Question

At $300K$, the standard enthalpies of formation of $C_6{H}_{5}COOH\left(s\right),C{O}_2\left(g\right)$ and $H_{2}O\left(l\right)$ are $-408,-393$ and $-286$ $kJ{mol}^{-1}$ respectively. Calculate the heat of combustion of benzoic acid at (i) constant pressure and (ii) constant volume.
$(R=8.31J{mol}^{-1}{K}^{-1})$

• A. $\Delta H=-4891kJ{mol}^{-1};\Delta U=-3199.75kJ{mol}^{-1}$
• B. $\Delta H=-3751kJ{mol}^{-1};\Delta U=-1000.75kJ{mol}^{-1}$
• C. $\Delta H=-4501kJ{mol}^{-1};\Delta U=-3199.75kJ{mol}^{-1}$
• D. $\Delta H=-3201kJ{mol}^{-1};\Delta U=-3199.75kJ{mol}^{-1}$

Answer 1

The correct option is D $\Delta H=-3201kJ{mol}^{-1};\Delta U=-3199.75kJ{mol}^{-1}$

Solution:- (D) $\Delta H=-3201KJ/mol;\Delta U=-3199.75KJ/mol$

$C_6{H}_{5}COOH+\frac{13}{2}{O}_2⟶7C{O}_2+3H_{2}O$

$q\left(p\right)$ and $q\left(v\right)$ be the heat of combustion of benzoic acid at constant pressure and constant volume respectively.

$q\left(p\right)=\Delta H=[7\times (-393)+3(-286)]-[(-408)-0]$

$\Rightarrow \Delta H=-3609+408=-3201KJ/mol$

$\Delta n=7-\frac{13}{2}=0.5$

$q\left(v\right)=\Delta E=\Delta H-\Delta nRT=-3201-(0.5\times 8.31\times {10}^{-3}\times 300)=-3199.75KJ/mol$