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(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^o = -2.303 \, RT \, \log \, K_{eq}$ Giv

Vedclass · Accepted Answer

$-5.527 \, kJ \, mol^{-1}$

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(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by: $\Delta G^o = -2.303 \, RT \, \log \, K_{eq}$ Given: $R = 8 \, J \, K^{-1} \, mol^{-1}$,$T = 300 \, K$,$K_{eq} = 10$ Substituting the values: $\Delta G^o = -2.303 	imes 8 	imes 300 	imes \log(10) \, J \, mol^{-1}$ Since $\log(10) = 1$: $\Delta G^o = -2.303 	imes 8 	imes 300 \, J \, mol^{-1} = -5527.2 \, J \, mol^{-1}$ Converting to $kJ \, mol^{-1}$: $\Delta G^o = -5.527 \, kJ \, mol^{-1}$

If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$

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