Question

The reaction of cyanamide, \(NH_2CN(s)\), with dioxygen was carried out in a bomb calorimeter, and \(∆U\) was found to be \(–742.7\ kJ\ mol^{–1}\) at \(298\ K\). Calculate enthalpy change for the reaction at \(298\ K\).
\(NH_2CN(g)+\frac 32O_2(g)→N_2(g)+CO_2(g)+H_2O(l)\)

Answer 1

Given reaction (at 298 K):

NH2CN(s) + 3/2O2(g) -> N2(g) + CO2(g) + H2O(l)

\[ \Delta U = -742.7\ \text{kJ mol}^{-1} \quad \text{at } 298\ \text{K} \]

Relation between ΔH and ΔU (at constant T):

\[ \Delta H = \Delta U + \Delta n_g \, R T \] where \(\Delta n_g\) = (moles of gaseous products) − (moles of gaseous reactants).

Step 1: Calculate Δn_g

Gaseous species:

• Reactants: {3}{2} {mol of } O2(g)
• Products: N2(g) (1 mol) + CO2(g) (1 mol) = 2 mol gas

\[ \Delta n_g = 2 - \tfrac{3}{2} = \tfrac{1}{2} = 0.5 \]

Step 2: Use ΔH = ΔU + Δn_gRT

Use \(R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\), \(T = 298\ \text{K}\):

\[ \Delta H = -742.7\ \text{kJ mol}^{-1} + 0.5 \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 298\ \text{K} \] Convert the correction term to kJ: \[ 0.5 \times 8.314 \times 298 \approx 1238.7\ \text{J mol}^{-1} \approx 1.24\ \text{kJ mol}^{-1} \] \[ \Delta H \approx -742.7\ \text{kJ mol}^{-1} + 1.24\ \text{kJ mol}^{-1} \approx -741.5\ \text{kJ mol}^{-1} \]

The enthalpy change for the reaction at 298 K is approximately \[ \boxed{\Delta H \approx -7.42 \times 10^{2}\ \text{kJ mol}^{-1}} \] (about \(-741.5\ \text{kJ mol}^{-1}\)).