(Solved): I would like you to calculate the Gibbs free energy for ATP hydrolysis in cells. The reaction is AT ...

I would like you to calculate the Gibbs free energy for ATP hydrolysis in cells. The reaction is ATP + $$\mathrm{H}2\mathrm{O}$$ goes to $$\mathrm{ADP}+$$ phosphate Let's assume these cellular concentrations $$\begin{array}{l}\text{ ATP }=2\mathrm{mM}(0.002\mathrm{M})\\ \text{ ADP }=0.2\mathrm{mM}(0.0002\mathrm{M})\\ {\mathrm{H}}_2\mathrm{O}=45\mathrm{M}\end{array}$$ phosphate $$=20\mathrm{mM}(0.02\mathrm{M})$$ Recall that the Gibbs free energy is equal to the standard free energy $$+\mathrm{RT}$$ In (concentrations of products/concentrations of substrates; ie, [ADP] $$\times [$$ phosphate $$]/[\mathrm{ATP}]\times \left[{\mathrm{H}}_2\mathrm{O}\right]$$ ) When calculating this let's use the following take the natural log $$(\ln )$$ of the ratio of concentrations, use $$RT=0.59\mathrm{kcal}/\mathrm{mol}$$ and use $$?7.3$$ $$\mathrm{kcal}/\mathrm{mol}$$ for the standard free energy of ATP hydrolysis. Report your value in $$\mathrm{kcal}/\mathrm{mol}$$