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* \ln K_(eq)\\\\\Delta G^o=-RT* \ln ((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* \ln ((C_(in))/(C_(out)))`

`\Delta G^o=-(8.314J/K.mol)* (310K)* \ln ((10mmol.dm^(3-))/(140mmol.dm^(3-)))`

`\Delta G^o=6.80* 10^(3)J/mol=6.80kJ/mol`

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