Question

Hydrazine, N2H4, is an extremely unstable compound and has been used as liquid rocket fuel. Using the following thermochemical information, write an equation describing the heat of reaction of liquid hydrazine and hydrogen peroxide, H2O2 in terms of A, B and C:

N2H4 (l) + 2 H2O2 (l)  N2 (g) + 4 H2O (l)

N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (l) ΔH1 = A

H2 (g) + 2 O2 (g) → H2O(l) ΔH2 = B

H2 (g) + O2 (g) → H2O2 (l) ΔH3 = C

Assuming A= -622 kJ; B= -285.8 kJ; C = -187.8 kJ what is the value of the enthalpy change for the reaction shown above?

Answer 1

Given reactions are

N2H4 (l) + 2 H2O2 (l) ------> N2 (g) + 4 H2O (l) : H = ? ---(1)

N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (l) : ΔH1 = A = -622 kJ   ----(2)

H2 (g) + (1/2) O2 (g) → H2O(l) : ΔH2 = B = -285.8 kJ   -------(3)

H2 (g) + O2 (g) → H2O2 (l): ΔH3 = C = -187.8 kJ     -----(4)

(1) = (2) + [2*(3)] [2* reverse of (4)]

So H = H₁ + [2*H₂]+[2*(-H₃ )]

            = -622 + [2*(-285.8)] +[2*(-(-187.8))]

             = -818 kJ

Therefore the value of the enthalpy change for the reaction shown above is -818 kJ