1. Thermochemistry The study of energy
Chapter 6

2. Kinetic Energy Ek The energy of an object in motion Ek= ½ mv2
Mass need to be in kg
A car moving at 40 mph has a greater kinetic energy than a car moving at 20 mph

3. Types of kinetic energy
Work: energy used to cause an object with mass to move
Heat: energy used to cause the temperature of an object to increase

4. Potential Energy Stored energy in an object by virtue of its position.
Units of energy:
Joule J,
1 Calorie Cal = J

5. 5. 1 The Nature of Energy
Energy is capacity to do work or to produce heat.
Thermochemistry is the study of heat change in chemical reactions.

6. Energy is the capacity to do work
Radiant energy comes from the sun and is earth’s primary energy source
Thermal energy is the energy associated with the random motion of atoms and molecules
Chemical energy is the energy stored within the bonds of chemical substances
Nuclear energy is the energy stored within the collection of neutrons and protons in the atom
Potential energy is the energy available by virtue of an object’s position
6.1

7. Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that are at different temperatures.
Temperature is a measure of the thermal energy
Said another way temperature is a measure of random motion (KE) of particles.
Temperature (K) = Thermal Energy (J)
6.2

8. Physics vs. Chemistry
Chemical energy: is the energy stored within the bonds of chemical substances (potential)
Thermal energy: is the energy associated with the random motion of atoms and molecules (Kinetic molecular theory) (kinetic)
In this chapter we will discuss the transfer of these types of energy.

9. Energy Systems
System: portion we single out to study typically the chemical
Ex: reactants products.
Surroundings: everything else
Ex: Reaction vessel, environment

11. Types of Systems

12. C B A

13. Exothermic
A reaction that results in the evolution of heat. Thus heat flows out of the system
Exo = out = heat loss from system = energy loss = -heat

14. An Exothermic Reaction
2H2 (g) + O2 (g) H2O (l) + energy
H2O (g) H2O (l) + energy

16. Endothermic
A reaction that absorbs heat from their surroundings. Thus heat flows into a system.
Endo = In to = heat gain to system = energy gain = + heat

17. An Endothermic Reaction
energy + 2HgO (s) Hg (l) + O2 (g)
energy + H2O (s) H2O (l)

19. Work, work, work Recall that energy
Can be transferred in the form of motion and heat
Work = force x distance
Units
f = Newton, N
d = meter, m
W = N*m or Joule, J

20. Force is any kind of push or pull exerted on an object.
Ex: gravity, mechanical, electrostatic
Distance is how far the object moved as a result of the force applied.

21. Work in Biology
Think about water moving from the ground up the trunk of a tree. What part of the system if any undergoes a change in potential energy? Is work done in the process?

22. Guided Questions What is changing location ?
Does this change involve potential energy?

23. Water moves up the trunk against the force of gravity
Water moves up the trunk against the force of gravity. Thus the potential energy of the water does change
Work is movement of a mass over a distance against an opposing force.

25. Homework
Chang: pg 255
1, 9, 10
****BL:Pg 188
1, 2, 3, 6, 9, 11,

26. 5.2 First law of thermodynamics
Aka: The law of conservation of energy
Energy is neither created nor destroyed, thus energy is converted from one form to another.
Thus the energy of the universe is constant
Universe = System + Surroundings

27. Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings

28. The total amount of energy contained by the water in a reservoir is constant.
A. At the top of the dam, the energy is potential (EP).
B-C. As the water falls over the dam, its velocity increases, and potential energy is converted into kinetic energy (EK).
D. At the bottom of the dam, the kinetic energy gained by the water is largely converted into heat and sound as the water dashes against the rocks.

29. Internal Energy
The energy (E) of a system can be defined as the sum of the kinetic and potential energies of all of the particles in a system.
Internal energy can be changed by a flow of HEAT, WORK, or BOTH

30. Change in a system energy
E = Efinal - Einitial
E > 0 = system gained energy
E < 0 = system lost energy to the surroundings
Enthalpy can not be negative, think about it like a bank account (EX pg 235)

31. Relating heat to work E = q + w
E = change in system’s internal energy
q = heat
w = work

32. Signs signs every where are signs

33. Write a figure description for this figure so that a student reading this text would understand.
Identify key terms in your description.

