Question

At $300K$ the standard enthalpies of formation of ${C_6}{H_5}COO{H_{(s)}},C{O_{2(g)}}$ and ${H_2}{O_{(l)}}$ are $ - 408, - 393$ and $ - 286{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ respectively. Calculate the heat of combustion of benzoic acid at constant volume
A.$ + 3201{\text{ kJ}}$
B.$ + 3199.75{\text{ kJ}}$
C.$ - 3201{\text{ kJ}}$
D.${\text{ - 3199}}{\text{.75 kJ}}$

Answer 1

Hint:The heat of combustion of benzoic acid can be calculated using the reaction of the combustion of one mole of benzoic acid. The enthalpy of a reaction can be determined by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products. If a certain amount of heat is absorbed during the course of the reaction, the enthalpy of the reaction is said to be positive. If a certain amount of heat is released during the course of the reaction, the enthalpy of the reaction is said to be negative.

Formula used: $\Delta H = \Delta U + \Delta {n_g}RT$
Where, $\Delta H$ is the heat of combustion of benzoic acid
$\Delta U$ is the change in internal energy of the system during the process of combustion.
And, $\Delta {n_g}$ is a change in the number of moles of gaseous substances as we move from reactants to products.

Complete step by step answer:
The chemical reaction for the combustion of benzoic acid can be written as
${C_6}{H_5}COOH + \dfrac{{15}}{2}{O_2} \to 7C{O_2} + 3{H_2}O$
Now we know that the enthalpy of a reaction is given by the formula, ${\Delta _r}H = {\left( {{\Delta _f}H} \right)_{{\text{products}}}} - {\left( {{\Delta _f}H} \right)_{{\text{reactants}}}}$
Substituting the values we can find the enthalpy of combustion of benzoic acid as
$\Delta H = 7\Delta {H_f}\left( {C{O_2}} \right) + 3\Delta {H_f}\left( {{H_2}O} \right) - \Delta {H_f}\left( {{C_6}{H_5}COOH} \right)$
Substituting the values:
$\Delta H = 7 \times \left( { - 393} \right) + 3 \times ( - 286) - \left( { - 408} \right)$
Thus, $\Delta H = - 3201{\text{ kJ}}$
We know that the exchange of heat at constant volume is equal to the change in internal energy of the system. So we can write the formula:
$\Delta H = \Delta U + \Delta {n_g}RT$
Solving for $\Delta U$, we get, $\Delta U = \Delta H - \Delta {n_g}RT$
$ \Rightarrow \Delta U = - 3201 + \left( {7 - \dfrac{{15}}{2}} \right) \times \dfrac{{8.314}}{{1000}} \times 300$
$\therefore \Delta U = - 3199.75{\text{ kJ}}$

Thus, the correct answer is D.

Note:
The enthalpy of formation of any element present in its standard state is always zero. In this reaction oxygen is present as a dioxygen molecule in the gaseous state. Hence, its enthalpy of formation is zero.