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Standard entropies of $X_2$, $Y_2$ and $XY_3$ are 60, 40 and 50 J $K^{-1}mol^{-1}$ respectively. For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \rightleftharpoons XY_3$, $\Delta H = -30$ kJ, to be at equilibrium, the temperature should be
A
500 K
B
750 K
C
1000 K
D
1250 K
Detailed Solution
$\Delta S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants}$
$\Delta S^\circ = S^\circ(XY_3) - \left[\frac{1}{2}S^\circ(X_2) + \frac{3}{2}S^\circ(Y_2)\right]$
$\Delta S^\circ = 50 - \left[\frac{1}{2}(60) + \frac{3}{2}(40)\right] = 50 - (30 + 60) = -40$ J $K^{-1}mol^{-1}$
At equilibrium $\Delta G = \Delta H - T\Delta S = 0$, so $T = \frac{\Delta H}{\Delta S}$
$T = \frac{-30\times10^3\ J\,mol^{-1}}{-40\ J\,K^{-1}mol^{-1}}$
T = 750 K
$\Delta S^\circ = S^\circ(XY_3) - \left[\frac{1}{2}S^\circ(X_2) + \frac{3}{2}S^\circ(Y_2)\right]$
$\Delta S^\circ = 50 - \left[\frac{1}{2}(60) + \frac{3}{2}(40)\right] = 50 - (30 + 60) = -40$ J $K^{-1}mol^{-1}$
At equilibrium $\Delta G = \Delta H - T\Delta S = 0$, so $T = \frac{\Delta H}{\Delta S}$
$T = \frac{-30\times10^3\ J\,mol^{-1}}{-40\ J\,K^{-1}mol^{-1}}$
T = 750 K