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Question: What is the heat of formation of \({C_6}{H_6}\), give that the heat of combustion of benzene, carbon...

What is the heat of formation of C6H6{C_6}{H_6}, give that the heat of combustion of benzene, carbon and hydrogen are −782,  −94 - 782,\; - 94 and −68  K - 68\;{\rm{K}} respectively.
A.+14  K.Cal + 14\;{\rm{K}}{\rm{.Cal}}
B.−14  K.Cal - 14\;{\rm{K}}{\rm{.Cal}}
C.+28  K.Cal + 28\;{\rm{K}}{\rm{.Cal}}
D.−28  K.Cal - 28\;{\rm{K}}{\rm{.Cal}}

Explanation

Solution

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 C6H6{C_6}{H_6}.

Complete step by step answer:
Given:
The heat of combustion of benzene, carbon and hydrogen
ΔH(C6H6)=−782  K.Cal\Delta H\left( {{C_6}{H_6}} \right) = - 782\;{\rm{K}}{\rm{.Cal}}
ΔH(C)=−94  K.Cal\Delta H\left( C \right) = - 94\;{\rm{K}}{\rm{.Cal}}
ΔH(H2)=−68  K.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.
ΔHf=∑ΔHf(P)−∑ΔHf(R)\Delta {H_f} = \sum {\Delta {H_f}} (P) - \sum {\Delta {H_f}} (R)
Where,
ΔHf\Delta {H_f} is an enthalpy change of formation.
ΔHf(P)\Delta {H_f}(P) is heat of formation of products.
ΔHf(R)\Delta {H_f}(R) is heat of formation of reactants.
We have to calculate the heat of formation of C6H6{C_6}{H_6} (ΔHf)\left( {\Delta {H_f}} \right) ,
6C+3H2→C6H66C + 3{H_2} \to {C_6}{H_6} (i)
C6H6+152O2→6CO2+3H2O{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 ΔH(C6H6)=−782  K.Cal\Delta H\left( {{C_6}{H_6}} \right) = - 782\;{\rm{K}}{\rm{.Cal}}.
C+O2→CO2C + {O_2} \to C{O_2} (iii)
The heat of combustion of carbon is ΔH(C)=−94  K.Cal\Delta H\left( C \right) = - 94\;{\rm{K}}{\rm{.Cal}}
H2+12O2→H2O{H_2} + \dfrac{1}{2}{O_2} \to {H_2}O (iv)
The heat of combustion of hydrogen is ΔH(H2)=−68  K.Cal\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 6CO2+3H2O6C{O_2} + 3{H_2}O which is subtracted by heat of combustion of benzene.
ΔHf=6[ΔH(C)]+3[ΔH(H2)]−[ΔH(C6H6)] =6(−94)+3(−68)−(−782) =−564−204+782 =+14  K.Cal\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 C6H6{C_6}{H_6} is +14  K.Cal + 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.