Calculate Delta G^circ for the reaction, CH_4 | vedclass

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Question

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

Vedclass · Accepted Answer

$-98.716 	ext{ kJ mol}^{-1}$

Vedclass · Answer

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^\circ = -RT \ln K_p$.Given: $R = 8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$, $T = 298 	ext{ K}$, $K_p = 2 	imes 10^{17}$.$\Delta G^\circ = -8.314 	imes 298 	imes \ln(2 	imes 10^{17})$.$\Delta G^\circ = -8.314 	imes 298 	imes (\ln 2 + \ln 10^{17})$.$\Delta G^\circ = -8.314 	imes 298 	imes (0.693 + 17 	imes 2.303)$.$\Delta G^\circ = -8.314 	imes 298 	imes (0.693 + 39.151) = -8.314 	imes 298 	imes 39.844$.$\Delta G^\circ \approx -98716 	ext{ J mol}^{-1} = -98.716 	ext{ kJ mol}^{-1}$.

Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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