Standard entropy of X_2, Y_2 and XY_3 are 60, | vedclass

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(C) For a reaction to be at equilibrium,$\Delta G = 0$.Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.For the reactio

Vedclass · Accepted Answer

$750$

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(C) For a reaction to be at equilibrium,$\Delta G = 0$.Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.For the reaction $\frac{1}{2} X_2 + \frac{3}{2} Y_2 	o XY_3$,$\Delta H = -30 \ kJ = -30000 \ J$.Calculating $\Delta S$ for the reaction:$\Delta S = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$$\Delta S = 50 - [\frac{1}{2} 	imes 60 + \frac{3}{2} 	imes 40] \ J \ K^{-1} \ mol^{-1}$$\Delta S = 50 - [30 + 60] = -40 \ J \ K^{-1} \ mol^{-1}$.At equilibrium,$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \ J}{-40 \ J \ K^{-1}} = 750 \ K$.

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

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