Question

The enthalpy of combustion of benzene from the following data will be:
i) $6C\left( s \right) + 3{H_2}\left( g \right) \to {C_6}{H_6}\left( l \right);\;\;\;\;\;\Delta H = + 45.9\;kJ$
ii) ${H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right);\;\;\;\;\;\Delta H = - 285.9\;kJ$
iii) $C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);\;\;\;\;\;\Delta H = - 393.5\;kJ$
A. $ + 3172.8\;kJ$
B. $ - 1549.2\;kJ$
C. $ - 3172.8\;kJ$
D. $ - 3264.6\;kJ$

Answer 1

Hint: We have the Hess’s law of constant heat summation for those reactions which take place in multiple steps or can be expressed as so.

Complete step by step answer:
From thermodynamics, we can define the enthalpy of formation of a compound as the enthalpy change accompanying the formation of one mole of it from its constituents. We denote the enthalpy of formation with ${\Delta _f}H$ .
We can also define the enthalpy of combustion of a compound as the enthalpy change accompanying the combustion of one mole of it. We denote the enthalpy of formation with ${\Delta _c}H$ .
So, it is evident that we are provided with the enthalpy of formation of benzene, water and carbon dioxide which are also involved in the combustion reaction of benzene. It will become clearer with the following chemical equation for the combustion of benzene:
${C_6}{H_6}\left( l \right) + \dfrac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( l \right)$
So, we can express the above reaction by using the given ones as follows:
We have to take the reverse reaction of $6C\left( s \right) + 3{H_2}\left( g \right) \to {C_6}{H_6}\left( l \right);\;\;\;\;\;\Delta H = + 45.9\;kJ$ because we need benzene to be the reactant: ${C_6}{H_6}\left( l \right) \to 6C\left( s \right) + 3{H_2}\left( g \right);\;\;\;\;\;\Delta H = - 45.9\;kJ$. The sign for enthalpy would also be changed as reverse of an endothermic reaction would be exothermic.
We can multiply ${H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right);\;\;\;\;\;\Delta H = - 285.9\;kJ$ by three as we require three moles of water as product: $3{H_2}\left( g \right) + \dfrac{3}{2}{O_2}\left( g \right) \to 3{H_2}O\left( l \right);\;\;\;\;\;\Delta H = - \left( {285.9 \times 3} \right)\;kJ$. The enthalpy is an extensive property so it would also change accordingly.
Now, we can multiply $C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);\;\;\;\;\;\Delta H = - 393.5\;kJ$ by six as we require six moles of carbon dioxide as product: $6C\left( s \right) + 6{O_2}\left( g \right) \to 6C{O_2}\left( g \right);\;\;\;\;\;\Delta H = - \left( {393.5 \times 6} \right)\;kJ$.
Now, we will use the Hess’s law of constant heat summation which can be stated as in case a reaction occur in multiple steps then we can determine the reaction enthalpy as the sum of the reaction enthalpies of the involved reactions given that all are at the same temperature. So, let’s write the combustion reaction of benzene as the sum of the formation reactions as follows:

We can now calculate the enthalpy of combustion as follows:
\[\begin{array}{c}
{\Delta _c}H = \Delta {H_1} + \Delta {H_2} + \Delta {H_3}\\
 = \left( { - 45.9\;kJ} \right) + \left\{ { - \left( {285.9 \times 3} \right)\;kJ} \right\} + \left\{ { - \left( {393.5 \times 6} \right)\;kJ} \right\}\\
 = - {\rm{3264}}{\rm{.6}}\;kJ
\end{array}\]
Hence, the enthalpy of combustion of benzene is \[ - {\rm{3264}}{\rm{.6}}\;kJ\]

Hence, the correct option is (D).

Note:
We have to use the balanced reactions and be careful with the stoichiometry while adding the reactions.