1. Principles of Reactivity: Energy and Chemical Reactions
AP Notes Chapter 5
Principles of Reactivity:
Energy and Chemical Reactions

2. study of energy transfer or heat flow
Thermodynamics
study of energy transfer or heat flow
Energy
Kinetic Energy
Thermal - Heat
Mechanical
Electrical
Sound
Potential Energy
Chemical
Gravitational
Electrostatic

3. Energy is... The ability to do work. Conserved. made of heat and work.
Work is a force acting over a distance.
W=F x d
Power is work done over time.
P=W t
Heat is energy transferred between objects because of temperature difference.

4. First Law of Thermodynamics
Law of conservation of energy
Total energy of the universe is constant
Temperature and heat
Heat is not temperature
Thermal energy = particle motion
Total thermal energy is sum of all a materials individual energies

5. The Universe is divided into two halves.
the system and the surroundings.
The system is the part you are concerned with.
The surroundings are the rest.
q into system = -q from surroundings

6. System: region of space where process occurs
Surroundings: region of space around system

7. Thermodynamic Properties
1. State functions
-depend on nature of process NOT method
-path independent
-denoted by capital letters
E → state function

8. Thermodynamic Properties
2. Path functions
-depends on how process is carried out
-path dependent
-denoted by lower case
q & w → path functions

9. Work →w
w = -  (PV)
w = - P  V - V  P
Constant P wP = - P  V
Constant V wV = - V  P

10. Work →w
w = - P  V
assume ideal piston
(i.e. no heat loss)
w =  E
 E = - P  V
in isothermics
w = -q

11. PV = nRT
R=0.0821L atm/mol K
In isothermic conditions
-q = w = -nRT
R=8.314 J/mol K

12. DU = q + w
In isothermics DU = 0 so q=-w
DU = q - PDV
In constant V or constant P
w=0 so DU = q

13. Calorie - Amount of heat needed to raise the temperature of 1 gram H20, 1 degree centigrade
Nutritional calorie [Calorie]= 1 Kcal
SI unit → Joule
1 cal = J
1000cal = 1Kcal=4.184KJ

14. Direction of energy flow
Every energy measurement has three parts.
A unit ( Joules or calories).
A number how many.
and a sign to tell direction.
negative - exothermic
positive- endothermic
No change in energy - isothermic

15. Factors Determining Amount of Heat
Amount of material
Temperature change
Heat capacity

16. Specific Heat
Amount of heat needed to raise the temperature of 1 gram of material 1 degree centigrade
Heat = q
q = m Cp T
where Cp = Heat capacity
T = T2 – T1

17. CP (J/mol.oC)
H2O(s) =
H2O(l) =
H2O(g) =

18. What amount of heat is needed to just boil away 100
What amount of heat is needed to just boil away 100. mL of water in a typical lab?

19. Surroundings
System
Energy
DE <0

20. Exothermic reactions release energy to the surroundings.
q < 0 (-)
Exothermic

21. Heat
Potential energy

22. Surroundings
System
Energy
DE >0

23. Endothermic reactions absorb energy from the surroundings.
q > 0 (+)
Endothermic

24. Heat
Potential energy

25. 1. How much heat is absorbed when 10
1. How much heat is absorbed when 10.0 g of H20 are heated from 200C to 250C?

26. 2. How much heat is needed to completely vaporize 10
2. How much heat is needed to completely vaporize 10.0 g of H2O at its boiling point?

27. Heat of Vaporization
Hvap(H2O) = kJ/mol
Heat of Fusion
Hfus(H2O) = 6.01 kJ/mol

28. Energy processes:
Within a phase
q=mHv or mHf
Between phases
q=mCpT

30. 3. How much total heat is used to raise the temperature of 100
3. How much total heat is used to raise the temperature of g H2O from-15o to 125oC?

31. Some rules for heat and work
Heat given off is negative.
Heat absorbed is positive.
Work done by system on surroundings is positive.
Work done on system by surroundings is negative.
Thermodynamics- The study of energy and the changes it undergoes.

