Standard entropy of X 2 , Y 2 and X Y 2 are 60,40 and 50 JK − 1 mol − 1 , respectively. For the reaction, 2 1 X 2 + 2 3 Y 2 ⟶ X Y 3 , Δ H = − 30 kJ , to be at equilibrium, the temperat

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Standard entropy of X 2 ​ , Y 2 ​ and X Y 2 ​ are 60,40 and 50 JK − 1 mol − 1 , respectively. For the reaction, 2 1 ​ X 2 ​ + 2 3 ​ Y 2 ​ ⟶ X Y 3 ​ , Δ H = − 30 kJ , to be at equilibrium, the temperature will be

TestHub · Accepted Answer

Δ S = S ( X Y 3 ​ ) − 2 1 ​ S ( X 2 ​ ) − 2 3 ​ S ( Y 2 ​ ) = 50 − 30 − 60 = − 40 J mol − 1 K − 1 Δ H = − 30 kJ = − 3000 J Δ G = Δ H − T Δ S At equilibrium, Δ G = 0 T = Δ S Δ H ​ = − 40 − 3000 ​ = 750 K

Chemistry - Thermodynamics - II Question with Solution | TestHub

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