Question

What is the maximum amount of non-pV work that can be done by the reaction

2H2 + O2 ->2H2O

if DG(H2O)= -2237.13 kJ/mol, and DG(H2) = DG (O2) = 0?

Answer 1

Solution :-

2H2 + O2   --- > 2H2O

Using the given standard free energy values of the reactant and product we can find the free energy change for the reaction

The free energy change of the reaction is non PV work

Delta G rxn = sum of delta Go product - sum of delta Go reactant

                     = [H2O*2] – [(H2*2)(O2*1)]

                     = [-2237.13 kJ/mol *2 mol ] – [(0*2)+(0*1)]

                     = -4474.26 kJ

Therefore the amount of non PV work done by the reaction is -4474.26 kJ