Given, for reaction (i) ‌‌H2O(ℓ)→H+((aq. )+OH−((aq. ); ∆Hr=57.32‌kJ (ii) H2(g)+‌
1
2
O2(g)→H2O(ℓ); ∆Hr=−286.20‌kJ For reaction (i) ∆Hr=∆H∘‌f(H+.aq)+ ∆H∘‌f(OH−.aq)−∆H∘‌f(H2O,ℓ) 57.32=0+∆Hf∘(OH−1,aq)− ∆H∘‌f(H2O,ℓ)‌‌...(iii) For reaction (ii) ∆Hr=∆H∘‌f(H2O,ℓ)−∆H∘‌f(H2,g)− ‌
1
2
∆H∘‌f(O2,g) −286.20=∆H∘‌f(H2O,ℓ) On replacing this value in equ. ( iii ) we have ‌57.32=∆H∘‌f(OH−.aq)−(−286.20) ‌∆H∘‌f=−286.20+57.32 ‌=−228.88‌kJ