Question

Given the following data:

HNO3(l) → 1/2N2(g) + 3/2O2(g) + 1/2H2(g)	ΔH° = 174.1 kJ
2N2(g) + 5O2(g) → 2N2O5(g)	ΔH° = 28.4 kJ
H2(g) + 1/2O2(g) → H2O(l)	ΔH° = -285.8 kJ

Calculate ΔH° for the reaction:
2HNO₃(l) → N₂O₅(g) + H₂O(l)
Note that you should be able to answer this one without needing to use any additional information from the thermo table.

Answer 1

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

2N₂(g) + 5O₂(g) → 2N₂O₅(g)    ; ΔH°₂ = 28.4 kJ                        -------(2)

H₂(g) + 1/2O₂(g) → H₂O(l)     ; ΔH°₃ = -285.8 kJ                        ------(3)

2HNO₃(l) → N₂O₅(g) + H₂O(l)   : ΔH° = ?                               -------(4)

Eqn (4) can be obtained from remaining equations as follows:

Eqn(1) = [2xEqn(1)] + [(1/2)xEqn(2)] +Eqn(3)

So ΔH° = [2x ΔH°₁ ] + [ (1/2)x ΔH°₂ ] + ΔH°₃

            = [2x(174.1 kJ)] + [(1/2)x(28.4 kJ)] + (-285.8 kJ)

            = +76.6 kJ