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

Question

Quizplus · Accepted Answer

The positive value of ΔH indicates that the reaction is endothermic, meaning it absorbs heat. According to Le Chatelier's principle, increasing the temperature of an endothermic reaction favors the forward reaction, while decreasing the temperature favors the reverse reaction. Therefore, at high temperatures, the equilibrium constant will shift towards the product side. As a result, the equilibrium constant will be greater than 1 at temperatures above 1042 K. At temperatures below this, the equilibrium constant will be less than 1.

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

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