4.46) is one of the most useful relationships in the whole of physical chemistry, because it links a thermodynamic quantity to a measurable equilibrium constant. A large negative \(\Delta G^\circ\) corresponds to a large \(K\) and a reaction that goes nearly to completion; a positive \(\Delta G^\circ\) corresponds to a \(K\) less than one and a reaction that barely proceeds.
4.12 Summary
Scope of thermodynamics
- Thermodynamics is concerned with the energy changes accompanying physical and chemical transformations. It gives no information about the rates of these processes or the mechanisms involved.
Basic terms
- The part of the universe chosen for study is the system; everything outside it is the surroundings. Together they make up the universe.
- An open system exchanges both energy and matter, a closed system exchanges energy but not matter, and an isolated system exchanges neither.
- An extensive property depends on the amount of matter (mass, volume, internal energy, heat capacity, number of moles). An intensive property does not (pressure, temperature, surface tension, viscosity, melting point, boiling point, specific heat).
- A state function depends only on the state of the system and not on the path taken to reach it. A path function depends on the path; work \((W)\) and heat \((Q)\) are path functions.
- A system is in thermodynamic equilibrium when its state functions do not vary with time.
Types of process
- Isothermal: \(\Delta T = 0\) and \(\Delta U = 0\). Isobaric: \(\Delta P = 0\). Isochoric: \(\Delta V = 0\). Adiabatic: \(Q = 0\).
- A reversible process is one in which the driving and opposing forces differ by an infinitesimal amount, which can be reversed by an infinitesimal change in conditions, which proceeds infinitely slowly through an infinite number of steps, and in which the system attains mechanical equilibrium with the surroundings at the end of every step.
Heat, work and their signs
- Energy entering the system is positive; energy leaving it is negative. Thus \(+Q\) heat absorbed by the system, \(-Q\) heat released by the system, \(+W\) work done on the system, \(-W\) work done by the system.
- Pressure–volume work: \(W = -P_{ext}\,\Delta V = -P_{ext}(V_2 - V_1)\).
- In a free expansion into vacuum, \(P_{ext} = 0\), therefore \(W = 0\).
- \(1\ \text{J} = 1\ \text{kg m}^2\text{ s}^{-2} = 1\ \text{Pa m}^3\) and \(1\ \text{dm}^3\ \text{bar} = 100\ \text{J}\).
Maximum work
- Work is maximum when the opposing pressure is infinitesimally smaller than the pressure of the gas, that is, when the expansion is reversible.
- For an isothermal reversible expansion of \(n\) moles of an ideal gas,
\[W_{max} = -2.303\,nRT\,\log_{10}\frac{V_2}{V_1} = -2.303\,nRT\,\log_{10}\frac{P_1}{P_2}\]
Internal energy and the first law
- Internal energy \((U)\) is the total kinetic and potential energy of the particles of a system. It is a state function and an extensive property; only \(\Delta U = U_2 - U_1\) can be measured.
- First law of thermodynamics: energy can neither be created nor destroyed, only converted from one form into another.
\[\Delta U = Q + W, \qquad dU = dQ + dW\]
- Applied to particular processes: isothermal \(W = -Q\); adiabatic \(-\Delta U = -W\); isochoric \(\Delta U = Q_v\); isobaric \(Q_p = \Delta U + P_{ext}\,\Delta V\).
Enthalpy
- \(H = U + PV\), and \(\Delta H = H_2 - H_1\). Enthalpy is a state function.
- At constant pressure, \(\Delta H = Q_p\) — the enthalpy change is the heat exchanged at constant pressure.
- For reactions of solids and liquids \(\Delta H = \Delta U\); for reactions involving gases
\[\Delta H = \Delta U + \Delta n_g RT\]
where \(\Delta n_g\) is the difference between the moles of gaseous products and gaseous reactants.
- Work done in a chemical reaction: \(W = -\Delta n_g RT\). If \(n_2 > n_1\), work is done by the system; if \(n_1 > n_2\), work is done on the system; if \(n_1 = n_2\), no \(PV\) work is done.
