Topic 13 · page 11 of 12

Entropy and K

The temperature dependence of K comes from entropy. Only the surroundings term, minus delta H over T, changes with temperature, and ΔS total = R ln K turns that change into a change in K.

15 min read13.8(i)13.8(ii)

Read first: Temperature changes K

Temperature and total entropy 13.8(i)

From the entropy topic, the total entropy change of a reaction is the sum of the entropy change of the system and that of the surroundings:

\Delta S_\text{total} = \Delta S_\text{system} + \Delta S_\text{surroundings} \qquad \Delta S_\text{surroundings} = -\frac{\Delta H}{T}

Only one term depends on temperature. $\Delta S_\text{system}$ is treated as constant, and $\Delta S_\text{surroundings}$ contains the T. Raising the temperature shrinks the size of $\Delta H / T$ , so the surroundings matter less.

Exothermic reaction: $\Delta H$ is negative, so $-\Delta H/T$ is positive. It gets smaller as T rises, so $\Delta S_\text{total}$ falls.
Endothermic reaction: $\Delta H$ is positive, so $-\Delta H/T$ is negative. Its size shrinks as T rises, so it becomes less negative, and $\Delta S_\text{total}$ rises.

Worked exampleTotal entropy for the Haber process

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , with $\Delta H$ = −92 kJ mol⁻¹ and $\Delta S_\text{system}$ = −198.8 J K⁻¹ mol⁻¹. Calculate $\Delta S_\text{total}$ at 298 K and at 700 K.

Convert: $\Delta H$ = −92 000 J mol⁻¹.
At 298 K: $\Delta S_\text{surroundings}$ = +92 000 ÷ 298 = +308.7. $\Delta S_\text{total}$ = −198.8 + 308.7 = +109.9 J K⁻¹ mol⁻¹.
At 700 K: $\Delta S_\text{surroundings}$ = +92 000 ÷ 700 = +131.4. $\Delta S_\text{total}$ = −198.8 + 131.4 = −67.4 J K⁻¹ mol⁻¹.
$\Delta S_\text{total}$ fell from positive to negative as the temperature rose, as expected for an exothermic reaction.

Worked exampleTotal entropy for N₂O₄

$\ce{N2O4(g) <=> 2NO2(g)}$ , with $\Delta H$ = +57 kJ mol⁻¹ and $\Delta S_\text{system}$ = +176 J K⁻¹ mol⁻¹ (illustrative values, close to data-book figures).

At 298 K: $\Delta S_\text{surroundings}$ = −57 000 ÷ 298 = −191.3. $\Delta S_\text{total}$ = 176.0 − 191.3 = −15.3 J K⁻¹ mol⁻¹.
At 700 K: $\Delta S_\text{surroundings}$ = −57 000 ÷ 700 = −81.4. $\Delta S_\text{total}$ = 176.0 − 81.4 = +94.6 J K⁻¹ mol⁻¹.
$\Delta S_\text{total}$ rose, as expected for an endothermic reaction.

Four bar charts of entropy changes in J per K per mol. Haber at 298 K: system minus 198.8, surroundings plus 308.7, total plus 109.9. Haber at 700 K: system minus 198.8, surroundings plus 131.4, total minus 67.4. N2O4 at 298 K: system plus 176.0, surroundings minus 191.3, total minus 15.3. N2O4 at 700 K: system plus 176.0, surroundings minus 81.4, total plus 94.6. — The system bar is the same at both temperatures. The surroundings bar shrinks as T rises, and the total bar follows it.

Total entropy and K 13.8(ii)

The total entropy change is linked to the equilibrium constant by

\Delta S_\text{total} = R \ln K

where R = 8.31 J K⁻¹ mol⁻¹ and K is the equilibrium constant at that temperature, used as a plain number. Rearranged:

\ln K = \frac{\Delta S_\text{total}}{R} \qquad K = e^{\,\Delta S_\text{total}/R}

Three things follow, with no calculator needed.

$\Delta S_\text{total}$ positive means ln K is positive, so K is greater than 1. Products are favoured.
$\Delta S_\text{total}$ zero means ln K = 0, so K = 1.
$\Delta S_\text{total}$ negative means K is less than 1. Reactants are favoured.

Because ln K moves in the same direction as $\Delta S_\text{total}$ , a bigger $\Delta S_\text{total}$ means a bigger K. Combine this with the previous section:

MethodCalculating K from the total entropy change

Find $\Delta S_\text{total}$ in J K⁻¹ mol⁻¹. Convert $\Delta H$ from kJ to J before dividing by T.
Divide by R = 8.31 to get ln K.
Take the exponential, $K = e^{\ln K}$ , using the $e^x$ button.

Worked exampleK for the Haber process at two temperatures

Use the $\Delta S_\text{total}$ values above.

298 K: ln K = +109.9 ÷ 8.31 = 13.23, so K = e^13.23 = 5.6 × 10⁵ (units atm⁻²).
700 K: ln K = −67.4 ÷ 8.31 = −8.11, so K = e^−8.11 = 3.0 × 10⁻⁴ (units atm⁻²).

