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Question: At \(300K\) the standard enthalpies of formation of \({C_6}{H_5}COO{H_{(s)}},C{O_{2(g)}}\) and \({H_...

At 300K300K the standard enthalpies of formation of C6H5COOH(s),CO2(g){C_6}{H_5}COO{H_{(s)}},C{O_{2(g)}} and H2O(l){H_2}{O_{(l)}} are −408,−393 - 408, - 393 and −286 kJ mol−1 - 286{\text{ kJ mo}}{{\text{l}}^{ - 1}} respectively. Calculate the heat of combustion of benzoic acid at constant volume
A.+3201 kJ + 3201{\text{ kJ}}
B.+3199.75 kJ + 3199.75{\text{ kJ}}
C.−3201 kJ - 3201{\text{ kJ}}
D. - 3199.75 kJ{\text{ - 3199}}{\text{.75 kJ}}

Explanation

Solution

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: ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
Where, ΔH\Delta H is the heat of combustion of benzoic acid
ΔU\Delta U is the change in internal energy of the system during the process of combustion.
And, Δng\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
C6H5COOH+152O2→7CO2+3H2O{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, ΔrH=(ΔfH)products−(ΔfH)reactants{\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
ΔH=7ΔHf(CO2)+3ΔHf(H2O)−ΔHf(C6H5COOH)\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:
ΔH=7×(−393)+3×(−286)−(−408)\Delta H = 7 \times \left( { - 393} \right) + 3 \times ( - 286) - \left( { - 408} \right)
Thus, ΔH=−3201 kJ\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:
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
Solving for ΔU\Delta U, we get, ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta {n_g}RT
⇒ΔU=−3201+(7−152)×8.3141000×300\Rightarrow \Delta U = - 3201 + \left( {7 - \dfrac{{15}}{2}} \right) \times \dfrac{{8.314}}{{1000}} \times 300
∴ΔU=−3199.75 kJ\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.