Standard entropies of X2, Y2, and XY3 are 60, 40, and 50 J K⁻¹ mol⁻¹ respectively. For the reaction 1/2 X2 + 3/2 Y2 → XY3, ΔH = -30 kJ, to be at equilibrium, the temperature should be:

Question

Accepted Answer

1000K. To find the equilibrium temperature, use ΔG = ΔH - TΔS = 0. Calculate ΔS = S(products) - S(reactants) = 50 - (1/2 × 60 + 3/2 × 40) = -40 J K⁻¹ mol⁻¹. Set ΔH = TΔS: -30,000 J = T × -40 J K⁻¹, giving T = 750 K.

Standard entropies of X2, Y2, and XY3 are 60, 40, and 50 J K⁻¹ mol⁻¹ respectively. For the reaction 1/2 X2 + 3/2 Y2 → XY3, ΔH = -30 kJ, to be at equilibrium, the temperature should be:

Solution

Related NEET Chemistry questions