Gibbs Energy — Spontaneity and the Effect of Temperature

Why the Total Entropy Criterion Is Awkward to Use

The second law settles direction: a change is spontaneous when the total entropy of the universe increases.

$$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$$

The criterion is correct, and it is also inconvenient. Only the first term belongs to the reaction being studied. The second belongs to everything outside the flask, and a chemist working at a bench has no way of measuring the entropy of a laboratory, a water bath and the air above it. Every spontaneity question would need two calculations, one of them about a region with no fixed boundary.

There is a way out, and it rests on a single observation. The surroundings receive whatever heat the system releases. At constant pressure that heat is the enthalpy change of the reaction, with its sign flipped:

$$\Delta H_{surr} = -\Delta H_{sys}$$

The surroundings are large, so their temperature does not change while they take in that heat, and the exchange is effectively reversible for them. Their entropy change is then

$$\Delta S_{surr} = \frac{\Delta H_{surr}}{T} = -\frac{\Delta H_{sys}}{T}$$

That expression carries no property of the surroundings in it. It is written entirely in terms of $\Delta H$ of the system and the temperature. So the surroundings can be eliminated from the criterion altogether, and what is left is a test on the system alone. Carrying out that elimination is what produces the Gibbs energy.

Key Point: The entropy change of the surroundings at constant $T$ and $p$ is $\Delta S_{surr} = -\Delta H_{sys}/T$. This single substitution converts a criterion about the universe into a criterion about the system.

Deriving G = H - TS

Start from the second law and put the substitution in.

$$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} = \Delta S_{sys} - \frac{\Delta H_{sys}}{T}$$

Multiply every term by $T$, which is positive, so no inequality is disturbed:

$$T\Delta S_{total} = T\Delta S_{sys} - \Delta H_{sys}$$

For a spontaneous change $\Delta S_{total} > 0$, and therefore

$$T\Delta S_{sys} - \Delta H_{sys} > 0$$

Now multiply through by $-1$. Multiplying an inequality by a negative number reverses it, and this reversal is the reason the final criterion reads "less than zero" rather than "greater than zero":

$$\Delta H_{sys} - T\Delta S_{sys} < 0$$

The left-hand side is a combination of system properties only, and it deserves a name. Define a new function

$$G = H - TS$$

$G$ is built from $H$, $T$ and $S$, all of which are state functions or state variables, so Gibbs energy $G$ is itself a state function. It has units of energy, and since $H$ and $S$ are extensive, $G$ is extensive too.

For a change at constant temperature,

$$\Delta G = \Delta H - T\Delta S$$

and the inequality above becomes simply $\Delta G < 0$. The subscript "sys" is dropped from here on; every symbol in the Gibbs equation refers to the system.

Key Point (Definition): Gibbs energy is $G = H - TS$, an extensive state function. At constant temperature its change is $\Delta G = \Delta H - T\Delta S$, and $\Delta G = -T\Delta S_{total}$.

That last identity is worth keeping. Since $T\Delta S_{total} = -(\Delta H - T\Delta S)$, the Gibbs energy change is nothing more than the total entropy change of the universe multiplied by $-T$. The two criteria are the same statement in different clothing: $\Delta S_{total}$ positive and $\Delta G$ negative always agree, because $T$ is positive.

The Sign of Delta G

The criterion applies at constant temperature and constant pressure, which is how almost every reaction in a school or college laboratory is run.

A positive $\Delta G$ never means "nothing can happen". It means the reverse reaction is the spontaneous one, and that the forward reaction can only be driven by an outside agency. Electrolysis of water has $\Delta G > 0$ and runs perfectly well as long as the power supply is on.

$\Delta G = 0$ deserves care. It does not mean the system is dead. A beaker of ice and water at $273\ \mathrm{K}$ has ice melting and water freezing at matched rates, and $\Delta G$ for $\mathrm{H_2O(s)} \rightleftharpoons \mathrm{H_2O(l)}$ is zero at that temperature. Equilibrium is the state at which $G$ has reached its minimum, so no further change in either direction can lower it.

