Q.8. The reaction of cyanamide, NH2CNs, 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.

NH2CNg+32O2gN2g+CO2g+H2Ol

NEETprep Answer:
Enthalpy change for a reaction (∆H) is given by the expression,
∆H = ∆U + ∆ngRT
Where,
∆U = change in internal energy
ng = change in number of moles
For the given reaction,
∆ng = ∑ng (products) – ∑ng (reactants)
= (2 – 2.5) moles
∆ng = –0.5 moles
And,
∆U = –742.7 kJ mol-1
T = 298 K
R = 8.314 × 10–3 kJ mol-1 K

Substituting the values in the expression of ∆H:
∆H = (–742.7 kJ mol-1) + (–0.5 mol) (298 K) (8.314 × 10-3 kJ mol-1 K-1) = –742.7 – 1.2
∆H = –743.9 kJ mol-1