Glucose makes up about 0.10 by mass of human blood

Asked by admin @ 30/10/2021 in Chemistry viewed by 260 People

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.

Answered by admin @ 30/10/2021

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.

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