Chapter 15, Problem 17P

Interpretation Introduction

(a)

Interpretation:

The values of $\Delta G°\text{'}$ and the equilibrium constant, $K_{{}_{\text{eq}}}^{\text{'}}$ at the temperature of $25°\text{C}$ for the reaction between ATP and pyruvate are to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K_{\text{eq}}$. The equilibrium constant is independent of the initial amount of the reactant and product.

Answer to Problem 17P

The values of $\Delta G°\text{'}$ and the equilibrium constant, $K_{{}_{\text{eq}}}^{\text{'}}$ at the temperature of $25°\text{C}$ ifor the reaction between ATP and pyruvate are $+31.4\text{kJ}{\text{mol}}^{-1}$ and $3.01\times {10}^{-6}$ respectively.

Explanation of Solution

The given reversible reaction is,

$\text{ATP}+\text{pyruvate}\rightleftharpoons \text{phosphoenolpyruvate}+\text{ADP}$.

The standard Gibbs free energy for the conversion of ATP to ADP is $\Delta G°\text{'}=-30.8\text{kJ}{\text{mol}}^{-1}$.

The given standard Gibbs free energy for the conversion of pyruvate to phosphoenolpyruvate is $\Delta G°\text{'}=-61.9\text{kJ}{\text{mol}}^{-1}$.

Thus, the change in the standard Gibbs free energy for the conversions is calculated as follows.

$\begin{array}{c}\text{Change}\text{in}\Delta G°\text{'}=\Delta G°\text{'}\text{ for}\text{ATP}\text{to}\text{ADP}-\Delta G°\text{'}\text{ for}\text{pyruvate}\text{to}\text{phosphoenolpyruvate}\\ \Delta G°\text{'}=-30.5\text{kJ}{\text{mol}}^{-1}-\left(-61.9\text{kJ}{\text{mol}}^{-1}\right)\\ =+31.4\text{kJ}{\text{mol}}^{-1}\end{array}$

The value of universal gas constant is $8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}$.

The standard temperature is $25°\text{C}$.

The conversion of degrees Celsius into Kelvin is done as,

$0°\text{C}=\text{273}\text{K}$

Thus, the given temperature becomes,

$\begin{array}{c}25°\text{C}=25+\text{273}\text{K}\\ =\text{298}\text{K}\end{array}$

The value of $K_{{}_{\text{eq}}}^{\text{'}}$ is calculated by the expression,

$K_{{}_{\text{eq}}}^{\text{'}}=e^{\frac{-\Delta G°\text{'}}{RT}}$

Where,

• $\Delta G°\text{'}$ is the Gibbs free energy.
• $R$ is the universal gas constant.
• $T$ is the temperature.

The calculated value of $R\times T$ is $8.315\times {10}^{-3}\text{kJ}{\text{mol}}^{-1}{\text{K}}^{-1}\times \text{298}\text{K}=2.47\text{kJ}{\text{mol}}^{-1}$

Substitute the values of Gibbs free energy for the conversions and $R\times T$ in the above expression.

$\begin{array}{c}{K}_{{}_{\text{eq}}}^{\text{'}}=e^{\frac{-31.4}{2.47}}\\ =3.01\times {10}^{-6}\end{array}$

Thus, the value of the Gibbs free energy and for the equilibrium constants for the given reaction are $+31.4\text{kJ}{\text{mol}}^{-1}$ and $3.01\times {10}^{-6}$ respectively.

Interpretation Introduction

(b)

Interpretation:

The equilibrium ration for the conversion of of pyruvate to phosphoenolpyruvate is to be stated.

Concept introduction:

When any reaction is at equilibrium then a constant expresses a relationship between the reactant side and the product side. This constant is known as equilibrium constant. It is denoted by $K_{\text{eq}}$. The equilibrium constant is independent of the initial amount of the reactant and product.

Answer to Problem 17P

The equilibrium ration for the conversion of of pyruvate to phosphoenolpyruvate is $3.28\times {10}^4$.

Explanation of Solution

The given ratio for the conversion of ATP to ADP is $K_{{}_{\text{eq}}}^{\text{'}}\frac{\left[\text{ADP}\right]}{\left[\text{ATP}\right]}=10$.

The calculated value of $K_{{}_{\text{eq}}}^{\text{'}}$ is $3.01\times {10}^{-6}$.

The expression equilibrium constant, $K_{{}_{\text{eq}}}^{\text{'}}$ for the given reaction is written as follows.

$K_{{}_{\text{eq}}}^{\text{'}}=\frac{\left[\text{Phosphoenolpyruvate}\right]\left[\text{ADP}\right]}{\left[\text{Pyruvate}\right]\left[\text{ATP}\right]}$

The reversed expression equilibrium constant, $K_{{}_{\text{eq}}}^{\text{'}}$ for the given reaction is written as follows.

$\frac{1}{K_{{}_{\text{eq}}}^{\text{'}}}=\frac{\left[\text{Pyruvate}\right]\left[\text{ATP}\right]}{\left[\text{Phosphoenolpyruvate}\right]\left[\text{ADP}\right]}$

Substitute the values of $K_{{}_{\text{eq}}}^{\text{'}}$ and ratio for the conversion of ATP to ADP in the above expression.

$\begin{array}{c}\frac{1}{3.01\times {10}^{-6}}=\frac{\left[\text{Pyruvate}\right]}{\left[\text{Phosphoenolpyruvate}\right]}\times 10\\ \frac{\left[\text{Pyruvate}\right]}{\left[\text{Phosphoenolpyruvate}\right]}=3.28\times {10}^4\end{array}$

Thus, the equilibrium ration for the conversion of pyruvate to phosphoenolpyruvate is $3.28\times {10}^4$.