Introduction to Thermochemistry by Canonigo

This slide presentation discusses the idea of thermodynamics in chemistry.

1. UDM PRAYER
Almighty God, creator and giver of life
We give You thanks and vow to seek Your light
Through the knowledge that we gain and impart with our peers
In our University

Grant us forgiveness for the sins we confess
And give us strength to discern what is best
Four ourselves, and more so, for the benefit of others

Teach us to have grateful hearts and enlightened minds
As we learn and share our talents to the community

For the greater good and a brighter future
Of the City of Manila and our country

All these for your greater glory Amen

2. CHEMISTRY 101 LESSON
GENERAL
CHEMISTRY
2
THERMOCHEMISTRY

3. CLASS
ATTENDANCE
RAISE TO SELECTIVE MEMBER TO REPORT
THIER ATTENDANCE EACH GROUP

4. BEFORE WE
START!
1. Arrange your respective chairs
2. Clean your areas
3. Listen attentively

5. CHAPTER 1 : THE
ACTIVITY
Starting this class we will start out
jorney of knowledge
by remember our past of what we
benn learned before we move on to
the next chapter.

6. ACTIVITY : POTENTIAL
OR KINETIC?
Direction:
Identify the pictures if shows PE
(Potential Energy) or KE (Kinetic
Energy)

8. KINETIC ENERGY/KE

10. POTENTIAL ENERGY/PE

12. KINETIC ENERGY / KE

14. KINETIC ENERGY

16. POTENTIAL ENERGY / PE

18. POTENTIAL ENERGY / PE

19. Can Energy capacity to
produce a heat?
GUIDE QUESTIONS
What are the ways
in which energy can
occur?

20. ACTIVITY : ACTIVITY:
WORDS BY NUMBERS
Direction:
Guessing the correct word by using clues of
mathematics function (First Clue) to know
the first letter or last and their definition
(Second Clue)

21. The first clue : first letter 2x + 2 ; where x=7

22. The first clue : first letter 2x + 2 ; where x=7
second Clue : "Feels like summer when i'm with you."

23. The first clue : first letter 2x + 2 ; where x=7
second Clue : "Feels like summer when i'm with you."
"Pantropiko"

24. The First letter : 10x - 15 ; where x=2

ITEM NUMBER 1

25. The First letter : 10x - 15 ; where x=2
Second clue : Reaction or physical changes in
which the system loses heat.

ITEM NUMBER 1

26. The First letter : 10x - 15 ; where x=2
Second clue : Reaction or physical changes in
which the system loses heat.

ITEM NUMBER 1
"EXOTHERMIC"

27. First Clue : Last letter : 6y - 5 ; where y=5

ITEM NUMBER 2

28. First Clue : Last letter : 6y - 5 ; where y=5
Second Clue : Ability to do "Work"

ITEM NUMBER 2

29. First Clue : Last letter : 6y - 5 ; where y=5
Second Clue : Ability to do "Work"

ITEM NUMBER 2
"ENERGY"

30. First Clue : First Letter : 3x + 10 ; where x=3

ITEM NUMBER 3

31. First Clue : First Letter : 3x + 10 ; where x=3
Second Clue : Everything that a system can
exhange energy with or connected to it.

ITEM NUMBER 3

32. First Clue : First Letter : 3x + 10 ; where x=3
Second Clue : Everything that a system can
exhange energy with or connected to it.

ITEM NUMBER 3
"Surroundings"

33. First clue : First letter : 2x - 3 ; where x=4

ITEM NUMBER 4

34. First clue : First letter : 2x - 3 ; where x=4
Second clue : Reaction or physical changes in
which system absorbs heat

ITEM NUMBER 4

35. First clue : First letter : 2x - 3 ; where x=4
Second clue : Reaction or physical changes in
which system absorbs heat

ITEM NUMBER 4
"Endothermic"

36. First Clue First letter : 2y - 16 ; where y=12

ITEM NUMBER 5

37. First Clue First letter : 2y - 16 ; where y=12
Second Clue : It is a form of energy transfer
between two objects as a result between two
objects as a result of their differences in
temperature.

ITEM NUMBER 5

38. First Clue First letter : 2y - 16 ; where y=12
Second Clue : It is a form of energy transfer
between two objects as a result between two
objects as a result of their differences in
temperature.

