AP Chemistry

Unit 6.6

INTRODUCTION TO ENTHALPY OF REACTION

Unit 6.6

Enduring Understanding:

• Changes in a substance’s properties or change into a different substance requires an exchange of energy

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Learning Objective:

• Calculate the heat, q, absorbed or released by a system undergoing a chemical reaction in relationship to the amount of the reacting substance in moles and the molar enthalpy of reaction

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Enthalpy Change of a Reaction

• Enthalpy change of a reaction:
• The amount of heat energy released (negative values) or absorbed (positive values) by a chemical reaction at constant pressure
• ΔH_rxn
• Represents the difference in enthalpy between the reactants and products
• The amount of heat generated or absorbed depends on the amounts of reactants and products
• Units:
• kJ/mol_rxn or J/mol_rxn

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Enthalpy Change of a Reaction

• The magnitude of ΔH_rxn reflects the stoichiometric amounts of reactants and products for the reaction as written
• For example:

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• These reactions (endothermic) show that 180 kJ of heat is absorbed when one mole of each reactant produces 2 moles of product

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

Or

N₂(g) + O₂(g) + 180 kJ → 2NO(g)

Enthalpy Change of a Reaction

• We can, therefore, use the heat as a conversion factor:

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• Example:
• How much heat is required to react 2.4 moles of N₂(g) with excess O₂(g) to form NO(g)?

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

Enthalpy Change of a Reaction

• We can, therefore, use the heat as a conversion factor:

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• Example:
• How much heat is required to react 2.4 moles of N₂(g) with excess O₂(g) to form NO(g)?

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

2.4 moles N₂

X 180 kJ = 430 kJ

1 mole N₂

Enthalpy Change of a Reaction

• We can, therefore, use the heat as a conversion factor:

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• Example:
• If 25 kJ of heat was absorbed in the reaction, how many moles of NO were produced?

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

Enthalpy Change of a Reaction

• We can, therefore, use the heat as a conversion factor:

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• Example:
• If 25 kJ of heat was absorbed in the reaction, how many moles of NO were produced?

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

25 kJ

X 2 moles NO = 0.28 moles NO

180 kJ

Enthalpy Change of a Reaction

• We can, therefore, use the heat as a conversion factor:

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• Example:
• If 25 kJ of heat was absorbed in the reaction, how many moles of NO were produced?

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N₂(g) + O₂(g) → 2NO(g) ΔH = +180 kJ/mol_rxn

25 kJ

X 2 moles NO = 0.28 moles NO

180 kJ

NOTE: At constant pressure ΔH = q

Enthalpy Change of a Reaction

• If you want to know why ΔH = q:

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Practice: I Do

1. If you were to react 7.25 moles of H₂S with 9.34 moles of O₂, how much heat would be released?

2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔH= -1120 kJ/mol_rxn

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Practice: I Do

• If you were to react 7.25 moles of H₂S with 9.34 moles of O₂, how much heat would be released?

2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔH= -1120 kJ/mol_rxn

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7.25 moles H₂S x -1120 kJ = -4060 kJ

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2 moles H₂S

Practice: I Do

• If you were to react 7.25 moles of H₂S with 9.34 moles of O₂, how much heat would be released?

2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔH= -1120 kJ/mol_rxn

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7.25 moles H₂S x -1120 kJ = -4060 kJ

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9.34 moles O₂ x -1120 kJ = -3490 kJ

2 moles H₂S

3 moles O₂

Practice: I Do

• If you were to react 7.25 moles of H₂S with 9.34 moles of O₂, how much heat would be released?

2H₂S(g) + 3O₂(g) → 2H₂O(l) + 2SO₂(g) ΔH= -1120 kJ/mol_rxn

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7.25 moles H₂S x -1120 kJ = -4060 kJ

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9.34 moles O₂ x -11120 kJ = -3490 kJ

2 moles H₂S

3 moles O₂

-3490 kJ is the amount of heat that is released; O₂ is the limiting reactant