Worked exampleK for N₂O₄ at two temperatures

298 K: ln K = −15.3 ÷ 8.31 = −1.84, so K = 0.16 (units atm).
700 K: ln K = +94.6 ÷ 8.31 = 11.38, so K = 8.8 × 10⁴ (units atm).

These are the figures in the table on page 10. The units of $K_\text{p}$ are put back afterwards, from the expression. The relationship uses K as a number.

The same story in one line 13.8(ii)

Substitute $\Delta S_\text{total} = \Delta S_\text{system} - \Delta H / T$ into $\Delta S_\text{total} = R \ln K$ :

\ln K = \frac{\Delta S_\text{system}}{R} - \frac{\Delta H}{RT}

For an exothermic reaction ( $\Delta H$ negative) the last term is positive and shrinks as T rises, so ln K falls. For an endothermic reaction it is negative and shrinks towards zero, so ln K rises. Both are the same argument as before.

When $\Delta S_\text{total}$ = 0, K = 1. That is the temperature at which the reaction becomes feasible in the entropy topic.

Worked exampleThe temperature at which K = 1

Find the temperature at which $K$ = 1 for $\ce{N2O4(g) <=> 2NO2(g)}$ .

K = 1 means ln K = 0, so $\Delta S_\text{total}$ = 0, so $\Delta S_\text{system} = \Delta H / T$ .
T = $\Delta H / \Delta S_\text{system}$ = 57 000 ÷ 176 = 324 K (51 °C).
The reaction is endothermic, so K rises with temperature: above 324 K, K > 1, and below it K < 1. This matches the table: 0.16 at 298 K, 56 at 400 K.

Check yourself

$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ has $\Delta H$ = −196 kJ mol⁻¹ and $\Delta S_\text{system}$ = −188 J K⁻¹ mol⁻¹. Calculate $\Delta S_\text{total}$ at 298 K and at 700 K, and state the effect of the temperature rise on $\Delta S_\text{total}$ .
Use your 700 K value from question 1 to calculate K at 700 K.
For $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ , $\Delta H$ = +178 kJ mol⁻¹ and $\Delta S_\text{system}$ = +160.4 J K⁻¹ mol⁻¹. Calculate $\Delta S_\text{total}$ and K at 1000 K.
Use $\Delta S_\text{total}$ = R ln K to explain why K for an endothermic reaction increases with temperature.
A reaction has K = 4.0 × 10⁻³ at one temperature. Calculate $\Delta S_\text{total}$ and say what it tells you about the position of equilibrium.
Find the temperature at which K = 1 for the Haber process, given $\Delta H$ = −92 kJ mol⁻¹ and $\Delta S_\text{system}$ = −198.8 J K⁻¹ mol⁻¹. Say whether K is greater or less than 1 at 600 K.

Answers cover these until you have tried the questions

At 298 K: $\Delta S_\text{surroundings}$ = +196 000 ÷ 298 = +657.7, so $\Delta S_\text{total}$ = −188 + 657.7 = +469.7 J K⁻¹ mol⁻¹ (+470). At 700 K: $\Delta S_\text{surroundings}$ = +196 000 ÷ 700 = +280.0, so $\Delta S_\text{total}$ = −188 + 280.0 = +92.0 J K⁻¹ mol⁻¹. The reaction is exothermic, so $\Delta S_\text{total}$ decreases as the temperature rises.
ln K = 92.0 ÷ 8.31 = 11.07, so K = e^11.07 = 6.4 × 10⁴ (atm⁻¹).
$\Delta S_\text{surroundings}$ = −178 000 ÷ 1000 = −178.0. $\Delta S_\text{total}$ = 160.4 − 178.0 = −17.6 J K⁻¹ mol⁻¹. ln K = −17.6 ÷ 8.31 = −2.12, so K = e^−2.12 = 0.12 (atm).
The reaction is endothermic, so $\Delta S_\text{surroundings}$ = −ΔH/T is negative. As T rises its size falls, so it becomes less negative. $\Delta S_\text{system}$ is constant, so $\Delta S_\text{total}$ increases. Since $\Delta S_\text{total}$ = R ln K, ln K increases, so K increases.
$\Delta S_\text{total}$ = 8.31 × ln(4.0 × 10⁻³) = 8.31 × (−5.52) = −45.9 J K⁻¹ mol⁻¹. It is negative and K is less than 1, so the equilibrium mixture contains mostly reactants.
$\Delta S_\text{total}$ = 0, so T = ΔH ÷ Δ $S_\text{system}$ = −92 000 ÷ −198.8 = 463 K (190 °C). The reaction is exothermic, so K falls as T rises. At 600 K, which is above 463 K, K is less than 1 (at 600 K, $\Delta S_\text{total}$ = −198.8 + 153.3 = −45.5 J K⁻¹ mol⁻¹, so K ≈ 4 × 10⁻³).