The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO (g) + NO2 (g) → N2O3 (g) ΔH o rxn = - 39.8 kJ (2) NO (g) + NO2 (g) + O2 (g) → N2O5 (g) ΔH o rxn = - 112.5 kJ (3) 2NO2 (g) → N2O4 (g) ΔH o rxn = - 57.2 kJ (4) 2NO (g) + O2 (g) → 2NO2 (g) ΔH o rxn = - 114.2 kJ (5) N2O5 (s) → N2O5 (g) ΔH o subl = 54.1 kJ Calculate the heat of reaction for N2O3 (g) + N2O5 (s) → 2N2O4 (g) ΔH = kJ

The heat of reaction for N₂O₃ (g) + N₂O₅ (s) → 2 N₂O₄ (g) is ΔH = - 22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO (g) + NO₂ (g) → N₂O₃ (g) ΔH o rxn = - 39.8 kJ

(2) NO (g) + NO₂ (g) + O₂ (g) → N₂O₅ (g) ΔH o rxn = - 112.5 kJ

(3) 2 NO₂ (g) → N₂O₄ (g) ΔH o rxn = - 57.2 kJ

(4) 2 NO (g) + O₂ (g) → 2 NO₂ (g) ΔH o rxn = - 114.2 kJ

(5) N₂O₅ (s) → N₂O₅ (g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃ (g) + N₂O₅ (s) → 2 N₂O₄ (g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

If the chemical equation is inverted, the symbol of ΔH is also reversed. If the coefficients are multiplied, multiply ΔH by the same factor. If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃ (g) + N₂O₅ (s) → 2 N₂O₄ (g) you must do the following:

Multiply equation (3) by 2

(3) 2*[2 NO₂ (g) → N₂O₄ (g) ] ΔH o rxn = - 57.2 kJ*2

4 NO₂ (g) → 2 N₂O₄ (g) ΔH o rxn = - 114.4 kJ

Reverse equations (1) and (2)

(1) N₂O₃ (g) → NO (g) + NO₂ (g) ΔH o rxn = 39.8 kJ

(2) N₂O₅ (g) → NO (g) + NO₂ (g) + O₂ (g) ΔH o rxn = 112.5 kJ

Equations (4) and (5) are maintained as stated.

(4) 2 NO (g) + O₂ (g) → 2 NO₂ (g) ΔH o rxn = - 114.2 kJ

(5) N₂O₅ (s) → N₂O₅ (g) ΔH o subl = 54.1 kJ

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH = - 114.4 kJ + 39.8 kJ + 112.5 kJ - 114.2 kJ + 54.1 kJ

ΔH = - 22.2 kJ

The heat of reaction for N₂O₃ (g) + N₂O₅ (s) → 2 N₂O₄ (g) is ΔH = - 22.2 kJ