Consider the following data : Heat of formation of $$CO_{2}$$(g)= -393.5 kJ $$mol^{-1}$$ Heat of formation of $$H_{2}0$$(l) = - 286.0 kJ $$mol^{-1}$$ Heat of combustion of benzene =-3267.0 kJ $$mol^{-1}$$ The heat of formation of benzene is _ . (Nearest integer)

We need to find the heat of formation of benzene from combustion data.

$$ C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) $$

By Hess’s law: $$\Delta H_{\text{combustion}} = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})$$. Therefore,

$$ \Delta H_{\text{comb}} = [6\Delta H_f(CO_2) + 3\Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + \tfrac{15}{2}\Delta H_f(O_2)] $$

Since $$O_2$$ is an element in its standard state, $$\Delta H_f(O_2) = 0$$. Rearranging gives

$$ \Delta H_f(C_6H_6) = 6\Delta H_f(CO_2) + 3\Delta H_f(H_2O) - \Delta H_{\text{comb}} $$

Substituting the numerical values $$\Delta H_f(CO_2) = -393.5\ \text{kJ/mol}$$, $$\Delta H_f(H_2O) = -286.0\ \text{kJ/mol}$$, and $$\Delta H_{\text{comb}} = -3267.0\ \text{kJ/mol}$$ yields:

$$ 6 \times (-393.5) = -2361.0\ \text{kJ} $$

$$ 3 \times (-286.0) = -858.0\ \text{kJ} $$

$$ \Delta H_f(C_6H_6) = -2361.0 + (-858.0) - (-3267.0) $$

$$ = -3219.0 + 3267.0 = 48.0\ \text{kJ/mol} $$

The answer is 48 kJ/mol.