Chapter 10, Problem 1E

Summary Introduction

To calculate: The free energy change for glucose to enter into cells at the given concentration.

Concept introduction: The free energy change for moving a substance across a membrane depends upon the concentrations on both sides of the membrane. In case of ions, the free energy change depends upon the membrane potential. A random distribution of molecules is an energy-rich condition because when all the concentrations are equal, the free energy change is minimized. Free energy change is denoted as ΔG.

Explanation of Solution

Explanation:

The extracellular concentration of glucose is 5mM = 0.005M.

The intracellular concentration of glucose is 3mM = 0.003M.

The free energy change $\text{ΔG = RT}\text{ln}\frac{{\left[\text{glucose}\right]}_{\text{in}}}{{\left[\text{glucose}\right]}_{\text{out}}}$ (1)

Here,

R is the gas constant = 8.3145 J · K^-1· mol^-1.

T is the temperature = 298K.

Substitute the given values in Equation (1).

$\begin{array}{c}\text{ΔG =}\left(\text{8}\text{.3145 J }\cdot {\text{K}}^{\text{-1}}\cdot {\text{ mol}}^{\text{-1}}\right)\left(\text{298K}\right)\text{ln}\frac{\left[\text{0}\text{.003}\right]}{\left[\text{0}\text{.005}\right]}\\ =\left(\text{8}\text{.3145 J }\cdot {\text{K}}^{\text{-1}}\cdot {\text{ mol}}^{\text{-1}}\right)\left(\text{298K}\right)\text{ln(0}\text{.6)}\\ \text{=}\left(\text{8}\text{.3145 J }\cdot {\text{K}}^{\text{-1}}\cdot {\text{ mol}}^{\text{-1}}\right)\left(\text{298K}\right)(-0.51082)\\ \text{= -1270 J }\cdot {\text{mol}}^{\text{-1}}\text{= }-1.27kJ\times mo{l}^{-1}\end{array}$

Conclusion

Conclusion:

The free energy change for glucose to enter into cells at the given concentration is -1.27kJ · mol^-1.