Question

The reaction of cyanamide, $N{H}_{2}CN\left(s\right)$ with oxygen was run in a bomb calorimeter and $\Delta U$ was found to be $-742.24kJmo{l}^{-1}$.The magnitude of $\Delta H_{298}$ for the reaction
$N{H}_{2}CN\left(s\right)+\frac{3}{2}\left(g\right)\to N_2\left(g\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$ is ______ kJ. (Rounded off to the nearest integer)

[Assume ideal gases and $R=8.314Jmo{l}^{-1}{K}^{-1}$]

Answer 1

$N{H}_{2}CN\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to N_2\left(g\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$

$\because$ $\Delta H=\Delta U+\Delta n_{g}RT$

$\Delta n_g=2-\frac{3}{2}=0.5$

Assume that the $\Delta U$ is given at the same temperature.
$\Delta H_{298}=-742.24kJ+\frac{0.5\times 8.314\times 298}{1000}kJ$

= - 742.24 + 1.24
= -741 kJ
$|\Delta H_{298}|=741kJ$