1. Chapter 5 Thermochemistry

2. Energy Energy is the ability to do work or transfer heat.
Thermodynamics, which is the study of energy and its transformations.
Specifically, thermochemistry is the study of chemical reactions and the energy changes that involve heat.

3. Chemical Energy is Mainly Potential Energy
The most important form of potential energy in molecules is electrostatic potential energy, Eel:
Reminder: the unit of energy commonly used is the Joule:

4. Attraction between Ions
Electrostatic attraction is seen between oppositely charged ions.
Energy is released when chemical bonds are formed; energy is consumed when chemical bonds are broken.
Think of the energy associated with chopping a log

5. First Law of Thermodynamics
Energy can be converted from one form to another, but it is neither created nor destroyed.
To heat your home, chemical energy needs to be converted to heat.
Sunlight is converted to chemical energy in green plants.
There are MANY more examples of conversion of energy.

6. Definitions: System and Surroundings
The portion of the universe that we single out to study is called the system (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder, piston and everything beyond).

7. Types of Systems Note that this is a CLOSED system.
Open System: a region of the universe being studied that can exchange heat AND mass with its surroundings.
Closed System: a region of the universe being studied that can ONLY exchange heat with its surroundings (NOT mass)
Isolated System: a region of the universe that can NOT exchange heat or mass with its surroundings
Note that this is a CLOSED system.

8. Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we use E to represent it.
We generally don’t know E, only how it CHANGES (ΔE).

9. Internal Energy ΔE = Efinal − Einitial
By definition, the change in internal energy, ΔE, is the final energy of the system minus the initial energy of the system:
ΔE = Efinal − Einitial

10. Changes in Internal Energy
IF: ΔE > 0, Efinal > Einitial
the system absorbed energy from the surroundings.

11. Changes in Internal Energy
IF: ΔE < 0, Efinal < Einitial
the system released energy to the surroundings.

12. Thermodynamic Quantities Have Three Parts
A number
A unit
A sign
Note about the sign:
A positive ΔE results when the system gains energy from the surroundings.
A negative ΔE results when the system loses energy to the surroundings.

13. Changes in Internal Energy
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is, ΔE = q + w.

14. ΔE, q, w, and Their Signs

15. Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic.

16. Exchange of Heat between System and Surroundings
When heat is released by the system into the surroundings, the process is exothermic.

17. State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.
However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.
In the system below, the water could have reached room temperature from either direction.

18. State Functions Therefore, internal energy is a state function.
It depends only on the present state of the system, not on the path by which the system arrived at that state.
And so, ΔE depends only on Einitial and Efinal.

19. State Functions However, q and w are not state functions.
Whether the battery is shorted out or is discharged by running the fan, its ΔE is the same, but q and w are different in the two cases.
What are some other state functions? E, P, t, V

20. Work
Usually the only work done by chemical or physical change is the mechanical work associated with a change in volume of gas.
We know: w = F x d = F x delta h
Since P = F/A we can write F = P x A
Plugging in for F above we have w = P x A x delta h
But we also know delta V = A x delta h which can be rearranged to delta H = delta V/A
Plugging in delta H for w above we have w = P x A x deltaV/A which simplifies to w = P x delta V
Since the system is doing work we give w a negative sign. Thus w = - P x delta V
Plugging in for A in F = P x A we get F = P x So F = P x A x d = P x A x delta H = P x delta V not taking into account the sign.

21. Work
We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PΔV
The work is NEGATIVE because it is work done BY the system.

22. Enthalpy
If a process takes place at constant pressure (and we usually work at atmospheric pressure) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy (H) of the system.
Enthalpy is the internal energy plus the product of pressure and volume:
We define enthalpy as H = E + PV
H = E + PV

23. Enthalpy
When the system changes at constant pressure, the change in enthalpy, ΔH, is
ΔH = Δ (E + PV)
This can be written
ΔH = ΔE + PΔV

24. Enthalpy
Since ΔE = q + w and w = −PΔV, we can substitute these into the enthalpy expression:
ΔH = ΔE + PΔV
ΔH = (q + w) − w
ΔH = q
So, at constant pressure, the change in enthalpy is the heat gained or lost.

25. Endothermic and Exothermic
A process is endothermic when ΔH is positive.
A process is exothermic when ΔH is negative.

