Standard entropies of X_2, Y_2 and XY_5 are 7 | vedclass

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(B) For the reaction: $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightleftharpoons XY_5$ $\Delta S_{rxn}^0 = S^0(XY_5) - [\frac{1}{2} S^0(X_2) + \frac{5}{2}

Vedclass · Accepted Answer

$700$

Vedclass · Answer

(B) For the reaction: $\frac{1}{2} X_2 + \frac{5}{2} Y_2 ightleftharpoons XY_5$ $\Delta S_{rxn}^0 = S^0(XY_5) - [\frac{1}{2} S^0(X_2) + \frac{5}{2} S^0(Y_2)]$ $\Delta S_{rxn}^0 = 110 - [(\frac{1}{2} 	imes 70) + (\frac{5}{2} 	imes 50)] = 110 - [35 + 125] = 110 - 160 = -50 \ J \ K^{-1} \ mol^{-1}$ At equilibrium,$\Delta G^0 = 0$,so $\Delta H^0 = T \Delta S^0$ Given $\Delta H^0 = -35 \ kJ \ mol^{-1} = -35000 \ J \ mol^{-1}$ $T = \frac{\Delta H^0}{\Delta S^0} = \frac{-35000}{-50} = 700 \ K$

Standard entropies of $X_2, Y_2$ and $XY_5$ are $70, 50$ and $110 \ J \ K^{-1} \ mol^{-1}$ respectively. The temperature in Kelvin at which the reaction $\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5$ with $\Delta H = -35 \ kJ \ mol^{-1}$ will be at equilibrium is . . . . . . (Nearest integer).

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