Given $Δ_{f} H^{\circ} (H X_{2} O (g)) = - 242 kJ mo l^{- 1}$ , bond enthalpy of $H - H = 436$ and of $O = O = 498 kJ mo l^{- 1}$ . Select the correct statement(s).

AThe average $O - H$ bond enthalpy in water is $463.5 kJ mo l^{- 1}$ .
BThe enthalpy of atomisation of $H X_{2} O (g)$ (to $2 H (g) + O (g)$ ) is $927 kJ mo l^{- 1}$ .
C $Δ H$ for $2 H X_{2} O (g) 2 H X_{2} (g) + O X_{2} (g)$ is $+ 484 kJ$ .
D $Δ H$ for $2 H X_{2} (g) + O X_{2} (g) 2 H X_{2} O (g)$ is $- 242 kJ$ .

Answer: A, B, C

For

H X_{2} O

Δ_{f} H = BE (H - H) + \frac{1}{2} BE (O = O) - 2 BE (O - H)

, so

2 BE (O - H) = 436 + 249 + 242 = 927

and the average

BE = 463.5

. The atomisation enthalpy is

927

. Doubling and reversing the formation reaction gives

+ 484

for decomposition of

2

mol, while forming

2

mol of steam from elements gives

- 484

, not

- 242