The sign of $\Delta G$ says nothing whatever about rate, exactly as the sign of $\Delta S_{total}$ said nothing about rate. Diamond turning into graphite has a negative $\Delta G$ at room temperature and takes geological time.

[JEE/NEET] If a question gives $\Delta S_{total}$ and asks for $\Delta G$, or the reverse, use $\Delta G = -T\Delta S_{total}$ directly rather than rebuilding the whole calculation.

Delta G as the Maximum Useful Work

Read the Gibbs equation as a division of the energy released by a reaction.

$$\Delta H = \Delta G + T\Delta S$$

$\Delta H$ is the total energy the reaction makes available at constant pressure. The term $T\Delta S$ is the portion that has to be spent on the entropy requirement of the change; it is dispersed as heat and cannot be harnessed. What remains, $\Delta G$, is the part that can be taken out as useful work.

Key Point: At constant $T$ and $p$, $-\Delta G$ equals the maximum non-expansion (useful) work obtainable from a change. This is why $G$ was long called the free energy: it is the energy free to do useful work.

"Non-expansion" is the operative word. A reaction that releases gas does pressure-volume work on the atmosphere simply by pushing it back, and that work is not available for anything else. Electrical work from a cell, the work of driving a motor, the work of pumping ions across a membrane are all non-expansion work, and $-\Delta G$ is their ceiling.

The word maximum matters as well. The full $-\Delta G$ is delivered only if the change is carried out reversibly, meaning through a continuous sequence of near-equilibrium states. A hydrogen fuel cell drawing a small current approaches it; burning the same hydrogen in air delivers none of it as useful work, because all of the energy leaves as heat.

For $\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}$, $\Delta G^{\circ} = -237.1\ \mathrm{kJ\,mol^{-1}}$. One mole of hydrogen consumed in a fuel cell can deliver at most $237.1\ \mathrm{kJ}$ of electrical work, whatever the design of the cell. Any real cell delivers less.

The Four Cases: How Temperature Enters

Temperature appears in the Gibbs equation only as a multiplier on $\Delta S$. That single position is enough to make temperature decisive, because $T\Delta S$ grows as $T$ rises while $\Delta H$ stays roughly fixed. Whether temperature helps or hurts depends on the signs of $\Delta H$ and $\Delta S$, and there are four combinations.

The first two rows need no arithmetic. If $\Delta H$ is negative and $\Delta S$ positive, then $-T\Delta S$ is also negative and both terms push $\Delta G$ down; the reaction goes at any temperature. The decomposition of hydrogen peroxide, $2\,\mathrm{H_2O_2(l)} \rightarrow 2\,\mathrm{H_2O(l)} + \mathrm{O_2(g)}$, is of this kind. If $\Delta H$ is positive and $\Delta S$ negative, both terms push $\Delta G$ up and no temperature can rescue the reaction; ozone forming from oxygen with heat absorbed and gas moles falling is such a case.

The last two rows are where exam questions live, because there the two terms fight.

Both negative. The enthalpy term favours the reaction and the entropy term opposes it. At low temperature $T\Delta S$ is small and $\Delta H$ wins, so the reaction goes. Raise the temperature and $T\lvert\Delta S\rvert$ eventually exceeds $\lvert\Delta H\rvert$, and the reaction stops. Ammonia synthesis, $\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$, has $\Delta H^{\circ} = -92.4\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S^{\circ} = -198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$: thermodynamically it prefers a cold reactor, which is precisely why the industrial process needs a catalyst to get any speed at a temperature low enough for a decent yield.

Both positive. The entropy term favours the reaction and the enthalpy term opposes it. Heating makes $T\Delta S$ large enough to overcome $\Delta H$. Limestone decomposing, $\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$, refuses to go at room temperature and runs in a lime kiln above about $1110\ \mathrm{K}$. Melting, boiling and sublimation all sit in this row too, which is why every solid has a temperature above which it melts.

