Chapter 16, Problem 10P

(a)

Summary Introduction

To determine: The equilibrium constant for the reaction at 25°C.

Introduction:

The citric acid cycle is a series of various types of chemical reactions. In the last step of citric acid cycle, malate undergoes dehydrogenation to produce oxaloacetate, which is essential for the entry of acetyl CoA into the cycle to start second round of the cycle.

Explanation of Solution

Explanation:

The conversion of malate into oxaloacetate can be written in equation form as:

$\text{L-Malate}+{\text{NAD}}^{+}\to \text{Oxaloacetate}+\text{NADH}+{\text{H}}^{+}$

Standard free energy change or $\Delta G{\text{'}}^{\circ }$ for this reaction is 30kJ/mol. R is a gas constant and its value is 8.314 J/mol, and T is the standard temperature at 278K. The equilibrium constant $\left(K{\text{'}}_{eq}\right)$ can be calculated by the formula,

$\ln \text{K}{\text{'}}_{\text{eq}}=\frac{\Delta G{\text{'}}^{\circ }}{-\text{RT}}$                                                                                                    (1)

Here, $\Delta G{\text{'}}^{\circ }$ is the standard free energy change. $\text{K}{\text{'}}_{\text{eq}}$ is the equilibrium constant. R is the universal gas constant and T is the absolute temperature.

Substitute the value in equation (1):

$\begin{array}{c}\ln \text{K}{\text{'}}_{\text{eq}}=\frac{\Delta G{\text{'}}^{\circ }}{-\text{RT}}\\ =-\frac{\left(30k\text{J/mol}\right)}{\left(2.48k\text{J/mol}\right)}\\ =-12.1\\ =-0.121\times {10}^2\end{array}$

To calculate the value of $\text{K}{\text{'}}_{\text{eq}}$, take antilog of -12.1.

$\begin{array}{c}\ln \text{K}{\text{'}}_{\text{eq}}=-12.1\\ \text{K}{\text{'}}_{\text{eq}}=5.6\times {10}^{-6}\end{array}$

Conclusion

Conclusion:

The equilibrium constant for the conversion of malate into oxaloacetate at standard room temperature is $\underset{\_}{5.6\times 10^{-6}}$.

(b)

Summary Introduction

To determine: Oxaloacetate concentration at pH 7.

Introduction:

At equilibrium, the concentration of the reactant and the product are same. It is the state of balance where all the forces are canceled out by equal and opposing force. Equilibrium of the substances is dependent on the concentration of reactants and products.

Explanation of Solution

Explanation:

At pH 7, $\text{K}{\text{'}}_{\text{eq}}$ can be calculated as:

$\text{K}{\text{'}}_{\text{eq}}=\frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{L-malate}\right]\left[{\text{NAD}}^{\text{+}}\right]}$                                                                         (2)

Here, $\text{K}{\text{'}}_{\text{eq}}$ is the equilibrium constant. [Oxaloacetate] is the concentration of oxaloacetate. [NADH] is the concentration of NADH. [L-malate] is the concentration of L- malate. [NAD⁺] is the concentration of NAD⁺. Thus, the values given and calculated are:

$\text{K}{\text{'}}_{\text{eq}}=5.6\times {10}^{-6}$

$\begin{array}{c}\frac{\left[{\text{NAD}}^{+}\right]}{\left[\text{NADH}\right]}=10\\ \frac{\left[\text{NADH}\right]}{\left[{\text{NAD}}^{+}\right]}=\frac{1}{10}\\ \left[\text{L-malate}\right]=0.20\text{mM}\\ \text{=2}\times {\text{10}}^{-2}\text{M}\end{array}$

Substitute the values in equation (2):

$\begin{array}{c}\text{K}{\text{'}}_{\text{eq}}=\frac{\left[\text{Oxaloacetate}\right]\left[\text{NADH}\right]}{\left[\text{L-malate}\right]\left[{\text{NAD}}^{\text{+}}\right]}\\ 5.6\times {10}^{-6}=\frac{\left[\text{Oxaloacetate}\right]}{2.0\times {10}^{-2}\text{M}\times \text{10}}\\ \left[\text{Oxaloacetate}\right]=5.6\times {10}^{-6}\times 2.0\times {10}^{-2}\text{M}\times \text{10}\\ =1.12\times {10}^{-8}\text{M}\end{array}$

Conclusion

Conclusion:

The concentration of oxaloacetate in pH 7 is $\underset{\_}{1.12\times 10^{-8}M}$.

(c)

Summary Introduction

To determine: The number of oxaloacetate molecules in the rat liver mitochondrion.

Introduction:

Total number of molecules can be obtained by multiplying the number of moles and the Avogadro number. Avogadro number can be defined as the particles or molecules of the substance present in particular amount of the substance. Avogadro number value is $6.022\times {10}^{23}\text{molecules/mole}$.

Explanation of Solution

Explanation:

Given,

Diameter of mitochondrion 2.0 µm, radius of mitochondrion 1.0 µm, [Oxaloacetate] is $1.12\times {10}^{-8}\text{M}\text{.}$

$\text{Volume of mitochondria}=\frac{4}{3}{\text{πr}}^{\text{3}}$                                                                (3)

Here, r is the radius of mitochondria which is 1 µm.

$\begin{array}{c}r=1\mu \text{m}\times \frac{{10}^{-2}\text{cm}}{{10}^{-6}\mu \text{m}}\\ =1\times {10}^{-4}\text{cm}\end{array}$

Constant $\pi =3.14$

Substitute the values in equation (3).

$\begin{array}{c}\text{Volume of mitochondria}=\frac{4}{3}\times \text{3}\text{.14}\times {\left({10}^{-4}\text{cm}\right)}^3\\ =4.18\times {10}^{-12}{\text{cm}}^3\end{array}$

1L of solution contains $1.12\times {10}^{-8}\text{M}$ of oxaloacetate. Thus,

$\begin{array}{l}1\text{L}=1000{\text{cm}}^3\\ 1000{\text{cm}}^3=1.1\times {10}^{-8}\text{M}\\ \text{4}\text{.19}\times {\text{10}}^{-12}{\text{cm}}^3=4.61\times {10}^{-23}\text{M}\end{array}$

Avogadro number is the number of particles in per unit volume, the value of Avogadro number is constant $6.023\times {10}^{23}{\text{M}}^{-1}.$ Total number of molecules can be obtained by multiplying the number of molecules with the Avogadro number.

$\text{Total}\text{number}\text{of}\text{molecules}=\text{Number}\text{of moles}\times \text{Avogadro}\text{number}$          (4)

Substitute the values in equation (4).

$\begin{array}{c}\text{Total}\text{number}\text{of}\text{molecules}=\text{Number}\text{of moles}\times \text{Avogadro}\text{number}\\ \text{= 4}\text{.61}\times {\text{10}}^{-23}\text{M}\times 6.023\times {10}^{23}{\text{M}}^{-1}\\ =27.77\approx 28\end{array}$

Conclusion

Conclusion:

The number of molecules of oxaloacetate in a single rat liver mitochondrion is $\underset{\_}{28}$.