Enthalpies of physical transformations
- Enthalpy of fusion \((\Delta_{fus}H)\): one mole of solid converted to liquid at constant temperature and pressure; for ice, \(+6.01\ \text{kJ mol}^{-1}\) at 0 °C.
- Enthalpy of vaporization \((\Delta_{vap}H)\): one mole of liquid converted to vapour; for water, \(+40\ \text{kJ mol}^{-1}\) at 100 °C and \(+44\ \text{kJ mol}^{-1}\) at 25 °C.
- Enthalpy of sublimation \((\Delta_{sub}H)\): one mole of solid converted directly to vapour. Since enthalpy is a state function, \(\Delta_{sub}H = \Delta_{fus}H + \Delta_{vap}H\).
- Enthalpy of ionization \((\Delta_{ion}H)\): removal of an electron from one mole of gaseous atoms; for Na, \(494\ \text{kJ mol}^{-1}\). Electron gain enthalpy \((\Delta_{eg}H)\): one mole of gaseous atoms accepting an electron; for Cl, \(-349\ \text{kJ mol}^{-1}\).
- Enthalpy of atomization \((\Delta_{atom}H)\): dissociation of one mole of a gaseous substance into atoms.
- Enthalpy of solution \((\Delta_{soln}H)\): one mole of substance dissolved in a specified amount of solvent, and
\[\Delta_{soln}H = \Delta_L H + \Delta_{hyd}H\]
where the crystal lattice enthalpy \(\Delta_L H\) is always positive and the hydration enthalpy \(\Delta_{hyd}H\) is always negative.
Thermochemistry
- \(\Delta_r H = \sum H_{\text{products}} - \sum H_{\text{reactants}}\), with each substance in a definite physical state and in the amounts given by the balanced equation.
- \(\Delta_r H\) positive means the reaction is endothermic; \(\Delta_r H\) negative means it is exothermic.
- The standard state of a substance is its most stable form at 1 bar and 298 K, or 1 M concentration for a species in solution.
- A thermochemical equation specifies the balanced equation, the physical states, the number of moles, and the value and sign of \(\Delta_r H^\circ\). Reversing the equation reverses the sign of \(\Delta_r H^\circ\); multiplying or dividing the coefficients multiplies or divides \(\Delta_r H^\circ\) by the same factor.
- Standard enthalpy of formation \((\Delta_f H^\circ)\): one mole of a compound formed from its elements, all in standard states. For an element, \(\Delta_f H^\circ = 0\). Hence
\[\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})\]
- Standard enthalpy of combustion \((\Delta_c H^\circ)\): one mole of a substance in its standard state completely oxidised.
- Bond enthalpy is the enthalpy change required to break a particular covalent bond in one mole of gaseous molecules. For polyatomic molecules an average bond enthalpy is used, and
\[\Delta_r H^\circ = \sum \Delta H^\circ(\text{reactant}) - \sum \Delta H^\circ(\text{product})\]
- Hess’s law of constant heat summation: the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes of the individual steps. It follows directly from enthalpy being a state function, and it allows enthalpy changes of reactions that cannot be carried out to be calculated indirectly.
Spontaneity, entropy and Gibbs energy
- A spontaneous process occurs of its own accord, proceeds in one direction only, may be fast or slow, and continues until equilibrium is reached.
- Exothermicity alone is not a sufficient criterion of spontaneity — melting of ice and dissolution of NaCl are endothermic yet spontaneous.
- Entropy \((S)\) measures molecular disorder or randomness. It is a state function, and
\[\Delta S = \frac{Q_{rev}}{T}, \qquad \text{unit } \text{J K}^{-1}\]
Entropy increases from solid to liquid to gas, and on dissolution or dissociation.
- Second law of thermodynamics: the total entropy of a system and its surroundings increases in a spontaneous process, \(\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0\). At equilibrium \(\Delta S_{total} = 0\); if \(\Delta S_{total} < 0\) the process is nonspontaneous.
