At 298 K, if the standard Gibbs energy chang | vedclass

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Question

(A) The relationship between standard Gibbs energy change and equilibrium constant is given by the equation: $\Delta_r G^{\ominus} = -RT \ln K_p$.Conv

Vedclass · Accepted Answer

$+20.15$

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(A) The relationship between standard Gibbs energy change and equilibrium constant is given by the equation: $\Delta_r G^{\ominus} = -RT \ln K_p$.Converting natural logarithm to base $10$: $\Delta_r G^{\ominus} = -2.303 RT \log_{10} K_p$.Given $\Delta_r G^{\ominus} = -115 \ kJ \ mol^{-1} = -115000 \ J \ mol^{-1}$, $R = 8.314 \ J \ K^{-1} \ mol^{-1}$, and $T = 298 \ K$.Substituting the values: $-115000 = -2.303 	imes 8.314 	imes 298 	imes \log_{10} K_p$.$\log_{10} K_p = \frac{115000}{2.303 	imes 8.314 	imes 298}$.$\log_{10} K_p = \frac{115000}{5705.84} \approx 20.15$.

At $298 \ K$, if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$, the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

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