Calculate ∆H for the reaction (j)

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

Target equation is N₂H₄ + O₂ ==> N₂ + 2H₂O

Use Hess' Law to find ∆H of this reaction:

N2H4 + H2O ==> 2NH3 + 1/2O2 ∆H = +143 kJ (reversed #3 and changed sign)

2NH3 + 3N2O ==> 4N2 + 3H2O ∆H = -1010 kJ (added #1)

3N2H4 + 3H2O ==> 3N2O + 9H2 ∆H = +951 kJ (reversed #2 and multiplied by 3)

9H2 + 4.5O2 ==> 9H2O ∆H = -2574 kJ (multiplied #4 by 9)

Add them all up and cancel like terms on opposite sides to get:

4N2H4 + 4O2 ==> 4N2 + 8H2O ∆H = -2490

But this is for FOUR times the original equation, so we divided everything by 4 to get...

N2H4 + O2 ==> N2 + 2H2O ∆H = -623 kJ for the reaction as written