Hess's Law of Heat Summation

Question

Liquid hydrazine, N2H4(l), is a rocket fuel. It combusts in oxygen gas to form nitrogen gas and liquid water:

N2H4(l) + O2(g) -> N2(g) + 2H2O(l)

Use the following thermochemical equations to calculate the enthalpy change for the combustion of liquid hydrazine:

2NH3(g) + 3N2O(g) -> 4N2(g) + 3H2O(l) ΔH = -1010kJ

N2O(g) + 3H2(g) -> N2H4(l) + H2O(l) ΔH = -317kJ

2NH3(g) + 0.5O2(g) -> N2H4(l) + H2O(l) ΔH = -143kJ

H2(g) + 0.5O2(g) -> H2O(l) ΔH = -286kJ

Answer 1

There may be other mathematical manipulations other than using fractions, but this is my way of approaching this problem dealing with Hess' Law.

TARGET EQUATION: N2H4(l) + O2(g) -> N2(g) + 2H2O(l)

Given:

(1): 2NH3(g) + 3N2O(g) -> 4N2(g) + 3H2O(l) ΔH = -1010kJ

(2): N2O(g) + 3H2(g) -> N2H4(l) + H2O(l) ΔH = -317kJ

(3): 2NH3(g) + 0.5O2(g) -> N2H4(l) + H2O(l) ΔH = -143kJ

(4): H2(g) + 0.5O2(g) -> H2O(l) ΔH = -286kJ

Procedure:

(1)÷4: 0.5 NH3 + 3/4 N2O ==> N2 + 3/4 H2O ... ∆H = -1010/4 = -252.5 kJ

rev(3)÷4: 1/4 N2H4 + 1/4 H2O ==> 0.5 NH3 + 0.125 O2 ... ∆H = +143/4 = +35.8 kJ

rev(2)x3/4: 3/4 N2H4 + 3/4 H2O ==> 3/4 N2O + 2.25 H2 ... ∆H = +317 x 3/4 = +237.8 kJ

(4)x2.25: 2.25 H2 + 1.125 O2 ==> 2.25 H2O ... ∆H = -286 x 2.25 = -643.5 kJ

_____________________________________________ add them/combine and/or cancel like terms to get...

N2H4(l) + O2(g) -> N2(g) + 2H2O(l) TARGET EQUATION

∆H = -252.5 + 35.8 + 237.8 -643.5 = -622.4 kJ