34. Work (w) W > 0 = work is done ON the system ( + W)
W < 0 = work is done BY the system on the surroundings.
(-W)

35. Prove it….
Write an example that illustrates work being done by a system
Then write a sentence illustrating work being done on a system

36. Heat (q) q > 0 = heat is added to the system. (+ q endothermic)
q < 0 = heat is released from the system (-q exothermic)

37. Prove it ….
Write an example that illustrates heat being added to a system
Then write a sentence illustrating heat being released form a system

38. Example:
Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done by the system.

39. Answer E = q + w E = 15.6 + (- 1.4) = 14.2 q = + 15.6 (endothermic)
W = kJ (work is done by the system)
E = q + w
E = (- 1.4) = 14.2

40. State Function
A Property of a function that is determined by specifying its condition or its present state
The value of a state function depends only on the present state of the system - not how it arrived there
Energy IS A STATE FUNCTION
Enthalpy IS A STATE FUNCTION
Temperature IS A STATE FUNCTION
Heat IS NOT A STATE FUNCTION
Work IS NOT A STATE FUNCTION

41. Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

42. Example
A B C
It does not mater whether I heated the water (C  B) or cooled that water (A  B) to get it to the temperature. Internal energy at point B would be the same regardless.

43. Change in energy ΔE
ΔE is a state function because it is independent of pathway.
(all we care about is energy at Ei and energy at Ef.
Heat (q) and work (w) are not state functions!!!!!

44. Example

45. 5.2 Homework Chang: pg 255 11,12,17,18 ***BL: Pg 189
17, 19, 21, 22, 25, 26

46. 5.3 Enthalpy Enthalpy (H):
Accounts for the heat flow in chemical changes occurring at constant pressure when no forms of work other than P, V work
*** state function****
Results from a change in P (atm) or V (L) of the system.

47. What is Pressure volume work
Build up of gas that causes a piston to lift against the force of gravity.
Work = -P ΔV

49. Enthalpy : H = E + PV E = internal energy
Surroundings
Surroundings
HEAT
HEAT
System
System
ΔH >0
Endothermic
ΔH <0
Exothermic

50. Don’t forget you can manipulate these equations
Enthalpy = H = E + PV
E = H  PV
H = E + PV

51. Calculating Work (w) work = force  distance
since pressure = force / area
work = pressure  volume
wsystem = PV
Note: P = pressure of the external environment.

52. Example:
Calculate the work associated with the expansion of gas from 46 L to
64 L at a constant external pressure of 15 atm.
W = -PΔV

53. Answer w = -PΔV P = 15 atm V = 64L – 46L = 18L w = -15atm (18L)
= L*atm
To convert to J J/1 L*atm

54. More Equations
We can analyze enthalpy (H) by looking at heat (q) at a constant pressure (note the subscript):
H = qp
We can analyze energy (E) by looking at heat (q) at a constant volume (note the subscript):
E = qv

55. 5.3 Homework
Chang: 15, 16,
BL: Pg 190
#’s 27 A and C only, 29 skip A do B, 30
HINT ON #30 write a balance chemical reaction first, don’t over think it.

56. 5.4 Enthalpy Changes
Enthalpy: The heat content of a chemical system is called the enthalpy (symbol: H)
The enthalpy change (Δ H) is the amount of heat released or absorbed when a chemical reaction occurs at constant pressure.

57. More enthalpy
Enthalpy, along with the pressure and volume of a system, is a state function (property of a system that depends only on its state, not how it arrived at its present state).

58. Enthalpy of reactions Hrxn = Hproducts – Hreactants
To calculate the enthalpy of a reaction we need to subtract the final enthalpy from the initial enthalpy or
Hrxn = Hproducts – Hreactants

59. What is the enthalpy of a rxn?
The heat of the reaction ΔH or ΔHrxn that occurs from the change of products to reactants.
Balanced equations that show an enthalpy change are called thermochemical equations.
2H2 (g) + O2 (g)  2H2O (g) ΔH = kJ

60. Enthalpy Diagrams
We can use diagrams to illustrate the change in enthalpy of a reaction.
Exothermic = - ΔH value
Endothermic = + ΔH value

61. Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
H2O (s) H2O (l)
DH = 6.01 kJ
6.4

62. Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)
DH = kJ
6.4

63. 2H2 (g) + O2 (g)  2H2O (g) ΔH = kJ ΔHrxn = H (products) – H (reactants) kJ + - (90.83 kJ) = kJ

64. Reaction time
Reaction time

66. Guidelines for thermochemical reactions
1. Enthalpy is an extensive property
Ex: combustions of 1 mol = ΔH -40 kJ
combustions of 2 mol = ΔH -80 kJ
2. Enthalpy of a forward reaction is equal to (but opposite in sign) of enthalpy of the reverse reaction
2H2 (g) + O2 (g)  2H2O (g) ΔH = kJ
2H2O (g)  2H2 (g) + O2 (g) ΔH = kJ

67. 3. Enthalpy of the reaction depends on the state of the reactants and the products.
Producing H2O (l) or H2O (g) would change the enthalpy value for a thermochemical equation
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ

68. Example Problem
How much heat is released when 4.50g of methane gas is burned in a constant-pressure system according the equation below?
CH4 (g) + 2O2(g)  CO2 (g) + 2H2O (l) ΔH = -890 kJ

69. Steps Use mole ratios to establish a proportion
CH4 (g) + 2O2(g)  CO2 (g) + 2H2O (l) ΔH = -890 kJ
Use mole ratios to establish a proportion
for every 1mol CH4 (g) burned -890 kJ of heat is produced
2. Use dimensional analysis to convert g to mol to heat. (since we know heat production from 1 mole of CH4)
4.50g CH4 1mol CH kJ = kJ
16.0g CH mol CH4

70. Homework Chang: BL: Pg: 190 #’s 31, 33, 36, 37, 40
Read 5.5 Calorimetry

71. Calorimetry Measures heat flow
Calorimeter: used to measure the exchange of heat that accompanies chemical reactions.

