Question

19) Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K.

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ?

ΔG°f (kJ/mol) -110.9, 87.6, 51.3, -237.1

Answer 1

Given that

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ?

ΔG°f (kJ/mol) -110.9, 87.6, 51.3, -237.1

2 HNO₃ (aq) + NO(g) ........> 3 NO ₂ (g) + H₂O(l)

First calculate the free energy change as follows:

ΔG^o = ΔG^o products * ΔG^o reactants

ΔG^o = 3(ΔG^o_NO2) + ΔG^o_H2O - 2(ΔG^o_HNO3) - ΔG^o_NO

          = 3(51.3) - 237.1 - 2(-110.9) - (87.6)

          = 51 kJ

We know that;

ΔG^o = - RT ln K

ln K = - (51 kJ) / (8.314*10^-3 kJ mol^-1 K^-1) (298 K)

     = -20.58

   So, K = e^-20.58

= 1.15*10^-9