Question

What is the heat of formation of ${C_6}{H_6}$, give that the heat of combustion of benzene, carbon and hydrogen are $ - 782,\; - 94$ and $ - 68\;{\rm{K}}$ respectively.
A.$ + 14\;{\rm{K}}{\rm{.Cal}}$
B.$ - 14\;{\rm{K}}{\rm{.Cal}}$
C.$ + 28\;{\rm{K}}{\rm{.Cal}}$
D.$ - 28\;{\rm{K}}{\rm{.Cal}}$

Answer 1

Hint:
We know that the heat of formation is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituents elements given that each element should be in its normal physical state. Now that we have given the heat of combustion of benzene, carbon and hydrogen with the help of that we are able to calculate the heat of formation of ${C_6}{H_6}$.

Complete step by step answer:
Given:
The heat of combustion of benzene, carbon and hydrogen
$\Delta H\left( {{C_6}{H_6}} \right) = - 782\;{\rm{K}}{\rm{.Cal}}$
$\Delta H\left( C \right) = - 94\;{\rm{K}}{\rm{.Cal}}$
$\Delta H\left( {{H_2}} \right) = - 68\;{\rm{K}}{\rm{.Cal}}$
The standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.
\[\Delta {H_f} = \sum {\Delta {H_f}} (P) - \sum {\Delta {H_f}} (R)\]
Where,
$\Delta {H_f}$ is an enthalpy change of formation.
$\Delta {H_f}(P)$ is heat of formation of products.
$\Delta {H_f}(R)$ is heat of formation of reactants.
We have to calculate the heat of formation of ${C_6}{H_6}$$\left( {\Delta {H_f}} \right)$ ,
$6C + 3{H_2} \to {C_6}{H_6}$ (i)
${C_6}{H_6} + \dfrac{{15}}{2}{O_2} \to 6C{O_2} + 3{H_2}O$ (ii)
The heat of combustion of benzene is $\Delta H\left( {{C_6}{H_6}} \right) = - 782\;{\rm{K}}{\rm{.Cal}}$.
$C + {O_2} \to C{O_2}$ (iii)
The heat of combustion of carbon is $\Delta H\left( C \right) = - 94\;{\rm{K}}{\rm{.Cal}}$
${H_2} + \dfrac{1}{2}{O_2} \to {H_2}O$ (iv)
The heat of combustion of hydrogen is $\Delta H\left( {{H_2}} \right) = - 68\;{\rm{K}}{\rm{.Cal}}$
Now, according to the formula of heat of formation will be sum of heat combustion of carbon dioxide and water along with their stoichiometric coefficients $6C{O_2} + 3{H_2}O$ which is subtracted by heat of combustion of benzene.
$\Delta {H_f} = 6\left[ {\Delta H\left( C \right)} \right] + 3\left[ {\Delta H\left( {{H_2}} \right)} \right] - \left[ {\Delta H\left( {{C_6}{H_6}} \right)} \right]\\
 = 6\left( { - 94} \right) + 3\left( { - 68} \right) - \left( { - 782} \right)\\
 = - 564 - 204 + 782\\
 = + 14\;{\rm{K}}{\rm{.Cal}}$
Therefore, Option A is correct.
The heat of formation of ${C_6}{H_6}$ is $ + 14\;{\rm{K}}{\rm{.Cal}}$.

Note:Take care of the signs and values should be taken according to the chemical reactions and further multiplied with their respective stoichiometric coefficients. Stoichiometric coefficients are multiplied because the heat of formation is the amount of heat absorbed or evolved, when ‘one mole’ of a compound is formed.