Topic 12 · page 7 of 17

Balance and temperature

Feasibility is a balance between the entropy change of the system and that of the surroundings. Temperature shifts the balance, because the surroundings term is −ΔH / T and shrinks as T rises.

14 min read12.9(i)12.9(ii)12.10

Read first: Surroundings and total entropy

The balance 12.9(i)

$\Delta S_\text{total}$ has two parts, and each can be positive or negative. The reaction goes when the sum is positive. Returning to the cold reaction of page 1:

The system has a large positive $\Delta S_\text{system}$ (3 mol of solid make gas, liquid and dissolved ions).
The reaction is endothermic, so $\Delta S_\text{surroundings}$ is negative.
At room temperature the first outweighs the second, so $\Delta S_\text{total}$ is positive and the reaction goes.

Temperature changes the balance 12.9(ii)

Only one term depends on temperature: $\Delta S_\text{surroundings} = -\Delta H / T$ . As T rises, $\Delta H / T$ gets smaller, so the surroundings matter less. Whichever sign $\Delta S_\text{surroundings}$ has, its size shrinks as the temperature rises, and $\Delta S_\text{system}$ gets more say.

The decomposition of calcium carbonate shows this. It has $\Delta H$ = +178 kJ mol⁻¹ and $\Delta S_\text{system}$ = +160.4 J K⁻¹ mol⁻¹. We assume both stay constant with temperature.

A graph of entropy change in J per K per mol against temperature from 300 K to 1500 K. The entropy change of the system is a horizontal dashed line at plus 160.4. The entropy change of the surroundings is a curve rising from about minus 600 at 300 K towards zero. The total is the sum, a curve crossing zero at 1110 K. Left of that the label says not feasible, right of it feasible. — Calcium carbonate decomposition. The surroundings term (orange) shrinks as T rises; the total (teal) crosses zero at 1110 K and the reaction becomes feasible above it.

Worked exampleCalcium carbonate at two temperatures

At 298 K: $\Delta S_\text{surroundings}$ = −178 000 ÷ 298 = −597.3. $\Delta S_\text{total}$ = +160.4 − 597.3 = −437 J K⁻¹ mol⁻¹. Not feasible: limestone is stable in a cliff.
At 1300 K: $\Delta S_\text{surroundings}$ = −178 000 ÷ 1300 = −136.9. $\Delta S_\text{total}$ = +160.4 − 136.9 = +23 J K⁻¹ mol⁻¹. Feasible: limestone decomposes in a lime kiln.

One negative term is fine 12.10

Each term can be positive or negative, so four combinations are possible. Find the sign of $\Delta H$ and the sign of $\Delta S_\text{system}$ and the outcome follows.

$\Delta H$	$\Delta S_\text{system}$	$\Delta S_\text{surroundings}$	Feasible?	Example
negative	positive	positive	at all temperatures	$\ce{2H2O2(l) -> 2H2O(l) + O2(g)}$
negative	negative	positive	at low temperature only	$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ (below 463 K)
positive	positive	negative	at high temperature only	$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ (above 1110 K)
positive	negative	negative	never	$\ce{3O2(g) -> 2O3(g)}$

The two middle rows are the interesting ones. In row 2 the reaction goes because the surroundings gain more entropy than the system loses, but only while the temperature is low enough for $-\Delta H / T$ to be big. In row 3 the reaction goes because the system gains more than the surroundings lose, but only when the temperature is high enough for $-\Delta H / T$ to be small.

Check yourself

State, for each combination, whether a reaction is feasible at all temperatures, only at low temperature, only at high temperature, or never: (a) $\Delta H$ negative, $\Delta S_\text{system}$ positive; (b) $\Delta H$ positive, $\Delta S_\text{system}$ positive; (c) $\Delta H$ negative, $\Delta S_\text{system}$ negative; (d) $\Delta H$ positive, $\Delta S_\text{system}$ negative.
Explain how an endothermic reaction can be spontaneous at room temperature.
A reaction has $\Delta H$ = +30 kJ mol⁻¹ and $\Delta S_\text{system}$ = +120 J K⁻¹ mol⁻¹ (illustrative values). Calculate $\Delta S_\text{total}$ at 298 K and at 400 K.
A reaction has $\Delta H$ = −120 kJ mol⁻¹ and $\Delta S_\text{system}$ = −250 J K⁻¹ mol⁻¹ (illustrative values). Calculate $\Delta S_\text{total}$ at 298 K and at 600 K, and comment.
Explain, in terms of $\Delta S_\text{surroundings}$ , why raising the temperature helps an endothermic reaction with a positive $\Delta S_\text{system}$ .
A student says: " $\Delta S_\text{system}$ is negative, so the reaction cannot occur." Comment.

Answers cover these until you have tried the questions

(a) at all temperatures (both terms positive). (b) only at high temperature (the surroundings term is negative and shrinks as T rises). (c) only at low temperature (the surroundings term is positive but shrinks as T rises). (d) never (both terms negative).
$\Delta S_\text{system}$ is positive and larger than the negative $\Delta S_\text{surroundings}$ (which is $-\Delta H / T$ for the endothermic $\Delta H$ ), so $\Delta S_\text{total}$ is positive and the reaction is feasible.
At 298 K: $\Delta S_\text{surroundings}$ = −30 000 ÷ 298 = −100.7. $\Delta S_\text{total}$ = 120 − 100.7 = +19 J K⁻¹ mol⁻¹. At 400 K: $\Delta S_\text{surroundings}$ = −30 000 ÷ 400 = −75. $\Delta S_\text{total}$ = 120 − 75 = +45 J K⁻¹ mol⁻¹. Feasible at both temperatures, and more so when hotter.
At 298 K: $\Delta S_\text{surroundings}$ = 120 000 ÷ 298 = +402.7. $\Delta S_\text{total}$ = −250 + 402.7 = +153 J K⁻¹ mol⁻¹ (feasible). At 600 K: $\Delta S_\text{surroundings}$ = 120 000 ÷ 600 = +200. $\Delta S_\text{total}$ = −250 + 200 = −50 J K⁻¹ mol⁻¹ (not feasible). The reaction is feasible only at low temperature.
For an endothermic reaction $\Delta S_\text{surroundings}$ = −ΔH ÷ T is negative. As T rises the size of this negative term falls, so $\Delta S_\text{system}$ outweighs it more easily and $\Delta S_\text{total}$ becomes positive.
Wrong: a reaction can occur if $\Delta S_\text{total}$ is positive. If the reaction is exothermic, $\Delta S_\text{surroundings}$ can be positive and large enough to outweigh the negative $\Delta S_\text{system}$ (as in ammonia synthesis at 298 K).