Clicker 1

  • Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction 2 NO(g) + O2(g) → 2 NO2(g) ΔH°rxn = ?
    • Given:
      • N2(g) + O2(g) → 2 NO(g) ΔH°rxn = +183 kJ
      • 1/2 N2(g) + O2(g) → NO2(g) ΔH°rxn = +33 kJ
    • A) -150. kJ
    • B) -117 kJ
    • C) -333 kJ
    • D) +115 kJ
    • E) +238 kJ

Standard Conditions and Standard Enthalpy of Formation (ΔHf)

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  • The standard state is the state of a material at a defined set of conditions.
    • Pure gas at exactly 1 atm pressure
    • Pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
      • Usually 25 °C
    • Substance in a solution with concentration 1 M
  • The standard enthalpy change, ΔH°, is the enthalpy change when all reactants and products are in their standard states.
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  • The standard enthalpy of formation, ΔH°f, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements.
    • The elements must be in their standard states.
    • The ΔHf° for a pure element in its standard state = 0 kJ/mol.

Table of Standard Enthalpies (ΔHf)

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Practice Problem: Standard Enthalpies of Formation (ΔHf)

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  • Write the appropriate equations for the heats of formation of MgCO3(s) and C6H12O6(s)

Clicker 2

  • The standard enthalpy of formation (ΔHo f) for potassium chloride is the enthalpy change for the reaction:
    • A) K(g) + ½Cl2(g) è KCl(g)
    • B) K+(g) + Cl- (g) è KCl(s)
    • C) 2K(s) + Cl2(g) è 2KCl(s)
    • D) K(s) + ½Cl2(g) è KCl(s)
    • E) K+(g) + Cl- (g) è KCl(g)

Calculating Standard Enthalpy Change for a Reaction

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  • Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products.
  • The ΔH° for the reaction is then the sum of the ΔHf° for the component reactions. ΔH° reaction = ∑nΔHf °(products) − ∑nΔHf °(reactants) ∑ means sum. n is the stoichiometric coefficient of the reaction.

CH4(g) + 2 O2(g) → CO2(g) + 2H2O(g)

  • C(s, graphite) + 2 H2(g) → CH4(g) ΔHf°= − 74.6 kJ/mol CH4
  • C(s, graphite) + O2(g) → CO2(g) ΔHf°= −393.5 kJ/mol CO2
  • H2(g) + ½ O2(g) → H2O(g) ΔHf° = −241.8 kJ/mol H2O

CH4(g) + 2 O2(g) → CO2(g) + 2H2O(g)

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Practice Problem: Standard Enthalpies of Formation (ΔHf )

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  • What is the change in enthalpy for the reaction 4NH3(g) + 5O2(g) è 4NO(g) + 6H2O(g)

Clicker 3

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  • Use the ΔH°f information provided to calculate ΔH°rxn for the following SO2Cl2(g) + 2 H2O(l) → 2 HCl(g) + H2SO4(l) ΔH°rxn = ?
    • ΔH° f (kJ/mol)
    • SO2Cl2(g) -364
    • H2O(l) -286
    • HCl(g) -92
    • H2SO4(l) -814
  • A) -256 kJ
  • B) +161 kJ
  • C) -62 kJ
  • D) +800. kJ
  • E) -422 kJ

Ionic Bonding and the Crystal Lattice

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  • The extra energy that is released comes from the formation of a structure in which every cation is surrounded by anions, and vice versa.
  • This structure is called a crystal lattice.
  • The crystal lattice is held together by the electrostatic attraction of the cations for all the surrounding anions.
  • The crystal lattice maximizes the attractions between cations and anions, leading to the most stable arrangement.

Lattice Energy

  • The extra stability that accompanies the formation of the crystal lattice is measured as the lattice energy.
  • The lattice energy is the energy released when the solid crystal forms from separate ions in the gas state.
    • Always exothermic
    • Hard to measure directly, but can be calculated from knowledge of other processes
  • Lattice energy depends directly on the size of charges and inversely on distance between ions.

Determining Lattice Energy: The Born–Haber Cycle

  • The Born–Haber cycle is a hypothetical series of reactions that represents the formation of an ionic compound from its constituent elements.
  • The reactions are chosen so that the change in enthalpy of each reaction is known except for the last one, which is the lattice energy.

Born–Haber Cycle and Hess’s Law

  • Use Hess’s law to add up enthalpy changes of other reactions to determine the lattice energy.
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  • The force of attraction between charged particles is inversely proportional to the distance between them.
  • Larger ions mean the center of positive charge (nucleus of the cation) is farther away from the negative charge (electrons of the anion).
    • Larger ion = weaker attraction
    • Weaker attraction = smaller lattice energy

Lattice Energy versus Ion Size

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  • The force of attraction between oppositely charged particles is directly proportional to the product of the charges.
  • Larger charge means the ions are more strongly attracted.
    • Larger charge = stronger attraction
    • Stronger attraction = larger lattice energy
  • The force of attraction between oppositely charged particles is directly proportional to the product of the charges.
  • Larger charge means the ions are more strongly attracted.
    • Larger charge = stronger attraction
    • Stronger attraction = larger lattice energy
  • Of the two factors, ion charge is generally more important.

Which compound should have the largest lattice energy?

Ch 11

  • What is a Gas?

Vocab

Term Definition
bond energy amount of energy it takes to break one mole of a bond in a compound
standard state state of a material at a defined set of conditions
standard enthalpy change (ΔH°) the enthalpy change when all reactants and products are in their standard states
standard enthalpy of formation (ΔH°f) the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements
Born–Haber cycle a hypothetical series of reactions that represents the formation of an ionic compound from its constituent elements