Standard entropy of X_{2}, Y_{2} and XY_{3} are 60, 40 and 50 J K^{–1} mol^{–1}, respectively. For the reaction, 1/2 X_{2} + 3/2 Y_{2} → XY_{3}, ΔH = –30 kJ, to be at equilibrium, the temperature will be

- 500 K
- 750 K
- 1000 K
- 1250 K

**Answer**

1/2X_{2} + 3/2Y_{2} → XY_{3}

ΔS°_{reaction} = ΔS°_{products} - ΔS°_{reactants }

= 50 – 60/2 - 3*40/2

= -40 J/mol

ΔG = ΔH – TΔS

ΔH = -30 kJ, ΔS = -40J/(K mol)

At equilibrium, ΔG = 0

T = ΔH/ΔS = 750 K

**The correct option is B.**