1. CHAPTER 10
Chemical Reactions
10.4 Chemical Reactions and Energy

2. There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)

3. Chapter 4.2 Chemical Reactions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions

4. Chapter 4.2 Chemical Reactions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Photosynthesis is an endothermic reaction: energy is absorbed

5. Chapter 4.2 Chemical Reactions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Photosynthesis is an endothermic reaction: energy is absorbed
Cellular respiration is an exothermic reaction: energy is released

6. Properties of Solutions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions

7. Energy is absorbed from the surroundings so the pack feels cold
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Energy is absorbed from the surroundings so the pack feels cold

8. Energy is released into the surroundings so the pack feels hot
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Energy is released into the surroundings so the pack feels hot

9. Properties of Solutions
There are three key components to a chemical reaction:
Reactants
Products
Energy (in or out)
Chapter 4.2
Chemical Reactions
Chapter 9.3
Properties of Solutions
Change in enthalpy
enthalpy: the amount of energy that is released or absorbed during a chemical reaction

10. Enthalpy change (∆H, J/mole)
Reaction
Exothermic
Endothermic
Energy
is released
is absorbed
Enthalpy change (∆H, J/mole)
is a negative number
∆H < 0
is a positive number
∆H > 0

11. Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ

12. Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ
Reactants Products Energy
thermochemical equation: the equation that gives the chemical reaction and the energy information of the reaction.

13. Enthalpy calculations
Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ
The reverse chemical reaction involves the same amount of energy, but the energy flow is reversed (“in” instead of “out”):
CO2(g) C(s) + O2(g) ∆H = kJ

14. Enthalpy calculations
Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ
1 mole mole mole
The combustion of twice as much carbon releases twice as much energy:
2C(s) + 2O2(g) CO2(g) ∆H = –787.0 kJ
2 moles moles moles

15. Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ
2 moles /2 moles mole

16. Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ
2 moles /2 moles mole
Rewrite the chemical equation using coefficients with the smallest whole numbers possible

17. Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ
2 moles /2 moles mole
x 2
4Fe(s) + 3O2(g) Fe2O3(s) ∆H = ?
4 moles moles moles
What is the enthalpy change for this reaction?

18. Enthalpy calculations
Chemical equation for the formation of rust:
2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ
2 moles /2 moles mole
x 2
x 2
4Fe(s) + 3O2(g) Fe2O3(s) ∆H = –1,648.4 kJ
4 moles moles moles

19. Enthalpy of formation Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ
This is also the chemical equation for the formation of CO2.
∆Hreaction = ∆Hformation of CO2 = –393.5 kJ

20. Enthalpy of formation Chemical equation for the combustion of carbon:
C(s) + O2(g) CO2(g) ∆H = –393.5 kJ
This is also the chemical equation for the formation of CO2.
∆Hreaction = ∆Hformation of CO2 = –393.5 kJ
∆Hf (CO2) = –393.5 kJ/mole
The formation of 1 mole of CO2 releases kJ of energy

21. Enthalpy of formation
Enthalpies of formation of some common substances

22. Enthalpy of formation
Enthalpies of formation of some common substances
Knowing these values and the following equation, you can calculate unknown enthalpy values:

23. Enthalpy of formation Let’s see an example!
Enthalpies of formation of some common substances
Let’s see an example!
Knowing these values and the following equation, you can calculate unknown enthalpy values:

24. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.

25. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
Asked: ∆Hf(glucose) = ?

26. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
Asked: ∆Hf(glucose) = ?
Given:
C6H12O6(s) + 6O2(g) CO2(g) + 6H2O(g) ∆H = –2,808 kJ
From the table of enthalpies of formation
Note: In this problem we have to use the enthalpy of formation of gaseous water which is different from the enthalpy of formation of liquid water.
In the text solved problem 313 the wrong enthalpy of formation is used.

27. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
Asked: ∆Hf(glucose) = ?
Given:
C6H12O6(s) + 6O2(g) CO2(g) + 6H2O(g) ∆H = –2,808 kJ
From the table of enthalpies of formation
Relationships:

28. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
C6H12O6(s) + 6O2(g) CO2(g) + 6H2O(g) ∆H = –2,808 kJ
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ

29. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
C6H12O6(s) + 6O2(g) CO2(g) + 6H2O(g) ∆H = –2,808 kJ
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products
Relationships:
10.4 Chemical Reactions and Energy

30. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

31. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products
Remember to multiply by the coefficients!

32. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

33. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

34. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

35. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

36. Enthalpy calculations
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
reactants
products

37. Enthalpy calculations
The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.
Formation of glucose
6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ
1 mole
Asked: ∆Hf(glucose) = ?
Answer: ∆Hf(glucose) = –1,004 kJ/mole

38. A B A B +
∆H = X

39. The reverse reaction changes the sign of ∆HB A B +
∆H = X A B A B +
∆H = –X
The reverse reaction changes the sign of ∆H

40. If three times more substances are involved, ∆H is three times greater+
∆H = X
x 3 A B A B
∆H = 3X +
If three times more substances are involved, ∆H is three times greater

41. ∆H(reaction) = ∆Hf (products) – ∆Hf (reactants)B A B +
∆H = X
∆H(reaction) = ∆Hf (products) – ∆Hf (reactants) A B A B
∆H(reaction) =
∆Hf
– ∆Hf + ∆Hf

42. Energy profile
Thermochemical equation:
Reactants Products ∆H = … kJ

43. also have stored energy
Energy profile
Thermochemical equation:
Reactants Products ∆H = … kJ
have stored energy
also have stored energy
Energy flow
during the reaction

44. We can graph the change in energy as the reaction takes place
Energy profile
Thermochemical equation:
Reactants Products ∆H = … kJ
have stored energy
also have stored energy
Energy flow
during the reaction
We can graph the change in energy as the reaction takes place

45. We can graph the change in energy as the reaction takes place
Energy profile
Thermochemical equation:
Reactants Products ∆H = … kJ
have stored energy
also have stored energy
Energy flow
during the reaction
Energy
We can graph the change in energy as the reaction takes place
Progress of reaction

46. also have stored energy
Energy profile
Thermochemical equation:
Reactants Products ∆H = … kJ
have stored energy
also have stored energy
Energy flow
during the reaction

47. The reaction cannot start without this initial input of energy
Energy profile
Activation energy
The reaction cannot start without this initial input of energy

48. Energy profile Combustion of carbon: C(s) + O2(g) CO2(g)
Wood does not spontaneously light itself up on fire

49. Energy profile Reaction of sodium in water:
Na(s) + H2O(l) NaOH + H2(g)
Sodium reacts with water immediately (and violently) upon contact

50. Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2).

51. Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A
∆H1

52. Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A
∆H1 A B
∆H2

53. Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A
∆H1 A B
∆H2 B P
∆H3

54. Hess’s law
Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A
∆H1 A B
∆H2 B P
∆H3 R P
∆H4

55. Hess’s law ∆H1 ∆H2 ∆H3 ∆H4 Hess’s law: ∆H4 = ∆H1 + ∆H2 + ∆H3 R A A B BP
∆H3 R P
∆H4

56. Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite.

57. Hess’s law
Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite.
Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Relationships: Hess’s law

58. Hess’s law
Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite.
Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Relationships: Hess’s law
Strategy: Create a path that leads from Cgr to Cd.

59. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Cgr(s) is a reactant in:
Cgr(s) Cd(s)

60. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Cgr(s) is a reactant in:
Cgr(s) Cd(s)
Cgr(s) is also a reactant in:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole

61. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Cd(s) is a product in:
Cgr(s) Cd(s)

62. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Cd(s) is a product in:
Cgr(s) Cd(s)
Cd(s) is a reactant in:
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole

63. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Cd(s) is a product in:
Cgr(s) Cd(s)
Cd(s) is a reactant in:
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the reverse reaction so that Cd(s) is a product, and adjust DH:
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole

64. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole

65. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole

66. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole

67. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole
Cgr(s) Cd(s) ∆H = (– ) kJ/mole

68. Hess’s law Asked: Cgr(s) Cd(s) ∆H = ?
Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole
Write the sum of the two equations:
Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole
CO2(g) Cd(s) + O2(g) ∆H = kJ/mole
Cgr(s) Cd(s) ∆H = +1.9 kJ/mole

69. Energy profile of a reaction
Hess’s law