Chapter 13, Problem 13P

(a)

Summary Introduction

To calculate: The ratio $\left[\text{Y}\right]/\left[\text{X}\right]$ at equilibrium.

Introduction:

The ATP refers to adenosine triphosphate. A complex organic compound cell produces during the process of cellular respiration. Cellular respiration is the process by which energy is produced in a cell. Energy is released from the breakdown of glucose in cellular respiration. This energy is then used by the cell for performing various life processes. Mitochondria are considered as the powerhouses of the cell, which synthesizes energy in the form of ATP.

Explanation of Solution

Calculation:

The equilibrium constant is equal to the ratio of $[{\text{Y}}_{\text{eq}}]/[{\text{X}}_{\text{eq}}]$.

The value of gas constant R is 0.0083 kJ/mol K. The equilibrium constant is calculated as follows at the give temperature 293 K:

$\begin{array}{c}(\Delta G^{\prime }°)=-RT\ln ({K^{\prime }}_{\text{eq}})\\ \ln ({K^{\prime }}_{\text{eq}})=\frac{(\Delta G^{\prime }°)}{-RT}\end{array}$

$\begin{array}{c}\ln ({K^{\prime }}_{\text{eq}})=\frac{(20.0\text{kJ}/\text{mol})}{(-0.0083\text{kJ}/\text{mol}\cdot \text{K})(293\text{K})}\\ =\frac{20.0}{-2.48}\\ =-8.07\end{array}$

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=e^{(-8.07)}\\ =3.12\times {10}^{-4}\end{array}$

Conclusion

Conclusion:

The ratio $\left[\text{Y}\right]/\left[\text{X}\right]$ at equilibrium is $\underset{\_}{3\times 10^{-4}}$.

(b)

Summary Introduction

To calculate: The ratio of $\left[\text{Y}\right]/\left[\text{X}\right]$ at equilibrium for the given  reaction provided the  temperature is 25°C and the equilibrium concentrations of ATP, ADP, and Pi are 1M..

Introduction:

The ATP refers to adenosine triphosphate. A complex organic compound cell produces during the process of cellular respiration. Cellular respiration is the process by which energy is produced in a cell. Energy is released from the breakdown of glucose in cellular respiration. This energy is then used by the cell for performing various life processes. Mitochondria are considered as the powerhouses of the cell, which synthesizes energy in the form of ATP..

Explanation of Solution

Calculation:

The two given chemical reactions having their energy change $(\Delta G^{\prime }°)$ are as follows:

$\begin{array}{l}\begin{array}{cc}\text{X}\to \text{Y}& (\Delta G^{\prime }°)=+20.0\text{kJ}/\text{mol}\\ \text{ATP}+{\text{H}}_{\text{2}}\text{O}\to \text{ADP}+\text{Pi}& (\Delta G^{\prime }°)=-30.5\text{kJ}/\text{mol}\end{array}\\ \underset{\_}{}\\ \\ \begin{array}{cc}\text{X}+\text{ATP}\to \text{Y}+\text{ADP}+\text{Pi}& (\Delta G^{\prime }°)=-10.5\text{kJ}/\text{mol}\end{array}\end{array}$

The value of $({K^{\prime }}_{\text{eq}})$ is calculated as follows:

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=\frac{[{\text{Y}}_{\text{eq}}][{\text{Pi}}_{\text{eq}}][{\text{ADP}}_{\text{eq}}]}{[{\text{X}}_{\text{eq}}][{\text{ATP}}_{\text{eq}}]}\\ =\frac{[{\text{Y}}_{\text{eq}}][1\text{M}][1\text{M}]}{[{\text{X}}_{\text{eq}}][1\text{M}]}\\ =\frac{[{\text{Y}}_{\text{eq}}]}{[{\text{X}}_{\text{eq}}]}\end{array}$

The value of gas constant R is 0.0083 kJ/mol K. The equilibrium constant is calculated as follows at the give temperature 293 K.:

$\begin{array}{c}(\Delta G^{\prime }°)=-RT\ln ({K^{\prime }}_{\text{eq}})\\ \ln ({K^{\prime }}_{\text{eq}})=\frac{(\Delta G^{\prime }°)}{-RT}\end{array}$

$\begin{array}{c}\ln ({K^{\prime }}_{\text{eq}})=\frac{(-10.5\text{kJ}/\text{mol})}{(-0.0083\text{kJ}/\text{mol}\cdot \text{K})(293\text{K})}\\ =4.23\end{array}$

$\begin{array}{c}({K^{\prime }}_{\text{eq}})=e^{(4.23)}\\ =68.7\end{array}$

Conclusion

Conclusion:

The ratio of $\left[\text{Y}\right]/\left[\text{X}\right]$ at equilibrium for the given reaction is $\underset{\_}{68.7}$.

(c)

Summary Introduction

To calculate: The ratio of $\left[\text{Y}\right]/\left[\text{X}\right]$ for the ATP-coupled reaction under physiological conditions.

Introduction:

The ATP refers to adenosine triphosphate. A complex organic compound cell produces during the process of cellular respiration. Cellular respiration is the process by which energy is produced in a cell. Energy is released from the breakdown of glucose in cellular respiration. This energy is then used by the cell for performing various life processes. Mitochondria are considered as the powerhouses of the cell, which synthesizes energy in the form of ATP.

Explanation of Solution

Calculation:

The value of $({K^{\prime }}_{\text{eq}})$ is calculated as follows:

$({K^{\prime }}_{\text{eq}})=\frac{[{\text{Y}}_{\text{eq}}][{\text{Pi}}_{\text{eq}}][{\text{ADP}}_{\text{eq}}]}{[{\text{X}}_{\text{eq}}][{\text{ATP}}_{\text{eq}}]}$

This above equation can be modified as following because the ratio $[\text{Y}]/[\text{X}]$ is equal to equilibrium constant.

$\frac{[\text{Y}]}{[\text{X}]}=\frac{({K^{\prime }}_{\text{eq}})[{\text{ATP}}_{\text{eq}}]}{[{\text{Pi}}_{\text{eq}}][{\text{ADP}}_{\text{eq}}]}$

$\begin{array}{c}\frac{[\text{Y}]}{[\text{X}]}=\frac{({K^{\prime }}_{\text{eq}})[{\text{ATP}}_{\text{eq}}]}{[{\text{Pi}}_{\text{eq}}][{\text{ADP}}_{\text{eq}}]}\\ =\frac{[69.1\text{M}][0.00805\text{M}]}{[0.00805\text{M}][\text{0}\text{.00093}\text{M}]}\\ =\frac{[0.556255]}{[0.0000074865]}\\ =7.4\times {10}^4\end{array}$

Conclusion

Conclusion:

The ratio of $\left[\text{Y}\right]/\left[\text{X}\right]$ for the ATP-coupled reaction under physiological conditions is $\underset{\_}{7.4\times 10^4}$.