There are two routes to a large enough $T\Delta S$ in this row. Either $\Delta S$ is small and $T$ must be pushed very high, or $\Delta S$ is large and a modest temperature will do. Reactions that release a gas belong to the second kind.

One warning about reading the table. The words low and high are relative to the particular reaction, not to any fixed scale. For a reaction with a small $\lvert\Delta H\rvert$ and a large $\Delta S$, the crossover may sit below room temperature, so ordinary laboratory conditions already count as high. For limestone the same word means above a thousand kelvin. Only the arithmetic fixes the boundary.

[JEE/NEET] Reading the signs off a balanced equation is usually enough to place a reaction in the table. A rise in the number of gaseous moles means $\Delta S$ is positive, a fall means negative, and whether heat is given out or taken in fixes $\Delta H$.

The Crossover Temperature

In the two rows where the terms fight, there is one temperature at which they exactly cancel. Setting $\Delta G = 0$,

$$0 = \Delta H - T\Delta S \qquad \Rightarrow \qquad T = \frac{\Delta H}{\Delta S}$$

Key Point: $T = \Delta H / \Delta S$ is the temperature at which $\Delta G$ changes sign. Below it one term dominates, above it the other does, and at it the system is at equilibrium.

Read the Gibbs equation as a straight line in $T$: $\Delta G = \Delta H - \Delta S \cdot T$ has intercept $\Delta H$ on the $\Delta G$ axis and slope $-\Delta S$. The crossover temperature is where that line cuts the $T$ axis. When $\Delta S$ is positive the line slopes down, so the reaction turns spontaneous above the crossing. When $\Delta S$ is negative the line slopes up, so the reaction turns non-spontaneous above the crossing.

The formula only means something when $\Delta H$ and $\Delta S$ have the same sign; if they have opposite signs the quotient is negative, and a negative Kelvin temperature is the arithmetic telling you the sign never changes.

The unit trap. Tabulated $\Delta H$ is almost always in $\mathrm{kJ\,mol^{-1}}$ and tabulated $\Delta S$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$. These cannot be subtracted or divided as they stand. Convert first, every single time.

$$\Delta H = 178.3\ \mathrm{kJ\,mol^{-1}} = 178300\ \mathrm{J\,mol^{-1}}$$

A student who divides $178.3$ by $160.6$ gets $1.11\ \mathrm{K}$ instead of $1110\ \mathrm{K}$, out by a factor of a thousand and absurd on its face. The same slip in a $\Delta G$ calculation gives a number in the tens of thousands of kilojoules. Two habits stop it: write the unit next to every number as you substitute, and glance at the size of the answer before moving on.

[Board] Working in joules throughout and converting the result to kilojoules at the end is safer than converting $\Delta S$, because it leaves only one conversion to remember.

Question 1: Gibbs energy for ammonia synthesis

For $\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$, $\Delta_r H^{\circ} = -92.4\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S^{\circ} = -198.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Find $\Delta_r G^{\circ}$ at $298\ \mathrm{K}$ and say whether the reaction is spontaneous.

Answer:

The two quantities are in different units, so I convert the enthalpy to joules first.

$$\Delta_r H^{\circ} = -92.4\ \mathrm{kJ\,mol^{-1}} = -92400\ \mathrm{J\,mol^{-1}}$$

Now I substitute into the Gibbs equation.

$$\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ} = -92400 - (298)(-198.7)\ \mathrm{J\,mol^{-1}}$$

The second term has two minus signs, so it comes back positive.

$$\Delta_r G^{\circ} = -92400 + 59213 = -33187\ \mathrm{J\,mol^{-1}}$$

That is $-33.2\ \mathrm{kJ\,mol^{-1}}$, which is negative, so the reaction is spontaneous at $298\ \mathrm{K}$.

