Why do some changes happen by themselves and others never do? Energy alone cannot decide: ice melts on a warm day although melting absorbs heat. Entropy, the second law and the Gibbs energy $G=H-TS$ give the full answer, and $\Delta_{r}G^{\circ}=-RT\ln K$ links it to the position of equilibrium.
Leave an ice cube on a table in a warm room and it melts. Nobody needs to push it; the change simply happens. Yet melting is endothermic — the ice absorbs heat from the room. The cold pack of the chapter opening is also endothermic, and so is the evaporation of sweat. On the other hand, the reverse changes — water freezing into ice on a warm table, a dissolved salt crystallising out of its solution while the solution warms up — never happen by themselves. Something other than the release of heat decides the direction of natural change.
That something is entropy, a measure of how widely matter and energy are spread out. The second law of thermodynamics says that every spontaneous change increases the total entropy of the universe. For chemists, who work at constant temperature and pressure, the second law is most useful in the form of the Gibbs energy: a process is spontaneous when $\Delta G=\Delta H-T\Delta S$ is negative. This single criterion explains why ice melts above $0\ ^{\circ}\mathrm{C}$ and freezes below it, why limestone decomposes only in a hot kiln, and — through $\Delta_{r}G^{\circ}=-RT\ln K$ — how far a reaction goes before it reaches equilibrium, the subject of the next chapter.
Spontaneous Processes
📘 Definition 5.19 — Spontaneous process
A spontaneous process is one that, once started, proceeds by itself without continuing help from an outside agency — water flowing downhill, heat flowing from a hot body to a cold one, a gas spreading into a vacuum, iron rusting in moist air. A process that does not occur unless it is driven continuously from outside (pumping water uphill, electrolysis of water) is non-spontaneous. The reverse of a spontaneous process is non-spontaneous.
⚠️ "Spontaneous" does not mean "fast"
Thermodynamics tells us whether a change can happen, not how quickly. A mixture of hydrogen and oxygen at room temperature is thermodynamically set to form water, yet it can stand for years until a spark starts the reaction; diamond is unstable with respect to graphite at room conditions, but the change is immeasurably slow. Speed is the subject of chemical kinetics (Class 12).
Many spontaneous reactions are exothermic — combustion, neutralisation, rusting — and for a long time chemists believed that a negative $\Delta H$ was the criterion. It is not. Melting of ice above $0\ ^{\circ}\mathrm{C}$, evaporation of water, dissolving ammonium nitrate and the decomposition of limestone at $1200\ \mathrm{K}$ are all endothermic and spontaneous, and a gas expands into a vacuum with $\Delta H=0$. What these changes have in common is that matter, or energy, becomes more spread out.
Entropy and the Second Law
📘 Definition 5.20 — Entropy
Entropy $S$ is a state function that measures the degree of disorder or randomness of a system — more precisely, how widely its energy and particles are spread out among the available arrangements. It is extensive, with unit $\mathrm{J\,K^{-1}}$ (molar entropy: $\mathrm{J\,K^{-1}\,mol^{-1}}$). For a change carried out reversibly at temperature $T$, with heat $q_{rev}$ absorbed by the system,
$$\Delta S=\frac{q_{rev}}{T}.$$
For a phase change at the transition temperature, $\Delta_{trs}S=\dfrac{\Delta_{trs}H}{T_{trs}}$ (for example $\Delta_{vap}S=\Delta_{vap}H/T_{b}$).
Entropy rises when a solid melts or a liquid boils (molecules gain freedom to move), when a substance is heated (energy spreads over more levels), when a solid dissolves, when gases mix, and when a reaction increases the number of moles of gas. Dividing by $T$ makes sense: a given amount of heat causes a bigger relative increase in disorder in a cold system than in an already hot, disordered one.
📐 Law 5.9 — Second law of thermodynamics
In any spontaneous process the total entropy of the system and its surroundings increases; at equilibrium it is constant:
$$\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}\gt0\ \ \text{(spontaneous)},\qquad \Delta S_{total}=0\ \ \text{(equilibrium)}$$
For an isolated system, which has no surroundings to exchange anything with, $\Delta S_{sys}\gt0$ for every spontaneous change. When a process at constant $T$ and $p$ transfers heat $q_{p}=\Delta H_{sys}$ to or from large surroundings, the surroundings absorb $-\Delta H_{sys}$ reversibly, so $\Delta S_{surr}=-\dfrac{\Delta H_{sys}}{T}$.
