Standard entropies of $X_2$, $Y_2$ and $XY_3$ are $60$, $40$ and $50 ext{ J K}^{-1} ext{mol}^{-1}$ respect

Question

Standard entropies of $X_2$, $Y_2$ and $XY_3$ are $60$, $40$ and $50 	ext{ J K}^{-1} 	ext{mol}^{-1}$ respectively. For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 ightleftharpoons XY_3$; $\Delta H = -30 	ext{ kJ}$ to be at equilibrium, the temperature should be:

Accepted Answer

For the reaction: $\frac{1}{2}X_2 + \frac{3}{2}Y_2 ightleftharpoons XY_3$
The entropy change of the reaction ($\Delta S$) is:
$\Delta S = S(XY_3) - \left[ \frac{1}{2}S(X_2) + \frac{3}{2}S(Y_2) ight]$
$\Delta S = 50 - \left[ \frac{1}{2}(60) + \frac{3}{2}(40) ight]$
$\Delta S = 50 - [30 + 60] = 50 - 90 = -40 	ext{ J K}^{-1} 	ext{mol}^{-1}$
At equilibrium, the Gibbs free energy change $\Delta G = 0$.
Since $\Delta G = \Delta H - T\Delta S$, we have:
$\Delta H = T\Delta S \implies T = \frac{\Delta H}{\Delta S}$
Given $\Delta H = -30 	ext{ kJ} = -30000 	ext{ J}$
$T = \frac{-30000 	ext{ J}}{-40 	ext{ J K}^{-1}} = 750 	ext{ K}$

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