- Gibbs energy \(G = H - TS\) expresses the criterion of spontaneity using properties of the system alone:
\[\Delta G = \Delta H - T\Delta S, \qquad \Delta G = -T\,\Delta S_{total}\]
\(\Delta G < 0\) spontaneous, \(\Delta G > 0\) nonspontaneous, \(\Delta G = 0\) at equilibrium.
- Effect of the signs of \(\Delta H\) and \(\Delta S\): both negative — spontaneous at low temperatures only; both positive — spontaneous at high temperatures only; \(\Delta H\) negative with \(\Delta S\) positive — spontaneous at all temperatures; \(\Delta H\) positive with \(\Delta S\) negative — nonspontaneous at all temperatures.
- Temperature of equilibrium, the cross-over from spontaneous to nonspontaneous behaviour, is \(T = \dfrac{\Delta H}{\Delta S}\).
- Relating Gibbs energy to equilibrium:
\[\Delta G = \Delta G^\circ + RT \ln Q\]
and at equilibrium, where \(\Delta G = 0\) and \(Q\) becomes \(K\),
\[\Delta G^\circ = -RT \ln K = -2.303\,RT \log_{10} K\]
Flowchart
FC 4.8 – The logic of the chapter: define the system and its surroundings → identify the process type (isothermal, isobaric, isochoric, adiabatic, reversible) → apply the first law ΔU = Q + W to find energy changes → branch: constant volume → heat measured is Q_v = ΔU; constant pressure → heat measured is Q_p = ΔH → obtain ΔH from enthalpies of formation, bond enthalpies, or Hess’s law → find ΔS for the change → combine as ΔG = ΔH – TΔS → branch: ΔG negative → spontaneous; ΔG positive → nonspontaneous; ΔG = 0 → equilibrium, and ΔG° = –2.303 RT log₁₀K.
4.13 Objective Questions
- The correct thermodynamic conditions for the spontaneous reaction at all temperatures are
- \(\Delta H < 0\) and \(\Delta S > 0\)
- \(\Delta H > 0\) and \(\Delta S < 0\)
- \(\Delta H < 0\) and \(\Delta S < 0\)
- \(\Delta H < 0\) and \(\Delta S = 0\)
- A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, \(\Delta U\) of the gas will be
- \(-500\) J
- \(+500\) J
- \(-1013\) J
- \(+1013\) J
- In which of the following does the entropy of the system decrease?
- Crystallization of liquid into solid
- Temperature of a crystalline solid is increased from 0 K to 115 K
- \(\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{H}\text{(g)}\)
- \(2\,\mathrm{NaHCO_3}\text{(s)} \rightarrow \mathrm{Na_2CO_3}\text{(s)} + \mathrm{CO_2}\text{(g)} + \mathrm{H_2O}\text{(g)}\)
- The enthalpy of formation for all elements in their standard states is
- unity
- zero
- less than zero
- different for different elements
- Which of the following reactions is exothermic?
- \(\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{H}\text{(g)}\)
- \(\mathrm{C}\text{(s)} \rightarrow \mathrm{C}\text{(g)}\)
- \(2\,\mathrm{Cl}\text{(g)} \rightarrow \mathrm{Cl_2}\text{(g)}\)
- \(\mathrm{H_2O}\text{(s)} \rightarrow \mathrm{H_2O}\text{(l)}\)
- 6.24 g of ethanol are vaporized by supplying 5.89 kJ of heat. The enthalpy of vaporization of ethanol will be
- \(43.4\ \text{kJ mol}^{-1}\)
- \(60.2\ \text{kJ mol}^{-1}\)
- \(38.9\ \text{kJ mol}^{-1}\)
- \(20.4\ \text{kJ mol}^{-1}\)
- If the standard enthalpy of formation of methanol is \(-238.9\ \text{kJ mol}^{-1}\), then the entropy change of the surroundings will be
- \(-801.7\ \text{J K}^{-1}\)
- \(801.7\ \text{J K}^{-1}\)
- \(0.8017\ \text{J K}^{-1}\)
- \(-0.8017\ \text{J K}^{-1}\)
- Which of the following are not state functions? 1. \(Q + W\) 2. \(Q\) 3. \(W\) 4. \(H - TS\)
- 1, 2 and 3
- 2 and 3
- 1 and 4
- 2, 3 and 4
- For the vaporization of water at 1 bar, \(\Delta H = 40.63\ \text{kJ mol}^{-1}\) and \(\Delta S = 108.8\ \text{J K}^{-1}\text{mol}^{-1}\). At what temperature is \(\Delta G = 0\)?
