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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?

User Amitav Roy
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1 Answer

3 votes

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

User Ikashnitsky
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