1.4.3 (HL)—Spontaneity

Condition	Meaning
ΔG < 0	spontaneous in the stated direction
ΔG = 0	equilibrium
ΔG > 0	non-spontaneous in the stated direction

For a temperature-dependent reaction, find the boundary by setting ΔG° = 0 in ΔG° = ΔH° − TΔS°, then use the signs of ΔH° and ΔS° to decide which temperature range is spontaneous.

Use the signs before calculating: ΔH < 0 and ΔS > 0 is favourable at all temperatures, while ΔH > 0 and ΔS < 0 is not; matching signs create a temperature threshold. At T = ΔH/ΔS, first put ΔH and ΔS in consistent units, then test one temperature on each side rather than guessing the inequality.

Worked threshold example: $\ce{CH4(g) + H2O(g) -> 3H2(g) + CO(g)}$ has $\Delta H^\circ=205\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S^\circ=216\,\mathrm{J\,K^{-1}\,mol^{-1}}=0.216\,\mathrm{kJ\,K^{-1}\,mol^{-1}}$ . At the boundary, $0=205-T(0.216)$ , so $T=205/0.216=949\,\mathrm{K}$ . Because both changes are positive, the forward reaction is spontaneous above 949 K and non-spontaneous below it, assuming $\Delta H^\circ$ and $\Delta S^\circ$ are approximately constant.

Representative question

[Maximum number: 2]

Calculate the temperature at which this reaction is no longer spontaneous.

Use your answer to part (a)(i) and section 1 of the data booklet.
The standard entropy change of this reaction is $\Delta S^{\ominus}=-233 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ .
If you did not obtain an answer to part (a)(i) then use the value $-80.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , although this is not the correct answer.

$\begin{aligned} \Delta S^{\theta}=-233 \times 10^{-3}<\mathrm{kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \gg \boldsymbol{O} \boldsymbol{R} \Delta H^{\theta}=-312000<\mathrm{J} \mathrm{~mol}^{-1}> \text { « } \Delta G^{\theta}=\Delta H^{\theta}-T \Delta S^{\theta}> \text { « } 0=(-312)-\mathrm{T}\left(-233 \times 10^{-3}\right) » \boldsymbol{O} \boldsymbol{R}<0=(-312000)-\mathrm{T}(-233) » \mathrm{~T}=1340 \text { «K } 0 \text { or above» } \end{aligned}$

Award [2] for correct final answer. If the alternative data is used the answer is 343 K .

Do not award ECF for M2 if the answer is a negative Kelvin temperature.

Retrieve the route: identify complete or incomplete combustion products, compare fuels and biofuels, balance fuel-cell half-equations, calculate ΔS° and ΔG°, then use ΔG, Q and K to reason about spontaneity and equilibrium.

Check products before balancing, evidence before evaluation, oxidation versus reduction, kelvin and unit consistency, the sign of ΔG, and whether Q is below, equal to, or above K.

1.4.3 (HL)—Spontaneity

Spontaneity and Temperature

Finding a Spontaneity Boundary

Assessment in practice

Combustion and Thermodynamics Summary

Question 1