Find the enthapy formation.

Question

Find the enthapy formation of NO_(g)

N2(g) + 3H2(g)   → 2 NH3(g)              DHº= – 91,80 kJ/mol

4NH3(g) + 5O2(g)  →4NO(g) + 6H2O(g)      DHº= – 906,20 kJ/mol

H2(g) + ½ O2(g) →H2O(g)               DHº= – 241,80 kJ/mol

Answer 1

This is a problem involving the use of Hess' Law.

Target equation: N₂ + O₂ ==> 2NO

Given:

(1). N2(g) + 3H2(g)  → 2 NH3(g)             ∆Hº= – 91,80 kJ/mol

(2). 4NH3(g) + 5O2(g)  →4NO(g) + 6H2O(g) ∆Hº= – 906,20 kJ/mol

(3). H2(g) + ½ O2(g) →H2O(g)              ∆Hº= – 241,80 kJ/mol

Procedure:

copy(1): N2(g) + 3H2(g)  ==> 2 NH3(g) ... ∆H = -91.80 kJ/mol

copy (2)x1/2: 2NH3 + 5/2 O2 ==> 2NO + 3H2O ... ∆H = 1/2 x -906.2 kJ/mol = -453 kJ/mol

reverse (3)x3: 3H2O ==> 3H2 + 3/2 O2 ... ∆H = 3 x +241.8 kJ/mol = +725.4 kJ/mol

Add them all to get ...

N2(g) + 3H2(g) + 2NH3(g) + 5/2 O2(g) + 3H2O(g) ==> 2 NH3(g) + 2NO(g) + 3H2O(g) + 3H2(g) + 3/2 O2(g)

Combine/cancel like terms to get ...

N2(g) + O2(g) ==> 2NO(g) ... Target equation

∆H = -91.80 + (-453) + 725.4 = 181 kJ/mol