72. How it works
Reaction using known quantities of reactants is conducted in an insulated vessel that is submerge in a known qty of water.
The heat created from the reaction will increase the temperature of the water surrounding the vessel.
The amount of heat emitted can be calculated using the total heat capacity of the calorimeter and its contents.

73. Heat Capacity
The temperature change experienced by an object when it absorbs a certain amount of heat energy.
The heat required to raise the temperature of a substance by 1K or 1ºC
q = CΔT
q = heat energy released or absorbed by the rxn (J)
C = heat capacity (J/K)
ΔT = change in temperature (K)
NOTE: when a sample gains heat (+q) ΔT is positive
when a sample loses heat (-q) ΔT is negative

74. A note on temperature scale
A temperature change in Kelvin is equal in magnitude to the temperature change in degrees Celsius.
In the following calculations we are looking at the value for the quantity of the ΔT not an actual Temp reading.
Thus 1K = 1ºC

75. Molar Heat Capacity
The heat capacity of one mole of a substance is called its molar heat capacity.
If the molar heat capacity for 1 mol =20 J/mol*K
Then the molar heat capacity for 10 moles = 20(10)
200 J/mol*K (20 J/mol*K for each mole present in the reaction)

76. Specific Heat The heat capacity of 1 gram of a substance.
The heat required to raise 1 gram of a substance by 1K or 1ºC
q = sm ΔT
q= heat energy released or absorbed by the rxn (J)
s = specific heat (J/g*K)
m = mass of solution (g)
ΔT = change in temperature in (K)

77. Enthalpy and Specific Heat
FYI: When reactions occur at constant pressure
ΔH = q
Thus enthalpy change (ΔH) is equal to heat (q)

78. Dt = tfinal – tinitial = 50C – 940C = -890C
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = J/g • 0C
Dt = tfinal – tinitial = 50C – 940C = -890C
q = msDt
= 869 g x J/g • 0C x –890C
= -34,000 J
6.5

79. Specific Heat
Specific heat of a substance can be determined by measuring the temperature change that a known mass of a substances undergoes when it gains or loses a specific quantity of heat.
Specific heat = qty of heat transferred (g of substance) x (temp change)

80. Example
209J is required to increase the temp of 50.0 g of water by 1.00K.
Thus the specific heat of water is
4.18 J/gK
s = J
(50.0g) (1.00 K)

81. Example:
How much heat in kJ is required to increase the temperature if 150 g of water from 25°C to 42°C. Specific heat of water is 4.18 J/g°C. (Memorize specific heat water)

82. Answer
q = msΔT
q = 150 (4.18)(42-25)
q = kJ

83. Molar Heat Capacity Example
If you have 1 mol of water with a specific heat of 4.18 J/g*K, the molar heat capacity of water is.
4.18 J/g*K (18.0g) = 75.2J/mol*K
1mol
What if you have 26 moles?

84. **Constant Pressure Calorimetry **
Fig 5.18 pg 170
At constant pressure (as in a solution) ΔH can be measured directly.

85. IF we assume that no heat is lost
IF we assume that no heat is lost. then the heat gained by the solution is lost from the heated metal ball.
Heat lost = Heat gained
qsol = (s) x (g of solution) x ΔT = -qrxn
ΔT > 0 means the rxn is exothermic qrxn < 0
Metal ball

86. Example
When a student mixes 50 mL of 1.0M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter the temperature of the resulting 100 mL solution increases from 21ºC to 27.5 ºC. Calculate the enthalpy change for the reaction assuming that the calorimeter loses only a negligible amount of heat. The density of the solution is 1.0 g/mL with a specific heat of
4.18 J/g*K

87. q = smΔT m = 100mL (1.0 g) = 100 g 1 mL q = (4.18 J/gK)(100 g)(6.5)
50 mL 1.0 M NaOH
100 mL
ΔT = = 6.5
S = J/g*K
D = 1.0 g/mL
50 mL 1.0 M HCl + =
q = smΔT
m = 100mL (1.0 g) = 100 g
1 mL
q = (4.18 J/gK)(100 g)(6.5)
= 2.7 x103 J
Now we must go back to the problem and determine if the heat (q) is endothermic (+) or exothermic (-)
Because the temperature increases heat is being released and the reaction is exothermic
FINAL ANSWER q = = x103 J