- 273.4 K
- 393.4 K
- 373.4 K
- 293.4 K
- Bond enthalpies of H–H, Cl–Cl and H–Cl bonds are \(434\ \text{kJ mol}^{-1}\), \(242\ \text{kJ mol}^{-1}\) and \(431\ \text{kJ mol}^{-1}\), respectively. The enthalpy of formation of HCl is
- \(245\ \text{kJ mol}^{-1}\)
- \(-93\ \text{kJ mol}^{-1}\)
- \(-245\ \text{kJ mol}^{-1}\)
- \(93\ \text{kJ mol}^{-1}\)
4.14 Check Your Concepts
A sealed glass ampoule containing a liquid medicine is warmed in a water bath. Classify it as an open, closed or isolated system, and justify your answer in terms of what crosses the boundary.
Molar volume is obtained by dividing volume by the number of moles, and both of these are extensive properties. Explain why molar volume is nevertheless an intensive property.
Assertion: A reversible process can never be carried out in a laboratory. Reason: A reversible process proceeds infinitely slowly through an infinite number of steps. State whether the assertion and the reason are correct, and whether the reason correctly explains the assertion.
In Problem 4.8 the same gas was taken from the same initial state to the same final state by three different routes, and three different values of work were obtained. Explain what this tells you about the nature of \(W\), and state one quantity in the same problem that would have been identical along all three routes.
Why does the external pressure \(P_{ext}\), and not the pressure of the gas itself, appear in the expression \(W = -P_{ext}\Delta V\)?
Why does a natural logarithm appear in the expression for the maximum work of an isothermal reversible expansion, when no logarithm appears in the expression for work against a constant external pressure?
A bicycle pump becomes noticeably warm when air is pumped quickly into a tyre. Identify the type of process involved, write the form the first law takes for it, and explain the rise in temperature.
Heat is described as a path function, and yet \(Q_v\) and \(Q_p\) are described as state functions. Resolve this apparent contradiction.
For each of the following, state whether work is done on the system, by the system, or not at all, giving \(\Delta n_g\) in each case: (i) \(\mathrm{NH_4Cl}\text{(s)} \rightarrow \mathrm{NH_3}\text{(g)} + \mathrm{HCl}\text{(g)}\), (ii) \(\mathrm{N_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{NO}\text{(g)}\), (iii) \(2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{SO_3}\text{(g)}\).
Why is the standard state of carbon taken as graphite rather than as diamond, when both are pure forms of the element?
Explain why the standard enthalpy of formation of an element in its standard state is taken as zero. Is this a measured result or a convention?
The enthalpy of sublimation of ice at 0 °C equals the sum of its enthalpy of fusion and its enthalpy of vaporization at the same temperature. Which property of enthalpy makes this necessarily true?
The two O–H bonds in a water molecule are identical, and yet the energies required to break them one after the other are 499 kJ and 428 kJ. Explain why the two values differ, and state what quantity chemists tabulate instead.
