1. Chapter 17 - Thermochemistry Heat and Chemical Change

2. Energy and Heat
Thermochemistry - concerned with heat changes that occur during chemical reactions.

3. Energy and Heat
Energy - capacity for doing work or supplying heat
weightless, odorless, tasteless
if within the chemical substances - called chemical potential energy

4. Energy and Heat
Gasoline contains a significant amount of chemical potential energy

5. Energy and Heat
Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.
only changes can be detected!
flows from warmer  cooler object

6. Exothermic and Endothermic Processes
Essentially all chemical reactions, and changes in physical state, involve either:
release of heat, or
absorption of heat

7. Exothermic and Endothermic
In studying heat changes, think of defining these two parts:
the system - the part of the universe on which you focus your attention
the surroundings - includes everything else in the universe

8. Exothermic and Endothermic
Together, the system and it’s surroundings constitute the universe
Thermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa.
Figure 17.2, page 506

9. Law of Conservation of Energy
The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.
All the energy is accounted for as work, stored energy, or heat.

10. Endothermic
Fig. 17.2a, p heat flowing into a system from it’s surroundings:
defined as positive
q has a positive value
called endothermic
system gains heat as the surroundings cool down

11. Endothermic

12. Exothermic
Fig. 17.2b, p heat flowing out of a system into it’s surroundings:
defined as negative
q has a negative value
called exothermic
system loses heat as the surroundings heat up

13. Exothermic

14. Exothermic and Endothermic
Every reaction has an energy change associated with it
Exothermic reactions release energy, usually in the form of heat.
Endothermic reactions absorb energy
Energy is stored in bonds between atoms

15. Heat Capacity and Specific Heat
A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.

16. Heat Capacity and Specific Heat
Used except when referring to food
a Calorie, written with a capital C, always refers to the energy in food
1 Calorie = 1 kilocalorie = 1000 cal.

17. Heat Capacity and Specific Heat
The calorie is also related to the joule, the SI unit of heat and energy
named after James Prescott Joule
4.184 J = 1 cal

18. Heat Capacity and Specific Heat
Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 oC

19. Heat Capacity and Specific Heat
Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)
often called simply “Specific Heat”
Note Table 17.1, page 508

20. Heat Capacity and Specific Heat
Water has a HUGE value, compared to other chemicals

21. Heat Capacity and Specific Heat
For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC)
Thus, for water:
it takes a long time to heat up, and
it takes a long time to cool off!
Water is used as a coolant!

22. Heat Capacity and Specific Heat
q = mass (g) x T x C
heat abbreviated as “q”
T = change in temperature
C = Specific Heat
Units are either J/(g oC) or cal/(g oC)

23. q = m x T x C C = q m x T Example
When 435J of heat is added to 3.4 g of olive oil at 21 °C, the temperature increases to 85 °C. What is the specific heat of olive oil?
q = m x T x C
C = q
m x T

24. C = q
m x T
C = J
3.4g x (85°C - 21°C)
C = 2 J/g°C

25. Calorimetry
Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.
The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter

26. Calorimetry
Foam cups are excellent heat insulators, and are commonly used as simple calorimeters
For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system.

27. Calorimeter

28. Calorimetry
Changes in enthalpy = H
q = H These terms will be used interchangeably in this textbook.

29. q = H = m x C x T Calorimetry Thus,
H is negative for an exothermic reaction.
H is positive for an endothermic reaction.

30. Calorimetry
Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system

31. Bomb Calorimeter

32. Calorimeter Problem
50ml of HCl
(in H2O)
Ti= 22.5°C

33. Calorimeter Problem
Add 50ml of NaOH
(in H2O)
Ti= 22.5°C

34. Calorimeter Problem
Chemical
Reaction

35. Calorimeter Problem
100ml of HCl and NaOH
(in H2O)
Tf= 26°C

36. Calorimeter Problem 1ml of H2O = 1g 100ml of H2O = 100g m = 100g
Ti= 22.5°C
Tf= 26°C
CH20= 4.18J/g°C

37. H = 1460J = 1.46kJ Calorimeter Problem H = m x C x T
H = m x C x (Tf –Ti)
= 100g x 4.18J/g°C x (26°C °C)
H = 1460J = 1.46kJ

38. C + O2 ® CO2
+ 395 kJ
Energy
Reactants
Products
C + O2
395kJ
C O2

39. In terms of bonds Breaking this bond will require energy.C O O C O
Breaking this bond will require energy. C O O C
Making these bonds gives you energy.
In this case making the bonds gives you more energy than breaking them.

40. Exothermic
The products are lower in energy than the reactants
Releases energy

41. CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products

42. The products are higher in energy than the reactants Absorbs energy
Endothermic
The products are higher in energy than the reactants
Absorbs energy

43. An equation that includes energy is called a thermochemical equation
Chemistry Happens in
MOLES
An equation that includes energy is called a thermochemical equation
CH4+2O2®CO2+2H2O kJ

44. CH4+2O2®CO2+2H2O kJ
1 mole of CH4 releases 802.2kJ of energy.
When you make kJ you also make 2 moles of water.

45. Thermochemical Equations
A heat of reaction is the heat change for the equation, exactly as written
The physical state of reactants and products must also be given.
Standard conditions for the reaction is kPa (1 atm.) and 25 oC

46. ΔH = 514 kJ CH4+2O2 ® CO2 + 2 H2O + 802.2 kJ 16.05g CH4 1 mol CH4
If grams of CH4 are burned completely, how much heat will be produced?
16.05g CH4
1 mol CH4
1mol CH4
802.2 kJ
10.3g CH4
ΔH = 514 kJ