26. Enthalpy of Reaction
The change in enthalpy, ΔH, is the enthalpy of the products minus the enthalpy of the reactants:
ΔHrxn = Hproducts − Hreactants

27. Enthalpy of Reaction
This quantity, ΔHrxn, is called the enthalpy of reaction, or the heat of reaction.

28. The Truth about Enthalpy
Enthalpy is an extensive property.
The amount of DH is proportional to the amount of reactant consumed.
The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to ΔH for the reverse reaction.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = – kJ
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g) H = kJ
The enthalpy change for a reaction depends on the states of the reactants and the products.
2H2O(l)  2H2O(g) DH = +88 kJ

29. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure ΔH through calorimetry, the measurement of heat flow.
The instrument used to measure heat flow is called a calorimeter.

30. Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1 °C) is its heat capacity. If the amount of the substance heated is one gram, it is the specific heat. If the amount is one mole, it is the molar heat capacity.
Thus heat capacity is heat per degree and specific heat is heat per degree per gram

31. Heat Capacity q = C Δt q = amount of heat transferred (J)
C = heat capacity (J/oC)
Δt = change in temperature (oC)
Heat capacity is an extensive property and depends on the amount of substance present

32. Heat Capacity
A copper mug has a heat capacity of 77.5 J/˚C. After adding hot water to the mug, the temperature of the mug changed from 25.0 ˚C to 85.0 ˚C. How much heat did the mug absorb from the water?
q = CΔt = (77.5 J/˚C)(85.0 oC – 20.0 oC)
q = (77.5 J/˚C)(60.0 oC)
q = J

33. Specific Heat
Specific heat is an intensive property that does not depend on the amount of substance
units of J / goC
s = specific heat
Cs = heat capacity
m = mass of substance
q = heat transferred
Δt= temperature change
Different substances and phases of matter have different specific heats
Some text books use the symbol s to signify specific heat.
C = q/delta t
More often Cs is written as q m delta t 33

34. Specific Heat The specific heat of water is 4.184 J / goC.
How much heat is required to heat 20.0 g of water from 15 oC to 25 oC?
q = mCsΔt = (20.0 g)(4.184 J / goC)(25 oC – 15 oC)
q = J
q = heat transferred
m = mass of substance
Cs = specific heat
Δt = change in temperature 34

35. Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter, the heat change for the system can be found by measuring the heat change for the water in the calorimeter.
The specific heat for water is J/g∙K.
We can calculate ΔH for the reaction with this equation:
qsoln = Cs × msoln × ΔT = –qrxn
Remember delta H = q

36. Example: Specific Heat
Calculate the specific heat of a metal if it takes 235 J to raise the temperature of a g sample by 2.53 °C.
The symbol q is used to mean “quantity of heat”.

37. Example
A 0.50 g sample of NH4NO3 is dissolved in 25.0 g of water in a coffee-cup calorimeter and the water temperature decreases from to 24.32oC. If we assume that the heat absorbed in the reaction comes from water, find q for the system.
NH4NO3 (s)  NH4+ (aq) + NO3- (aq)

38. Constant Pressure Calorimetry
g Al (s) was heated to 97.2 C then dumped into g H2O at 21.3 C. The temperature of the H2O rose to 24.2 C. What is the specific heat of Al (s)?

39. Constant Pressure Calorimetry
Heat lost by the system is equal to the heat gained by the surroundings
Heat lost by the Al (s) is equal to the heat gained by the H2O ()
-qmet = qw = g x J/g.oC x 2.9oC
Typically delta H is given as kJ/mol
-qmet = qw = 988 J
qmet = -988 J

40. Constant Pressure Calorimetry
g Al (s) was heated to 97.2 C then dumped into g H2O at 21.3 C. The temperature of the H2O rose to 24.2 C. What is the specific heat of Al (s) ?
-qmet = 988 J
qmet = -988 J = g x Cs x (24.2C-97.2C)
qmet = -988 J = g x Cs x (-73C)
Cs = J/g·C

41. Hess’s Law
ΔH is well known for many reactions, and it is inconvenient to measure ΔH for every reaction in which we are interested.
However, we can calculate ΔH using published ΔH values and the properties of enthalpy.

42. Hess’s Law
Hess’s law: If a reaction is carried out in a series of steps, ΔH for the overall reaction equals the sum of the enthalpy changes for the individual steps.
Because H is a state function, for a particular set of reactants and products, ΔH is the same whether the reaction takes place in one step or in a series of steps.

