For a certain reaction at 300 K,K=10,then De | vedclass

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(D) The relationship between standard Gibbs free energy change and equilibrium constant is given by $\Delta G^{\circ} = -RT \ln(K)$.Given $R = 8.314 \

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$57$

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(D) The relationship between standard Gibbs free energy change and equilibrium constant is given by $\Delta G^{\circ} = -RT \ln(K)$.Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $K = 10$.$\Delta G^{\circ} = -8.314 	imes 300 	imes \ln(10)$$\Delta G^{\circ} = -8.314 	imes 300 	imes 2.303 \ J \ mol^{-1}$$\Delta G^{\circ} = -5744.14 \ J \ mol^{-1}$Converting to $kJ \ mol^{-1}$ by dividing by $1000$:$\Delta G^{\circ} = -5.744 \ kJ \ mol^{-1}$Expressing in terms of $	imes 10^{-1} \ kJ \ mol^{-1}$:$-5.744 \ kJ \ mol^{-1} = -57.44 	imes 10^{-1} \ kJ \ mol^{-1}$.Rounding to the nearest integer value as per the options,we get $57$.

For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

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