Question

The enthalpy changes at 298K in successive breaking of O-H bond of water are H2O(g)→H(g) + OH(g) :∆H =498 k j mol-1and OH(g) →H(g) + O(g) : ∆H = 428 k j mol-1the bond enthalpy of O-H is(a) 498 k j mol-1 (b) 428 k j mol-1 (c) 70 k j mol-1 (d) 463 k j mol-1

Answer 1

Answer:

The correct answer is option d.

Explanation:

$H_2O(g)\rightarrow H(g) + OH(g),\Delta H_{1,OH}=498 kJ/mol$..[1]

$OH(g)\rightarrow H(g) + O(g),\Delta H_{2,OH}=428 kJ/mol$...[2]

The bond enthalpy of O-H bond will be the average both both enthalies required in [1] and [2].

$\Delta H_{O-H}=\frac{\Delta H_{1,OH}+\Delta H_{2,OH}}{2}$

$=\frac{498 kJ/mol+428 kJ/mol}{2}=463 kJ/mol$

The bond enthalpy of O-H is 463 kJ/mol.