Chapter 7 Making and Breaking of Bonds

1
Chapter 7Making and Breaking of Bonds
Chemical reactions are characterized by the
making and breaking of chemical bonds. One
possible consequence of a chemical reaction is a
transfer of energy from the system to the
surroundings.
2
Energy

• Hydrocarbons
• Carbohydrates
• reaction with oxygen to release energy.

3
Energy

• Kinetic Energy
• Potential Energy
• Transfer of energy
• Conversion of energy
• Thermodynamics

4
Energy

• Energy is released when making a chemical bond.
• Energy is consumed when breaking a chemical bond.

5
Heat

• "Heat is energy in transit." page 263.
• No energy transferred means no heat interaction.

6
Heat

• Heat has units of energy.
• Heat is not the same as temperature.
• In order for a heat interaction to occur between
  two systems, they must have different
  temperatures.
• This heat interaction can form the basis for a
  definition of temperature!

7
Heat and the Kinetic Molecular Theory

• System
• Surroundings
• Boundary

Figure 7.1
8
The First Law of Thermodynamics

• Energy is conserved.
• No exceptions have been observed.
• Yet.
• That's why it's called a law.

9
The First Law of Thermodynamics

• Energy can be exchanged between the system and
  the surroundings.
• Energy cannot appear or disappear.
• Energy entering or leaving a system can do so
  under two forms
• heat (q)
• work (w)

10
The First Law of Thermodynamics

• Energy transfers may occur under two separate
  special conditions
• system held at constant volume, or
• system held at constant pressure.

11
The First Law of Thermodynamics

• These special conditions have a profound effect
  on the form (q or w) in which energy can be
  transferred.
• Chemistry is typically done at constant pressure.
  Why?

12
The First Law of Thermodynamics

• When energy is exchanged in the form of heat with
  a system held at constant pressure, the heat
  energy, qP, is described with a new term,
  enthalpy (H).

13
The First Law of Thermodynamics

• In chemistry, the term enthalpy of reaction (?H)
  is used.
• It says how much energy is released or consumed
  in the form of heat if the reaction occurs under
  the condition of constant pressure.

14
The First Law of Thermodynamics
Table 7.2
15
State Functions

• State of a system
• P, T, n, V,
• Extensive properties of a system depend on the
  size of the system.
• Intensive properties of a system are independent
  of the system size.
• State functions are independent of a systems
  history.

16
State Functions

• H is a state function
• So are T, P, V, E, ...

17
The Enthalpy of a System

• Enthalpy, H, is a state function.
• Calorimetry
• Measures heat interactions associated with
  chemical and physical changes.

Figure 7.5
18
The Enthalpy of a System

• Calorimeters like the one shown measure ?E.
• A little math is used to determine ?H once ?E is
  measured.

19
Specific Heat

• Consider the following

Figure 7.6
20
Specific Heat

• Water and mercury have different specific heats.
• Specific heat is the energy required to raise the
  temperature of one gram of material one degree
  Celsius.
• units
• cal/C g
• J/C g

21
Specific Heat

• molar heat capacity
• specific heat molecular weight
• quantity of heat required to raise 1 mole of a
  substance 1 degree Celsius.

22
Specific Heat
Table 7.3
23
Enthalpies of Reaction

• ?H heat of reaction at constant P.
• may be gt0 endothermic reaction
• may be lt0 exothermic reaction

24
Enthalpies of Reaction

• Enthalpy of reaction (?H) is an extensive
  property. It changes with the stoichiometric
  coefficients.
• 2H2O(g) O2(g) ? 2H2O(g) ?H -483.64 kJ
• 4H2O(g) 2O2(g) ? 4H2O(g) ?H -967.28 kJ

25
Enthalpies of Reaction

• The sign of ?H changes for a reverse reaction.
• 2H2O(g) ? 2H2(g) O2(g) ?H 483.64 kJ
• 2H2(g) O2(g) ? 2H2O(g) ?H -483.64 kJ

26
Enthalpies of Reaction

• The heat released for any amount of product or
  reactant can be determined from a single
  thermochemical equation.
• 2H2(g) O2(g) ? 2H2O(g) ?H -483.64 kJ
• How much heat will be released when 50 grams of
  H2(g) are consumed?

27
Enthalpies of Reaction
28
Enthalpy as a State Function

• H is a state function.
• ?H Hfinal - Hinitial
• independent of path
• a. convert reactant into atoms
• b. make products from same atoms
• c. no extra or missing atoms!

29
Enthalpy as a State Function

• Reactions do not proceed this way.
• It doesn't matter when calculating ?H.
• H is a state function.

30
Standard-State Enthalpies of Reaction

• For a reaction carried out at a pressure of 1
  bar, ?H ?H.
• ?H is called the standard-state enthalpy of
  reaction.
• 1 bar is part of the definition of standard
  conditions.
• Most tabulated enthalpies of reaction are ?H.

31
Calculating Enthalpies of Reaction

• Easier to calculate ?H than to measure it.
• These are calculations, not estimates.
• Break all the reactant bonds.
• Form all the product bonds.
• ?H the difference in energy between these two
  processes.

32
Enthalpies of Atom Combination

• N(g) 3 H(g) ? NH3(g)
• Notice these are not the elemental forms.
• If these 4 atoms combine to produce one ammonia
  molecule, 1171.76 kJ of energy will be released.

