Question

At $300$K the standard enthalpies of formation of $C_6{H}_{5}COO{H}_{\left(s\right)},C{O}_{2\left(g\right)}$ and $H_2{O}_{\left(l\right)}$ are $-408,-393$ and $-286$ kJ $mo{l}^{-1}$ respectively. Calculate the heat of combustion of benzoic acid at constant volume.

• A. $+3201$ kJ
• B. $+3199.75$ kJ
• C. $-3201$ kJ
• D. $-3199.75$ kJ

Answer 1

Answer: (D)

$-3199.75$ kJ

Heat at constant volume, is equal to the $\Delta U$ of the system.

$C_6{H}_{5}COOH+\frac{15}{2}{O}_2\to 7C{O}_2+3H_{2}O$

$\Delta H_{comb}=7\Delta H_f\left(C{O}_2\right)+3\Delta H_f\left(H_{2}O\right)-\Delta H_f\left(C_6{H}_{5}COOH\right)$

$=7\times (-393)+3(-286)-(-408)$

$=-3201kJ$

$\Delta n_g=7-\frac{15}{2}=\frac{-1}{2}$

$\Delta n_{g}RT=\frac{-1}{2}\times \frac{8.314}{1000}\times 300\approx -1.25KJ$

$\therefore \Delta H=\Delta U+\Delta n_{g}RT\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT=-3199.75KJ$