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Anastasy [175]
3 years ago
12

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O

2(g)→NO2(g), ΔfH∘A=33.2 kJ mol−1 12N2(g)+12O2(g)→NO(g), ΔfH∘B=90.2 kJ mol−1
Chemistry
2 answers:
Alisiya [41]3 years ago
7 0
We apply Hess's Law and obtain:
2ΔHf(B) + ΔH(reaction) = 2ΔHf(A)
ΔH(reaction) = 2(33.2) - 2(90.2)
ΔH(reaction) = -114 kJ / mol
Alika [10]3 years ago
6 0

The enthalpy of the reaction : <u>114 kJ/mol</u>

<h3>Further explanation </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat/heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0

<em>The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero. </em>

From the problem can be known

∆Hf ° NO₂ = 33.2 kJ / mol

∆Hf ° NO = 90.2 kJ / mol, so that

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

∆H ° rxn = 2.∆Hf ° NO₂ - 2. ∆Hf ° NO  (O₂ not included)

∆H ° rxn = 2.33.2 - 2.90.2

∆H ° rxn = 66.4 - 180.4

∆H ° rxn = -114 kJ / mol

<h3>Learn more </h3>

Delta H solution

brainly.com/question/10600048

an exothermic reaction

brainly.com/question/1831525

as endothermic or exothermic

brainly.com/question/11419458

an exothermic dissolving process

brainly.com/question/10541336

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