📐 Law 5.10 — Third law of thermodynamics
The entropy of a perfectly crystalline substance approaches zero as the temperature approaches absolute zero. This gives entropy an absolute zero point, so absolute (standard molar) entropies $S^{\circ}$ can be tabulated — unlike $H$, whose absolute value is unknown. Note that $S^{\circ}$ of an element at $298\ \mathrm{K}$ is not zero (contrast $\Delta_{f}H^{\circ}$). For a reaction, $\Delta_{r}S^{\circ}=\sum\nu S^{\circ}(\text{products})-\sum\nu S^{\circ}(\text{reactants})$.
Substance
$\mathrm{C(graphite)}$
$\mathrm{CaO(s)}$
$\mathrm{H_{2}O(l)}$
$\mathrm{CaCO_{3}(s)}$
$\mathrm{H_{2}(g)}$
$\mathrm{H_{2}O(g)}$
$\mathrm{N_{2}(g)}$
$\mathrm{NH_{3}(g)}$
$\mathrm{O_{2}(g)}$
$\mathrm{CO_{2}(g)}$
$S^{\circ}$ / $\mathrm{J\,K^{-1}\,mol^{-1}}$
5.7
39.8
69.9
92.9
130.7
188.8
191.6
192.5
205.1
213.7
Standard molar entropies at $298\ \mathrm{K}$. Solids have the smallest values and gases the largest; liquid and gaseous water differ by about $119\ \mathrm{J\,K^{-1}\,mol^{-1}}$.
✏️ Example 1 — Predicting the sign of ΔS
Predict whether the entropy of the system increases or decreases: (a) $\mathrm{H_{2}O(l)\rightarrow H_{2}O(g)}$; (b) $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow 2SO_{3}(g)}$; (c) $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$; (d) $\mathrm{NaCl(s)\rightarrow Na^{+}(aq)+Cl^{-}(aq)}$; (e) cooling a sample of nitrogen from $300\ \mathrm{K}$ to $200\ \mathrm{K}$; (f) $\mathrm{N_{2}(g)+O_{2}(g)\rightarrow 2NO(g)}$.
💡 Show solution▶
(a), (b) (a) Increases: liquid → gas. (b) Decreases: 3 mol of gas → 2 mol of gas.
(c), (d) (c) Increases: a gas is formed from a solid. (d) Increases: the ordered lattice breaks up into mobile ions (the ions also order some water molecules around them, but for NaCl the net change is positive).
(e) Decreases: lower temperature, less thermal motion.
(f) Hard to predict: 2 mol of gas → 2 mol of gas; $\Delta S$ is small (in fact about $+25\ \mathrm{J\,K^{-1}\,mol^{-1}}$). The mole-count rule only works when $\Delta n_{g}\ne0$.
✏️ Example 2 — Entropy of phase changes
(a) Find the entropy of vaporisation of water at its normal boiling point ($\Delta_{vap}H=40.66\ \mathrm{kJ\,mol^{-1}}$, $T_{b}=373\ \mathrm{K}$) and the entropy of fusion of ice ($\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$, $T_{m}=273\ \mathrm{K}$). (b) Explain why $\Delta_{vap}S$ is so much larger.
💡 Show solution▶
(a) At the transition temperature the change is reversible: $\Delta_{vap}S=\dfrac{40\,660\ \mathrm{J\,mol^{-1}}}{373\ \mathrm{K}}=109.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$; $\Delta_{fus}S=\dfrac{6010}{273}=22.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$.
(b) Melting only loosens the ordered crystal into a liquid in which molecules are still in contact; vaporisation spreads them through a volume over a thousand times larger.
Units Remember to convert kJ to J: entropies are tabulated in $\mathrm{J\,K^{-1}\,mol^{-1}}$, enthalpies in $\mathrm{kJ\,mol^{-1}}$.
✏️ Example 3 — The second law in action
For $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$ at $298\ \mathrm{K}$, $\Delta_{r}H^{\circ}=-92.2\ \mathrm{kJ\,mol^{-1}}$. (a) Find $\Delta_{r}S^{\circ}$ from the table. (b) Find $\Delta S_{surr}$ and $\Delta S_{total}$, and decide whether the reaction is spontaneous under standard conditions.
💡 Show solution▶
(a) $\Delta_{r}S^{\circ}=2(192.5)-[191.6+3(130.7)]=385.0-583.7=-198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$ — negative, as expected for 4 mol gas → 2 mol gas.
(b) surroundings $\Delta S_{surr}=-\dfrac{\Delta H}{T}=\dfrac{+92\,200}{298}=+309.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$: the heat released spreads into the surroundings.
(b) total $\Delta S_{total}=-198.7+309.4=+110.7\ \mathrm{J\,K^{-1}\,mol^{-1}}\gt0$: spontaneous at $298\ \mathrm{K}$. The system becomes more ordered, but the surroundings gain more entropy than the system loses.
