91Ó°ÊÓ

Hydrolysis is a sort of breakdown reaction in which one of the reactants is water, while the other reactant is often broken chemical bonds with water.

(a)

The reaction of ATP hydrolysis is as following –

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

The equilibrium constant for this reaction can be expressed as –

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

$∆G_{rxn}^{°}=R·T·InK$

Since,

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

Assuming the standard conditions role="math" localid="1663819432369" $\left(1atm,298K\right)$ –

$$\begin{gathered} InK=\frac{-30.5×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}}×298K \\ InK=12.31044 \\ K=222001.50 \\ =2.22·{10}^5 \end{gathered}$$

Therefore, the value is obtained as $2.22·{10}^5$.

(b)

The reaction of dehydration-condensation to form glucose phosphate is as following –

${\text{Glucose + HPO}}_{\text{4}}^{\text{2 -}}{\text{+H}}^{\text{+}}⇌{\text{[glucose phosphate]}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$

The equilibrium constant for this reaction can be expressed as –

$\text{K =}\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]·\left[{\text{[glucose phosphate]}}^{\text{-}}\right]}{\left[{\text{HPOH}}_{\text{4}}^{\text{2 -}}\right]·\left[{\text{H}}^{\text{+}}\right]·\text{[Glucose]}}$

Assuming the standard conditions $(1atm,298K)$ –

$$\begin{gathered} InK=\frac{13.80×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}·298K} \\ \text{lnK = - 5.56996} \\ \text{K = 3.811}·{\text{10}}^{\text{- 3}} \end{gathered}$$

Therefore, the value is obtained as $3.811·{10}^{-3}$.

(c)

The coupling reaction of ATP with glucose is as following –

${\text{Glucose + ATP}}^{\text{4 -}}⇌{\text{[glucose phosphate]}}^{\text{-}}{\text{+ ADP}}^{\text{3 -}}$

The equilibrium constant for this reaction can be expressed as –

$K=\frac{\left[AD{P}^{3-}\right]×\left[{\left[glu\cos e-phosphate\right]}^{-}\right]}{\left[AT{P}^{4-}\right]×\left[glu\cos e\right]}$

Assuming the standard conditions $\left(1atm,298K\right)$ –

$$\begin{gathered} InK=\frac{-16.70×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}·298K} \\ InK=6.74047K \\ =845.958 \\ =8.46·{10}^2 \end{gathered}$$

Therefore, the value is obtained as$8.46·{10}^2$.

(d)

Upon increase in temperature to${\text{37}}^{\text{o}}\text{C = 310 K}$,all$K$can be recalculated.

For Hydrolysis of ATP –

$$\begin{gathered} InK=\frac{-30.5×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}×310K} \\ \text{lnK = 11.834} \\ \text{K = 1,378}·{\text{10}}^{\text{5}} \end{gathered}$$

Thus, the change in$K$is –

$$\begin{gathered} ∆K=\text{1.378}·{\text{10}}^{\text{5}}\text{- 2.22}·{\text{10}}^{\text{5}} \\ ∆K=\text{- 8.42}·{\text{10}}^{\text{4}} \end{gathered}$$

For Dehydration Condensation to Form Glucose Phosphate –

$$\begin{gathered} InK=\frac{13.80×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}·310K} \\ InK=-\text{5.354} \\ K=\text{4.728}·{\text{10}}^{\text{- 3}} \end{gathered}$$

Thus, the change in$K$is –

$$\begin{gathered} ∆K=\text{4.728}·{\text{10}}^{\text{- 3}}\text{- 3.811}·{\text{10}}^{\text{- 3}} \\ ∆K=\text{9.17}·{\text{10}}^{\text{- 4}} \end{gathered}$$

For coupled reaction between ATP and Glucose –

$$\begin{gathered} InK=\frac{-16.70×{10}^3{\displaystyle \frac{J}{mol}}}{-8.314{\displaystyle \frac{J}{mol·K}}·310K} \\ InK=\text{6.47955 K} \\ =\text{651.676} \end{gathered}$$

Thus, the change in$K$is –

$$\begin{gathered} ∆K=\text{6.52}·{\text{10}}^{\text{2}}\text{- 8.46}·{\text{10}}^{\text{2}} \\ \text{∆K=-1.94}·{\text{10}}^{\text{2}} \end{gathered}$$

Therefore, the values are obtained as$\text{- 8.42}·{\text{10}}^{\text{4}}\text{, 9.17}·{\text{10}}^{\text{- 4}}$ and $\text{- 1.94}·{\text{10}}^{\text{2}}$.