Deciding Whether a Reaction Is Spontaneous at a New Temperature
An industrial team studies the contact process step SO₂(g) + ½O₂(g) ⇌ SO₃(g). At 298 K the standard enthalpy change is ΔH° = −99 kJ mol⁻¹ and the standard entropy change is ΔS° = −94 J K⁻¹ mol⁻¹. The team wants to run the reaction at 700 K to get a useful rate, but a junior chemist argues that because ΔH° is negative the reaction must be spontaneous at all temperatures, so heating is harmless. The reactor also operates at high pressure, and the team must decide whether temperature or pressure is the better lever for improving equilibrium yield.
1. Calculate ΔG° at 298 K and at 700 K, and state whether the forward reaction is spontaneous under standard conditions at each temperature.
Read the worked answer
Use ΔG° = ΔH° − TΔS°. Convert ΔS° to kJ: −94 J K⁻¹ mol⁻¹ = −0.094 kJ K⁻¹ mol⁻¹.
At 298 K: ΔG° = −99 − (298 × −0.094) = −99 + 28.0 = −71 kJ mol⁻¹. Negative, so spontaneous.
At 700 K: ΔG° = −99 − (700 × −0.094) = −99 + 65.8 = −33 kJ mol⁻¹. Still negative, so still spontaneous under standard conditions, but much less favourable.
The junior chemist is wrong in principle: a negative ΔH° does not guarantee spontaneity at all temperatures. Here ΔS° is negative, so the −TΔS° term becomes increasingly positive as T rises. The reaction becomes non-spontaneous when T > ΔH°/ΔS° = 99/0.094 ≈ 1053 K. At 700 K it is spontaneous but with a smaller driving force.
2. Use Le Chatelier's principle and the sign of ΔS° to decide whether increasing temperature or increasing pressure better improves the equilibrium yield of SO₃.
Read the worked answer
The forward reaction is exothermic (ΔH° negative), so increasing temperature shifts the equilibrium to the left, lowering the yield of SO₃. This is consistent with ΔS° being negative: the forward reaction reduces the number of gas molecules (1.5 mol → 1 mol), so it is entropically disfavoured and heat opposes it.
Increasing pressure shifts the equilibrium toward the side with fewer gas moles, i.e. toward SO₃, increasing yield. So pressure is the better lever for yield; temperature is used to achieve an acceptable rate, with a compromise temperature rather than the highest possible temperature.
The two levers act in opposite directions on yield, so the team should raise pressure and choose a moderate temperature that balances rate against equilibrium conversion.