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — entropy change when an ideal gas expands
Not required by the CBSE syllabus. Entropy and the criteria for spontaneity are listed by JEE Main, NEET and JEE Advanced; this calculation appears mainly in JEE problems.
For isothermal expansion of $n$ mol of ideal gas from $V_{1}$ to $V_{2}$, the reversible path has $q_{rev}=-w_{rev}=nRT\ln\tfrac{V_{2}}{V_{1}}$ (Rule 5.2), so $\Delta S_{sys}=\dfrac{q_{rev}}{T}=nR\ln\dfrac{V_{2}}{V_{1}}$; doubling the volume of $1\ \mathrm{mol}$ gives $8.314\ln2=+5.76\ \mathrm{J\,K^{-1}}$. Because $S$ is a state function, the same $\Delta S_{sys}$ applies to free expansion into a vacuum between the same states. There $q=0$, so $\Delta S_{surr}=0$ and $\Delta S_{total}=+5.76\ \mathrm{J\,K^{-1}}\gt0$: the free expansion is spontaneous, while in the reversible expansion the surroundings lose exactly $5.76\ \mathrm{J\,K^{-1}}$ and $\Delta S_{total}=0$.
Interactive 1 — ΔG = ΔH − TΔS Sign Map and Crossover Temperature
Choose a process or set $\Delta H$ and $\Delta S$ yourself (both assumed constant with temperature). The graph shows $\Delta G=\Delta H-T\Delta S$ as a straight line against $T$: it starts at $\Delta H$ and has slope $-\Delta S$. Where the line lies in the green region the process is spontaneous. The small map on the right shows which of the four sign cases you are in. Drag the temperature, or press ▶ Play to sweep it from $0$ to $1500\ \mathrm{K}$.
——298 K
ΔH—
TΔS—
ΔG = ΔH − TΔS—
crossover T = ΔH/ΔS—
Gibbs Energy and Spontaneity
The second law needs the entropy of the surroundings as well as of the system — awkward, because we usually care only about the system. At constant temperature and pressure, $\Delta S_{surr}=-\Delta H_{sys}/T$, so
$$\Delta S_{total}=\Delta S_{sys}-\frac{\Delta H_{sys}}{T}\quad\Longrightarrow\quad -T\Delta S_{total}=\Delta H_{sys}-T\Delta S_{sys}.$$
The right-hand side contains only properties of the system. J. Willard Gibbs gave it a name.
📘 Definition 5.21 — Gibbs energy
The Gibbs energy (Gibbs free energy) of a system is
$$G=H-TS.$$
It is a state function and an extensive property. For a change at constant temperature,
$$\Delta G=\Delta H-T\Delta S\qquad\text{and}\qquad \Delta G=-T\Delta S_{total}\ \ (\text{constant }T,\,p).$$
$-\Delta G$ is the maximum useful (non-expansion) work that the process can deliver, for example as electrical work in a cell (Class 12).
📐 Rule 5.11 — The Gibbs energy criterion and the four sign cases
At constant temperature and pressure: $\;\Delta G\lt0$ spontaneous; $\;\Delta G=0$ at equilibrium; $\;\Delta G\gt0$ non-spontaneous (the reverse process is spontaneous). With $\Delta H$ and $\Delta S$ roughly constant over the range of temperature:
The temperature at which the sign changes is the crossover temperature $T=\Delta H/\Delta S$ (with $\Delta H$ and $\Delta S$ of the same sign), where $\Delta G=0$.
⚠️ Units and signs in ΔG = ΔH − TΔS
$\Delta H$ is usually in $\mathrm{kJ\,mol^{-1}}$ and $\Delta S$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$: divide $\Delta S$ by 1000 before multiplying by $T$. Use kelvin, never °C. And keep the sign of $\Delta S$: for $\Delta S=-198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$, $-T\Delta S$ is positive.
✏️ Example 4 — When does limestone decompose?
For $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$, $\Delta_{r}H^{\circ}=+178.3\ \mathrm{kJ\,mol^{-1}}$ (Section 5.3). (a) Find $\Delta_{r}S^{\circ}$ from the table. (b) Calculate $\Delta_{r}G^{\circ}$ at $298\ \mathrm{K}$ and at $1200\ \mathrm{K}$. (c) Find the temperature above which the decomposition is spontaneous under standard conditions (assume $\Delta H$ and $\Delta S$ do not change with temperature).
(b) 298 K $\Delta_{r}G^{\circ}=178.3-298(0.1606)=178.3-47.9=+130.4\ \mathrm{kJ\,mol^{-1}}$: not spontaneous — limestone is stable at room temperature.
(b) 1200 K $\Delta_{r}G^{\circ}=178.3-1200(0.1606)=178.3-192.7=-14.4\ \mathrm{kJ\,mol^{-1}}$: spontaneous.
