Hess's Laws Problem

Question

Given the following equations:

N2(g) + O2(g) --> 2NO(g) = +190.0kj

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

2H2(g) + O2(g) --> 2H2O(g) = -483.7kj

Calculate the enthalpy for the reaction:

4NH3(g) + 5O2(g) --> 4NO(g) + 6H20(g)

Answer 1

4NH3(g) + 5O2(g) --> 4NO(g) + 6H20(g) .. TARGET EQUATION

Given:

eq. 1: N2(g) + O2(g) --> 2NO(g) = +190.0kj

eq. 2: N2(g) +3H2(g) --> 2NH3(g) = -91.8kj

eq. 3: 2H2(g) + O2(g) --> 2H2O(g) = -483.7kj

Procedure:

reverse eq.2 and multiply by 2: 4NH3(g) ==> 2N2(g) + 6 H2(g) .. +91.8 x 2 = +183.6 kJ

copy eq.1 and multiply by 2: 2N2(g) + 2O2(g) ==> 4NO(g) .. +190.0 x 2 = + 380 kJ

copy eq.3 and multiply by 3: 6H2(g) + 3O2(g) ==> 6H2O(g) .. -483.7 x 3 = -1451.1 kJ

Add all to get ...

4NH3(g) + 2N2(g) + 2O2(g) + 6H2(g) + 3O2(g) ==> 2N2(g) + 6H2(g) + 4NO(g) + 6H2O(g)

Combine/cancel like terms to end up with ...

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

Enthalpy of reaction = 183.6 + 380 - 1451.1 = -887.5 kJ