Ans: $\Delta_r G^{\circ} = -33.2\ \mathrm{kJ\,mol^{-1}}$; spontaneous at $298\ \mathrm{K}$. Watch out: Both signs are negative here, so the entropy term works against the reaction. Dropping one minus sign gives $-151.6\ \mathrm{kJ\,mol^{-1}}$, which is a common wrong answer.

Question 2: Rusting of iron, two ways

For $4\,\mathrm{Fe(s)} + 3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{Fe_2O_3(s)}$ at $298\ \mathrm{K}$, $\Delta_r H = -1648\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S = -549.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Find $\Delta_r G$, and confirm it against the total entropy change.

Answer:

Converting the enthalpy, $\Delta_r H = -1648 \times 10^{3}\ \mathrm{J\,mol^{-1}}$.

$$\Delta_r G = -1648000 - (298)(-549.4) = -1648000 + 163721 = -1.484 \times 10^{6}\ \mathrm{J\,mol^{-1}}$$

So $\Delta_r G = -1484\ \mathrm{kJ\,mol^{-1}}$, strongly negative, and rust forms spontaneously even though the system loses entropy.

For the cross-check I work out the entropy change of the surroundings.

$$\Delta S_{surr} = -\frac{\Delta_r H}{T} = \frac{1648000}{298} = +5530\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

$$\Delta S_{total} = -549.4 + 5530 = +4980.6\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

$$\Delta G = -T\Delta S_{total} = -(298)(4980.6) = -1.484 \times 10^{6}\ \mathrm{J\,mol^{-1}}$$

The two routes agree exactly, as they must.

Ans: $\Delta_r G = -1484\ \mathrm{kJ\,mol^{-1}}$; the huge positive $\Delta S_{surr}$ from the heat released is what makes the change spontaneous. Watch out: A negative $\Delta_r S$ for the system is not a reason to call a reaction non-spontaneous. Here the surroundings gain far more entropy than the system loses.

Question 3: Sorting reactions into the four cases

Classify each reaction by the signs of $\Delta H$ and $\Delta S$ and state its temperature behaviour: (i) $2\,\mathrm{H_2O_2(l)} \rightarrow 2\,\mathrm{H_2O(l)} + \mathrm{O_2(g)}$, exothermic, gas produced; (ii) $3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{O_3(g)}$, endothermic, gas moles fall; (iii) $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(s)}$; (iv) $\mathrm{NH_4Cl(s)} \rightarrow \mathrm{NH_3(g)} + \mathrm{HCl(g)}$, endothermic.

Answer:

(i) $\Delta H$ negative and $\Delta S$ positive, since one mole of gas appears where there was none. Both terms drive $\Delta G$ negative, so it is spontaneous at all temperatures.

(ii) $\Delta H$ positive and $\Delta S$ negative, since three moles of gas become two. Both terms drive $\Delta G$ positive, so it is non-spontaneous at all temperatures.

(iii) Freezing releases heat, so $\Delta H$ is negative, and a liquid becoming a solid loses disorder, so $\Delta S$ is negative. Both negative means spontaneous only at low temperature — below $273\ \mathrm{K}$ at $1\ \mathrm{bar}$.

(iv) $\Delta H$ positive, and one solid gives two moles of gas so $\Delta S$ is strongly positive. Both positive means spontaneous only at high temperature, which matches solid ammonium chloride subliming when heated.

Ans: (i) all $T$; (ii) no $T$; (iii) low $T$ only; (iv) high $T$ only.

Question 4: Crossover temperature for limestone

For $\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$, $\Delta_r H^{\circ} = +178.3\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S^{\circ} = +160.6\ \mathrm{J\,K^{-1}\,mol^{-1}}$, both taken as independent of temperature. Find $\Delta_r G^{\circ}$ at $298\ \mathrm{K}$ and the temperature above which the decomposition becomes spontaneous.

Answer:

First the value at room temperature, with the enthalpy converted to joules.

$$\Delta_r G^{\circ} = 178300 - (298)(160.6) = 178300 - 47859 = +130441\ \mathrm{J\,mol^{-1}}$$

That is $+130.4\ \mathrm{kJ\,mol^{-1}}$, positive, so limestone does not decompose at room temperature.