(c) $T=\dfrac{\Delta H}{\Delta S}=\dfrac{178.3}{0.1606}=1.11\times10^{3}\ \mathrm{K}$ (about $837\ ^{\circ}\mathrm{C}$). Lime kilns run at about $900$–$1000\ ^{\circ}\mathrm{C}$.
✏️ Example 5 — Using the four cases
Decide whether each process is spontaneous at all temperatures, at none, or only above or below some temperature; where there is a crossover, find it. (a) $\Delta H=-50\ \mathrm{kJ\,mol^{-1}}$, $\Delta S=-100\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (b) $\Delta H=+30\ \mathrm{kJ\,mol^{-1}}$, $\Delta S=+80\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (c) $\Delta H=-120\ \mathrm{kJ\,mol^{-1}}$, $\Delta S=+40\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (d) $\Delta H=+285\ \mathrm{kJ}$, $\Delta S=-137\ \mathrm{J\,K^{-1}}$ for $\mathrm{3O_{2}(g)\rightarrow 2O_{3}(g)}$. (e) Show that ice melts at $273\ \mathrm{K}$ using $\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{fus}S=22.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$.
💡 Show solution▶
(a) Both negative: spontaneous below $T=\dfrac{-50\,000}{-100}=500\ \mathrm{K}$. Check at $300\ \mathrm{K}$: $-50+30=-20\ \mathrm{kJ}$ ✓; at $800\ \mathrm{K}$: $-50+80=+30\ \mathrm{kJ}$ (non-spontaneous).
(b) Both positive: spontaneous above $T=30\,000/80=375\ \mathrm{K}$.
(c), (d) (c) $\Delta H\lt0$, $\Delta S\gt0$: $\Delta G\lt0$ at every temperature. (d) $\Delta H\gt0$, $\Delta S\lt0$: $\Delta G\gt0$ at every temperature — ozone forms only when energy is supplied (UV light, electric discharge).
(e) $T=\dfrac{6010}{22.0}=273\ \mathrm{K}$: at $0\ ^{\circ}\mathrm{C}$ ice and water are in equilibrium ($\Delta G=0$); above it $\Delta G\lt0$ for melting, below it $\Delta G\gt0$ (freezing is spontaneous).
Gibbs Energy and Equilibrium
A reaction mixture does not run all the way to products or stay entirely as reactants: it settles at an equilibrium composition, described by an equilibrium constant $K$ (Chapter 6). As the reaction proceeds, the Gibbs energy of the mixture falls until it reaches a minimum; there $\Delta_{r}G=0$ and the system is at equilibrium. The standard Gibbs energy change $\Delta_{r}G^{\circ}$ — for converting reactants entirely into products, all in their standard states — decides where that minimum lies.
📐 Rule 5.12 — Standard Gibbs energy and the equilibrium constant
$$\Delta_{r}G^{\circ}=-RT\ln K=-2.303\,RT\log K$$
with $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ and $\Delta_{r}G^{\circ}$ in $\mathrm{J\,mol^{-1}}$. Equivalently $K=e^{-\Delta_{r}G^{\circ}/RT}$. Also $\Delta_{r}G^{\circ}=\Delta_{r}H^{\circ}-T\Delta_{r}S^{\circ}$.
$\Delta_{r}G^{\circ}\lt0$ ⇔ $K\gt1$: products are favoured at equilibrium.
$\Delta_{r}G^{\circ}=0$ ⇔ $K=1$.
$\Delta_{r}G^{\circ}\gt0$ ⇔ $K\lt1$: reactants are favoured (but some product always forms).
Because of the exponential, a change of only about $5.7\ \mathrm{kJ\,mol^{-1}}$ in $\Delta_{r}G^{\circ}$ at $298\ \mathrm{K}$ changes $K$ by a factor of 10.
✏️ Example 6 — From ΔG° to K and back
(a) Using Example 3, calculate $\Delta_{r}G^{\circ}$ and $K$ at $298\ \mathrm{K}$ for $\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g)}$. (b) A reaction has $K=10$ at $300\ \mathrm{K}$. Find $\Delta_{r}G^{\circ}$. (c) At about what temperature does $\Delta_{r}G^{\circ}$ for ammonia synthesis become zero (assume constant $\Delta H$, $\Delta S$)? What does this mean for $K$ there?
(a) K $\log K=\dfrac{-\Delta_{r}G^{\circ}}{2.303RT}=\dfrac{33\,000}{2.303\times8.314\times298}=5.78$, so $K=10^{5.78}=6.1\times10^{5}$: ammonia is strongly favoured at equilibrium at $298\ \mathrm{K}$ (but the reaction is very slow without a catalyst).
