Question
Calculate the enthalpy change, ΔrH, for the following reaction,
4 NH3 (g) + 5 O2(g) → 4 NO (g) + 6 H2O (g)
given the thermochemical equations below.
N2 (g) + O2 (g) → 2 NO (g) ΔrH° = +181 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔrH° = 91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) ΔrH° = 484 kJ
A. 127 kJ
B. 395 kJ
C. 906 kJ
D. 211 kJ
E. +1270 kJ
4 NH3 (g) + 5 O2(g) → 4 NO (g) + 6 H2O (g)
given the thermochemical equations below.
N2 (g) + O2 (g) → 2 NO (g) ΔrH° = +181 kJ
N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔrH° = 91.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (g) ΔrH° = 484 kJ
A. 127 kJ
B. 395 kJ
C. 906 kJ
D. 211 kJ
E. +1270 kJ