Consider the reversible decomposition of a solid X ( s ) into two gaseous products Y ( g ) and Z ( g ) : X ( s ) ⇌ Y ( g ) + Z ( g ) . The standard enthalpy change for this reaction is Δ H ∘ = + 90 kJ

Question

Consider the reversible decomposition of a solid $X (s)$ into two gaseous products $Y (g)$ and $Z (g)$ : $X (s) ⇌ Y (g) + Z (g)$ .

The standard enthalpy change for this reaction is $Δ H^{\circ} = + 90 kJ mol^{- 1}$ and the standard entropy change is $Δ S^{\circ} = + 250 J K^{- 1} mol^{- 1}$ at $298 K$ .

Assume $Δ H^{\circ}$ and $Δ S^{\circ}$ are temperature independent.

Which of the following statement(s) is/are correct? (Use $R = 8.314 J K^{- 1} mol^{- 1}$ )

TestHub · Accepted Answer

The equilibrium temperature is T e q ​ = Δ S ∘ Δ H ∘ ​ = 250 90000 ​ = 360 K . A. At 300 K , Δ G ∘ = 90000 − 300 × 250 = + 15000 J / m o l . K p ​ = e − Δ G ∘ / ( RT ) = e − 15000/ ( 8.314 × 300 ) ≈ 0.0024 . Q p ​ = P Y ​ P Z ​ = 0.1 × 0.1 = 0.01 . Since Q p ​ > K p ​ , the reaction proceeds in reverse. A is incorrect. B. At 400 K , Δ G ∘ = 90000 − 400 × 250 = − 10000 J / m o l . K p ​ = e − ( − 10000 ) / ( 8.314 × 400 ) ≈ 20.23 . At equilibrium, P Y ​ = P Z ​ = P , so P 2 = K p ​ ⟹ P = 20.23 ​ ≈ 4.5 a t m . B is incorrect. C. At 298 K , Δ G ∘ = 90000 − 298 × 250 = + 15500 J / m o l . For spontaneity, Δ G = Δ G ∘ + RT ln Q p ​ < 0 . Let P Y ​ = P Z ​ = P . 15500 + 8.314 × 298 ln ( P 2 ) < 0 ⟹ P < e − 3.128 ≈ 0.0438 a t m . Since 0.0438 < 0.05 , C is correct. D. The reaction X ( s ) ⇌ Y ( g ) + Z ( g ) increases the moles of gas ( Δ n g ​ = + 2 ). Increasing volume shifts equilibrium towards more moles of gas (products). D is correct.

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