Match the following.
| Column A | Column B |
|---|---|
| a. \(\Delta_{fus}H\) | i. \(-2.303\,RT \log_{10} K\) |
| b. \(\Delta S\) | ii. \(\Delta U + \Delta n_g RT\) |
| c. \(\Delta H\) | iii. \(Q_{rev}/T\) |
| d. \(\Delta G^\circ\) | iv. Solid converted to liquid at constant temperature |
Diamond is thermodynamically less stable than graphite at ordinary temperature and pressure, so the conversion of diamond into graphite is spontaneous. Explain why diamonds nevertheless survive unchanged for centuries, and state which key point about spontaneity this illustrates.
Common salt dissolves in water even though the process absorbs heat. Explain, using both enthalpy and entropy, why this happens, and state why exothermicity alone cannot be used as a criterion of spontaneity.
Predict whether \(\Delta S\) is positive, negative or close to zero for each change, giving a one-line reason: (i) \(\mathrm{I_2}\text{(s)} \rightarrow \mathrm{I_2}\text{(g)}\), (ii) \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}\), (iii) \(\mathrm{H_2}\text{(g)} + \mathrm{Cl_2}\text{(g)} \rightarrow 2\,\mathrm{HCl}\text{(g)}\), (iv) freezing of water at \(-5\) °C.
A reaction has \(\Delta H\) negative and \(\Delta S\) negative. State how its spontaneity changes as the temperature is raised, and explain why the temperature has this effect by referring to the equation \(\Delta G = \Delta H - T\Delta S\).
Assertion: For a reaction at equilibrium, \(\Delta G = 0\) but \(\Delta G^\circ\) is generally not zero. Reason: \(\Delta G^\circ\) refers to reactants and products in their standard states, which is generally not an equilibrium composition. State whether the assertion and reason are correct, and whether the reason explains the assertion.
The industrial synthesis of ammonia is exothermic and proceeds with a decrease in the number of moles of gas. State what temperature thermodynamics would favour, explain why industrial plants nevertheless operate at a raised temperature, and name the branch of chemistry that supplies the missing part of the picture.
Comment on the statement: no work is involved in an expansion of gas in vacuum.
State the first law of thermodynamics.
What is enthalpy of fusion?
What is the standard state of a substance?
State whether \(\Delta S\) is positive, negative or zero for the reaction \(2\,\mathrm{H}\text{(g)} \rightarrow \mathrm{H_2}\text{(g)}\). Explain.
State the second law of thermodynamics in terms of entropy.
If the enthalpy change of a reaction is \(\Delta H\), how will you calculate the entropy of the surroundings?
Comment on the spontaneity of reactions for which \(\Delta H\) is positive and \(\Delta S\) is negative.
Obtain the relationship between \(\Delta G^\circ\) of a reaction and the equilibrium constant.
What is entropy? Give its units.
How will you calculate reaction enthalpy from data on bond enthalpies?
What is the standard enthalpy of combustion? Give an example.
What is the enthalpy of atomization? Give an example.
Obtain the expression for work done in a chemical reaction.
Derive the expression for \(PV\) work.
What are intensive properties? Explain why density is an intensive property.
Derive the expression for the maximum work.
Obtain the relationship between \(\Delta H\) and \(\Delta U\) for gas phase reactions.
State Hess’s law of constant heat summation. Illustrate with an example. State its applications.
Although \(\Delta S\) for the formation of two moles of water from \(\mathrm{H_2}\) and \(\mathrm{O_2}\) is \(-327\ \text{J K}^{-1}\), it is spontaneous. Explain. (Given \(\Delta H\) for the reaction is \(-572\) kJ.)
Obtain the relation between \(\Delta G\) and \(\Delta S_{total}\). Comment on the spontaneity of the reaction.
The following are some processes occurring in nature: a river originates in a mountain and flows towards the sea; after proper incubation for 21 days a chicken egg hatches and a baby chick comes out. List out some more processes you come across in nature. Identify the processes that are in accordance with the second law of thermodynamics and those which are against it.
4.15 Practice With Numericals
(Use \(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\), \(1\ \text{dm}^3\ \text{bar} = 100\ \text{J}\), and standard molar masses wherever required.)