Both $\Delta H$ and $\Delta S$ are positive, so this is the case that turns spontaneous on heating. The crossover is where $\Delta_r G^{\circ} = 0$.

$$T = \frac{\Delta_r H^{\circ}}{\Delta_r S^{\circ}} = \frac{178300\ \mathrm{J\,mol^{-1}}}{160.6\ \mathrm{J\,K^{-1}\,mol^{-1}}} = 1110\ \mathrm{K}$$

Above $1110\ \mathrm{K}$ the term $T\Delta S$ exceeds $178300\ \mathrm{J\,mol^{-1}}$ and $\Delta_r G^{\circ}$ turns negative. At $1110\ \mathrm{K}$ itself the solid and its products are in equilibrium. A lime kiln is run above this temperature for exactly this reason.

Ans: $\Delta_r G^{\circ} = +130.4\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$; spontaneous above $T = 1110\ \mathrm{K}$. Watch out: Dividing $178.3$ by $160.6$ without converting gives $1.11\ \mathrm{K}$. Any crossover temperature that comes out near absolute zero, or in the millions, is a unit error.

Question 5: The temperature above which dinitrogen tetroxide dissociates

$\mathrm{N_2O_4(g)} \rightarrow 2\,\mathrm{NO_2(g)}$ has $\Delta_r H^{\circ} = +57.2\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S^{\circ} = +175.8\ \mathrm{J\,K^{-1}\,mol^{-1}}$. Is it spontaneous at $298\ \mathrm{K}$? Find the temperature above which it is.

Answer:

$$\Delta_r G^{\circ} = 57200 - (298)(175.8) = 57200 - 52388 = +4812\ \mathrm{J\,mol^{-1}}$$

So $\Delta_r G^{\circ} = +4.8\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$, just positive, so the dissociation is not spontaneous under standard conditions, though only barely.

$$T = \frac{57200}{175.8} = 325.4\ \mathrm{K}$$

Above about $325\ \mathrm{K}$, some $52\ {}^\circ\mathrm{C}$, the entropy term takes over and $\mathrm{N_2O_4}$ dissociates spontaneously. This is why a sealed tube of the pale gas darkens when warmed in a water bath: more brown $\mathrm{NO_2}$ is formed.

Ans: not spontaneous at $298\ \mathrm{K}$ ($\Delta_r G^{\circ} = +4.8\ \mathrm{kJ\,mol^{-1}}$); spontaneous above $325.4\ \mathrm{K}$. Watch out: A small positive $\Delta G$ means the crossover temperature is close by. Do not read "non-spontaneous" as "impossible under any conditions".

Question 6: Ice and water at the melting point

The enthalpy of fusion of ice is $6.00\ \mathrm{kJ\,mol^{-1}}$ and ice and water are in equilibrium at $273\ \mathrm{K}$ under $1\ \mathrm{bar}$. Find $\Delta S_{fus}$. Then say what happens to $\Delta G$ for melting at $263\ \mathrm{K}$ and at $283\ \mathrm{K}$.

Answer:

Equilibrium means $\Delta G = 0$ for $\mathrm{H_2O(s)} \rightarrow \mathrm{H_2O(l)}$ at $273\ \mathrm{K}$.

$$0 = \Delta H_{fus} - T\Delta S_{fus} \qquad \Rightarrow \qquad \Delta S_{fus} = \frac{6000\ \mathrm{J\,mol^{-1}}}{273\ \mathrm{K}} = 21.98\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

Melting has both $\Delta H$ and $\Delta S$ positive, so it is the high-temperature case and $273\ \mathrm{K}$ is its crossover.

At $263\ \mathrm{K}$: $\Delta G = 6000 - (263)(21.98) = 6000 - 5781 = +219\ \mathrm{J\,mol^{-1}}$, positive, so ice does not melt and water freezes instead.

