Standard entropy of $$X_2, Y_2$$ and $$XY_3$$ are 60, 40 and 50 $$JK^{-1}\ mol^{-1}$$, respectively. For the reaction, $$\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3,\ \Delta H = -30\ kJ$$, to be at equilibrium, the temperature will be

The balanced chemical equation is
$$\frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightarrow XY_3$$

1. Calculate the standard entropy change $$\Delta S^{\circ}$$.

Standard molar entropies (in $$J\,K^{-1}\,mol^{-1}$$):
$$S^{\circ}(X_2)=60,\; S^{\circ}(Y_2)=40,\; S^{\circ}(XY_3)=50$$

Products: one mole of $$XY_3$$ contributes $$50$$.
Reactants: $$\tfrac12$$ mol $$X_2$$ gives $$\tfrac12\times60 = 30$$ and $$\tfrac32$$ mol $$Y_2$$ gives $$\tfrac32\times40 = 60$$.
Total for reactants $$= 30 + 60 = 90$$.

Hence
$$\Delta S^{\circ} = 50 - 90 = -40\;J\,K^{-1}\,mol^{-1}$$

2. The standard enthalpy change is given: $$\Delta H^{\circ} = -30\;kJ = -30000\;J$$.

3. At equilibrium under standard conditions, $$\Delta G^{\circ}=0$$ and
$$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$$

Setting $$\Delta G^{\circ}=0$$:
$$0 = \Delta H^{\circ} - T\Delta S^{\circ}$$
$$T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$$

Insert the values (both in joule units):
$$T = \frac{-30000}{-40} = 750\;K$$

Therefore, the temperature at which the reaction is at equilibrium is 750 K.

Option C which is: 750 K