Calculate ∆H for the reaction (n)

Question

Calculate ∆H for the reaction

N₂H₄(l) + O₂(g) → N₂(g) + 2 H₂O(l)

given the following data:

Equation	∆H(kJ)
2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(l)	-1010
N2O(g) + 3 H2(g) → N2H4(l) + H2O(l)	-317
2 NH3(g) + ½ O2(g) → N2H4(l) + H2O(l)	-143
H2(g) + ½ O2(g) → H2O(l)	-286

∆H= kJ

Answer 1

Hess is a mess!

The strategy is to work backward from the desired equation. Clearly we need N₂H₄(l) on the left side, so generate a new equation by subtracting Eqn.1 -Eqn. 3.

2 NH₃(g) + 3 N₂O(g) → 4 N₂(g) + 3 H₂O(l)

N₂H₄(l) + H₂O(l) → 2 NH₃(g) + ½ O₂(g)

3 N₂O(g) + N₂H₄(l) → 4 N₂(g) + 2 H₂O(l) + ½ O₂(g) (∆H(kJ) = -867)

by adding this reaction to the negative (opposite) of the desired reaction will give us an idea of what we need.

3 N₂O(g) + N₂H₄(l) → 4 N₂(g) + 2 H₂O(l) + ½ O₂(g) (opposite reaction of what we need)

N₂(g) + 2 H₂O(l) → N₂H₄(l) + O₂(g), so adding them

3 N₂O(g) → 3 N₂(g) + 3/2 O₂(g), and dividing by 3

N₂O(g) → N₂(g) + ½ O₂(g).

We have determined what reaction we need, now we need the enthalpy of it.

We need to add this rxn to the one we got from (Eqn.1 - Eqn. 3), but how do we generate it?

Eliminate NH₃(g) to give a new reaction. (Eqn. 1 + Eqn. 2 - Eqn 3).

4 N₂O(g) + 3 H₂(g) → 4 N₂(g) + ½ O₂(g) +3 H₂O(l) (∆H(kJ) = -1184)

Now eliminate water from the expression by adding 3x's the reverse of Eqn. 4.

3 H₂O(l) → 9 H₂(g) + 3/2 O₂(g) (∆H(kJ) = 858) and it gives.

4 N₂O(g) → 4 N₂(g) + 4/2 O₂(g) (∆H(kJ) = -326). This is equivalent to writing...

N₂O(g) → N₂(g) + ½ O₂(g) (∆H(kJ) = -326/4 = -81.5)

We now have all we need. Now add negative 3x of this to ...

3 N₂O(g) + N₂H₄(l) → 4 N₂(g) + 2 H₂O(l) + ½ O₂(g) (∆H(kJ) = -867)

3 N₂(g) + 3/2 O₂(g) → 3 N₂O(g) (∆H(kJ) = 81.5*3 = 244.5)

N₂H₄(l) + ½ O₂(g) → N₂(g) + 2 H₂O(l)

(∆H(kJ) = -622.5). The answer is -623 kJ accounting for 3 significant figures.