Chapter 13, Problem 4P

Interpretation Introduction

Interpretation:

The free-energy cost at $25°\text{C}$ of pumping ${\text{Ca}}^{2+}$ out of a cell is to be calculated.

Concept introduction:

The thermodynamic quantity that measures the free energy available in the system to work at a constant temperature is called Gibbs free energy. It is represented by $\text{G}\text{.}$ It determines the feasibility of the chemical reaction.

The expression to calculate the free-energy change is as follows:

$\Delta G=RT\ln \left(\frac{c_2}{c_1}\right)+ZF\Delta V$

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $c_2$ is the concentration outside of the cell
• $c_1$ is the concentration inside the cell.
• $Z$ is the electrical charge of the transported species.
• $F$ is the faraday constant.
• $\Delta V$ is the potential in volt across membrane.

Answer to Problem 4P

The free-energy cost at $25°\text{C}$ of pumping ${\text{Ca}}^{2+}$ out of a cell is $31.969\text{ kJ}\cdot {\text{mol}}^{-1}$.

Explanation of Solution

The expression to calculate the free-energy change is as follows:

$\Delta G=RT\ln \left(\frac{c_2}{c_1}\right)+ZF\Delta V$   ....... (1)

Where,

• $R$ is the gas constant.
• $T$ is the temperature.
• $c_2$ is the concentration outside of the cell
• $c_1$ is the concentration inside the cell.
• $Z$ is the electrical charge of the transported species.
• $F$ is the faraday constant.
• $\Delta V$ is the potential in volt across membrane.

The formula to convert $°\text{C}$ to Kelvin is:

$T\left(K\right)=T\left(°C\right)+273$

The value of $T\left(°C\right)$ is $\text{25 }°\text{C}$.

Substitute $T\left(°C\right)$ in the above expression.

$\begin{array}{c}T\left(K\right)=25+273\\ =298\text{ K}\end{array}$

The calcium ion $\left({\text{Ca}}^{2+}\right)$ moves out of a cell, so the potential in volt across membrane should be positive. The value of potential in volt across membrane is $60\text{ mV}\text{.}$

The value of $R$ is $8.314\text{J}{\text{mol}}^{-1}\cdot {\text{K}}^{-1}$.

The value of $T$ is $298\text{ K}$.

The value of $c_2$ is $1.5\text{ mM}$.

The value of $c_1$ is $0.4\text{ μM}$.

The value of $Z$ is 2.

The value of $F$ is $\text{96}\text{.5 kJ}\cdot {\text{mol}}^{-1}\cdot {\text{v}}^{-1}$.

The value of $\Delta V$ is $-60\text{ mV}$.

Substitute $R$, $T$, $c_2$, $c_1$, $Z$, $F$ and $\Delta V$ in the equation (1).

$\begin{array}{c}\Delta G=\left[\begin{array}{l}\left(8.314\text{J}{\text{mol}}^{-1}\cdot {\text{K}}^{-1}\right)\left(\frac{1\text{kJ}}{1000\text{J}}\right)\left(298\text{ K}\right)\ln \left(\frac{1.5\text{ mM}}{\left(0.4\text{ μM}\left(\frac{1\text{ mM}}{1000\text{ μM}}\right)\right)}\right)\\ +\left(2\right)\left(\text{96}\text{.5 kJ}\cdot {\text{mol}}^{-1}\cdot {\text{v}}^{-1}\right)\left(60\text{ mV}\left(\frac{1\text{V}}{1000\text{mV}}\right)\right)\end{array}\right]\\ =31.969\text{ kJ}\cdot {\text{mol}}^{-1}\end{array}$

Conclusion

The free-energy cost at $25°\text{C}$ of pumping ${\text{Ca}}^{2+}$ out of a cell is $31.969\text{ kJ}\cdot {\text{mol}}^{-1}$.