Standard entropies of X_2, Y_2 and XY_3 are 6 | vedclass

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(A) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 \rightleftharpoons XY_3$.First,calculate the entropy change of the reaction $\Delta S^{\circ}_{

Vedclass · Accepted Answer

$750$

Vedclass · Answer

(A) The reaction is $\frac{1}{2} X_2 + \frac{3}{2} Y_2 ightleftharpoons XY_3$.First,calculate the entropy change of the reaction $\Delta S^{\circ}_{rxn}$:$\Delta S^{\circ}_{rxn} = S^{\circ}(XY_3) - [\frac{1}{2} S^{\circ}(X_2) + \frac{3}{2} S^{\circ}(Y_2)]$$\Delta S^{\circ}_{rxn} = 50 - [(\frac{1}{2} 	imes 60) + (\frac{3}{2} 	imes 40)] \ J \ K^{-1} \ mol^{-1}$$\Delta S^{\circ}_{rxn} = 50 - [30 + 60] = 50 - 90 = -40 \ J \ K^{-1} \ mol^{-1}$.At equilibrium,$\Delta G = 0$,which implies $\Delta H = T \Delta S$.Given $\Delta H = -30 \ kJ = -30000 \ J$.$-30000 = T 	imes (-40)$.$T = \frac{-30000}{-40} = 750 \ K$.

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 ............. $K$.

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