For the reaction 2SO₃(g)$\rightarrow$2SO₂(g)+ O₂(g) $\Delta$H_r° = +198 kJ.mol^-1 and $\Delta$S_r° = 190 J.K^-1.mol^-1 at 298 K.The equilibrium constant for this reaction will be greater than 1 at A)all temperatures. B)temperatures above 1315 K. C)temperatures below 1042 K. D)no temperature. E)temperatures above 1042 K.

Question

Quizplus · Accepted Answer

The value of the enthalpy change for the reaction is +198 kJ, indicating that the reaction is endothermic. Therefore, an increase in temperature would favor the forward reaction to offset the added heat. The value of the entropy change for the reaction is positive (+190 J/K), indicating that the reaction proceeds with an increase in disorder or randomness. At temperatures above 1042 K, the increase in entropy would be expected to outweigh the increase in enthalpy, resulting in a positive value for the Gibbs free energy change and a Kc value greater than 1. Therefore, the equilibrium constant for this reaction will be greater than 1 at temperatures above 1042 K.

For the Reaction 2SO3(g) →\rightarrow→ 2SO2(g)+ O2(g) Δ\DeltaΔ Hr° = +198 KJ

For the Reaction
2SO₃(g) $\rightarrow$ 2SO₂(g)+ O₂(g) $\Delta$ H_r° = +198 KJ