Q102CP - Chemistry: Molecular Nature Of Matter And Change

Question

Energy from ATP hydrolysis drives many nonspontaneous cell reactions: 

$$\begin{gathered} \mathit{A}\mathit{T}{\mathit{P}}^{\mathbf{4}\mathbf{-}}\left(aq\right)\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(I\right)\mathbf{\rightleftharpoons } \\ \mathit{A}\mathit{D}{\mathit{P}}^{\mathbf{3}\mathbf{-}}\left(aq\right)\mathbf{+}\mathit{H}\mathit{P}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{∆}{\mathit{G}}^{\mathbf{\text{o'}}}\mathbf{=}\mathbf{-}\mathbf{30}\mathbf{.}\mathbf{5}\mathit{k}\mathit{J} \end{gathered}$$

Energy for the reverse process comes ultimately from glucose metabolism: 

${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{12}}}{\mathbf{\text{O}}}_{\mathbf{\text{6}}}{\mathbf{\text{(s) + 6O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{6CO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + 6H}}}_{\mathbf{\text{2}}}\mathbf{\text{O(l)}}$ 

(a) Find K for the hydrolysis of ATP at ${\mathbf{37}}^{\mathbf{°}}\mathit{C}$. 

(b) Find$\mathbf{∆}{\mathit{G}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{\text{o'}}}$ for metabolism of $\mathbf{1}$mol of glucose.

(c) How many moles of ATP can be produced by metabolism of $\mathbf{1}$mol of glucose? 

(d) If $\mathbf{36}$ mol of ATP is formed, what is the actual yield?

Step-by-Step Solution

Answer

a) At ${37}^{°}C$,$\text{K = 1.378}$ is the equilibrium constant for the ATP hydrolysis reaction.

b) For $1$mol of glucose metabolism, the $∆G_{rxn}^{°}$is$\text{ - 2878.992kJ}$ .

c) It is possible to manufacture $94.372$moles of ATP.

d) The ATP formation yield is just$\text{38.138\% }$ .

Step 1: Definition of Hydrolysis

Hydrolysis is a typical type of chemical reaction in which water is used to break down chemical bonds between two or more substances. The word hydrolysis comes from the Greek word hydro, which means water, and lysis, which means to break or unbind.

Step 2: Calculating K for the hydrolysis of ATP

(a) The ATP hydrolysis reaction is as follows:

${\text{ATP}}^{\text{4 - }}{\text{ + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{ADP}}^{\text{3 - }}{\text{ + HPO}}_{\text{4}}^{\text{ - }}{\text{ + H}}^{\text{ + }}$

This reaction's equilibrium constant can be written as:

$K=\frac{\left[AD{P}^{3-}\right]\times \left[HP{O}_4^{-}\right]\times \left[H^{+}\right]}{\left[AT{P}^{4-}\right]\times \left[H_{2}0\right]}$

Since

$$\begin{gathered} ∆G_{rxn}^{°}=-R\times T\times InK \\ InK=-\frac{∆G_{rxn}^{°}}{R\times T} \\ K=e^{InK} \end{gathered}$$

At the${\text{37}}^{\text{o}}\text{C = 310K}$ value,

$$\begin{gathered} InK=-\frac{-30.5\times {10}^3{\displaystyle \frac{J}{mol}}}{8.314{\displaystyle \frac{J}{mol\times K}}\times 310K} \\ InK=11.834 \\ K=1.378\times {10}^5 \end{gathered}$$

At${37}^{°}C$, the equilibrium constant for the ATP hydrolysis reaction is $K = 1.378 \times {10}^5$

Step 3: Calculating for metabolism

(b) We can use Appendix B and the equation below to get$∆G_{rxn}^{°\text{'}}$ :

$$\begin{gathered} {\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{(s) + 6O}}_{\text{2}}\text{(g)}\to {\text{6CO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O(l)} \\ ∆G_{rxn}^{°}=\stackrel{°}{a∆G_{products}^{°}}-\stackrel{°}{a}∆G_{reac\tan ts}^{°} \\ ∆G_{rxn}^{°}=\left[6mol\times \left(-394.40kJ/mol\right)+6mol\times \left(-273.192kJ/mol\right)\right]- \\ -\left[1mol\times \left(-910.56kJ/mol\right)+6mol\times \left(0.00kJ/mol\right)\right] \\ ∆G_{rxn}^{°}=-2878.992kJ \end{gathered}$$

For $1$mol of glucose metabolism, the is$∆G_{rxn}^{°} is -2878.992kJ$ .

Step 4: Determining the moles of ATP produced by metabolism

(c) Since we know the$∆G_{rxn}^{°}$ for $1$mol of ATP hydrolysis to ADP, we can assume that the reverse process - ATP synthesis - will have$∆G_{rxn}^{°}$ a of the opposite sign. Thus,

$∆G_{rxn}^{°}=30.5kJ$

Knowing how much energy it takes to manufacture one mol of ATP and how much energy mol of glucose metabolism produces (estimated in b),

$n\left(ATP\right)=\frac{∆G_{glu\cos e}^{°}}{∆G_{ATP}^{°}}$

It is important to note that the absolute value of energy is used.

$$\begin{gathered} n\left(ATP\right)=\frac{2878.992kJ}{30.5kJ/mol} \\ n\left(ATP\right)=94.392mol \end{gathered}$$

As a result, we can produce roughly $94.372$moles of ATP.

Step 5: Determining the actual yield

(d) The yield can be computed using the following formula:

$$\begin{gathered} Yield=\frac{n_{experimental}}{n_{theoretical}}\times 100\% \\ Yield=\frac{36mol}{94.392mol}\times 100\% \\ Yield=38.138\% \end{gathered}$$

The ATP formation yield is only$\text{38.138\% }$ .