For the reaction,A_{(g)} + B_{(g)} to C_{(g)} | vedclass

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(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$. Using the Gibbs free energy eq

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(C) Given: $\Delta H^o = -29.8 \, kJ \, mol^{-1}$,$\Delta S^o = -0.100 \, kJ \, K^{-1} \, mol^{-1}$,and $T = 298 \, K$. Using the Gibbs free energy equation: $\Delta G^o = \Delta H^o - T\Delta S^o$. Substituting the values: $\Delta G^o = -29.8 \, kJ \, mol^{-1} - (298 \, K 	imes -0.100 \, kJ \, K^{-1} \, mol^{-1})$. $\Delta G^o = -29.8 + 29.8 = 0 \, kJ \, mol^{-1}$. Since $\Delta G^o = -RT \ln K_{eq}$,and $\Delta G^o = 0$,we have $0 = -RT \ln K_{eq}$. This implies $\ln K_{eq} = 0$,which means $K_{eq} = e^0 = 1$.

For the reaction,$A_{(g)} + B_{(g)} \to C_{(g)} + D_{(g)}$,$\Delta H^o$ and $\Delta S^o$ are,respectively,$-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \, K$. The equilibrium constant for the reaction at $298 \, K$ is

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