Question

The enthalpy for the following reactions (ΔH^o) at 25^oC are given below

(i) 1/2 H₂(g) + 1/2 O₂(g) ----> OH(g) ΔH = -10.06 kcal

(ii) H₂(g) -----> 2H(g)  ΔH = 104.18 kcal

(iii) O₂(g) ------> 2O(g)   ΔH = 118.32 kcal

calculate the O - H bond energy in the hydroxyl radical. 

Answer 1

O - H(g) -----> O(g) + H(g)  ΔH = ?

On reversing Eq (i), we get