Question

Consider the following data: 
- Heat of formation of \( CO_2(g) \) = -393.5 kJ mol\(^{-1}\) 
- Heat of formation of \( H_2O(l) \) = -286.0 kJ mol\(^{-1}\) 
- Heat of combustion of benzene = -3267.0 kJ mol\(^{-1}\) 
The heat of formation of benzene is ……… kJ mol\(^{-1}\) (Nearest integer).

Hint:

Use Hess's law for enthalpy calculations by balancing the formation and combustion equations.

Answer 1

The balanced chemical equation for the combustion of benzene is: \[ C_6H_6 + \frac{15}{2} O_2 \rightarrow 6CO_2 + 3H_2O \] Applying Hess's law to determine the standard enthalpy of formation of benzene ($\Delta H_f(C_6H_6)$): \[ \Delta H_f(C_6H_6) = \Delta H_c - \left( 6\Delta H_f(CO_2) + 3\Delta H_f(H_2O) \right) \] Substituting the given values: \[ = -3267 - \left(6(-393.5) + 3(-286.0)\right) \] \[ = -3267 + 2361 + 858 \] The calculated standard enthalpy of formation for benzene is: \[ = -48.5 \approx 49 { kJ/mol} \]