At $283\ \mathrm{K}$: $\Delta G = 6000 - (283)(21.98) = 6000 - 6220 = -220\ \mathrm{J\,mol^{-1}}$, negative, so ice melts.

Ans: $\Delta S_{fus} = 21.98\ \mathrm{J\,K^{-1}\,mol^{-1}}$; $\Delta G = +219\ \mathrm{J\,mol^{-1}}$ at $263\ \mathrm{K}$ and $-220\ \mathrm{J\,mol^{-1}}$ at $283\ \mathrm{K}$.

Question 7: Entropy of vaporisation of water

Water boils at $373\ \mathrm{K}$ under $1\ \mathrm{bar}$ with $\Delta H_{vap} = 40.79\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta S_{vap}$, and explain why it is much larger than $\Delta S_{fus}$.

Answer:

At the boiling point liquid and vapour coexist, so $\Delta G = 0$ for the vaporisation.

$$\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b} = \frac{40790\ \mathrm{J\,mol^{-1}}}{373\ \mathrm{K}} = 109.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

That is about five times $\Delta S_{fus}$. Melting only frees molecules from fixed lattice sites while keeping them in contact; boiling separates them completely and gives each one the whole volume of the vessel to move in. The jump in disorder from liquid to gas is far bigger than the jump from solid to liquid.

Ans: $\Delta S_{vap} = 109.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$.

Question 8: Useful work from a fuel cell

$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}$ has $\Delta H^{\circ} = -285.8\ \mathrm{kJ\,mol^{-1}}$ and $\Delta G^{\circ} = -237.1\ \mathrm{kJ\,mol^{-1}}$ at $298\ \mathrm{K}$. What is the greatest electrical work obtainable per mole of hydrogen, and what happens to the rest of the energy?

Answer:

The maximum non-expansion work available at constant $T$ and $p$ is $-\Delta G$.

$$w_{useful,max} = -\Delta G^{\circ} = 237.1\ \mathrm{kJ}\ \text{per mole of}\ \mathrm{H_2}$$

The difference between the two quantities is the part tied up by the entropy requirement.

$$T\Delta S^{\circ} = \Delta H^{\circ} - \Delta G^{\circ} = -285.8 - (-237.1) = -48.7\ \mathrm{kJ\,mol^{-1}}$$

$\Delta S^{\circ}$ is negative here, since three half-moles of gas become a liquid, so the reaction must dump $48.7\ \mathrm{kJ}$ into the surroundings as heat to satisfy the second law. That heat cannot be converted into electrical work.

Ans: at most $237.1\ \mathrm{kJ}$ of electrical work per mole of $\mathrm{H_2}$; $48.7\ \mathrm{kJ}$ must leave as heat. Watch out: The ceiling is $-\Delta G$, not $-\Delta H$. Quoting $285.8\ \mathrm{kJ}$ as the available work ignores the entropy term entirely.

Question 9: Finding an entropy change from Delta G

A reaction at $298\ \mathrm{K}$ has $\Delta_r H = -28.4\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r G = -13.6\ \mathrm{kJ\,mol^{-1}}$. Find $\Delta_r S$ and say whether raising the temperature helps the reaction.

Answer:

I rearrange the Gibbs equation for $\Delta_r S$.

$$\Delta_r S = \frac{\Delta_r H - \Delta_r G}{T} = \frac{(-28400) - (-13600)}{298}\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

$$\Delta_r S = \frac{-14800}{298} = -49.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$$

Both $\Delta H$ and $\Delta S$ are negative, so this is the low-temperature case. Heating makes $-T\Delta S$ more positive and pushes $\Delta G$ up, so raising the temperature hurts. The reaction stops being spontaneous above

$$T = \frac{28400}{49.7} = 571\ \mathrm{K}$$

Ans: $\Delta_r S = -49.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$; heating hurts, and spontaneity is lost above about $571\ \mathrm{K}$.