(c) $T=\dfrac{92.2}{0.1987}=464\ \mathrm{K}$; there $K=1$. Above it $K\lt1$ — which is why the Haber process (about $700\ \mathrm{K}$) needs high pressure to get a good yield.
Interactive 2 — ΔG° and K: How Far Does a Reaction Go?
Set $\Delta_{r}G^{\circ}$ and the temperature. The graph plots $\log K=-\Delta_{r}G^{\circ}/(2.303RT)$ — a straight line whose slope gets gentler at higher $T$. The bar below shows the equilibrium position for a simple reaction $\mathrm{A\rightleftharpoons B}$, where the fraction of B at equilibrium is $K/(1+K)$. Press ▶ Play to sweep $\Delta_{r}G^{\circ}$ from $-40$ to $+40\ \mathrm{kJ\,mol^{-1}}$.
——
ΔrG°—
log K—
K—
fraction B (A ⇌ B)—
Exercises
I
Spontaneity and entropy — ideas.
1 What is a spontaneous process? Give two examples of spontaneous processes that are endothermic.
2 Is a spontaneous reaction necessarily fast? Give an example.
3 Predict the sign of $\Delta S_{sys}$: (a) $\mathrm{I_{2}(s)\rightarrow I_{2}(g)}$; (b) $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow 2SO_{3}(g)}$; (c) $\mathrm{2H_{2}O_{2}(l)\rightarrow 2H_{2}O(l)+O_{2}(g)}$; (d) $\mathrm{Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)}$.
4 Which has the higher entropy, and why: (a) $1\ \mathrm{mol}$ of $\mathrm{H_{2}O(l)}$ or $1\ \mathrm{mol}$ of $\mathrm{H_{2}O(g)}$ at $298\ \mathrm{K}$; (b) $1\ \mathrm{mol}$ of $\mathrm{O_{2}}$ at $300\ \mathrm{K}$ or at $600\ \mathrm{K}$ (same pressure); (c) diamond or graphite?
5 State the second law of thermodynamics in terms of entropy.
6 Explain why, at constant $T$ and $p$, $\Delta S_{surr}=-\Delta H_{sys}/T$.
7 State the third law of thermodynamics. Why can entropies, but not enthalpies, be tabulated as absolute values?
8 A gas expands into an evacuated flask with $\Delta H=0$. Why is the change spontaneous?
💡 Show solutions▶
1 A process that, once started, continues by itself without outside help. Endothermic examples: melting of ice above $0\ ^{\circ}\mathrm{C}$, evaporation of water, dissolving $\mathrm{NH_{4}NO_{3}}$ or KCl in water.
2 No. Diamond → graphite and the reaction of $\mathrm{H_{2}}$ with $\mathrm{O_{2}}$ at room temperature are spontaneous but extremely slow; thermodynamics says nothing about rate.
4 (a) The vapour: its molecules move freely through a much larger volume. (b) At $600\ \mathrm{K}$: more thermal energy spread over more levels. (c) Graphite ($5.7$ against $2.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$): its layers allow more vibrational freedom than the rigid diamond network.
5 In every spontaneous process the total entropy of the system and surroundings increases ($\Delta S_{total}\gt0$); at equilibrium $\Delta S_{total}=0$.
6 The heat released by the system, $-\Delta H_{sys}$, is absorbed by very large surroundings at constant $T$ — effectively reversibly — so $\Delta S_{surr}=q_{surr}/T=-\Delta H_{sys}/T$.
7 The entropy of a perfectly crystalline substance is zero at $0\ \mathrm{K}$. This fixes a natural zero for $S$, so absolute values can be measured from heat-capacity data; no such zero exists for $H$.
8 The gas molecules spread into a larger volume, so the entropy increases ($\Delta S\gt0$) while nothing happens in the surroundings: $\Delta S_{total}\gt0$.
II
Calculating entropy changes. Use the table of $S^{\circ}$ values.
1 Calculate $\Delta_{fus}S$ of ice at $273\ \mathrm{K}$ ($\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$).
2 Benzene boils at $353\ \mathrm{K}$ with $\Delta_{vap}H=30.8\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_{vap}S$.
3 Ethanol boils at $351\ \mathrm{K}$ and $\Delta_{vap}S=110\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Find $\Delta_{vap}H$.
4 Find $\Delta_{r}S^{\circ}$ for $\mathrm{2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)}$.
5 Find $\Delta_{r}S^{\circ}$ for $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$.
6 Find $\Delta_{r}S^{\circ}$ for $\mathrm{C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)}$ and explain why it is so small.
