Question

The reaction of cyanamide, NH₂CN_(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. 

Answer 1

Enthalpy change for a reaction (∆H) is given by the expression, 

 ∆H = ∆U + ∆n_gRT 

Where, 

∆U = change in internal energy 

∆n_g = change in number of moles 

For the given reaction, 

∆n_g = ∑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 ^–1

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