Question

At $300K$ the standard enthalpies of formation of $C_6{H}_{5}COOH\left(s\right),{CO}_2\left(g\right),H_{2}O\left(l\right)$ are $-408,-398$ and $-286kJ{mol}^{-1}$ respectively. Calculate the heat of combustion of benzoic acid at constant volume:

• A. $+3201kJ$
• B. $+3199.75kJ$
• C. $-3234.74kJ$
• D. $-3199.75kJ$

Answer 1

The correct option is C $-3234.74kJ$

${C_6{H}_{5}COOH}_{\left(s\right)}+\frac{15}{2}{O_2}_{\left(g\right)}⟶7{C{O}_2}_{\left(g\right)}+3{H_{2}O}_{\left(l\right)}$

$\Delta H=\sum a{\Delta }_f{H}_P-\sum b{\Delta }_f{H}_R$

$\Delta H=(7\times -398+3\times -286)-(-408-0)=-2751-858+408=-3236kJ/mol$

$\Delta n_g=7-\frac{15}{2}=-0.5$

$q_{\left(V\right)}=\Delta E=\Delta H-\Delta n_{g}RT$

whereas,

$q_{\left(V\right)}=?=$ heat of combustion at constant volume

$R=$ Molar gas constant $=8.31\times {10}^{-3}KJ/mol$

$T=$ Temperature $=300K\left[Given\right]$

$\therefore q_{\left(V\right)}=-3236-(-0.5\times 8.31\times {10}^{-3}\times 300)=-3234.754kJ/mol$