Question

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

• A. $415.6$
• B. $-452.46$
• C. $3260.2$
• D. $-3265.9$

Answer 1

The correct option is D $-3265.9$

Solution-

$C_6{H}_6\left(l\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

After balancing the equation,

$C_6{H}_6\left(l\right)+\frac{150}{2}{O}_2\left(g\right)\to 6C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

$△ng=6-\frac{15}{2}=-3/2$

$△U=-3263.9KJ/mol.$

$△H=△U+△ngRT$

$△H=-3263.9-\frac{3}{2}\times 8.314\times 298$

$△n=-3263900-1238.786$

$△n=-3265.9KJ/mol$