Question

The reaction of cyanamide, $N{H}_{2}C{N}_{\left(s\right)}$, with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7kJmo{l}^{-1}$ at 298 K. calculate enthalpy change for the reaction at 298 K.
$N{H}_{2}C{N}_{\left(g\right)}+\frac{3}{2}{O}_{2\left(g\right)}\to N_{2\left(g\right)}+C{O}_{2\left(g\right)}+H_2{O}_{\left(l\right)}$

Answer 1

Enthalpy change for a reaction $\left(\Delta H\right)$ is given by the expression,
$\Delta H=\Delta U+\Delta n_{g}RT$
Where,
$\Delta U$ = change in internal energy
$\Delta n_g$ = change in number of moles
For the given reaction,
$\Delta n_g=\sum n_g$ (products) - $\sum n_g$ (reactants)
= (2 - 2.5) moles
$\Delta n_g=–0.5$ moles
And,
$\Delta U=–742.7kJmo{l}^{-1}T=298KR=8.314\times 10–3kJmo{l}^{-1}{K}^{-1}$
Substituting the values in the expression of \Delta~H:
$\Delta H=(–742.7kJmo{l}^{-1})+(–0.5mol)\left(298K\right)(8.314\times {10}^{-3}kJmo{l}^{-1}{K}^{-1})=-742.7–1.2\Delta H=–743.9kJmo{l}^{-1}$