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(A) The standard Gibbs free energy change is given by $\Delta G^o = \Delta H^o - T\Delta S^o$. Given $\Delta H^o = 18451.2 \ cal$ and $\Delta S^o = 29

Vedclass · Accepted Answer

$6.94 	imes 10^{-8}$

Vedclass · Answer

(A) The standard Gibbs free energy change is given by $\Delta G^o = \Delta H^o - T\Delta S^o$. Given $\Delta H^o = 18451.2 \ cal$ and $\Delta S^o = 29.16 \ cal/K$ at $T = 298 \ K$. $\Delta G^o = 18451.2 - (298 	imes 29.16) = 18451.2 - 8689.68 = 9761.52 \ cal$. Using the relation $\Delta G^o = -RT \ln K_{eq}$,where $R = 1.987 \ cal/mol \cdot K$. $\ln K_{eq} = -\frac{\Delta G^o}{RT} = -\frac{9761.52}{1.987 	imes 298} = -\frac{9761.52}{592.126} \approx -16.485$. $K_{eq} = e^{-16.485} \approx 6.94 	imes 10^{-8}$.

Find the equilibrium constant for the reaction below at $298 \ K$. $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$ given that $\Delta H^o = 18.4512 \ kcal$ and $\Delta S^o = 29.16 \ cal/K$.

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