Answer:
[tex]6.14\cdot 10^{-6}[/tex]
Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
[tex]K_{eq} = \frac{[ADP][Pi]}{ATP}[/tex]
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:
[tex]\Delta G^o = -RT ln K_{eq}[/tex]
From here, rearrange the equation to solve for K:
[tex]K_{eq} = e^{-\frac{\Delta G^o}{RT}}[/tex]
Now we know from the initial equation that:
[tex]K_{eq} = \frac{[ADP][Pi]}{ATP}[/tex]
Let's express the ratio of ADP to ATP:
[tex]\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}[/tex]
Substitute the expression for K:
[tex]\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}[/tex]
Now we may use the values given to solve:
[tex]\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}[/tex]
