Question

Hydrazine, a component of rocket fuel, undergoes combustion to yield N₂ and H₂O.

N₂H₄ (l) + O₂ (g) → N₂ (g) + 2H₂O (l)

What is the enthalpy change of combustion of N₂H₄ (kJ/mole) Given Reaction ΔH/kJ 2NH₃ (g) + 3N₂O (g) → 4N₂ (g) + 3H₂O (l) - 1011 kJ N₂O (g) + 3H₂ (g) → N₂H₄ (l) + H₂O (l) - 317 kJ 4NH₃ (g) + O₂ (g) → 2N₂H₄ (l) + 2H₂O (l) - 286 kJ H₂(g) + $\frac{1}{2}$O₂ (g) → H₂O (l) - 285 kJ

• A. -620.5
• B. -622.75
• C. 1167.5
• D.
  • 622.75

Answer 1

Answer: (A)

-620.5 kJ/mole

Answer 2

The enthalpy change of combustion of N₂H₄ can be calculated using Hess's Law and the given reactions.

1. From Reaction 1: 2x + 3y = 156

2. From Reaction 3: z = 2x + 142

3. From Reaction 2: z = y - 32

Solving these equations, we find:

• x = -45.75
• y = 82.5
• z = 50.5 kJ

Then, the combustion reaction ΔH is calculated as:

ΔH = [2(–285)] – (50.5) = –620.5 kJ.