Question 10: A cold reaction that still happens

Solid $\mathrm{NH_4Cl}$ dissolving in water absorbs heat, with $\Delta H_{sol} = +15.0\ \mathrm{kJ\,mol^{-1}}$ and $\Delta S_{sol} = +75.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$ at $298\ \mathrm{K}$. Show that it dissolves spontaneously and identify which term is responsible.

Answer:

$$\Delta G_{sol} = 15000 - (298)(75.0) = 15000 - 22350 = -7350\ \mathrm{J\,mol^{-1}}$$

So $\Delta G_{sol} = -7.35\ \mathrm{kJ\,mol^{-1}}$, negative, and the salt dissolves on its own even though the test tube goes cold.

The enthalpy term is working against dissolution; it is the entropy term that carries the process. An ordered ionic lattice breaks up into ions moving freely through the solvent, which is a large gain in disorder, and at $298\ \mathrm{K}$ the product $T\Delta S$ is $22.35\ \mathrm{kJ\,mol^{-1}}$, comfortably larger than $15.0$.

Ans: $\Delta G_{sol} = -7.35\ \mathrm{kJ\,mol^{-1}}$; the positive $\Delta S$ term drives it. Watch out: The flask cooling is a sign that $\Delta H$ is positive, not that $\Delta G$ is.

Question 11: Spotting a unit error

For $2\,\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{SO_3(g)}$, $\Delta_r H^{\circ} = -198.4\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S^{\circ} = -187.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$. A student writes $\Delta_r G^{\circ} = -198.4 - (298)(-187.0) = +55527.6\ \mathrm{kJ\,mol^{-1}}$. Find the error and the correct value at $298\ \mathrm{K}$, then find the crossover temperature.

Answer:

The student subtracted a term in joules from a number in kilojoules. Every quantity has to be in the same unit before the subtraction, so I put the enthalpy into joules.

$$\Delta_r G^{\circ} = -198400 - (298)(-187.0) = -198400 + 55726 = -142674\ \mathrm{J\,mol^{-1}}$$

That is $-142.7\ \mathrm{kJ\,mol^{-1}}$, spontaneous at $298\ \mathrm{K}$. The student's answer of over fifty thousand kilojoules per mole is roughly a thousand times any ordinary reaction energy, which is the clue that a conversion was missed.

Both signs are negative, so the reaction is spontaneous only below its crossover.

$$T = \frac{198400}{187.0} = 1061\ \mathrm{K}$$

Above about $1061\ \mathrm{K}$ the oxidation of $\mathrm{SO_2}$ stops being spontaneous, which is why the contact process is run near $700\ \mathrm{K}$ with a catalyst rather than simply hotter.

Ans: $\Delta_r G^{\circ} = -142.7\ \mathrm{kJ\,mol^{-1}}$; spontaneous below $1061\ \mathrm{K}$.

Question 12: Minimum temperature for a reaction to run

A reaction has $\Delta_r H = +30.5\ \mathrm{kJ\,mol^{-1}}$ and $\Delta_r S = +66.0\ \mathrm{J\,K^{-1}\,mol^{-1}}$, both effectively constant. Will it proceed at $300\ \mathrm{K}$? Find the lowest temperature at which it will.

Answer:

$$\Delta_r G = 30500 - (300)(66.0) = 30500 - 19800 = +10700\ \mathrm{J\,mol^{-1}}$$

$\Delta_r G = +10.7\ \mathrm{kJ\,mol^{-1}}$ at $300\ \mathrm{K}$, so it does not proceed there.

Both quantities are positive, so heating helps. Setting $\Delta_r G = 0$,

$$T = \frac{30500}{66.0} = 462\ \mathrm{K}$$

At $462\ \mathrm{K}$ the system is at equilibrium; above it $\Delta_r G$ is negative and the reaction runs.

Ans: not spontaneous at $300\ \mathrm{K}$; spontaneous above $462\ \mathrm{K}$. Watch out: "Lowest temperature at which it is spontaneous" and "temperature at which $\Delta G = 0$" give the same number, but at that exact temperature the system is at equilibrium, not proceeding.