Calculate Delta_{r} G^{ominus} for the conver | vedclass

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(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln

Vedclass · Accepted Answer

$163 \, kJ \, mol^{-1}$

Vedclass · Answer

(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln K_{p} = -2.303 \, RT \log K_{p}$.Given values are $R = 8.314 \, J \, K^{-1} \, mol^{-1}$,$T = 298 \, K$,and $K_{p} = 2.47 	imes 10^{-29}$.Substituting these values into the equation:$\Delta_{r} G^{\ominus} = -2.303 	imes (8.314 \, J \, K^{-1} \, mol^{-1}) 	imes (298 \, K) 	imes \log(2.47 	imes 10^{-29})$.$\Delta_{r} G^{\ominus} = -2.303 	imes 8.314 	imes 298 	imes (\log 2.47 + \log 10^{-29})$.$\Delta_{r} G^{\ominus} = -5705.85 	imes (0.3927 - 29)$.$\Delta_{r} G^{\ominus} = -5705.85 	imes (-28.6073) \approx 163229 \, J \, mol^{-1}$.Converting to $kJ \, mol^{-1}$,we get $\Delta_{r} G^{\ominus} \approx 163 \, kJ \, mol^{-1}$.

Calculate $\Delta_{r} G^{\ominus}$ for the conversion of oxygen to ozone,$\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$ at $298 \, K$,if $K_{p}$ for this conversion is $2.47 \times 10^{-29}$.

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