1. Chapter 5 Thermochemistry
Lecture Presentation
Chapter 5 Thermochemistry
James F. Kirby
Quinnipiac University
Hamden, CT

2. Energy Energy is the ability to do work or transfer heat.
Energy used to cause an object that has mass to move is called work.
Work = force x distance
W = f x d
Energy used to cause the temperature of an object to rise is called heat.
Thermodynamics: the study of energy transformations
Thermochemistry applies the field to chemical reactions, specifically.

3. Kinetic Energy
Kinetic energy is energy an object possesses by virtue of its motion:

4. Potential Energy
Potential energy is energy an object possesses by virtue of its position or chemical composition.
The most important form of potential energy in molecules is electrostatic potential energy, Eel:
Q = charge
D = distance

5. Units of Energy 1 J = 1 kg·m2/s2
The SI unit of energy is the joule (J):
An older, non-SI unit is still in widespread use, the calorie (cal):
1 cal = J
(Note: this is not the same as the calorie of foods; the food calorie is 1 kcal!)
1 J = 1 kg·m2/s2

6. Definitions: System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder and piston).

7. Definitions: Work
Energy used to move an object over some distance is work:
w = F  d
where w is work, F is the force, and d is the distance over which the force is exerted.

8. Heat Energy can also be transferred as heat.
Heat flows from warmer objects to cooler objects.

9. Conversion of Energy Energy can be converted from one type to another.
The cyclist has potential energy as she sits on top of the hill.
As she coasts down the hill, her potential energy is converted to kinetic energy until the bottom, where the energy is converted to kinetic energy.

10. A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it.
What happens to the potential energy of the ball as it is raised?
What quantity of work, in J, is used to raise the ball?
After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground?
(Note: The force due to gravity is F = m × g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m ⁄ s2.)

11. Solve
(a) Because the ball is raised above the ground, its potential energy relative to the ground increases.
(b) The ball has a mass of 5.4 kg and is lifted 1.6 m. To calculate the work performed to raise the ball, we use Equation 5.3 and F = m × g for the force that is due to gravity:
W = F × d = m × g × d
= (5.4 kg)(9.8 m ⁄ s2)(1.6 m) = 85 kg-m2 ⁄ s2 = 85 J
Thus, the bowler has done 85 J of work to lift the ball to a height of 1.6 m.

12. (c) When the ball is dropped, its potential energy is converted to kinetic energy. We assume that the kinetic energy just before the ball hits the ground is equal to the work done in part (b), 85 J:
Ek = mv2 = 85 J = 85 kg-m2 ⁄ s2
We can now solve this equation for v:

13. First Law of Thermodynamics
Energy is neither created nor destroyed.
In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

14. Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

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

16. Changes in Internal Energy
If E > 0, Efinal > Einitial
Therefore, the system absorbed energy from the surroundings.
This energy change is called endergonic or endothermic.

17. Changes in Internal Energy
If E < 0, Efinal < Einitial
Therefore, the system released energy to the surroundings.
This energy change is called exergonic or exothermic.

18. 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, E = q + w.

19. E, q, w, and Their Signs

20. Example:
Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in the Figure and react to form a solid product C(s): A(g) + B(g) → C(s).
As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system?

21. Solution
Solve Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q = −1150 J and w = 480 kJ. Thus,
ΔE = q + w = (−1150 J) + (480 J) = −670 J
The negative value of ΔE tells us that a net quantity of 670 J of energy has been transferred from the system to the surroundings.

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

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

24. 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.

25. 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, E depends only on Einitial and Efinal.

26. State Functions However, q and w are not state functions.
Whether the battery is shorted out or is discharged by running the fan, its E is the same.
But q and w are different in the two cases.

27. Work
Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas).

28. Work
5.3
the work involved in the expansion or compression of gases is called Pressure-Volume Work (P-V work). When pressure is constant, the sign and magnitude of the work are given by:
w = −PV

29. W = -0 atm x 3.8 L = 0 L•atm = 0 joules
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?
w = -P DV
(a)
DV = 5.4 L – 1.6 L = 3.8 L
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = L•atm
w = L•atm x
101.3 J
1L•atm
= J
6.3

30. Enthalpy
If pressure is constant, we can account for heat flow during the process by measuring the enthalpy of the system.
Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV

31. Enthalpy
When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
This can be written
H = E + PV

32. Enthalpy
Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:
H = E + PV
H = (q + w) − w
H = q
So, at constant pressure, the change in enthalpy is the heat gained or lost.

