Consider the reaction : H2S(g) + \(\tfrac{3}{2}\) O2(g) → SO2(g) + H2O(g), ΔH° = 518.62 kJ What is the effect of increase of temperature and pressure on the reaction ?
Retardation and advancement of reaction, respectively
Le Chatelier's principle is applied twice, once for each change.
Effect of pressure. Count the gas moles on each side: \(1 + \tfrac{3}{2} = 2.5\) on the left against \(1 + 1 = 2\) on the right. The reaction proceeds with a decrease in the number of gas molecules, so raising the pressure shifts the equilibrium forward — an advancement.
Effect of temperature. The combustion of hydrogen sulphide is strongly exothermic; heat is a product. Raising the temperature therefore pushes the equilibrium backward — a retardation.
So on the chemically correct reading the answer is retardation with increasing temperature and advancement with increasing pressure.
The difficulty is that the stem prints \(\Delta H^{\circ} = +518.62\) kJ, a positive value, which would make the reaction endothermic. Taken literally that sign would reverse the temperature argument and give advancement in both cases. But the combustion of H2S to SO2 and water is unambiguously exothermic — the standard value is close to \(-518\) kJ mol-1 — so the printed sign is almost certainly a misprint for \(-518.62\) kJ.
It is worth noting that pressure affects the position of equilibrium here only because \(\Delta n_{gas} \ne 0\); had the moles balanced, pressure would have had no effect at all.
Note: the SPPU final answer key marked this question with code 9 ("No option is correct or the question is wrong"), consistent with the contradictory enthalpy sign. The answer recorded here follows the chemically correct exothermic reading and is flagged as key-disputed.
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