33
Enthalpies of Atom Combination

• N(g) 3 H(g) ? NH3(g)
• Reaction is called atom combination.
• Not meant to reflect actual mechanism.
• The enthalpy associated with it is called the
  enthalpy of atom combination, ?Hac.
• In the above example, ?Hac -1171.76 kJ.
• Why is it negative?

34
Enthalpies of Atom Combination

• The reverse reaction
• NH3(g) ? N(g) 3 H(g)
• is called atomization. The enthalpy change is
  called the enthalpy of atomization and in this
  example equals 1171.76 kJ.

35
Enthalpies of Atom Combination

• Can be used to study physical processes.
• How much heat is required to accomplish the
  following?
• CH3OH(l) ? CH3OH(g)

36
Enthalpies of Atom Combination

• Can be used to study physical processes
• How much heat is required to accomplish the
  following?
• CH3OH(l) ? CH3OH(g)
• ?Hac -2037.11 kJ for CH3OH(g)
• ?Hac -2075.11 kJ for CH3OH(l)

37
Enthalpies of Atom Combination

• Can be used to study physical processes
• CH3OH(l) ? CH3OH(g)

?H -2037.11 kJ - (-2075.11 kJ) 38.00 kJ
38
Enthalpies of Atom Combination

• Can be used to study chemical processes
• 2H2(g) O2(g) ? 2H2O(g) ?H ?

39
Enthalpies of Atom Combination

• Can be used to study chemical processes
• 2H2(g) O2(g) ? 2H2O(g)
• ?H2?Hac(H2O(g)) -2?Hac (H2(g))-1?Hac
  (O2(g) )

40
Enthalpies of Atom Combination

• Can be used to study chemical processes
• 2H2(g) O2(g) ? 2H2O(g)
• ?H2?Hac(H2O(g)) -2?Hac (H2(g))-1?Hac
  (O2(g) )
• From Appendix B.13
• ?H2(-926.29 kJ) -2(-435.30 kJ)-1(-498.340
  kJ)
• ?H -483.64 kJ

41
Enthalpies of Atom Combination

• Can be used to study chemical processes.
• 2H2(g) O2(g) ? 2H2O(g)
• Established this as exothermic.
• Calculated that 483.64 kJ of energy will be
  released for each mole of O2(g) consumed.

42
Using Enthalpies of Atom Combination to Probe
Chemical Reactions

• Isomers

43
Using Enthalpies of Atom Combination to Probe
Chemical Reactions

• Knowing ?Hac for each isomer allows for the
  calculation of the enthalpy change associated
  with the transformation

44
Using Enthalpies of Atom Combination to Probe
Chemical Reactions

• Knowing ?Hac gives insight into average bond
  strengths.
• These provide a microscopic interpretation of
  overall ?H for a reaction.
• endothermic or exothermic character
• magnitude of ?H

45
Using Enthalpies of Atom Combination to Probe
Chemical Reactions

• 4HF(g) SiO2(g) ? SiF4(g) 2H2O(g) ?H
  -103.4 kJ
• 4HCl(g) SiO2(g) ? SiCl4(g) 2H2O(g) ?H
  139.6 kJ
• Why the difference in sign?

46
Using Enthalpies of Atom Combination to Probe
Chemical Reactions

• 4HF(g) SiO2(g) ? SiF4(g) 2H2O(g) ?H
  -103.4 kJ
• 4HCl(g) SiO2(g) ? SiCl4(g) 2H2O(g) ?H
  139.6 kJ
• Why the difference in sign?
• The Si-F bond is stronger than the Si-Cl
  bond by an amount greater than the difference in
  bond strength between H-F and H-Cl shown on the
  next slide.

47
Bond Length and the Enthalpy of Atom Combination
Table 7.5
48
Bond Length and the Enthalpy of Atom Combination

• Longer bonds tend to be weaker bonds.
• Multiple bonds tend to be stronger than single
  bonds.
• This was also covered in section 4.8.

49
Hess's Law

• An alternative method for calculating ?H.
• Does not use ?Hac.
• Takes advantage of H being a state function.

50
Hess's Law

• Desired reactions are constructed from known
  reactions.
• ?H from known reactions combined in same way to
  calculate ?H for the desired reaction.
• Trial and error method!

51
Hess's Law

• C(s) H2O(g) ? CO(g) H2(g) ?H ?
• from
• C(s) ½O2(g) ? CO(g) ?H -110.53 kJ
• H2(g) ½O2(g) ? H2O(g) H2(g) ?H -214.82 kJ

52
Enthalpies of Formation

• Enthalpy of formation, ?Hf
• Combined in same way as ?Hac.
• Used to calculate ?H for a reaction.
• Tabulated in Appendix B.16.
• Don't mix ?Hf and ?Hac in a calculation.
• Use one set or the other.

53
Enthalpies of Formation

• Defined as enthalpy change associated with the
  formation of one mole of a substance under
  standard conditions (1 bar, T, ...) from the
  elements in their thermodynamically stable form
  at T.

54
Enthalpies of Formation

• At 25 C and 1 bar
• Oxygen is O2(g).
• Carbon is C(solid, graphite).
• The rest can be determined from B.16 by looking
  for the entry with ?Hf 0.
• Why is the entry with ?Hf 0 the
  thermodynamically stable form?