Question

Calculate the heat of formation of C₆H₆, given that the heats of combustion of benzene, carbon and hydrogen are 754, 94 and 68 kcal respectively.

Answer 1

Dear Student,

$$\begin{gathered} 6\mathrm{C}+3{\mathrm{H}}_2\to {\mathrm{C}}_6{\mathrm{H}}_6..\left(\mathrm{i}\right)∆{\mathrm{H}}_{\mathrm{f}}=? \\ {\mathrm{C}}_6{\mathrm{H}}_6+\frac{15}{2}{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}..\left(\mathrm{ii}\right)∆{\mathrm{H}}_1=754\mathrm{kcal} \\ \mathrm{C}+{\mathrm{O}}_2\to {\mathrm{CO}}_2...\left(\mathrm{iii}\right)∆{\mathrm{H}}_2=94\mathrm{kcal} \\ {\mathrm{H}}_2\hspace{0.17em}+\frac{1}{2}{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}...\left(\mathrm{iv}\right)∆{\mathrm{H}}_3=68\mathrm{kcal} \\ \mathrm{eq}.\left(\mathrm{i}\right)\mathrm{can}\mathrm{be}\mathrm{obtained}\mathrm{by}\mathrm{multipying}\mathrm{eq}.\left(\mathrm{iii}\right)\mathrm{by}6\mathrm{and}\mathrm{eq}.\left(\mathrm{iv}\right)\mathrm{by}3\mathrm{and}\mathrm{adding}\mathrm{both}\mathrm{the}\mathrm{equation}.\mathrm{eq}\left(\mathrm{ii}\right)\mathrm{is}\mathrm{then}\mathrm{substracted}\mathrm{from}\mathrm{sum}\mathrm{of}\mathrm{eq}\left(\mathrm{iii}\right)\mathrm{and}\mathrm{eq}\left(\mathrm{iv}\right)\mathrm{to}\mathrm{get}\mathrm{eq}\left(\mathrm{i}\right) \\ \mathrm{Therefore},∆{\mathrm{H}}_{\mathrm{f}}=(6\times ∆{\mathrm{H}}_2+3\times ∆{\mathrm{H}}_3)-∆{\mathrm{H}}_1 \\ ∆{\mathrm{H}}_{\mathrm{f}}=(6\times 94+3\times 68)-754=(564+204)-754=14\mathrm{kcal} \end{gathered}$$