Chemistry - Thermochemistry Question with Solution | TestHub
Study the following thermochemical equations:
(a) N2O(g) + 3H2(g) → H2O(l) + N2H4(l); ΔH = -75.56 kcal
(b) 4N2(g) + 3H2O(l) → 2NH3(g) + 3N2O(g); ΔH = +241.35 kcal
(c) N2H4(l) + H2O(l) → 2NH3(g) + 1/2 O2(g); ΔH = +34.18 kcal
(d) H2O(l) → H2(g) + 1/2 O2(g); ΔH = +68.32 kcal
The enthalpy of combustion (in kcal/mol) of liquid N2H4 is
Options:
Answer:
Solution:
From (d), Δf H(H2O,l) = -68.32. Let h = Δf H(N2H4), o = Δf H(N2O), and a = Δf H(NH3).
Using the given equations:
Equation (a): -68.32 + h - o = -75.56, which simplifies to h - o = -7.24.
Equation (c): 2a - h + 68.32 = 34.18, which simplifies to 2a - h = -34.14.
Equation (b): 2a + 3o + 204.96 = 241.35, which simplifies to 2a + 3o = 36.39.
Now, we solve this system of equations:
Subtracting the simplified Equation (c) from the simplified Equation (b):
(2a + 3o) - (2a - h) = 36.39 - (-34.14)
3o + h = 70.53
Substitute h = o - 7.24 (from simplified Equation (a)) into the above equation:
3o + (o - 7.24) = 70.53
4o = 77.77
o = 19.44
Now, find h:
h = o - 7.24 = 19.44 - 7.24 = 12.20
For the combustion of hydrazine:
N2H4(l) + O2(g) tarrow N2(g) + 2H2O(l)
The enthalpy change for this reaction (Δ H) is calculated as:
Δ H = 2 × Δf H(H2O,l) - Δf H(N2H4,l)
Δ H = 2(-68.32) - 12.20
Δ H = -136.64 - 12.20
Δ H = -148.84 kcal/mol
The answer is (c).
