The Ostwald process for the commercial production of nitric acid from ammonia an

Question

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H)

2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol

3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol

Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations (Water is also a product).

So i know the answer is this...

"Multiply the equations so that the coefficients of the N-products become the same as the N-educts of the following line.
4NH3(g)+5O2(g)------->4NO(g)+6H2O(g) -908 kj/mol (delta H) (x3)
2NO(g)+O2(g)----------->2NO2(g) -112 kj/mol (x 6)
3NO2(g)+H2O(l)-------->2HNO3(aq)+NO(g) -140 kj/mol (x 4)

12NH3(g)+15O2(g)------->12NO(g)+18H2O(...
12NO(g)+6O2(g)----------->12NO2(g)
12NO2(g)+4H2O(l)-------->8HNO3(aq)+4NO.."

What i want to know is why do we multiply the equations like that? Am i missing a concept? Because why can we just not add the eqautions together?

Explanation / Answer

The overall net reaction for the Ostwalds process for production of nitric acid is,

NH3 + 2O2 ---> HNO3 + H2O

Now, according to Hess's law, the overall change in enthalpy equals to the sum of changes in individual steps of reaction, thus,

4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)     -908 kJ/mol

Multiply second equation with 3,

6NO(g) + 3O2(g)-----------> 6NO2(g)     3 x -112 kj/mol

Multiply third equation with 2,

6NO2(g)+ 2H2O(l)--------> 4HNO3(aq)+ 2NO(g)     2 x -140 kj/mol

Add all three equations and divide by 4,

NH3(g) + 2O2(g) ---> HNO3(aq) + H2O    -1524 kJ/mol

Is your final answer

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