Calculate the ΔH for the reaction 4 NH3 (g) + 5 O2 (g) --> 4 NO (g) + 6 H20 (g), from the following data.

Question

1) N2 (g) + O2 (g) --> 2 NO (g) ΔH= -180.5 kJ

2) N2 (g) + 3 H2 (g) --> 2 NH3 (g) ΔH= -91.8 kJ

3) 2 H2 (g) + O2 (g) --> 2 H2O (g) ΔH= -483.6 kJ

Answer 1

This problem relies on Hess' Law:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(g) ... TARGET EQUATION

Given:

1) N2 (g) + O2 (g) --> 2 NO (g) ΔH= -180.5 kJ

2) N2 (g) + 3 H2 (g) --> 2 NH3 (g) ΔH= -91.8 kJ

3) 2 H2 (g) + O2 (g) --> 2 H2O (g) ΔH= -483.6 kJ

Process:

The idea is to rearrange the given equations so that we can combine them to get the target equation.

reverse(2) & x 2: 4NH3(g) ==> 2N2(g) + 6H2(g) ... ∆H = +183.6 kJ

copy(1) x 2: 2N2(g) + 2O2(g) ==> 4NO2(g) ... ∆H = -361 kJ

copy(3) x 3: 6H2(g) + 3O2(g) ==> 6H2O(g) ... ∆H = -1450.8

Add them up and combine or cancel appropriate items to get...

4NH3(g) + 2N2(g) + 2O2(g) + 6H2(g) + 3O2(g) ==> 2N2(g) + 6H2(g) + 4NO2(g) + 6H2O(g)

This leaves us with ...

4NH3(g) + 5O2(g) ==> 4NO2(g) + 6H2O(g) = TARGET EQUATION

∆H = 183.6 + (-361) + (-1450.8) = -1628.2 kJ