Question

For the combustion of 1 mol of liquid benzene at ${25}^{o}C$, the heat of reaction at constant pressure is given by, $C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$; $\Delta H_p=-780980calmo{l}^{-1}$.
What would be the heat of reaction at constant volume?

• A. $750985cal$
• B. $-700980cal$
• C. $-880090cal$
• D. $-780086cal$

Answer 1

The correct option is D $-780086cal$

Given that, $\Delta H_p=-780980cal/molT={25}^{o}C=25+273=298K$

we know $\Delta H=\Delta U+\Delta n_{g}RT$
$\Delta n_g=n_p-n_r=6-7.5=-1.5$

so, $\Delta H=\Delta U-1.5RT$
$\Delta U=\Delta H+1.5RT$
$\Delta U=-780980+1.5\times 2\times 298$
$\Delta U=-780086cal$
the heat of reaction at constant volume would be $-780086cal$