Question

2. Given the following data:
H2O(l) → H2(g) + 1/2O2(g) ΔH° = 285.8 kJ
2HNO3(l) → N2O5(g) + H2O(l) ΔH° = 76.6 kJ
2N2(g) + 5O2(g) → 2N2O5(g) ΔH° = 28.4 kJ

Calculate ΔH° for the reaction: 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l)
Note that you should be able to answer this one without needing to use any additional information from the thermo table.

I've attempted this question multiple times. I am able to get to the simplified eqaution broken down, but I am not able to match the answer numerically. Any concise and clear explanations would be greatly appreciated! I really need to understand the material prior to my exam!

Answer 1

H2(g) + 1/2O2(g) → H2O(l)    ΔH° = -285.8 kJ ---------- (1) (Reversing the equation)

N2O5(g) + H2O(l) → 2HNO3(l) ΔH° = -76.6 kJ -----------(2) (Reversing the equation)

Adding 1 and 2

H2(g) + 1/2O2(g)+N2O5(g) + →  2HNO3(l) ΔH° = (-76.6 kJ + (-285 kJ)) = -362.4 kJ ---------------(3)

N2(g) + 5/2O2(g) → N2O5(g) ΔH° = 14.2 kJ (Dividing by 2) -------------------------------------(4)

Adding 3 and 4

H2(g) + 3O2(g)+ N2(g) →  2HNO3(l) ΔH° = -348.2 kJ

Dividing through out by 2

1/2H2(g) + 3/2O2(g)+ 1/2N2(g)   → HNO3(l) ΔH° = -348.2/2 kJ = -174.1 kJ

ΔH° for the reaction: 1/2N2(g) + 3/2O2(g) + 1/2H2(g) → HNO3(l) is -174.1 kJ