Question

The reaction of cyanamide $\left(N{H}_{2}CN\right)$ with dioxygen was carried out in a bomb calorimeter and $\Delta U$ was found to be $-742.7{\text{kJmol}}^{-1}$ at $298K$. Calculate the enthalpy change for the reaction at $298K$
$N{H}_{2}CN\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to N_2\left(g\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$

• A. $-741.46{\text{kJ/mol}}^{-1}$
• B. $-743.9{\text{kJ/mol}}^{-1}$
• C. $+741.46{\text{kJ/mol}}^{-1}$
• D. $-743.9{\text{kJ/mol}}^{-1}$

Answer 1

The correct option is A $-741.46{\text{kJ/mol}}^{-1}$

$N{H}_{2}CN\left(s\right)+\frac{3}{2}{O}_2\left(g\right)\to N_2\left(g\right)+C{O}_2\left(g\right)+H_{2}O\left(l\right)$
$\Delta n_g=n_p-n_r=2-\frac{3}{2}=\frac{1}{2}=0.5\text{mol}$
$\Delta H=\Delta U+\Delta n_{g}RT$
$\Delta H=-742.7+(0.5\times 8.314\times {10}^{-3}\times 298)$
$\Delta H=(-742.7+1238.786\times {10}^{-3})=-741.46{\text{kJ mol}}^{-1}$