Standard entropies of X 2 , Y 2 and XY 3 are 60 , 40 and 50 J / K / mol respectively. At what temperature the following reaction will be at equilibrium

Step-by-step solution

So from this question we know that enthalpy for the reaction is - 30   KJ 1 2 X 2   +   3 2 Y 2   →   XY 3 ∆ H   =   - 30   KJ So we know from the concept of standard entropies is ∆ S   =   ∆ S P   -   ∆ S R ∆ S   =   ∆ S XY 3   - 3 2   ∆ S Y 2 - 1 2 ∆ S X 2 ∆ S   =   50   -   3 2 × 40 - 1 2 × 60 ∆ S   =   - 40   JK - 1 mol - 1