7 For the reaction in part 4, $\Delta_{r}H^{\circ}=-571.6\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta S_{surr}$ and $\Delta S_{total}$ at $298\ \mathrm{K}$. Is the reaction spontaneous although $\Delta S_{sys}$ is negative?
8 A system absorbs $2.00\ \mathrm{kJ}$ of heat reversibly at $400\ \mathrm{K}$. Find its entropy change.
6 $213.7-(5.7+205.1)=+2.9\ \mathrm{J\,K^{-1}\,mol^{-1}}$: one mole of gas replaces one mole of gas, and the solid contributes little.
7 $\Delta S_{surr}=571\,600/298=+1918\ \mathrm{J\,K^{-1}\,mol^{-1}}$; $\Delta S_{total}=1918-326.7=+1591\ \mathrm{J\,K^{-1}\,mol^{-1}}\gt0$: yes, spontaneous — the large heat release increases the entropy of the surroundings far more than the system loses.
Gibbs energy and spontaneity (assume $\Delta H$ and $\Delta S$ independent of temperature).
1 Define Gibbs energy and show that $\Delta G=-T\Delta S_{total}$ at constant $T$ and $p$.
2 A reaction has $\Delta H=-50\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=-0.100\ \mathrm{kJ\,K^{-1}\,mol^{-1}}$. Find $\Delta G$ at $300\ \mathrm{K}$ and at $800\ \mathrm{K}$.
3 For a reaction $\Delta H=+30\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=+80\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Above what temperature is it spontaneous?
4 Which sign combinations of $\Delta H$ and $\Delta S$ give a process that is (a) spontaneous at all temperatures; (b) never spontaneous?
5 Using Example 4, find $\Delta_{r}G^{\circ}$ for the decomposition of $\mathrm{CaCO_{3}}$ at $1000\ \mathrm{K}$.
6 Using $\Delta_{fus}H=6.01\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{fus}S=22.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$, find $\Delta G$ for melting ice at $263\ \mathrm{K}$, $273\ \mathrm{K}$ and $283\ \mathrm{K}$, and interpret.
7 For ammonia synthesis, $\Delta_{r}H^{\circ}=-92.2\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{r}S^{\circ}=-198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Find $\Delta_{r}G^{\circ}$ at $600\ \mathrm{K}$.
8 A reaction has $\Delta_{r}G^{\circ}=-200\ \mathrm{kJ\,mol^{-1}}$ but nothing seems to happen when the reactants are mixed. Suggest why.
💡 Show solutions▶
1 $G=H-TS$. $\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}=\Delta S_{sys}-\Delta H_{sys}/T$; multiply by $-T$: $-T\Delta S_{total}=\Delta H-T\Delta S=\Delta G$. So $\Delta G\lt0$ exactly when $\Delta S_{total}\gt0$.
8 The reaction is spontaneous but kinetically very slow (high activation energy); it may need heat, a spark or a catalyst to start.
IV
Gibbs energy and equilibrium. Use $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.
1 Write the relation between $\Delta_{r}G^{\circ}$ and $K$. What can you say about $K$ when $\Delta_{r}G^{\circ}\lt0$?
2 A reaction has $K=1.0\times10^{3}$ at $298\ \mathrm{K}$. Find $\Delta_{r}G^{\circ}$.
3 For a reaction at $298\ \mathrm{K}$, $\Delta_{r}G^{\circ}=+10.0\ \mathrm{kJ\,mol^{-1}}$. Find $K$.
4 Find $K$ at $298\ \mathrm{K}$ for a reaction with $\Delta_{r}G^{\circ}=-33.0\ \mathrm{kJ\,mol^{-1}}$.
5 What is $\Delta_{r}G^{\circ}$ for a reaction with $K=1$? Is the reaction "finished" at equilibrium in that case?
6 A reaction has $\Delta_{r}H^{\circ}=-40\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{r}S^{\circ}=-100\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Find $K$ at $300\ \mathrm{K}$ and at $500\ \mathrm{K}$. Explain the change.
💡 Show solutions▶
1 $\Delta_{r}G^{\circ}=-RT\ln K=-2.303RT\log K$; if $\Delta_{r}G^{\circ}\lt0$ then $\log K\gt0$, i.e. $K\gt1$.
5 $\Delta_{r}G^{\circ}=0$. No reaction is ever "finished": at equilibrium forward and reverse reactions continue at equal rates; with $K=1$ products and reactants are comparably favoured.