88. One step further
We solved for how much energy was produced from 100g of solution…
q = = x103 J
What if the question asked for heat produced from 1 mol HCl?
(0.05L HCl)(1.0 mol HCl) = 0.05 mol HCl in solution
1 L HCl
J/ (0.050 mol HCl) = J/mol of HCl

89. A coffee cup calorimeter initially contains 125 g of water at 24. 2°C
A coffee cup calorimeter initially contains 125 g of water at 24.2°C. Potassium bromide (10.5g) also at 24.2°C is added to the water, and after the KBr dissolves, the final temp is 21.1°C. Calculate the enthalpy change for dissolving the salt in J. Assume that the specific heat capacity of the solution is J/°Cg and that no heat transferred to the surroundings or to the calorimeter.

90. q = msΔT (think m total)
m = 125g H2O KBr = 135.5g total
q= (4.184)( )
q = J

91. *******Example***** A 46.2 g sample of copper is heated to
95.4 ºC and then placed into a calorimeter containing 75.0 g of water at 19.6 ºC. The final temp of the metal and water was
21.8 ºC. Calculate the specific heat of copper, assuming that all the heat lost by copper is gained by water.

92. Heat lost by copper = heat gained by the water
Assume:
Heat lost by copper = heat gained by the water
qlost = 46.2g(s)( ºC )
= (s)
qgained = 75.0 g(4.18)( ºC ) =
(s) = 689.7
S = 0.2 J/ºC *g

93. Bomb Calorimetry
Used to study combustion reactions
qrxn = -Ccal x ΔT

94. Homework Brown Lemay Pg 191-2 #’s
Sp heat/ capacity 43, 44, 45, 46, 47,
Calorimetry 49, 50

95. 6.3 Hess’s Law “Heat of summation”
states that if reactions are carried out in a series of steps, ΔH for the reaction will be equal to the sum of the enthalpy changes by the individual steps.

96. Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.
Goal: manipulate the steps of the equation (multiple, reverse) to cancel out terms to isolate the final reaction and calculate the new enthalpy (ΔH )

98. Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g) H = 540 kJ
-180 x 3 = -540

99. Solving Hess’s Law Problems
Given the following data, calculate the ΔH for the reaction:
H2S(g) O2(g)  H2SO4(l) ΔH =-706.5KJ
H2SO4(l)  SO3(g) + H2O(g) ΔH =184.5KJ
H2O(g)  H2O(l) ΔH=-99KJ ______________________________________
SO3(g) + H2O(l)  H2S(g) + 2O2(g) ΔH=

100. Solving Hess’s Law Problems
Given the following data, calculate the ΔH for the reaction:
C2H2 (g)+ 5/2 O2 ( g)  2CO2 (g) + H2O (l) ΔH = -1300kJ
C (s) + O2 (g)  CO2 (g) ΔH = kJ
H2 (g) + ½ O2 (g)  H2O (l) ΔH = kJ
2C (s) + H2 (g)  C2H2 (g)

101. __________________________________
C2H2 (g)+ 5/2 O2  2CO2 (g) + H2O (l) ΔH = -1300kJ
C (s) + O2 (g)  CO2 (g) ΔH = kJ
H2 (g) + ½ O2 (g)  H2O (l) ΔH = kJ
__________________________________
2C (s) + H2 (g)  C2H2 (g) ΔH =

102. Homework
BL packet
#’s 57, 59, 60, 61, 62

103. 5.7 Enthalpy of Formation
Change in enthalpy due to the formation of substances from stable forms of its component elements.
Standard enthalpy of formation (ΔHfº ) is the change in enthalpy for the reaction that forms 1 mol of the compound from its elements in their standard states.

104. Where to Find ΔHfº Values
Appendix C page 1041 Or
Table 5.3 page 177
NOTE THE STATE OF YOUR MOLECULE!!!

105. Where to Find ΔHfº Values
Appendix 3 pg A-8
NOTE THE STATE OF YOUR MOLECULE!!!

106. ΔHrxn = H (products) – H (reactants) multiply any ∆H by its coefficient

107. Example:
Using enthalpies of formation, calculate the standard change in enthalpy for the following reaction. Then determine if the reaction is endothermic or exothermic:
2Al (s) + Fe2O3 (s)  Al2O3 (s) + 2Fe (s)

108. ΔHrxn = H (products) – H (reactants)
2Al (s) + Fe2O3 (s)  Al2O3 (s) + 2Fe (s)
H =
( ) - ( )
ΔHrxn = -847 kJ VERY exothermic

109. Homework
#’s 67, 71, 72
(You will need to look up H in BL table)