Question

Nitrogen reacts with hydrogen to form ammonia according to the equation below.

mc023-1.jpg

Standard Enthalpies of Formation
Substance
DHf (kJ/mol)
C2 H2 (g)
–26.7
NH3 (g)
–46.19
HBr (g)
236.23
HCl (g)
–92.30
HF (g)
–268.6
HI (g)
25.9
NaCl (s)
–411.0

Based on the equation and the information in the table, what is the enthalpy of the reaction?
mc023-2.jpgHrxn = 46.19 kJ
mc023-3.jpgHrxn = –46.19 kJ
mc023-4.jpgHrxn = 92.38 kJ
mc023-5.jpgHrxn = –92.38 kJ

Answer 1

Balanced reaction: N2(g) + 3 H2(g) → 2 NH3(g)

ΔH°rxn = Σ nΔHf(products) − Σ nΔHf(reactants)
= 2(−46.19 kJ/mol) − [1(0) + 3(0)]
= −92.38 kJ

Answer: Hrxn = −92.38 kJ.