Question

Set up a Hess’s law cycle, and use the following information to calculate ΔH∘f for aqueous nitric acid, HNO3(aq). You will need to use fractional coefficients for some equations. 3NO2(g)+H2O(l)→2HNO3(aq)+NO(g) ΔH∘ = -137.3 kJ 2NO(g)+O2(g)→2NO2(g) ΔH∘ = -116.2 kJ 4NH3(g)+5O2(g)→4NO(g)+6H2O(l) ΔH∘ = -1165.2 kJ NH3(g) ΔH∘f = -46.1 kJ H2O(l) ΔH∘f = -285.8 kJ

Answer 1

Answer:

-207,4 kJ = ΔH∘fHNO₃(aq)

Explanation:

Ussing Hess's law it is possible to obtain the enthalpy of a reaction or a product by the sum of different half-reactions.

(1) 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g) ΔH° = -137.3 kJ

(2) 2NO(g) + O₂(g) → 2NO₂(g) ΔH° = -116.2 kJ

(3) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l) ΔH° = -1165.2 kJ

The sum of ²/₃ (1) + (2) gives:

⁴/₃NO(g) + ²/₃H₂O(l) + O₂(g) → ⁴/₃HNO₃(aq) ΔH° = ²/₃ (1) + (2) = -207,7 kJ

This reaction + ¹/₃ (3) produce:

⁴/₃NH₃(g) + ⁸/₃O₂(g) → + ⁴/₃H₂O(l) + ⁴/₃HNO₃(aq) ΔH° = -207,7 kJ +  ¹/₃ (3) = -596,1 kJ

This ΔH° is =  ⁴/₃ΔH∘fH₂O(l) + ⁴/₃ΔH∘fHNO₃(aq) - (⁴/₃ΔH∘fNH₃(g) + ⁸/₃ΔH∘fO₂(g))

As: NH₃(g) ΔH∘f = -46.1 kJ; H₂O(l) ΔH∘f = -285.8 kJ; O₂(g) ΔH∘f =0kJ:

-596,1 kJ = -⁴/₃285,8kJ +⁴/₃ΔH∘fHNO₃(aq)  - (-⁴/₃46,1kJ + 0 kJ)

-276,5 kJ = ⁴/₃ΔH∘fHNO₃(aq)

-207,4 kJ = ΔH∘fHNO₃(aq)

I hope it helps!