The value of Delta G^{ominus} for the phospho | vedclass

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(A) Given: $\Delta G^{\ominus} = 13.8 \, kJ \, mol^{-1} = 13.8 	imes 10^{3} \, J \, mol^{-1}$,$T = 298 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$.T

Vedclass · Accepted Answer

$3.81 	imes 10^{-3}$

Vedclass · Answer

(A) Given: $\Delta G^{\ominus} = 13.8 \, kJ \, mol^{-1} = 13.8 	imes 10^{3} \, J \, mol^{-1}$,$T = 298 \, K$,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$.The relationship between standard Gibbs free energy and equilibrium constant is given by: $\Delta G^{\ominus} = -RT \ln K_{c}$.Rearranging for $\ln K_{c}$: $\ln K_{c} = -\frac{\Delta G^{\ominus}}{RT}$.Substituting the values: $\ln K_{c} = -\frac{13.8 	imes 10^{3}}{8.314 	imes 298} = -\frac{13800}{2477.572} \approx -5.5698$.Calculating $K_{c}$: $K_{c} = e^{-5.5698} \approx 3.81 	imes 10^{-3}$.

The value of $\Delta G^{\ominus}$ for the phosphorylation of glucose in glycolysis is $13.8 \, kJ \, mol^{-1}$. Find the value of $K_{c}$ at $298 \, K$.

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