Question

One reaction that occurs in human metabolism is

$\begin{array}{l}{\mathbf{\text{HO}}}_{\mathbf{\text{2}}}{\mathbf{\text{CCH}}}_{\mathbf{\text{2}}}\underset{\underset{{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}}{\mathbf{\text{|}}}}{\mathbf{\text{C}}}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{CHCO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}{\mathbf{\text{+NH}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{N}}\stackrel{\stackrel{\mathbf{\text{O}}}{\mathbf{\parallel }}}{\mathbf{\text{C}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}\underset{\underset{{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}}{\mathbf{\text{|}}}}{\mathbf{\text{C}}}{\mathbf{\text{HCO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{\text{l}}\mathbf{\right)}\\ \mathbf{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Glutamic\hspace{0.17em}acid\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Glutamine}}\end{array}$

For this reaction $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{=}\mathbf{14}\mathit{k}\mathit{J}$ at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$

1. Calculate for this reaction at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$
2. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 84). Calculate $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$and K at ${\mathbf{25}}^{\mathbf{o}}\mathit{C}$for the following reaction:

$\mathbf{\text{Glutamic\hspace{0.17em}acid}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\text{+ATP}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\text{+NH}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{\text{Glutamine}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}\mathbf{\text{+ADP}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}\mathbf{\left(}\mathbf{\text{aq}}\mathbf{\right)}$

Short answer

1. The value for K is$K=3.52\times {10}^{-3}$
2. -16.5 kJ/mol is the value for $\Delta G^o.$

Step-by-step solution

The Calculation for k  in the solution

$\text{Glutamic\hspace{0.17em}acid}\left(aq\right)+N{H}_3\left(aq\right)\rightleftharpoons \text{Glutamine}\left(aq\right)+H_{2}O\left(l\right)$

$\begin{array}{l}\Delta G_R^0=-RT\ln K\\ K=e^{-\frac{\Delta G_R^0}{RT}}\\ =e^{-\frac{14kJ/mol}{0.008314kJ/mol-K\times 298K}}\\ K=3.52\times {10}^{-3}\end{array}$

Adding of the reaction forΔGo

$ATP\left(aq\right)+H_{2}O\left(l\right)\to ADP\left(aq\right)+H_{2}P{O}_4^{2-}$,$\Delta G_1^0=-30.5kJ/\text{mol}$

$\text{Glutamic\hspace{0.17em}acid}\left(aq\right)+N{H}_3\left(aq\right)\rightleftharpoons \text{Glutamine}\left(aq\right)+H_{2}O\left(l\right)$,$\Delta G_2^0=14.0kJ/\text{mol}$

By adding,

$\text{Glutamic\hspace{0.17em}acid}\left(aq\right)+ATP\left(aq\right)+N{H}_3\left(aq\right)\rightleftharpoons \text{Glutamine}\left(aq\right)+ADP\left(aq\right)+H_{2}P{O}_4^{2-}$

$\begin{array}{c}\Delta G_{}^0=\Delta G_1^0+\Delta G_2^0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-30.5+14.0\\ \Delta G_{}^0=-16.5kJ/\text{mol}\end{array}$

Conclusion

Therefore, the calculation value for K is $K=3.52\times {10}^{-3}$and -16.5 kJ/mol is the value for $\Delta G^o.$