Question

The distribution of Na+ ions across a typical biological membrane is 10 mmol dm-3 inside the cell and 140 mmol dm-3 outside the cell. At equilibrium the concentrations are equal. What is the Gibbs energy difference across the membrane at 37oC in units of kJ mol-1?

Answer 1

Answer :  The value of $\Delta G^o$ across the membrane is 6.80 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}\\\\\Delta G^o=-RT\times \ln (\frac{C_{in}}{C_{out}})$

where,

$\Delta G^o$ = standard Gibbs free energy  = ?

R = gas constant = 8.314 J/K.mol

T = temperature = $37^oC=273+37=310K$

$K_{eq}$  = equilibrium constant

$C_{in}$ = concentration inside the cell = $10mmol.dm^{3-}$

$C_{out}$ = concentration outside the cell = $140mmol.dm^{3-}$

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln (\frac{C_{in}}{C_{out}})$

$\Delta G^o=-(8.314J/K.mol)\times (310K)\times \ln (\frac{10mmol.dm^{3-}}{140mmol.dm^{3-}})$

$\Delta G^o=6.80\times 10^{3}J/mol=6.80kJ/mol$

Thus, the value of $\Delta G^o$ across the membrane is 6.80 kJ/mol