Thermochemistry

Question

6 NO(aq) + 4NH₃ (g) ----> 5N₂ (g) + 6H₂O Delta H_rxn = -1807

6 NO₂ (g) +8 NH₃(g) _---> 7 N₂ (g) +12 H₂O(g) Delta H_rxn = -2730

Calculate

3 NO(g) + N₂(g) + 3 H₂O(g) -----> 3NO₂ (g) + 2NH₃ (g) Delta H_{rxn =} ?

Answer 1

Since we need 3 NO2 on the left hand side, we start by reversing the second reaction and dividing by two to get:

(7/2) N2 + 6 H2O → 3 NO2 + 4 NH3 ΔH = 1365

The next step is to divide the first reaction by two to get:

3 NO + 2 NH3 → (5/2) N2 + 3 H2O ΔH = -903.5

When these two are added, and some terms occurring on both sides are canceled, the result is:

N2 + 3 H2O + 3 NO → 3 NO2 + 2 NH3 ΔH = 1365 - 903.5 = 461.5