Calculate the standard Gibbs energy change (D | vedclass

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(A) The formula for standard Gibbs energy change is $\Delta G^\circ = -RT \ln K_p$.Given: $R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$, $T = 298 	ext

Vedclass · Accepted Answer

-$100.1 	ext{ kJ mol}^{-1}$

Vedclass · Answer

(A) The formula for standard Gibbs energy change is $\Delta G^\circ = -RT \ln K_p$.Given: $R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$, $T = 298 	ext{ K}$, $K_p = 3.5 	imes 10^{17}$.Using the relation $\ln x = 2.303 \log_{10} x$:$\Delta G^\circ = -2.303 	imes R 	imes T 	imes \log_{10} K_p$.$\Delta G^\circ = -2.303 	imes 8.314 	imes 298 	imes \log_{10}(3.5 	imes 10^{17})$.$\log_{10}(3.5 	imes 10^{17}) = \log_{10}(3.5) + 17 \approx 0.544 + 17 = 17.544$.$\Delta G^\circ = -2.303 	imes 8.314 	imes 298 	imes 17.544 \approx -100135 	ext{ J mol}^{-1}$.Converting to kJ: $\Delta G^\circ \approx -100.1 	ext{ kJ mol}^{-1}$.

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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