47. CH4+2O2 ® CO2+2H2O kJ
How many liters of O2 at STP would be required to produce 23 kJ of heat?
22.4L x 2 = 44.8L produces 802.2kJ
23kJ
802.2kJ
= 0.029
0.029 x 44.8L =
VO2 = 1.30 L

48. CH4+2O2 ® CO2+2H2O kJ
How many grams of water would be produced with 506 kJ of heat?
2 mol of H2O produced with 802.2kJ
2 mol of H2O is 36g
506kJ
802.2kJ
x 36g
= 22.71g of H2O

49. Heat of Combustion
The heat from the reaction that completely burns 1 mole of a substance
Note Table 17.2 page 517

50. Summary, so far...

51. Enthalpy
The heat content a substance has at a given temperature and pressure
Can’t be measured directly because there is no set starting point
The reactants start with a heat content
The products end up with a heat content
So we can measure how much enthalpy changes

52. Enthalpy Symbol is H Change in enthalpy is DH (delta H)
If heat is released, the heat content of the products is lower
DH is negative (exothermic)
If heat is absorbed, the heat content of the products is higher
DH is positive (endothermic)

53. Energy
Change is down
DH is <0
Reactants
Products

54. Energy
Change is up
DH is > 0
Reactants
Products

55. Heat of Reaction
The heat that is released or absorbed in a chemical reaction
Equivalent to DH
C + O2(g) ® CO2(g) kJ
C + O2(g) ® CO2(g)
DH = kJ

56. Heat of Reaction
In thermochemical equation, it is important to indicate the physical state
H2(g) + 1/2O2 (g)® H2O(g) DH = kJ
H2(g) + 1/2O2 (g)® H2O(l) DH = kJ

57. Section 17.3 Heat in Changes of State

58. Heats of Fusion and Solidification
Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid

59. Heats of Fusion and Solidification
Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies

60. Heats of Fusion and Solidification
Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies
Thus, Hfus = -Hsolid
Note Table 17.3, page 522
Sample Problem 17-4, page 421

61. It takes 6.01kJ to melt 1 mol of ice.
Sample Problem
How many kJs of heat are required to melt a 10.0g popsicle at 0°C? Assume the popsicle has the same molar mass and heat of fusion as water.
It takes 6.01kJ to melt 1 mol of ice.
There are 18g in 1 mol of ice.
10g
18g/mol
x 6.01kJ/mol
= 3.34 kJ

62. Heats of Vaporization and Condensation
When liquids absorb heat at their boiling points, they become vapors.
Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.
Table 17.3, page 521

63. Heats of Vaporization and Condensation
Condensation is the opposite of vaporization.
Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses
Hvap = - Hcond

64. ΔHvaporization
ΔHfusion

65. Heats of Vaporization and Condensation
The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous
You can receive a scalding burn from steam when the heat of condensation is released!

66. Heats of Vaporization and Condensation
H20(l)  H20(g) m = 63.7g of liquid
Hvap = ?
63.7g
18g/mol
x 40.7kJ/mol
= 144kJ

67. Heat of Solution
Heat changes can also occur when a solute dissolves in a solvent.
Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance
Sodium hydroxide provides a good example of an exothermic molar heat of solution:

68. H2O(l)
NaOH(s)  Na1+(aq) + OH1-(aq)
Hsoln = kJ/mol
The heat is released as the ions separate and interact with water, releasing kJ of heat as Hsoln thus becoming so hot it steams!
Sample Problem 17-7, page 526

69. Section 17.4 Calculating Heat Changes

70. Hess’s Law
If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.
Called Hess’s law of heat summation
Example shown on page 529 for graphite and diamonds

71.  Diamond to Graphite C(s,diamond) C(s,graphite)
C(s,diamond) + O2  CO2 ΔH=-393.5kJ
C(s,graphite) + O2  CO2 ΔH=-395.4kJ
CO2  C(s,graphite) + O2 ΔH=395.45kJ

72. + C(s,diamond) + O2  CO2 ΔH=-393.5kJ
CO2  C(s,graphite) + O2 ΔH=395.45kJ +
C(s,diam)  C(s,graph) ΔH=-1.9kJ

73. If you turn an equation around, you change the sign:
Why Does It Work?
If you turn an equation around, you change the sign:
H2(g)+1/2 O2(g)®H2O(g) DH= kJ
H2O(g)®H2(g)+1/2 O2(g)DH = kJ
If you multiply the equation by a number, you multiply the heat by that number:
2H2O(g)®2H2(g)+O2(g) DH = kJ

74. Why does it work?
You make the products, so you need their heats of formation
You “unmake” the products so you have to subtract their heats.
How do you get good at this?

75. Standard Heats of Formation
The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions
Standard conditions: 25°C and 1 atm.
Symbol is ΔHf

76. Standard Heats of Formation
The standard heat of formation of an element = 0
This includes the diatomics
O2 N2 Cl2 etc……

77. What good are they?
Table 17.4, page 530 has standard heats of formation
The heat of a reaction can be calculated by:
subtracting the heats of formation of the reactants from the products
DHo
= (
Products) - (
Reactants)

78. CH4 (g) = - 74.86 kJ/mol O2(g) = 0 kJ/mol CO2(g) = - 393.5 kJ/mol
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
CH4 (g) = kJ/mol
O2(g) = 0 kJ/mol
CO2(g) = kJ/mol
H2O(g) = kJ/mol
DH= [ (-241.8)]-[ (0)]
DH= kJ

79. Examples
Sample Problem 17-8, page 531