Question

The combustion of one mole of benzene takes place at 298K and 1 atm. Enthalpy of combustion of benzene is 3267 KJ/mole. Calculate Enthalpy of formation of benzene. Given Enthalpy of formation of $C{O_2}$and ${H_2}O$is $- 393.5Kj/mol$ and $- 285.83Kj/mole$.

Answer 1

Since, given in the question is combustion of one mole of benzene takes place at 298K and 1 atm and the enthalpy of combustion of benzene i.e., $\Delta H$ is given and also enthalpies of formation of $C{O_2}$ and ${H_2}O$ is given. We know that,
$\Delta H_{rxn}^0 = \sum m\Delta H_f^0(products) - \sum n\Delta H_f^0(reactants)$ and for calculating the $\Delta H$ of benzene, we will substitute the values in the equation.

Complete step by step answer:
The equation formed is:
${C_6}{H_6}(Benzene) + \dfrac{{15}}{2}{O_2} \to 6C{O_2} + 3{H_2}O$
$\Delta {H_{rxn}} = - 3267{\text{ }}KJ/mole$
$C + {O_2} \to C{O_2}$ $\Delta H = - 393.5Kj/mole$
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$ $\Delta H = - 285.83Kj/mole$
$\Delta H = 6\Delta {H_f}C{O_2} + 3\Delta {H_f}{H_2}O - \Delta {H_f}{C_6}{H_6}$
We know that,
$\Delta H_{rxn}^0 = \sum m\Delta H_f^0(products) - \sum n\Delta H_f^0(reactants)$
$\Delta {H_f}(Benzene) = 6 \times ( - 393.5) + 3 \times ( - 285.8) + 3267$
$\Rightarrow - 3218.49 + 3267$
$\Delta {H_f}(Benzene) = 48.51KJ$

Hence, Enthalpy of formation of benzene is 48.51 KJ.
Note:
The magnitude of $\Delta H$ for a reaction will depend upon the physical states of the reactants and the products i.e., gas, liquid, solid, or the solution, then the pressure of any gases present, and the temperature at which the reaction is carried out. The enthalpies of formation that are measured under these conditions are known as standard enthalpies of formation ($\Delta H_f^0$). The enthalpy changes for the formation of 1 mole of a compound from its component elements when the component elements are each in their standard states, then the standard enthalpy of formation of any element in its most stable form is zero by definition.