Two moles of an ideal gas expand from 5.0 dm³ to 12.0 dm³ against a constant external pressure of 1.5 bar. Calculate the work done in dm³ bar and in joules, and state whether work is done on or by the system.
1.5 moles of an ideal gas are expanded isothermally and reversibly at 310 K from a volume of 2.0 dm³ to 8.0 dm³. Calculate the maximum work done.
Three moles of an ideal gas are compressed from 30 dm³ to 18 dm³ at a constant external pressure, and 2500 J of work is done on the gas. Calculate the external pressure in bar.
A system releases 45 kJ of heat to its surroundings while 28 kJ of work is done on it. Calculate \(\Delta U\) for the system.
For the reaction \(2\,\mathrm{CO}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{CO_2}\text{(g)}\), \(\Delta H = -566\) kJ at 298 K. Calculate \(\Delta U\) for the reaction.
Calculate the work done when one mole of \(\mathrm{CaCO_3}\) decomposes at 1000 K according to \(\mathrm{CaCO_3}\text{(s)} \rightarrow \mathrm{CaO}\text{(s)} + \mathrm{CO_2}\text{(g)}\), and state whether work is done on or by the system.
Calculate the total heat required to convert 54 g of ice at 0 °C into water at 50 °C. Given \(\Delta_{fus}H = 6.01\ \text{kJ mol}^{-1}\) at 0 °C and the specific heat of water \(= 4.18\ \text{J g}^{-1}\text{K}^{-1}\).
For potassium bromide, the crystal lattice enthalpy is \(+690\ \text{kJ mol}^{-1}\) and the enthalpy of hydration is \(-670\ \text{kJ mol}^{-1}\). Calculate its enthalpy of solution and state whether the solution will warm or cool on dissolution.
Calculate the standard enthalpy of the reaction \(2\,\mathrm{SO_2}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{SO_3}\text{(g)}\), given \(\Delta_f H^\circ(\mathrm{SO_2}) = -297\ \text{kJ mol}^{-1}\) and \(\Delta_f H^\circ(\mathrm{SO_3}) = -395\ \text{kJ mol}^{-1}\).
Calculate the standard enthalpy of combustion of ethanol, \(\mathrm{C_2H_5OH}\text{(l)}\), given \(\Delta_f H^\circ(\mathrm{C_2H_5OH}) = -277.7\ \text{kJ mol}^{-1}\), \(\Delta_f H^\circ(\mathrm{CO_2}) = -393.5\ \text{kJ mol}^{-1}\) and \(\Delta_f H^\circ(\mathrm{H_2O},\ \text{l}) = -285.8\ \text{kJ mol}^{-1}\). Write the balanced combustion equation first.
Using the bond enthalpies \(\Delta H^\circ(\mathrm{N \equiv N}) = 946\ \text{kJ mol}^{-1}\), \(\Delta H^\circ(\text{H–H}) = 436\ \text{kJ mol}^{-1}\) and \(\Delta H^\circ(\text{N–H}) = 391\ \text{kJ mol}^{-1}\), calculate \(\Delta_r H^\circ\) for \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}\).
Calculate the standard enthalpy of formation of \(\mathrm{C_2H_2}\text{(g)}\) from the following data: \(\Delta_c H^\circ(\mathrm{C_2H_2}) = -1300\) kJ; \(\mathrm{C}\text{(graphite)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)}\), \(\Delta H^\circ = -393.5\) kJ; \(\mathrm{H_2}\text{(g)} + \tfrac{1}{2}\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{H_2O}\text{(l)}\), \(\Delta H^\circ = -285.8\) kJ.
A reaction carried out at 298 K has \(\Delta H = -180\) kJ. Calculate the entropy change of the surroundings.
A liquid boils at 350 K and its enthalpy of vaporization at that temperature is \(30.8\ \text{kJ mol}^{-1}\). Calculate the entropy of vaporization of the liquid.
