Question

Calculate the free energy required to move 1 mol of ${\mathbf{K}}^{\mathbf{+}}$ ions from the outside of the cell (where$\left[K^{+}\right]\mathbf{=}\mathbf{4}\mathbf{mM}$ ) to the inside (where $\left[K^{+}\right]\mathbf{=}\mathbf{140}\mathbf{}\mathbf{mM}$) when the membrane potential is$\mathbf{-}\mathbf{70}\mathbf{}\mathbf{mV}$ and the temperature is ${\mathbf{37}}^{\mathbf{0}}\mathbf{}\mathbf{C}$.

Short answer

The free energy change to move the membrane potential is $2.4{\mathrm{kJmol}}^{-1}$.

Step-by-step solution

Introduction

The amount of free energy required to move a chemical over a membrane is determined by the concentrations on both sides. The membrane potentialdetermines the free energy change in the case of ions. When all the concentrations are identical, free energy change is reduced, making a random distribution of molecules an energy-rich state. stands for "free energy change."

Calculation of membrane potential.

The intracellular concentration of$\left[{\mathrm{K}}^{+}\right]$ is$0.004\mathrm{M}$

The extracellular concentration of$\left[{\mathrm{K}}^{+}\right]$ is$0.14\mathrm{M}$

The chemical potential difference is,

$∆\mathrm{G}=\mathrm{RTln}\frac{{\left[{\mathrm{K}}^{+}\right]}_{\mathrm{in}}}{{\left[{\mathrm{K}}^{+}\right]}_{\mathrm{out}}}+\mathrm{ZF}∆\mathrm{\phi }$

On substituting,

$$\begin{gathered} =\left(310\mathrm{K}\right)\ln \frac{\left[0.14\right]}{\left[0.004\right]}+\left(1\right)\left(96,485{\mathrm{JV}}^{-1}{\mathrm{mol}}^{-1}\right)\left(-0.070\mathrm{V}\right) \\ =9163{\mathrm{Jmol}}^{-1}-6754{\mathrm{Jmol}}^{-1} \\ =2.4{\mathrm{kJmol}}^{-1} \end{gathered}$$

Conclusion

Thus, the chemical potential difference when themembrane potential is $-70\mathrm{mV}=2.4{\mathrm{kJmol}}^{-1}$.