32. First Law of Thermodynamics
The energy of the universe is constant.
Law of conservation of energy.
q = heat
w = work
DE = q + w
Take the systems point of view to decide signs.

33. What is work? Work is a force acting over a distance. w= F x Dd
P = F/ area
d = V/area
w= (P x area) x D (V/area)= PDV
Work can be calculated by multiplying pressure by the change in volume at constant pressure.
units of liter - atm L-atm

34. Work needs a sign
If the volume of a gas increases, the system has done work on the surroundings.
work is negative
w = - PDV
Expanding work is negative.
Contracting, surroundings do work on the system w is positive.
1 L atm = J

35. Examples
What amount of work is done when 15 L of gas is expanded to 25 L at 2.4 atm pressure?
If J of heat are absorbed by the gas above. what is the change in energy?
How much heat would it take to change the gas without changing the internal energy of the gas?

36. Calorimetry Measuring heat. Use a calorimeter. Two kinds
Constant pressure calorimeter (called a coffee cup calorimeter)
heat capacity for a material, C is calculated
C= heat absorbed/ DT = DH/ DT
specific heat capacity = C/mass

37. Calorimetry molar heat capacity = C/moles
heat = specific heat x m x DT
heat = molar heat x moles x DT
Make the units work and you’ve done the problem right.
A coffee cup calorimeter measures DH.
An insulated cup, full of water.
The specific heat of water is 1 cal/gºC
Heat of reaction= DH = sh x mass x DT

39. Bomb Calorimeter

40. Calorimeter Constant heat capacity that is constant over a
temperature range
where
q(cal) = CC x TC
(CC = calorimeter constant)

41. 4. Determine the calorimeter
constant for a bomb calorimeter by mixing 50.0 mL of water at 25oC with 50.0 mL of water at 60.0 oC. The final temp. attained is 40oC.

42. 5. What is the molar heat of combustion of sulfur, if 2
5. What is the molar heat of combustion of sulfur, if 2.56 grams of solid flowers of sulfur are burned in excess oxygen in the bomb calorimeter in #4?

43. 6. Determine the molar heat of solution of LiOH
6. Determine the molar heat of solution of LiOH. HINT: Find the calorimeter constant first.

44. 7. The heat was absorbed by 815 g of water in calorimeter which had an initial temp of C. After equilibrium was reached, the final temperature was C.

45. Examples
The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.
A 46.2 g sample of copper is heated to 95.4ºC and then placed in a calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of both the water and the copper is 21.8ºC. What is the specific heat of copper?

46. Calorimetry Constant volume calorimeter is called a bomb calorimeter.
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
The heat capacity of the calorimeter is known and tested.
Since DV = 0, PDV = 0, DE = q

47. Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb
sample

48. Properties
intensive properties not related to the amount of substance.
density, specific heat, temperature.
Extensive property - does depend on the amount of stuff.
Heat capacity, mass, heat from a reaction.

49. Enthalpy abbreviated H H = E + PV (that’s the definition) PV = w
at constant pressure.
DH = DE + PDV
the heat at constant pressure qp can be calculated from
DE = qp + w = qp - PDV
qp = DE + P DV = DH

50. Hess’s Law Enthalpy is a state function.
It is independent of the path.
We can add equations to come up with the desired final product, and add the DH
Two rules
If the reaction is reversed the sign of DH is changed
If the reaction is multiplied, so is DH

51. Enthalpy and DHf
N2 + 2O2  2NO2
N2 + 2O2 + 68KJ/mol 2NO2
Which side is the heat written on? Why?
Look on your Heat Sheets!
Why is it only 34KJ/mol?

52. H (kJ)
2NO2
68 kJ N2
+ 2O2

53. O2
+ NO2
-112 kJ
H (kJ)
180 kJ
2NO2 N2
+ 2O2

54. O2
+ NO2
-112 kJ
H (kJ)
180 kJ
2NO2
68 kJ N2
+ 2O2

55. Hess’ Law Reactants → Products
The enthalpy change is the same whether the reaction occurs in one step or in a series of steps

56. Molar heat capacity

57. 8. Calculate the heat
needed to decompose
one mole of calcium
carbonate.

