Question

The combustions of benzene (l) gives $C{O}_2\left(g\right)$ and $H_{2}O\left(l\right)$. Given that heat of combustion of benzene at constant volume is $-3263.9kJmo{l}^{-1}$ at ${25}^{o}C$, heat of combustion (in $kJmo{l}^{-1}$) of benzene at constant pressure will be:
$(R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $+3260$
• B. $-3267.6$
• C. $+4152.6$
• D. $-452.46$

Answer 1

The correct option is B $-3267.6$

$C_6{H}_6\left(l\right)+\frac{15}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

$\Delta H=\Delta U+\Delta nRT$

$=-3263.9+\frac{(6-7.5)\times 8.314\times 298}{1000}$

$=-3263.9-3.7$

$=-3267.6kJ$

Hence, option $B$ is correct.