Relationship between free Energy and Equilibrium Constant

      ΔG = ΔG^o + RT InQ

Where Q is the reaction quotient

When equilibrium is attained, there is no further free energy change i.e. ΔG = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.

           ΔG^o = –RT In K_(eq) 

      or  ΔG^o = –2.303 RT log K_(eq)

In case of galvanic cells. Gibbs energy change ΔG is related to the electrical work done by the cell.

ΔG = -nFE_(cell) where n = no. of moles of electrons involved

                        F = the Faraday constant

                        E = emf of the cell

If reactants and products are in their standard states ΔG^o = –nFE^o_cell

Solved example: Calculate ΔG⁰ for conversion of oxygen to ozone 3/2 O₂ (g) ——> O₃ (g) at 298 K, if K_p for this conversion is 2.47 ´ 10^-29.

Solution: ΔG^o = –2.303 RT log K_p

Where R = 8.314 J/K mol, K_p = 2.47 ´ 10^-29, T = 298K

       ΔG⁰ = 16300 J/mol = 163 KJ/mol.

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