1. Thermochemistry Heat and Chemical Change
Charles Page High School
Dr. Stephen L. Cotton

2. Section 11.1 The Flow of Energy - Heat
OBJECTIVES:
Explain the relationship between energy and heat.

3. Section 11.1 The Flow of Energy - Heat
OBJECTIVES:
Distinguish between heat capacity and specific heat.

4. Energy and Heat
Thermochemistry - concerned with heat changes that occur during chemical reactions
Energy - capacity for doing work or supplying heat
weightless, odorless, tasteless
if within the chemical substances- called chemical potential energy

5. Energy and Heat
Gasoline contains a significant amount of chemical potential energy
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 Processes
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 Processes
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 11.3, page 294

9. Exothermic and Endothermic Processes
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. Exothermic and Endothermic Processes
Fig. 11.3a, 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. Exothermic and Endothermic Processes
Fig. 11.3b, 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

12. Exothermic and Endothermic
Fig. 11.4, page on the left, the system (the people) gain heat from it’s surroundings (the fire)
this is endothermic
On the right, the system (the body) cools as perspiration evaporates, and heat flows to the surroundings
this is exothermic

13. Exothemic 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

14. 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.
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.

15. 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
Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 oC

16. 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 11.2, page 296
Water has a HUGE value, compared to other chemicals

17. 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!
Note Figure 11.7, page 297

18. Heat Capacity and Specific Heat
To calculate, use the formula:
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)
Sample problem 11-1, page 299

19. Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES:
Construct equations that show the heat changes for chemical and physical processes.

20. Section 11.2 Measuring and Expressing Heat Changes
OBJECTIVES:
Calculate heat changes in chemical and physical processes.

21. 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

22. Calorimetry
Foam cups are excellent heat insulators, and are commonly used as simple calorimeters
Fig. 11.8, page 300
For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system

23. Calorimetry Changes in enthalpy = H
q = H These terms will be used interchangeably in this textbook
Thus, q = H = m x C x T
H is negative for an exothermic reaction
H is positive for an endothermic reaction (Note Table 11.3, p.301)

24. Calorimetry
Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system
Figure 11.9, page 301
Sample 11-2, page 302

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

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

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

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

29. Endothermic The products are higher in energy than the reactants
Absorbs energy
Note Figure 11.11, page 303

30. Chemistry Happens in MOLES
An equation that includes energy is called a thermochemical equation
CH4 + 2O2 ® CO2 + 2H2O kJ
1 mole of CH4 releases kJ of energy.
When you make kJ you also make 2 moles of water

31. 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

32. CH4 + 2 O2 ® CO2 + 2 H2O kJ
If grams of CH4 are burned completely, how much heat will be produced?
1 mol CH4
802.2 kJ
10. 3 g CH4
16.05 g CH4
1 mol CH4
= 514 kJ

33. How many grams of water would be produced with 506 kJ of heat?
CH4 + 2 O2 ® CO2 + 2 H2O kJ
How many liters of O2 at STP would be required to produce 23 kJ of heat?
How many grams of water would be produced with 506 kJ of heat?

34. Summary, so far...

35. 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

36. 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)

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

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

39. 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
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

40. Heat of Combustion
The heat from the reaction that completely burns 1 mole of a substance
Note Table 11.4, page 305

41. Section 11.3 Heat in Changes of State
OBJECTIVES:
Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.

42. Section 11.3 Heat in Changes of State
OBJECTIVES:
Calculate heat changes that occur during melting, freezing, boiling, and condensing.

43. 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
Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies

44. 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 11.5, page 308
Sample Problem 11-4, page 309

45. 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 11.5, page 308

46. 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

47. Heats of Vaporization and Condensation
Note Figure 11.5, page 310
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!

48. Heats of Vaporization and Condensation
H20(g)  H20(l) Hcond = kJ/mol
Sample Problem 11-5, page 311

49. 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:

50. Heat of Solution NaOH(s)  Na1+(aq) + OH1-(aq) Hsoln = - 445.1 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 11-6, page 313
H2O(l)

51. Section 11.4 Calculating Heat Changes
OBJECTIVES:
Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.

52. Section 11.4 Calculating Heat Changes
OBJECTIVES:
Calculate heat changes using standard heats of formation.

53. 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 314 for graphite and diamonds

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

55. 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?

56. 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
The standard heat of formation of an element = 0
This includes the diatomics

57. What good are they? DHo = ( Products) - ( Reactants)
Table 11.6, page 316 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)

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

59. Examples
Sample Problem 11-7, page 317