The standard Gibbs free energy change Delta G | vedclass

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Question

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K_{e

Vedclass · Accepted Answer

$4.7$

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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K_{eq}$.Given: $T = 25^{\circ} C = 298 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,and $K_{eq} = 0.15$.Substituting the values: $\Delta G^{\circ} = -8.314 	imes 298 	imes \ln(0.15)$.Using $\ln(0.15) \approx -1.897$:$\Delta G^{\circ} = -8.314 	imes 298 	imes (-1.897) \approx 4703 \ J \approx 4.7 \ kJ$.Alternatively,using $\log_{10}$:$\Delta G^{\circ} = -2.303 	imes 8.314 	imes 298 	imes \log_{10}(0.15) \approx -5705.8 	imes (-0.8239) \approx 4701 \ J \approx 4.7 \ kJ$.

The standard Gibbs free energy change $\Delta G^{\circ}$ at $25^{\circ} C$ for the dissociation of $N_2O_{4(g)}$ to $NO_{2(g)}$ is (given,equilibrium constant $K_{eq} = 0.15, R = 8.314 \ J \ K^{-1} \ mol^{-1}$) (in $kJ$)

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