Q99CP - Chemistry: Molecular Nature Of Matter And Change

Question

Find $\mathit{K}$for

(a) the hydrolysis of ATP, 

(b) the dehydration condensation to form glucose phosphate, and 

(c) the coupled reaction between ATP and glucose. 

(d) How does each  $\mathit{K}$ change when $\mathit{T}$  changes from ${\mathbf{25}}^{\mathbf{°}}\mathit{C}$  to ${\mathbf{37}}^{\mathbf{°}}\mathit{C}$ ?

Step-by-Step Solution

Answer

(a) The equilibrium constant for ATP dehydration reaction is$\text{K = 2.22}·{\text{10}}^{\text{5}}$  .

(b) The equilibrium constant for dehydration-condensation reaction is$\text{K = 3.811}·{\text{10}}^{\text{ - 3}}$  .

(c) The equilibrium constant for coupling reaction of ATP with glucose is $K=8.46\times {10}^2$ 

(d) The equilibrium constant for ATP dehydration reaction decreased by -$∆K=-8.42·{10}^4$ .

The equilibrium constant for dehydration-condensation reaction increased by $∆K=9.17·{10}^{-4}$ .

The equilibrium constant for coupling reaction of ATP with glucose decreased   .

Step 1: Concept Introduction

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.

Step 2: Hydrolysis of ATP

(a)

The reaction of ATP hydrolysis is as following –

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

The equilibrium constant for this reaction can be expressed as –

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

$∆G_{rxn}^{°}=R·T·InK$

Since,  

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

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

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

 

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

Step 3: Dehydration Condensation to Form Glucose Phosphate

(b)

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

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

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 $(1atm,298K)$  –

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

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

Step 4: Coupled Reaction between ATP and Glucose

(c)

The coupling reaction of ATP with glucose is as following –

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

The equilibrium constant for this reaction can be expressed as –

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

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

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

 

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

 

Step 5: Change in K when T changes

(d)

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

For Hydrolysis of ATP –

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

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 –

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

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 –

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

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 $\text{ - 1.94}·{\text{10}}^{\text{2}}$ .