6 $300\ \mathrm{K}$: $\Delta_{r}G^{\circ}=-40+30=-10\ \mathrm{kJ\,mol^{-1}}$, $K=e^{10\,000/(8.314\times300)}=e^{4.009}=55$. $500\ \mathrm{K}$: $\Delta_{r}G^{\circ}=-40+50=+10\ \mathrm{kJ\,mol^{-1}}$, $K=e^{-2.406}=0.090$. For an exothermic reaction with $\Delta S\lt0$, raising $T$ lowers $K$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET
E1.NEET A reaction is spontaneous at all temperatures if (A) $\Delta H\gt0$, $\Delta S\gt0$ (B) $\Delta H\lt0$, $\Delta S\gt0$ (C) $\Delta H\lt0$, $\Delta S\lt0$ (D) $\Delta H\gt0$, $\Delta S\lt0$
💡 Show solution▶
Idea $\Delta G=\Delta H-T\Delta S$ is negative for every $T\gt0$ only when $\Delta H\lt0$ and $-T\Delta S\lt0$, i.e. $\Delta S\gt0$.
Answer(B).
E2.JEE MainNumerical answer. For a process $\Delta H=+40.0\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=+125\ \mathrm{J\,K^{-1}\,mol^{-1}}$, both independent of temperature. The minimum temperature (in K) above which the process is spontaneous is ____.
💡 Show solution▶
Crossover $\Delta G=0$ at $T=\dfrac{\Delta H}{\Delta S}=\dfrac{40\,000}{125}$.
Answer320 K (spontaneous above it, since $\Delta S\gt0$).
E3.Board · 3 marks For $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g)}$, $\Delta_{r}H^{\circ}=-92.2\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_{r}S^{\circ}=-198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (a) Calculate $\Delta_{r}G^{\circ}$ at $298\ \mathrm{K}$ and predict whether the reaction is spontaneous. (b) Is it spontaneous at $600\ \mathrm{K}$ under standard conditions? (c) Explain the change in terms of the sign of $\Delta S$.
💡 Show solution▶
(a) 1 mark $\Delta_{r}G^{\circ}=-92.2-298(-0.1987)=-33.0\ \mathrm{kJ\,mol^{-1}}$: spontaneous.
(b) 1 mark $\Delta_{r}G^{\circ}=-92.2+600(0.1987)=+27.0\ \mathrm{kJ\,mol^{-1}}$: not spontaneous.
(c) 1 mark $\Delta S\lt0$, so the term $-T\Delta S$ is positive and grows with $T$; above $T=92.2/0.1987=464\ \mathrm{K}$ it outweighs the favourable $\Delta H$.
E4.JEE Main The equilibrium constant of a reaction at $300\ \mathrm{K}$ is $100$. Its $\Delta_{r}G^{\circ}$ is ($R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$) (A) $-11.5\ \mathrm{kJ\,mol^{-1}}$ (B) $+11.5\ \mathrm{kJ\,mol^{-1}}$ (C) $-5.74\ \mathrm{kJ\,mol^{-1}}$ (D) $-4.99\ \mathrm{kJ\,mol^{-1}}$
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Formula $\Delta_{r}G^{\circ}=-2.303RT\log K=-2.303\times8.314\times300\times2$.
Compute $=-11\,488\ \mathrm{J\,mol^{-1}}=-11.5\ \mathrm{kJ\,mol^{-1}}$. (C) is the value for $K=10$; (D) forgets the factor 2.303.
Answer(A).
E5.JEE AdvancedOne or more options correct. (A) For a spontaneous process in an isolated system, $\Delta S\gt0$. (B) At constant temperature and pressure, $\Delta G=0$ for a system at equilibrium. (C) The standard molar entropy of an element in its reference state at $298\ \mathrm{K}$ is zero. (D) Every exothermic reaction is spontaneous.
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(A) True: an isolated system has no surroundings to exchange with, so the second law applies to it alone.
(B) True: $\Delta G=-T\Delta S_{total}=0$ at equilibrium.
(C) False: the third law makes $S=0$ only for a perfect crystal at $0\ \mathrm{K}$; at $298\ \mathrm{K}$, $S^{\circ}(\mathrm{O_{2}})=205.1\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (It is $\Delta_{f}H^{\circ}$ that is zero.)
(D) False: with $\Delta S\lt0$ an exothermic reaction is non-spontaneous above $\Delta H/\Delta S$ (ammonia synthesis above $464\ \mathrm{K}$).
Answer(A), (B).
E6.Board · Case study In winter, ice on a pond melts on sunny days and the water refreezes at night. For $\mathrm{H_{2}O(s)\rightarrow H_{2}O(l)}$, $\Delta H=+6.01\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=+22.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$. (i) Calculate $\Delta G$ at $263\ \mathrm{K}$ and $283\ \mathrm{K}$ and say what happens at each temperature. (2 marks) (ii) Find the temperature at which ice and water are in equilibrium. (1 mark) (iii) Why is $\Delta S$ positive for melting? (1 mark)
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(i) $263\ \mathrm{K}$: $\Delta G=6.01-263(0.0220)=+0.22\ \mathrm{kJ\,mol^{-1}}$ — melting is non-spontaneous; water freezes. $283\ \mathrm{K}$: $6.01-283(0.0220)=-0.22\ \mathrm{kJ\,mol^{-1}}$ — ice melts.