43. Hess’s Law Example Calculate H for 2 C (s, gr) + H2(g)  C2H2(g)
Given the following:
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l ) H = – kJ
C(s, gr) + O2(g)  CO2(g) H = –393.5 kJ
H2(g) + ½O2(g)  H2O(l ) H = –285.8 kJ

44. Hess’s Law Example
2CO2(g) + H2O(l )  C2H2(g) + 5/2O2(g) H°rxn = – (– kJ) = kJ
2C(s, gr) + 2O2(g)  2CO2(g) H°rxn =(2–393.5 kJ) = –787.0 kJ
H2(g) + ½O2(g)  H2O(l ) H°rxn = –285.8 kJ
2CO2(g) + H2O(l ) + 2C(s, gr) + 2O2(g) + H2(g) + ½O2(g)  C2H2(g) + 5/2O2(g) + 2CO2(g) + H2O(l )
2C(s, gr) + H2(g)  C2H2(g) H°rxn = kJ

45. Enthalpies of Formation
An enthalpy of formation, ΔHf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

46. Standard Enthalpies of Formation
Standard enthalpies of formation, ΔHf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

47. Standard State
Most stable form and physical state of element at 1 atm (1 bar) and 25 °C (298 K)
Element
Standard state O
O2(g) C
C (s, gr) H
H2(g) Al
Al(s) Ne
Ne(g)
Note: All Hf° of elements in their standard states = 0
Forming element from itself.

48. Calculation of ΔH C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
Imagine this as occurring in three steps:
1) Decomposition of propane to the elements:
C3H8(g) C(graphite) + 4 H2(g)

49. Calculation of ΔH C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
Imagine this as occurring in three steps:
2) Formation of CO2:
3 C(graphite) + 3 O2(g) CO2(g)

50. Calculation of ΔH C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
Imagine this as occurring in three steps:
3) Formation of H2O:
4 H2(g) + 2 O2(g) H2O(l)

51. Calculation of ΔH C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
So, all steps look like this:
C3H8(g) C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) CO2(g)
4 H2(g) + 2 O2(g) H2O(l)

52. Calculation of ΔH C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
The sum of these equations is the overall equation!
C3H8(g) C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) CO2(g)
4 H2(g) + 2 O2(g) H2O(l)
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(l)

53. Calculation of ΔH We can use Hess’s law in this way:
ΔH = ΣnΔHf,products – ΣmΔHf°,reactants
where n and m are the stoichiometric coefficients.

54. Calculation of ΔH using Values from the Standard Enthalpy Table
C3H8(g) + 5 O2(g) CO2(g) + 4 H2O(l)
ΔH = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)]
= [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)]
= (− kJ) – (− kJ) = − kJ
Values for standard enthalpies of formation are found in Appendix C in your text.

55. Bond Enthalpy
The enthalpy associated with breaking one mole of a particular bond in a gaseous substance.
Consider this reaction: CH4 + Cl2  CH3Cl + HCl

56. Bond Enthalpy
The bond enthalpy is always positive because energy is required to break chemical bonds.
Energy is always released when a bond forms between gaseous fragments.
The greater the bond enthalpy, the stronger the bond.

57. Bond Enthalpies and Enthalpy of Reaction
ADD bond energy for ALL bonds made (+)
SUBTRACT bond energy for ALL bonds broken (–)
The result is an estimate of ΔH.
ΔHrxn = Σ (bond enthalpies – Σ (bond enthalpies
of bonds broken) of bonds formed)

58. Bond Enthalpies and Enthalpy of Reaction
So, we can predict whether a chemical reaction will be endothermic or exothermic using bond energies.
DHrxn =  (enthalpies of bonds broken)
-  (enthalpies of bonds formed)

59. Bond Enthalpies and Enthalpy of Reaction
Bonds broken C-H 413 kJ Cl-Cl 242 kJ Bonds formed C-Cl 328 kJ H-Cl 431 kJ
DHrxn =  (enthalpies of bonds broken)
-  (enthalpies of bonds formed)
DHrxn = (413 kJ kJ) – (328 kJ kJ) = -104 kJ

60. Bond Enthalpies and Enthalpy of Reaction
Bonds broken C-H 413 kJ Cl-Cl 242 kJ Bonds formed C-Cl 328 kJ H-Cl 431 kJ
CH4 + Cl2  CH3Cl + HCl
DHrxn =  (enthalpies of bonds broken)
-  (enthalpies of bonds formed)
DHrxn = (413 kJ kJ) – (328 kJ kJ) = -104 kJ