58. Standard Enthalpy
The enthalpy change for a reaction at standard conditions (25ºC, 1 atm , 1 M solutions)
Symbol DHº
When using Hess’s Law, work by adding the equations up to make it look like the answer.
The other parts will cancel out.

59. Example Given calculate DHº for this reaction DHº= -1300. kJ

60. Example Given +1300. kJ + 394 kJ + 286 kJ
calculate DHº for this reaction

61. Example Given +1300. kJ + 394 kJ + 286 kJ
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g)
+ 394 kJ
+ 286 kJ
calculate DHº for this reaction

62. Example Given + 2(394 kJ) + 286 kJ calculate DHº for this reaction
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g) 2 2 2
+ 2(394 kJ)
+ 286 kJ
calculate DHº for this reaction

63. Example Given + 788 kJ + 286 kJ calculate DHº for this reaction
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g) 2 4 2
+ 788 kJ 2
+ 286 kJ
calculate DHº for this reaction

64. Example Given + 788 kJ + 286 kJ calculate DHº for this reaction
1300.KJ + 2CO2(g) + H20(l)  C2H2 (g) + 5/2 O2 (g) 2 4 2
+ 788 kJ 2
+ 286 kJ
calculate DHº for this reaction

65. Example Given calculate DHº for this reaction 1300.KJ  C2H2 (g)
2C(s)  788KJ
H2(g)  286KJ
1300.KJ  788kJ + 286KJ
1300.KJ  1074KJ
calculate DHº for this reaction

66. Example Given calculate DHº for this reaction 1300.KJ  1074KJ 226KJ 
DHf = 1300KJ – 1074KJ
DHf = 226KJ (+sign shows Endothermic)
calculate DHº for this reaction
226KJ +

67. Example Now, You try it! Given DHº= +77.9kJ DHº= +495 kJ DHº= +435.9kJ
Calculate DHº for this reaction

68. Standard Enthalpies of Formation
Hess’s Law is much more useful if you know lots of reactions.
Made a table of standard heats of formation. The amount of heat needed to for 1 mole of a compound from its elements in their standard states.
Standard states are 1 atm, 1M and 25ºC
For an element it is 0
There is a table in Appendix 4 (pg A22)

69. Standard Enthalpies of Formation
Need to be able to write the equations.
What is the equation for the formation of NO2 ?
½N2 (g) + O2 (g) ® NO2 (g)
Have to make one mole to meet the definition.
Write the equation for the formation of methanol CH3OH.

70. Since we can manipulate the equations
We can use heats of formation to figure out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g) ® 2CO2 + 3H2O
which leads us to this rule.

71. Since we can manipulate the equations
We can use heats of formation to figure out the heat of reaction.
Lets do it with this equation.
C2H5OH +3O2(g) ® 2CO2 + 3H2O
which leads us to this rule.

72. 9. Calculate Hf of SO3(g) from the data given. Forms of Equations:
2SO2(g) + O2(g)  2SO3(g) H = kj or
2SO2(g) + O2(g)  2SO3(g) kj

73. 10. If the metabolism of glucose is combustion of C6H12O6(s), how much heat is produced by the metabolism of 1.00 g of glucose?

74. Additivity of Heats of Reaction
H = Enthalpy
H = E + PV
DH = DE + DPDV

75. at constant pressure:
E = qP + w
or E = qP - PV

76. now, at constant P:
VDP = 0
thus, H = E + PV
& if E = qP - PV
then H = qP

77. H = Hproducts - Hreactants
H = Enthalpy
a “state” function
cannot be measured
measure H
H = Hproducts - Hreactants

78. Standard Enthalpy of Formation
Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states

79. Thermodynamic Standard State
T = 25oC and P = 1 atm
Thermodynamic Data
Appendix L of text p. A31
reference point-
for ELEMENTS in standard state

80. 11. Combustion of Methane
HC = ?

81. Write equations and find ΔHf0 for methane,
carbon dioxide, & water

82. HC = ?