Determine the molar enthalpy (in kJ/mol) for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).

Question

Using the equations

N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

Determine the molar enthalpy (in kJ/mol) for the reaction 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g).

Answer 1

You want to rearrange the equations so that you can get the reactants/products to cancel out and leave you with the final equation. If we were to number your equations...

(1) N₂ (g) + O₂ (g) → 2 NO (g) ∆H° = 180.6 kJ/mol

(2) N₂ (g) + 3 H₂ (g) → 2 NH₃ (g) ∆H° = -91.8 kJ/mol

(3) 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.7 kJ/mol

If the final equation desired is:

4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

We will need to flip (2) and multiply it by 2 in order to get 4 NH₃ on the reactant side.

We will also need to multiply (1) by 2 to get 4 NO on the product side. If we were to do this, we would have:

(2, reversed & x2): 4 NH₃ (g) → 2 N₂ (g) + 6 H₂ (g)

(1, x2) 2 N₂ (g) + 2 O₂ (g) → 4 NO (g)

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Resulting (4): 4 NH₃ (g) +2 O₂ (g) → 6 H₂ (g) + 4 NO (g)

Since H₂ doesn't appear in the final equation, and we have six of them on the product side, we will need to also include (3) and multiply it by 3.

(4, above) 4 NH₃ (g) +2 O₂ (g) → 6 H₂ (g) + 4 NO (g)

(3, x3) 6 H₂ (g) + 3 O₂ (g) → 6 H₂O (g)

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Resulting (final): 4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

In order to find the total molar enthalpy, we will need to multiply by -1 for every rxn that was flipped (if you release 50 kJ/mol in the forward direction then you will absorb 50 kJ/mol in the reverse direction) and multiply by the factor that the rxn was increased by.

This means we have: (-2 * 91.8) + (2 * 180.6) + (3 * -483.7) kJ/mol