At 25 ° C , the enthalpy of the following processes are given: H 2 (   g ) + O 2 (   g ) → 2 OH ( g ) Δ H o = 78   kJ   mol - 1 H 2 (   g ) + 1 / 2 O 2 (   g ) → H 2 O ( g ) Δ H 0 = - 242   kJ   mol - 1 H 2 (   g ) → 2 H ( g ) Δ H o = 436   kJ   mol - 1 1 / 2 O 2 (   g ) → O ( g ) Δ H 0 = 249   kJ  

Step-by-step solution

2 H 2 O ( g ) → 2 H 2 (   g ) + O 2 (   g )    + ( 242 × 2 ) kJmol - 1 ....(1) H 2 (   g ) + O 2 (   g ) → 2 OH    + 78   kJ   mol - 1 ....(2) H 2 (   g ) → 2 H ( g )    + 436   kJ   mol - 1 ....(3) Using Hess's law of constant heat summation, adding equations(1), (2) and (3) and dividing the net result by 2 we get: ⇒ H 2 O ( g ) → H ( g ) + OH ( g )    998 × 1 2 = + 499   kJ