33. Endothermic and Exothermic
A process is endothermic when H is positive.
A process is exothermic when H is negative.

34. Example:
A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is L, and the final volume is L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? (1 L-atm = J)

35. Solve The volume change is
ΔV = Vfinal − Vinitial = L − L = L
Thus, the quantity of work is
w = −PΔV = −(1.35 atm)(0.730 L) = − L-atm
Converting L-atm to J, we have

36. Enthalpy of Reaction
5.4
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants

37. Enthalpy of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

38. The Truth about Enthalpy
Enthalpy is an extensive property.
H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
H for a reaction depends on the state of the products and the state of the reactants.

39. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ
6.4

40. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= 6470 kJ
6.4

41. C2H5OH(l) + 3O2→ (g) 2CO2(g) + 3H2O(l) ΔH = −555 kJ
Sample Exercise 5.4 Determining the Sign of ΔH
Indicate the sign of the enthalpy change, ΔH, in the following processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic:
(a) An ice cube melts;
(b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O.
Sample Exercise 5.5 Relating ΔH to Quantities of Reactants and Products
The complete combustion of ethanol, C2H5OH (FW = 46.0 g ⁄ mol), proceeds as follows:
C2H5OH(l) + 3O2→ (g) 2CO2(g) + 3H2O(l) ΔH = −555 kJ
What is the enthalpy change for combustion of 15.0 g
of ethanol?

42. 5.5
Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.
The instrument used to measure heat flow is called a calorimeter.

43. 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, usually given for one mole of the substance.

44. Heat Capacity and Specific Heat
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C).
Units:
J/g·˚C or J/g·K

45. Heat Capacity and Specific Heat
Specific heat, then, is

46. 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 well known (4.184 J/g∙K).
We can calculate H for the reaction with this equation:
q = m  Cs  T

47. Examples:
a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to 98 °C (near its boiling point)?
(b) What is the molar heat capacity of water?
2) (a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J ⁄ g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 °C.
(b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

48. Coffee-Cup Calorimetry Example:
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 to 27.5 °C. Calculate the enthalpy change for the reaction in kJ ⁄ mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g ⁄ mL, and that its specific heat is 4.18 J ⁄ g-K.

49. Coffee-Cup Calorimetry Example 2:
A lead pellet having a mass of g at 89.98˚C was placed in a constant-pressure calorimeter of negligible heat capacity containing mL of water. The water temperature rose from 22.50˚C to 23.17˚C. What is the specific heat of the lead pellet?

50. Bomb Calorimetry
Reactions can be carried out in a sealed “bomb” such as this one.
The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.
qrxn = – Ccal × ∆T

51. Bomb Calorimetry
Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.
For most reactions, the difference is very small.

52. 2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l)
The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2(g), and H2O(l):
2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l)
When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from to °C.
In a separate experiment the heat capacity of the calorimeter is measured to be kJ ⁄ °C. Calculate the heat of reaction for the combustion of a mole of CH6N2.

53. Hess’s Law
5.6
H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.
However, we can estimate H using published H values and the properties of enthalpy.

54. Hess’s Law
Hess’s law: If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.
Because H is a state function, the total enthalpy change depends only on the initial state (reactants) and the final state (products) of the reaction.

55. Enthalpies of Formation
5.7
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its constituent elements in their elemental forms.

56. Standard Enthalpies of Formation
Standard enthalpies of formation, ∆Hf°, are measured under standard conditions (25 ºC and 1.00 atm pressure).

57. Calculation of H 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)  3 C(graphite) + 4 H2(g)

58. Calculation of H 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) 3 CO2(g)

59. Calculation of H 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)  4 H2O(l)

60. Calculation of H C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
So, all steps look like this:
C3H8(g)  3 C(graphite) + 4 H2(g)
3 C(graphite) + 3 O2(g) 3 CO2(g)
4 H2(g) + 2 O2(g)  4 H2O(l)

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

62. Calculation of H We can use Hess’s law in this way:
H = nHf,products – mHf°,reactants
where n and m are the stoichiometric coefficients.

63. Calculation of H using Values from the Standard Enthalpy Table
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(− kJ) + 5(0 kJ)]
= [(− kJ) + (− kJ)] – [(− kJ) + (0 kJ)]
= (− kJ) – (− kJ) = − kJ

64. Energy in Foods
Most of the fuel in the food we eat comes from carbohydrates and fats.

65. Energy in Fuels
The vast majority of the energy consumed in this country comes from fossil fuels.

66. Other Energy Sources
Nuclear fission produces 8.5% of the U.S. energy needs.
Renewable energy sources, like solar, wind, geothermal, hydroelectric, and biomass sources produce 7.4% of the U.S. energy needs.