For the reaction frac{1}{2}X_2 + frac{3}{2}Y_ | vedclass

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(B) The change in entropy for the reaction is given by: $\Delta S = \sum \Delta S_{	ext{products}} - \sum \Delta S_{	ext{reactants}}$$\Delta S = \De

Vedclass · Accepted Answer

$750$

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(B) The change in entropy for the reaction is given by: $\Delta S = \sum \Delta S_{	ext{products}} - \sum \Delta S_{	ext{reactants}}$$\Delta S = \Delta S_{XY_3} - [\frac{1}{2} \Delta S_{X_2} + \frac{3}{2} \Delta S_{Y_2}]$$\Delta S = 50 - [\frac{60}{2} + \frac{3 	imes 40}{2}] \ J/mol \cdot K$$\Delta S = 50 - [30 + 60] = 50 - 90 = -40 \ J/mol \cdot K = -0.04 \ kJ/mol \cdot K$At equilibrium,$\Delta G = 0$,so $\Delta G = \Delta H - T \Delta S = 0$,which implies $T = \frac{\Delta H}{\Delta S}$.$T = \frac{-30 \ kJ/mol}{-0.04 \ kJ/mol \cdot K} = 750 \ K$.

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

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