Chapter 20, Problem 20.99P

(a)

Interpretation Introduction

Interpretation:

For the hydrolysis of ATP, the equilibrium constant $K$ should be founded at $298K$.

Concept introduction:

Adenosine triphosphate ATP: The main job of ATP is to store energy and release it when cell is in need of energy.

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$,

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^o\text{+}\text{RT}\ln \left(\text{K}\right)\\ \left(or\right)\\ {\text{ΔG}}^o=-\text{RT}\ln \left(\text{K}\right)\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$ is standard free energy values.

Answer to Problem 20.99P

The hydrolysis of ATP is equilbrium constant value is $\underset{\_}{K=2.22\times 10^5}.$

Explanation of Solution

The chemical equation for hydrolysis of ATP is,

$\left(1\right).AT{P}^{4-}\left(aq\right)+H_{2}O\left(l\right)\underset{}{\rightleftharpoons }AD{P}^{3-}\left(aq\right)+HP{O}_4^{2-}\left(aq\right)+H^{+}\left(aq\right)\Delta G^o=-30.5kJ$

Free energy equation is,

  $\begin{array}{c}\text{ΔG=0=}{\text{ΔG}}^{\text{o}}+\text{RT}\ln \left(\text{K}\right)\\ \\ {\text{ΔG}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)-------\left[1\right]\\ \\ {\text{ΔG}}^{\text{o}}\text{=-30}\text{.5}\text{kJ}\\ \\ \text{R=}\text{8}\text{.314}\text{J/mol}\times \text{K}\\ \\ \text{T=}\text{273}\text{K+25}\text{K}\text{=298}\text{K}\text{.}\end{array}$

Consider the reaction (1) hydrolysis of ATP

Calculation for equilibrium constant $\text{K}$

We know that equilibrium equation,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)$

Rearrange the above equation,

  $\begin{array}{c}\ln \text{K}=-\frac{{\text{ΔG}}^{\text{o}}}{\text{RT}}=\left(\frac{-30.5\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(298\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=12.3104\\ \\ \text{K=}{\text{e}}^{12.3104}\\ \\ \text{K=}2.2199\times {10}^5\text{=}2.22\times 10^5.\end{array}$

The hydrolysis of ATP is equilbrium constant value is $\underset{\_}{K=2.22\times 10^5}.$

(b)

Interpretation Introduction

Interpretation:

For the dehydration condensation to from glucose phosphate, the equilibrium constant $K$ should be calculated at $298K$.

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

    $\text{ΔG}\text{=}{{\displaystyle \sum {\text{nΔG}}_{\text{f}}}}^{\text{°}}\text{(products)-}{{\displaystyle \sum {\text{mΔG}}_{\text{f}}}}^{\text{°}}\text{(reactants)}$

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^o\text{+}\text{RT}\ln \left(\text{K}\right)\\ \left(or\right)\\ {\text{ΔG}}^o=-\text{RT}\ln \left(\text{K}\right)\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$ is standard free energy values.

Answer to Problem 20.99P

The ATPs, dehydration equilbrium is $\underset{\_}{K=3.81\times 10^{-3}}.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$\left(2\right)\text{.}\text{Glucose}+HP{O}_4^{2-}\left(aq\right)\underset{}{\rightleftharpoons }{\left[\text{Glucose}\text{phosphate}\right]}^{-}+H_{2}O\left(l\right)\Delta G^o=13.8kJ$

Free energy equation is,

  $\begin{array}{c}\text{ΔG=0=}{\text{ΔG}}^{\text{o}}+\text{RT}\ln \left(\text{K}\right)\\ \\ {\text{ΔG}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)-------\left[1\right]\\ \\ {\text{ΔG}}^{\text{o}}\text{=}\text{13}\text{.8}\text{kJ}\\ \\ \text{R=}\text{8}\text{.314}\text{J/mol}\times \text{K}\\ \\ \text{T=}\text{273}\text{K+25}\text{K}\text{=298}\text{K}\text{.}\end{array}$

Consider the reaction (2) dehydration of ATP

Calculation for equilibrium constant $\text{K}$

We know that equilibrium equation,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)$

Rearrange the above equation,

  $\begin{array}{c}\ln \text{K}=-\frac{{\text{ΔG}}^{\text{o}}}{\text{RT}}=\left(\frac{13.8\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(298\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=-5.569969\\ \\ \text{K=}{\text{e}}^{-5.569969}\\ \\ \text{K=}3.8105985\times {10}^{-3}\\ \\ \text{K=}3.81\times 10^{-3}.\end{array}$

Hence, the ATPs, dehydration equilbrium is $\underset{\_}{K=3.81\times 10^{-3}}.$

(c)

Interpretation Introduction

Interpretation:

For the coupled reaction between ATP and glucose chemical equilibrium constant $K$ should be calculated at $298K$.

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

    $\text{ΔG}\text{=}{{\displaystyle \sum {\text{nΔG}}_{\text{f}}}}^{\text{°}}\text{(products)-}{{\displaystyle \sum {\text{mΔG}}_{\text{f}}}}^{\text{°}}\text{(reactants)}$

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^o\text{+}\text{RT}\ln \left(\text{K}\right)\\ \left(or\right)\\ {\text{ΔG}}^o=-\text{RT}\ln \left(\text{K}\right)\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$ is standard free energy values.

