1. Chapter 5 Thermochemistry
Jessica Contino

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.
Energy used to cause the temperature of an object to rise is called heat.

3. Potential Energy
Potential energy is energy an object possesses by virtue of its position or chemical composition.

4. Kinetic Energy
Kinetic energy is energy an object possesses by virtue of its motion. 1 2
KE =  mv2

5. Units of Energy kg m2 1 J = 1  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
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. Solution Sample Exercise 5.1 Describing and Calculating Energy Changes
A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 feet) and then drops the ball back to the ground. (a) What happens to the potential energy of the bowling ball as it is raised from the ground? (b) What quantity of work, in J, is used to raise the ball? (c) After the ball is dropped, it gains kinetic energy. If we assume that all of the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the speed of the ball at the instant 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.)
Solution
Analyze: We need to relate the potential energy of the bowling ball to its position relative to the ground. We then need to establish the relationship between work and the change in potential energy of the ball. Finally, we need to connect the change in potential energy when the ball is dropped with the kinetic energy attained by the ball.
Plan: We can calculate the work done in lifting the ball by using Equation 5.3: w = F  d. The kinetic energy of the ball at the moment of impact equals its initial potential energy. We can use the kinetic energy and Equation 5.1 to calculate the speed, v, at impact.
Solve:
(a) Because the bowling ball is raised to a greater height above the ground, its potential energy increases.
(b) The ball has a mass of 5.4 kg, and it is lifted a distance of 1.6 m. To calculate the work performed to raise the ball, we use both Equation 5.3 and F = m  g for the force that is due to gravity:
© 2009, Prentice-Hall, Inc.

9. Solution (continued) Practice Exercise
Sample Exercise 5.1 Describing and Calculating Energy Changes
Solution (continued)
Thus, the bowler has done 85 J of work to lift the ball to a height of 1.6 m.
(c) When the ball is dropped, its potential energy is converted to kinetic energy. At the instant just before the ball hits the ground, we assume that the kinetic energy is equal to the work done in part (b), 85 J:
We can now solve this equation for v:
Check: Work must be done in part (b) to increase the potential energy of the ball, which is in accord with our experience. The units are appropriate in both parts (b) and (c). The work is in units of J and the speed in units of m/s. In part (c) we have carried an additional digit in the intermediate calculation involving the square root, but we report the final value to only two significant figures, as appropriate.
Comment: A speed of 1 m/s is roughly 2 mph, so the bowling ball has a speed greater than 10 mph upon impact.
What is the kinetic energy, in J, of (a) an Ar atom moving with a speed of 650 m/s, (b) a mole of Ar atoms moving with a speed of 650 m/s (Hint: 1 amu = 1.66  10–27 kg)
Answers: (a) 1.4  10–20 J, (b) 8.4  103 J
Practice Exercise
© 2009, Prentice-Hall, Inc.

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

11. Conversion of Energy Energy can be converted from one type to another.
For example, the cyclist above has potential energy as she sits on top of the hill.

12. Conversion of Energy
As she coasts down the hill, her potential energy is converted to kinetic energy.
At the bottom, all the potential energy she had at the top of the hill is now kinetic energy.

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
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.

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

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. Solution Practice Exercise
Sample Exercise 5.2 Relating Heat and Work to Changes of Internal Energy
Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement like that in Figure 5.3. Substances A and B react to form a solid product: 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?
Solution
Analyze: The question asks us to determine ΔE, given information about q and w.
Plan: We first determine the signs of q and w (Table 5.1) and then use Equation 5.5, ΔE = q + w, to calculate ΔE.
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,
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.
Comment: You can think of this change as a decrease of 670 J in the net value of the system’s energy bank account (hence the negative sign); 1150 J is withdrawn in the form of heat, while 480 J is deposited in the form of work. Notice that as the volume of the gases decreases, work is being done on the system by the surroundings, resulting in a deposit of energy.
Calculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.
Answer: +55 J
Practice Exercise
© 2009, Prentice-Hall, Inc.

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

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

23. State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

24. State Functions
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
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 = -PV

29. Enthalpy
If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, 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

30. 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

31. 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.

32. Endothermicity and Exothermicity
A process is endothermic when H is positive.

33. Endothermicity and Exothermicity
A process is endothermic when H is positive.
A process is exothermic when H is negative.

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

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

36. 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.

37. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

38. 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.

39. 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.

40. Heat Capacity and Specific Heat
Specific heat, then, is
Specific heat =
heat transferred
mass  temperature change
s = q
m  T

41. Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

42. Constant Pressure Calorimetry
Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation:
q = m  s  T

43. 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.

44. 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.

45. Hess’s Law
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.

46. Hess’s Law
Hess’s law states that “[i]f 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.”

47. Hess’s Law
Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

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

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

50. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
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)

51. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
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)

52. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
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)

53. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is:
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)

54. 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.

55. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] – [( kJ) + (0 kJ)] = ( kJ) – ( kJ) = kJ

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

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