Given the reactions
a.) n2(g)+o2(> 2no(g) delta h= -180.5
b.) n2(g) + 3h2(g) ——> 2nh3(g) delta h= -91.8
c.)2h2(g)+ o2(g) —-> 2h2o(g) delta h= -486.6

what is the delta h for the reaction below
4nh3(g)+5o2(> 4no(g)+6h2o(g)

The answer to your question is: ΔH = 1637.8

Explanation:

Hess' law: This law states that the enthalpy change can be calculated even if it is not calculated directly.

"if a chemical change takes place by several routes, the overall enthalpy change is the same regardless the route".

Process

A) N2(g)+O2(g)—->2NO(g)                              Δ H= -180.5

B) N2(g) + 3H2(g) ——> 2NH3(g)                    Δ H= -91.8

C)2H2(g)+ O2(g) —-> 2H2O(g)                       Δ H= -486.6

The result must be:

                                4NH3(g)+5O2(g)—->4NO(g)+6H2O(g)

Turn letter B and multiply it by 2

                               4NH3   ⇒  2N2  +  6H2       ΔH = 183.6

Multiply letter A by 2

                               2N2 + 2O2 ⇒ 4 NO             ΔH = -361  

Multiply letter C by 3

                              6H2 + 3O2 ⇒ 6H2O            ΔH = -1459.8

Finally we add the equations up and simplify then:

                            4NH3 + 5O2 ⇒ 4NO + 6 H2O

And we add the ΔH = 183.6 - 361 - 1459.8

                                 = -1637.8