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

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(B) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 	o XY_3$.First,calculate the change in entropy $\Delta S$ for the reaction:$\Delta S = S^{\cir

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(B) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 	o XY_3$.First,calculate the change in entropy $\Delta S$ 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] = 50 - [30 + 60] = 50 - 90 = -40 \, J \, K^{-1} \, mol^{-1}$.At equilibrium,the Gibbs free energy change $\Delta G = 0$.Since $\Delta G = \Delta H - T \Delta S$,at equilibrium,$\Delta H = T \Delta S$.Given $\Delta H = -30 \, kJ = -30000 \, J$.$T = \frac{\Delta H}{\Delta S} = \frac{-30000 \, J}{-40 \, J \, K^{-1}} = 750 \, K$.

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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