Question

The heat evolved in the combustion of benzene is given by
$C_6{H}_6\left(l\right)+7\frac{1}{2}{O}_2\left(g\right)\to 3H_{2}O\left(l\right)+6C{O}_2\left(g\right),\Delta H=-781.0Kcalmo{l}^{-1}$
When $156$ g of $C_6{H}_6$ is burnt in an open container, the amount of heat energy released will be

• A. $150.2kcalmo{l}^{-1}$
• B. $1562.0kcalmo{l}^{-1}$
• C. $6528.2kJmo{l}^{-1}$
• D. $2448.5kJmo{l}^{-1}$

Answer 1

Answer: (B), (C)

The correct options are
B $6528.2kJmo{l}^{-1}$
C $1562.0kcalmo{l}^{-1}$

As given,

$C_6{H}_6\left(l\right)+7\frac{1}{2}{O}_2\left(g\right)\to 3H_{2}O\left(l\right)+6C{O}_2\left(g\right),\Delta H=-781.0Kcalmo{l}^{-1}$

$78$ gm = 1 mol

$156/78$ = $2$ mol $\Delta H=-781.0\ast 2=-1562Kcalmo{l}^{-1}=-6528.2kJmo{l}^1$

(because 1 cal = 4.2 joule)