Calculate Delta G^{circ} (kJ/mol) at 127 ^{ | vedclass

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Question

(A) The formula for standard Gibbs free energy change is $\Delta G^{\circ} = -2.303 RT \log K_{eq}$.Given: $R = 8.314 \ J \cdot K^{-1} \cdot mol^{-1}

Vedclass · Accepted Answer

$-38.294$

Vedclass · Answer

(A) The formula for standard Gibbs free energy change is $\Delta G^{\circ} = -2.303 RT \log K_{eq}$.Given: $R = 8.314 \ J \cdot K^{-1} \cdot mol^{-1} = 8.314 	imes 10^{-3} \ kJ \cdot K^{-1} \cdot mol^{-1}$,$T = 127 + 273 = 400 \ K$,and $K_{eq} = 10^5$.Substituting the values:$\Delta G^{\circ} = -2.303 	imes (8.314 	imes 10^{-3} \ kJ \cdot K^{-1} \cdot mol^{-1}) 	imes (400 \ K) 	imes \log(10^5)$$\Delta G^{\circ} = -2.303 	imes 8.314 	imes 10^{-3} 	imes 400 	imes 5$$\Delta G^{\circ} = -2.303 	imes 3.3256 	imes 5$$\Delta G^{\circ} = -38.294 \ kJ/mol$.

Calculate $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

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