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(A) The relationship between standard Gibbs free energy change and equilibrium constant is given by the equation: $\Delta G^{\circ} = -2.303 \ RT \ \l

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$-5.74$

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(A) The relationship between standard Gibbs free energy change and equilibrium constant is given by the equation: $\Delta G^{\circ} = -2.303 \ RT \ \log \ K$. Given: $K = 10$,$T = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$. Substituting the values: $\Delta G^{\circ} = -2.303 	imes 8.314 \ J \ mol^{-1} \ K^{-1} 	imes 300 \ K 	imes \log(10)$. Since $\log(10) = 1$,we get: $\Delta G^{\circ} = -2.303 	imes 8.314 	imes 300 \ J \ mol^{-1}$. $\Delta G^{\circ} = -5744.14 \ J \ mol^{-1}$. Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} = -5.744 \ kJ \ mol^{-1}$. Thus,the correct option is $(A)$.

At $300 \ K$,the equilibrium constant for a reaction is $10$. The standard free energy change (in $kJ \ mol^{-1}$) for the reaction is

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