At 25°C, the enthalpy of the following processes are given:
H$$_2$$(g) + O$$_2$$(g) $$\to$$ 2OH(g) $$\Delta$$H$$^0$$ = 78 kJ mol$$^{-1}$$
H$$_2$$(g) + 1/2O$$_2$$(g) $$\to$$ H$$_2$$O(g) $$\Delta$$H$$^0$$ = -242 kJ mol$$^{-1}$$
H$$_2$$(g) $$\to$$ 2H(g) $$\Delta$$H$$^0$$ = 436 kJ mol$$^{-1}$$
1/2O$$_2$$(g) $$\to$$ O(g) $$\Delta$$H$$^0$$ = 249 kJ mol$$^{-1}$$
H$$_2$$O(g) $$\to$$ Hg + OHg $$\Delta$$H$$^0$$ = X kJmol$$^{-1}$$
What would be the value of X for the following reaction? (Nearest integer)

We are given the following thermochemical equations at 25 degrees C:

(1) $$H_2(g) + O_2(g) \rightarrow 2OH(g)$$, $$\Delta H_1^0 = 78$$ kJ/mol

(2) $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$$, $$\Delta H_2^0 = -242$$ kJ/mol

(3) $$H_2(g) \rightarrow 2H(g)$$, $$\Delta H_3^0 = 436$$ kJ/mol

(4) $$\frac{1}{2}O_2(g) \rightarrow O(g)$$, $$\Delta H_4^0 = 249$$ kJ/mol

We need to find $$X$$ for: $$H_2O(g) \rightarrow H(g) + OH(g)$$, $$\Delta H^0 = X$$ kJ/mol.

The atomization of water is $$H_2O(g) \rightarrow 2H(g) + O(g)$$. By Hess's Law, reversing reaction (2) and adding reactions (3) and (4):

$$\Delta H_{\text{atom}} = -\Delta H_2^0 + \Delta H_3^0 + \Delta H_4^0 = 242 + 436 + 249 = 927 \text{ kJ/mol}$$

Now, from reaction (1): $$H_2(g) + O_2(g) \rightarrow 2OH(g)$$, $$\Delta H_1^0 = 78$$ kJ/mol. We can write this as $$2H(g) + 2O(g) \rightarrow 2OH(g)$$, with enthalpy $$\Delta H_1^0 - \Delta H_3^0 - 2\Delta H_4^0$$:

$$\Delta H_{2OH} = 78 - 436 - 2(249) = 78 - 436 - 498 = -856 \text{ kJ/mol}$$

For one OH radical: $$H(g) + O(g) \rightarrow OH(g)$$, $$\Delta H = \frac{-856}{2} = -428$$ kJ/mol. So the bond formation energy of the O-H bond in OH is 428 kJ/mol.

The target reaction $$H_2O(g) \rightarrow H(g) + OH(g)$$ can be viewed as: first fully atomize water (costs 927 kJ/mol), then recombine one H and one O into OH (releases 428 kJ/mol).

Hence, $$X = 927 - 428 = 499$$ kJ/mol. So, the answer is $$499$$.