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.

(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.

(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.

(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.