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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°rxn = -39.8kj
(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH°rxn = -112.5kj
(3) 2NO₂(g) → N₂O₄(g) ΔH°rxn = -57.2kj
(4) 2NO(g) + O₂(g) → 2NO₂(g) ΔH°rxn = -114.2kj
(5) N₂O₅(s) → N₂O₅(g) ΔH°rxn = 54.1kj
calculate the standard enthalpy of reaction for
N₂O₃(g) + N₂O₅(g) → 2 N₂O₄(g)

Respuesta :

N2O3(g) + N2O5(s) 2 N2O4(g) has a heat of reaction of H = 22.2 kJ.

In order to accomplish the final chemical reaction from which you wish to obtain the reaction heat, we must first complete the intermediate steps.

N2O3(g) plus N2O5(s) = 2 N₂O₄ (g)

Equation (3)

should be multiplied by 2 4 NO2(g) 2 N2O4 (g) H orxn equals 114.4 kJ

Reverse the first and second equations: N2O3(g) = NO(g) + NO2 (g) H o rxn = 39.8 kJ N2O5(g), NO(g), NO2(g), and O2 (g) H orxn equals 112.5 kJ

The original form of equations (4) and (5) is kept.

O2(g) + 2 NO(g) = 2 NO2 (g)

H o rxn equals 114.2 kJ

Learn more about Enthalpies here-

https://brainly.com/question/13996238

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