At 25°C, the enthalpy of the following processes are given:
$$H_2(g) + O_2(g) \rightarrow 2OH(g) \quad \Delta H^o = 78 \text{ kJ mol}^{-1}$$
$$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g) \quad \Delta H^o = -242 \text{ kJ mol}^{-1}$$
$$H_2(g) \rightarrow 2H(g) \quad \Delta H^o = 436 \text{ kJ mol}^{-1}$$
$$\frac{1}{2}O_2(g) \rightarrow O(g) \quad \Delta H^o = 249 \text{ kJ mol}^{-1}$$
What would be the value of X for the following reaction? (Nearest integer)
$$H_2O(g) \rightarrow H(g) + OH(g) \quad \Delta H^o = X \text{ kJ mol}^{-1}$$

The required reaction is
$$H_2O(g) \rightarrow H(g) + OH(g) \qquad \Delta H^{\circ}=X$$

According to Hess’s law we can write
$$\Delta H^{\circ}= \bigl[\Delta H_f^{\circ}(H) + \Delta H_f^{\circ}(OH)\bigr] - \Delta H_f^{\circ}(H_2O)\quad -(1)$$
so we first calculate the standard enthalpies of formation of the radicals $$H(g)$$ and $$OH(g)$$ from the data supplied.

Step 1 : Enthalpy of formation of the hydrogen atom
The given dissociation of the hydrogen molecule is
$$H_2(g) \rightarrow 2H(g)\qquad \Delta H^{\circ}=436\text{ kJ mol}^{-1}$$
For one hydrogen atom the enthalpy change is half of this value:
$$\Delta H_f^{\circ}(H)=\frac{436}{2}=218\text{ kJ mol}^{-1}\quad -(2)$$

Step 2 : Enthalpy of formation of the hydroxyl radical
The given reaction is
$$H_2(g)+O_2(g)\rightarrow 2OH(g)\qquad \Delta H^{\circ}=78\text{ kJ mol}^{-1}$$
This creates two $$OH$$ radicals directly from the elements in their standard states, so
$$2\,\Delta H_f^{\circ}(OH)=78\text{ kJ}$$
$$\Rightarrow\; \Delta H_f^{\circ}(OH)=\frac{78}{2}=39\text{ kJ mol}^{-1}\quad -(3)$$

Step 3 : Enthalpy of formation of water vapour
The data table already gives
$$H_2(g)+\tfrac12O_2(g)\rightarrow H_2O(g)\qquad \Delta H^{\circ}=-242\text{ kJ mol}^{-1}$$
Hence
$$\Delta H_f^{\circ}(H_2O)=-242\text{ kJ mol}^{-1}\quad -(4)$$

Step 4 : Enthalpy of the required reaction
Substituting the values from $$(2)\!-\!(4)$$ in equation $$(1)$$:
$$\Delta H^{\circ}= \bigl[218 + 39\bigr] - (-242)$$
$$\Delta H^{\circ}=257 + 242$$
$$\Delta H^{\circ}=499\text{ kJ mol}^{-1}$$

Therefore, the nearest integer value of $$X$$ is $${\mathbf{499}}$$ kJ mol−1.