(ii) $\Delta G=0$: $T=\dfrac{6010}{22.0}=273\ \mathrm{K}$ ($0\ ^{\circ}\mathrm{C}$).
(iii) In the liquid the molecules are no longer fixed in an ordered lattice; they move and tumble, so disorder (the spread of energy and positions) increases.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Cement and lime production release roughly a twelfth of the world's $\mathrm{CO_{2}}$, much of it from the reaction $\mathrm{CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)}$ itself. "Calcium looping" proposes to run this reaction backwards and forwards to capture $\mathrm{CO_{2}}$ from power-station flue gas. At what temperatures must each step run, according to thermodynamics?
Why it matters. Pilot calcium-looping plants have operated in Spain, Germany, Taiwan and elsewhere; cement makers worldwide, including in India (the world's second-largest producer), are studying carbon capture to meet climate targets. The design temperatures follow directly from $\Delta G=\Delta H-T\Delta S$ and $\Delta_{r}G^{\circ}=-RT\ln K$.
What to do.
Using Example 4 ($\Delta_{r}H^{\circ}=+178.3\ \mathrm{kJ\,mol^{-1}}$, $\Delta_{r}S^{\circ}=+160.6\ \mathrm{J\,K^{-1}\,mol^{-1}}$), calculate $\Delta_{r}G^{\circ}$ at $900$, $1000$, $1100$, $1200$ and $1300\ \mathrm{K}$ (Interactive 1 helps).
For this reaction $K=p_{\mathrm{CO_{2}}}/(1\ \mathrm{bar})$ (solids do not appear in $K$; Chapter 6). Use Rule 5.12 to find the equilibrium pressure of $\mathrm{CO_{2}}$ at each temperature.
Flue gas contains about $0.15\ \mathrm{bar}$ of $\mathrm{CO_{2}}$. In the capture step (the "carbonator") CaO must take up $\mathrm{CO_{2}}$: roughly below what temperature is that possible? In the release step (the "calciner") pure $\mathrm{CO_{2}}$ at about $1\ \mathrm{bar}$ must be driven off: above what temperature?
Compare your answers with published operating temperatures of calcium-looping pilots (about $650\ ^{\circ}\mathrm{C}$ and $900$–$950\ ^{\circ}\mathrm{C}$). What could explain differences?
Tools: Rule 5.11, Rule 5.12, Interactive 1, a spreadsheet, a review article or agency report on calcium looping.
Hand in: A table of $T$, $\Delta_{r}G^{\circ}$ and $p_{\mathrm{CO_{2}}}$, the two estimated temperatures, and a short paragraph on whether the thermodynamics makes the scheme practical.
Going further. Why does assuming constant $\Delta H$ and $\Delta S$ become less reliable as the temperature range grows? Which data would you need to improve the estimate?
Instructor note: $\Delta_{r}G^{\circ}$ ($\mathrm{kJ\,mol^{-1}}$) $=178.3-0.1606T$: $33.8$, $17.7$, $1.6$, $-14.4$, $-30.5$; $p_{\mathrm{CO_{2}}}=e^{-\Delta G^{\circ}/RT}$: about $0.011$, $0.12$, $0.84$, $4.2$, $16.8\ \mathrm{bar}$. $p=0.15\ \mathrm{bar}$ when $\Delta_{r}G^{\circ}=-RT\ln0.15$, i.e. $T\approx1011\ \mathrm{K}$ (capture must run below this); $p=1\ \mathrm{bar}$ at the crossover $1110\ \mathrm{K}$ (release above). Real data give somewhat lower temperatures (the equilibrium $p_{\mathrm{CO_{2}}}$ reaches $1\ \mathrm{bar}$ near $1170\ \mathrm{K}$, and capture rates are fast enough at about $920\ \mathrm{K}$); the constant-$\Delta H$, $\Delta S$ approximation and kinetics explain the gap.
🧪 Self-check quiz
0/5 correct
Q1. In which change does the entropy of the system decrease?
Q2. A process has $\Delta H\gt0$ and $\Delta S\gt0$. It is
Q3. If $\Delta_{r}G^{\circ}\lt0$ for a reaction, its equilibrium constant is
Q4. According to the second law, a process is spontaneous when
Q5. For a process with $\Delta H=+20\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S=+50\ \mathrm{J\,K^{-1}\,mol^{-1}}$, the crossover temperature is