Question

Cyanamide (NH₂CN) is completely burnt in excess oxygen in a bomb calorimeter, ΔU was found to be −742.4 kJ mol⁻¹, calculate the enthalpy change of the reaction at 298 K.\[\ce{NH2CN_{(s)} + 3/2 O2_{(g)} -> N2_{(g)} + CO2_{(g)} + H2O_{(l)}}\] ΔH = ?

Answer 1

Given, T = 298 K; ∆U = −742.4 kJ mol⁻¹

∆H = ?

∆H = ∆U + ∆n_(g) RT

∆H = ∆U + (n_p – n_r) RT

∆H = `-742.4 + (2 - 3/2) xx 8.314 xx 10^-3 xx 298`

∆H = −742.4 + (0.5 × 8.314 × 10⁻³ × 298)

∆H = −742.4 + 1.24 

∆H = −741.16 kJ mol⁻¹