For a certain reaction \(\Delta H = -85\) kJ and \(\Delta S = -125\ \text{J K}^{-1}\). (a) Calculate \(\Delta G\) at 298 K and state whether the reaction is spontaneous. (b) Calculate the temperature at which the change over from spontaneous to nonspontaneous behaviour occurs.
The equilibrium constant of a reaction at 298 K is \(4.6 \times 10^{-3}\). Calculate \(\Delta G^\circ\) for the reaction and state whether reactants or products are favoured at equilibrium.
For the reaction \(\mathrm{N_2}\text{(g)} + 3\,\mathrm{H_2}\text{(g)} \rightarrow 2\,\mathrm{NH_3}\text{(g)}\) at 298 K, \(\Delta G^\circ = -33.0\ \text{kJ mol}^{-1}\). Calculate \(\Delta G\) when the partial pressures are \(P_{\mathrm{N_2}} = 2\) bar, \(P_{\mathrm{H_2}} = 1\) bar and \(P_{\mathrm{NH_3}} = 0.5\) bar, and comment on the direction in which the reaction will proceed.
How much heat is evolved when 12 g of CO reacts with \(\mathrm{NO_2}\)? The reaction is:
\[4\,\mathrm{CO}\text{(g)} + 2\,\mathrm{NO_2}\text{(g)} \rightarrow 4\,\mathrm{CO_2}\text{(g)} + \mathrm{N_2}\text{(g)}, \qquad \Delta_r H^\circ = -1200\ \text{kJ}\]
One mole of an ideal gas is compressed from 500 cm³ against a constant external pressure of \(1.2 \times 10^5\) Pa. The work involved in the process is 36.0 J. Calculate the final volume.
Calculate the maximum work when 24 g of \(\mathrm{O_2}\) are expanded isothermally and reversibly from a pressure of 1.6 bar to 1 bar at 298 K.
Calculate the work done in the decomposition of 132 g of \(\mathrm{NH_4NO_3}\) at 100 °C.
\[\mathrm{NH_4NO_3}\text{(s)} \rightarrow \mathrm{N_2O}\text{(g)} + 2\,\mathrm{H_2O}\text{(g)}\]
State whether work is done on the system or by the system.
- Calculate the standard enthalpy of the reaction
\[\mathrm{Fe_2O_3}\text{(s)} + 3\,\mathrm{CO}\text{(g)} \rightarrow 2\,\mathrm{Fe}\text{(s)} + 3\,\mathrm{CO_2}\text{(g)}\]
from the following data: \(\Delta_f H^\circ(\mathrm{Fe_2O_3}) = -824\ \text{kJ/mol}\), \(\Delta_f H^\circ(\mathrm{CO}) = -110\ \text{kJ/mol}\), \(\Delta_f H^\circ(\mathrm{CO_2}) = -393\ \text{kJ/mol}\)
For a certain reaction \(\Delta H^\circ = 219\) kJ and \(\Delta S^\circ = -21\) J/K. Determine whether the reaction is spontaneous or nonspontaneous.
Determine whether the following reaction is spontaneous under standard state conditions.
\[2\,\mathrm{H_2O}\text{(l)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{H_2O_2}\text{(l)}\]
if \(\Delta H^\circ = 196\) kJ, \(\Delta S^\circ = -126\) J/K. Does it have a cross-over temperature?
- Calculate \(\Delta U\) at 298 K for the reaction
\[\mathrm{C_2H_4}\text{(g)} + \mathrm{HCl}\text{(g)} \rightarrow \mathrm{C_2H_5Cl}\text{(g)}, \qquad \Delta H = -72.3\ \text{kJ}\]
How much \(PV\) work is done?
Calculate the work done during the synthesis of \(\mathrm{NH_3}\) in which the volume changes from 8.0 dm³ to 4.0 dm³ at a constant external pressure of 43 bar. In what direction does the work energy flow?
