Question

The concentration of glucose inside a cell is 0.12 mM. Outside the cell the concentration of glucose is 12.9 mM. Calculate the change in Gibbs free energy for the transport of 1.50 moles of glucose into the cell at 37°C.

Answer 1

Answer:

The value of he change in Gibbs free energy ΔG = - 18.083 KJ

Explanation:

Given data

The concentration of glucose inside a cell is (P) = 0.12 m M

The concentration of glucose outside a cell is (R) = 12.9 m M

No. of  moles = 1.5 moles

The change in Gibbs free energy

ΔG = RT ㏑$\frac{P}{R}$

ΔG = 8.314 × 310 ㏑$\frac{0.12}{12.9}$

ΔG = - 12.055 $\frac{J}{mole}$

Since No. of  moles = 1.5 moles

Therefore

ΔG = - 12.055 × 1.5

ΔG = - 18.083 KJ

This the value of he change in Gibbs free energy.