Calculate the enthalpy of the reaction 2no(g)+o2(g)→2no2(g) given the following reactions and enthalpies of formation: 12n2(g)+o2(g)→no2(g), δfh∘a=33.2 kj mol−1 12n2(g)+12o2(g)→no(g), δfh∘b=90.2 kj mol−1

The enthalpy of the reaction : 114 kJ/mol

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

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

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

Delta H solution

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an exothermic reaction

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as endothermic or exothermic

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an exothermic dissolving process

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