Answer to Problem 20.99P

Given coupled reaction equilbrium value is $\underset{\_}{K=8.46\times 10^2}.$

Explanation of Solution

The chemical equation for dehydration of ATP is,

$\left(3\right)\text{.}\text{Glucose}+AT{P}^{4-}\left(aq\right)\underset{}{\rightleftharpoons }{\left[\text{Glucose}\text{phosphate}\right]}^{-}+AD{P}^{3-}\left(aq\right)\Delta G^o=-16.7kJ$

Free energy equation is,

  $\begin{array}{c}\text{ΔG=0=}{\text{ΔG}}^{\text{o}}+\text{RT}\ln \left(\text{K}\right)\\ \\ {\text{ΔG}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)\\ \\ {\text{ΔG}}^{\text{o}}=-16.7\text{kJ/mol}\\ \\ {\text{ΔG}}^{\text{o}}\text{=}\text{-16}\text{.7}\text{kJ}\\ \\ \text{R=}\text{8}\text{.314}\text{J/mol}\times \text{K}\\ \\ \text{T=}\text{273}\text{K+25}\text{K}\text{=298}\text{K}\text{.}\end{array}$

Consider the coupled (3) reaction between ATP and glucose is,

The equilibrium equation,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)$

Rearrange the above equation to calculate $\text{K}$

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{-16.7\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(298\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=6.74047\\ \\ \text{K=}{\text{e}}^{6.74047}\text{=}8.45958\times {10}^2\text{=}8.46\times 10^2.\end{array}$

Hence, the coupled reaction equilbrium value is $\underset{\_}{K=8.46\times 10^2}.$

(d)

Interpretation Introduction

Interpretation:

Identify the change in $K$ when $T$ changes from ${20}^{o}Cto{37}^{o}C$ in each cases a,b and c 

Concept introduction:

Free energy change: The free energy change of a reaction is given by the subtraction of free energy changes of reactants from free energy changes of reactants.

    $\text{ΔG}\text{=}{{\displaystyle \sum {\text{nΔG}}_{\text{f}}}}^{\text{°}}\text{(products)-}{{\displaystyle \sum {\text{mΔG}}_{\text{f}}}}^{\text{°}}\text{(reactants)}$

Free energy change$\text{ΔG}$: change in the free energy takes place while reactant converts to product where both are in standard state. It depends on the equilibrium constant $\text{K}$

  $\begin{array}{l}\text{ΔG =}{\text{ΔG}}^o\text{+}\text{RT}\ln \left(\text{K}\right)\\ \left(or\right)\\ {\text{ΔG}}^o=-\text{RT}\ln \left(\text{K}\right)\end{array}$

  Where,

  $\text{T}$ is the temperature

  $\text{ΔG}$ is the free energy

  ${\text{ΔG}}^{\text{0}}$ is standard free energy values.

Answer to Problem 20.99P

Given each reactions equilibrium changes has $\text{K=}1.38\times 10^5,4.73\times 10^{-3}\text{and}6.52\times 10^2.$

Explanation of Solution

The chemical equation of ATP is,

$\begin{array}{l}\left(1\right).AT{P}^{4-}\left(aq\right)+H_{2}O\left(l\right)\underset{}{\rightleftharpoons }AD{P}^{3-}\left(aq\right)+HP{O}_4^{2-}\left(aq\right)+H^{+}\left(aq\right)\Delta G^o=-30.5kJ\\ \\ \left(2\right)\text{.}\text{Glucose}+HP{O}_4^{2-}\left(aq\right)\underset{}{\rightleftharpoons }{\left[\text{Glucose}\text{phosphate}\right]}^{-}+H_{2}O\left(l\right)\Delta G^o=13.8kJ\\ \\ \left(3\right)\text{.}\text{Glucose}+AT{P}^{4-}\left(aq\right)\underset{}{\rightleftharpoons }{\left[\text{Glucose}\text{phosphate}\right]}^{-}+AD{P}^{3-}\left(aq\right)\Delta G^o=-16.7kJ\end{array}$

The temperature changes from $20^{o}Cto37^{o}C$each of $K$changes.

The calculations at the new temperature $T=\left(273+37=310.0K\right)$

The equilibrium equation,

  ${\text{ΔG}}_{\text{rxn}}^{\text{o}}=-\text{RT}\ln \left(\text{K}\right)$

Case-1

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{-30.5\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(310\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=11.8339\\ \\ \text{K=}{\text{e}}^{11.8339}\text{=}1.37847\times {10}^5\text{=}1.38\times 10^5.\end{array}$

Case-2

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{13.8\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(310\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=-5.3543576\\ \\ \text{K=}{\text{e}}^{-5.3543576}\text{=}4.7275\times {10}^{-3}\text{=}4.73\times 10^{-3}.\end{array}$

Case-3

  $\begin{array}{c}\ln \text{K}=\frac{{\text{ΔG}}^{\text{o}}}{\text{-RT}}=\left(\frac{-16.7\text{kJ/mol}}{-\left(8.314\text{J/mol}\cdot \text{K}\right)\left(310\text{K}\right)}\right)\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ \\ \ln \text{K}=6.4795\\ \\ \text{K=}{\text{e}}^{6.4795}\text{=}6.51645\times {10}^2\text{=}6.52\times 10^2.\end{array}$

Hence, the each cases equilibrium changes has $\text{K=}1.38\times 10^5,4.73\times 10^{-3}\text{and}6.52\times 10^2.$