Calculate the amount of work done in (a) the oxidation of 1 mole of HCl(g) at 200 °C according to the reaction
\[4\,\mathrm{HCl}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow 2\,\mathrm{Cl_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(g)}\]
and (b) the decomposition of one mole of NO at 300 °C for the reaction
\[2\,\mathrm{NO}\text{(g)} \rightarrow \mathrm{N_2}\text{(g)} + \mathrm{O_2}\text{(g)}\]
- When 6.0 g of \(\mathrm{O_2}\) reacts with ClF as per
\[2\,\mathrm{ClF}\text{(g)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{Cl_2O}\text{(g)} + \mathrm{OF_2}\text{(g)}\]
the enthalpy change is 38.55 kJ. What is the standard enthalpy of the reaction?
- Calculate the standard enthalpy of formation of \(\mathrm{CH_3OH}\text{(l)}\) from the following data.
\(\mathrm{CH_3OH}\text{(l)} + \tfrac{3}{2}\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)} + 2\,\mathrm{H_2O}\text{(l)}, \qquad \Delta H^\circ = -726\ \text{kJ mol}^{-1}\)
\(\mathrm{C}\text{(graphite)} + \mathrm{O_2}\text{(g)} \rightarrow \mathrm{CO_2}\text{(g)}, \qquad \Delta_c H^\circ = -393\ \text{kJ mol}^{-1}\)
\(\mathrm{H_2}\text{(g)} + \tfrac{1}{2}\,\mathrm{O_2}\text{(g)} \rightarrow \mathrm{H_2O}\text{(l)}, \qquad \Delta_f H^\circ = -286\ \text{kJ mol}^{-1}\)
- Calculate \(\Delta H^\circ\) for the following reaction at 298 K.
\[\mathrm{H_2B_4O_7}\text{(s)} + \mathrm{H_2O}\text{(l)} \rightarrow 4\,\mathrm{HBO_2}\text{(aq)}\]
\(2\,\mathrm{H_3BO_3}\text{(aq)} \rightarrow \mathrm{B_2O_3}\text{(s)} + 3\,\mathrm{H_2O}\text{(l)}, \qquad \Delta H^\circ = 14.4\ \text{kJ mol}^{-1}\)
\(\mathrm{H_3BO_3}\text{(aq)} \rightarrow \mathrm{HBO_2}\text{(aq)} + \mathrm{H_2O}\text{(l)}, \qquad \Delta H^\circ = -0.02\ \text{kJ mol}^{-1}\)
\(\mathrm{H_2B_4O_7}\text{(s)} \rightarrow 2\,\mathrm{B_2O_3}\text{(s)} + \mathrm{H_2O}\text{(l)}, \qquad \Delta H^\circ = 17.3\ \text{kJ mol}^{-1}\)
Calculate the total heat required (a) to melt 180 g of ice at 0 °C, (b) to heat it to 100 °C, and then (c) to vapourise it at that temperature. Given \(\Delta_{fus}H^\circ(\text{ice}) = 6.01\ \text{kJ mol}^{-1}\) at 0 °C, \(\Delta_{vap}H^\circ(\mathrm{H_2O}) = 40.7\ \text{kJ mol}^{-1}\) at 100 °C, and the specific heat of water is \(4.18\ \text{J g}^{-1}\text{K}^{-1}\).
The enthalpy change for the reaction
\[\mathrm{C_2H_4}\text{(g)} + \mathrm{H_2}\text{(g)} \rightarrow \mathrm{C_2H_6}\text{(g)}\]
is \(-620\) J when 100 mL of ethylene and 100 mL of \(\mathrm{H_2}\) react at 1 bar pressure. Calculate the pressure–volume type of work and \(\Delta U\) for the reaction.
- Calculate the work done and comment on whether work is done on or by the system for the decomposition of 2 moles of \(\mathrm{NH_4NO_3}\) at 100 °C.
\[\mathrm{NH_4NO_3}\text{(s)} \rightarrow \mathrm{N_2O}\text{(g)} + 2\,\mathrm{H_2O}\text{(g)}\]