Question

Ammonia (NH3 (g), Delta.Hf = –45.9 kJ/mol ) reacts with oxygen to produce nitrogen and water (H2 O(g), Delta.Hf = –241.8 kJ/mol) according to the equation below.

4 upper N upper H subscript 3 (g) plus 3 upper O subscript 2 (g) right arrow 2 upper N subscript 2 (g) plus 6 upper H subscript 2 upper O (g).

What is the enthalpy of the reaction?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–3,572.1 kJ
–1,267.2 kJ
1,267.2 kJ
3,572.1 kJ

Answer 1

ΔHrxn = [2·ΔHf(N2) + 6·ΔHf(H2O)] − [4·ΔHf(NH3) + 3·ΔHf(O2)]
= [2·0 + 6(−241.8)] − [4(−45.9) + 3·0]
= −1450.8 − (−183.6)
= −1267.2 kJ

Answer: −1,267.2 kJ (exothermic).