In the glycolysis process,during the phosphorylation of glucose,the equilibrium constant at $298 \ K$ is $3.6 \times 10^{-3}$. Find the value of $\Delta G^{\Theta}$. What does this indicate? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
(A) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the formula: $\Delta G^{\Theta} = -RT \ln K$.
Substituting the given values: $\Delta G^{\Theta} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(3.6 \times 10^{-3})$.
$\Delta G^{\Theta} = -2477.572 \times (-5.6268) \approx 13940 \ J \ mol^{-1} = 13.94 \ kJ \ mol^{-1}$.
Since $\Delta G^{\Theta} > 0$,the reaction is non-spontaneous under standard conditions.
Substituting the given values: $\Delta G^{\Theta} = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(3.6 \times 10^{-3})$.
$\Delta G^{\Theta} = -2477.572 \times (-5.6268) \approx 13940 \ J \ mol^{-1} = 13.94 \ kJ \ mol^{-1}$.
Since $\Delta G^{\Theta} > 0$,the reaction is non-spontaneous under standard conditions.
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