The standard entropies of X_2,Y_2,and XY_3 ar | vedclass

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(C) The entropy change for the reaction is calculated as: $\Delta S^{\circ} = \sum \Delta S^{\circ}(	ext{products}) - \sum \Delta S^{\circ}(	ext{rea

Vedclass · Accepted Answer

$750$

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(C) The entropy change for the reaction is calculated as: $\Delta S^{\circ} = \sum \Delta S^{\circ}(	ext{products}) - \sum \Delta S^{\circ}(	ext{reactants})$$\Delta S^{\circ} = S^{\circ}(XY_3) - [\frac{1}{2}S^{\circ}(X_2) + \frac{3}{2}S^{\circ}(Y_2)]$$\Delta S^{\circ} = 50 - [\frac{1}{2} 	imes 60 + \frac{3}{2} 	imes 40] = 50 - [30 + 60] = -40 \ J \ K^{-1} \ mol^{-1}$At equilibrium,$\Delta G = 0$,so $\Delta G = \Delta H - T\Delta S = 0$,which implies $T = \frac{\Delta H}{\Delta S}$Given $\Delta H = -30 \ kJ = -30000 \ J$,we have $T = \frac{-30000 \ J}{-40 \ J \ K^{-1}} = 750 \ K$

$A$. Isothermal process	$i$. $q = \Delta U$
$B$. Adiabatic process	$ii$. $W = - P \times \Delta V$
$C$. Isobaric process	$iii$. $W = \Delta U$
$D$. Isochoric process	$iv$. $W = - nRT \ln \left(\frac{v_f}{v_i}\right)$

The standard entropies of $X_2$,$Y_2$,and $XY_3$ are $60$,$40$,and $50 \ J \ K^{-1} \ mol^{-1}$ respectively. For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ$ at equilibrium. What is the temperature of the system in $K$?

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$B$. Adiabatic process $ii$. $W = - P \times \Delta V$
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