ITEM NUMBER 5
"HEAT"

39. Why do you think
energy is significant to
our lives
GUIDE QUESTIONS

40. LESSON OBJECTIVES
Explain the energy changes
during chemical reaction
Distinguish between exothermic and
endothermic processes
Explain the first law of
thermodynamics
And the heat of reaction

41. CHAPTER 2:
DISCOVERING
This chapter we will
getting a knew ideas by
experience, you might get
excited for something
new

42. ACTIVITY : ENDOTHERMIC
OR EXOTHERMIC
Objectives : To determine
whether the dissolving of
some common solids in
water is exothermic or
endothermic

43. MgSO4 • 7H2O (Epsom Salt)
CaCl2 NH4Cl
Na2CO3 WATER • 250 mL Beaker •
Thermometer MATERIALS
01
02
03

44. Place 100mL of water into 250-mL beaker.
Immerse a thermometer in the water and
record the temperature if water Remove the thermometer and add about a
tablespoon of the solid to be tested. Stir for about
20 seconds Place the Thermometer into the
solution and record the temperature. Repeat the process for each of the solids to
be tested.
PROCEDURES

45. Energy Changes in Chemical Reaction
- Energy capacity to do work or to supply heat it
has either mass nor volume.

Energy has no concrete form, but it can only be
detected based on its effects. Energy causes
water in kettle to boil when subjected to heat.

46. -WHAT IS THERMOCHEMISTRY?
Thermochemistry- a study of energy during
chemical reaction and/or a change in phase
-Energy can be classified as potential energy and
kinetic energy. Potential is rest, Kinetic is energy
in motion

47. Open, Isolated and Closed System
Just like matter, energy is neither created or lost.
This principle is in accordance with the law of
conservation (the 1st law of thermodynamics)
This implies that the energy of the universe does
not change in amount but is only transformed.

48. System - is a region containing energy or
matter that is separated form its
surroundings
Sorroundings - everything that interacts
with the system

49. Three types of system
1. Open system - wherein transfer of matter and
energy occurs between the system and
surroundings
2. Closed system - only energy can transfer
between the system and the surroundings.
3. Isolated system - matter and energy cannot
transfer between the system and its surroundings.

50. Matter =
Energy =
Matter =
Energy =
Matter =
Energy =
open system
Closed System Isolated System
System System System
surroundings surroundings
surroundings

52. KE= 1/2 mv²
•The total energy is the summation of potential
and kinetic energy
•Energy Transformation
- Energy transfer may be in the form of heat or
work. Both have a very important with the
energy of the system and sorroundings.

53. A process or reaction involving the release the heat is
termed as an exothermic process, and those involve
the absorption of heat is defined as an endothermic
process.
Consider the direction of the flow of heat consider
the melting butter at room temperature, the butter is
the system, and the air is the surrounding. To the
melt of the butter, its temperature must increase;
therefore, heat must flow from the air to the butter.
The process is endothermic

54. Generally an endothermic process
favors an increase in temperature. On
the other hand an exothermic reaction
favors in decrease temperature.

55. First law of thermodynamics
- is the study of energy and its
transformations there are three laws
associated with thermodynamics.
the first law of thermodynamics states that
the energy of the universe is constant. This is
also reffered to as the law of conservation
that energy is neither created nor destroyed.

56. The total energy from the summation of potential and kinetic
energies is called the internal energy
Q= mc∆T
Where Q = heat absorbed or released, J
m= mass of substance, g
c= specific of heat of substance, J/g°C
∆T= change in Temperature
Work is defined as the force applied over a given distance. It's
energy transfer between and sorroundings due to a force acting
through a distance.

57. Work is defined as the force applied over
a given distance. It's energy transfer
between and sorroundings due to a force
acting through a distance.

58. Where ∆U = system is system of the equal heat
Q = heat
W = work is done
Heat absorbed (+Q) WORK is done on the
is positive system W(+)

Heat released (-Q) WORK is done by the
is negative system W (-)
∆U = Q - W

59. ∆U = Q - W
work is done
on the system
+W
heat is absorbed
+Q
-W
-Q
heat is released
work is done
by the system

60. ENTALPHY (HEAT OF REACTION)
The **heat of reaction**, also known as the **enthalpy
change of reaction** (\( \Delta H \)), is the amount of
heat absorbed or released during a chemical reaction at
constant pressure, and it is one of the key elements
studied in both chemistry and chemical engineering

61. ΔH
Suppose the gaseous molecules in the
cylinder are exerting pressure on the
piston, such that it moves to a distance
of ΔH. The applied pressure is taken as
the force applied per unit are
P = F
A
Heat transfer in cylinder equipped
with a movable piston
Where P is the pressure of the gas, A is the area of
the cylinder, and F is the Force Applied. Work, force
is applied over given distance (d). this can be
translated as
w = Fd, ΔH

62. ΔH
Substituting this equation, you now
have
Heat transfer in cylinder equipped
with a movable piston
w = PAΔH
but the volume of the cylinder (V) is equal to the
area of the piston. That change in volume (ΔV)
caused by a moving piston
ΔV= AΔH
integrating this equation 5.5, you now have
w= PΔV

63. Enthalphy, H, Is equal to the amount of heat flow in a
system with constant pressure Qp+ Because b∆E = Qp +
w, and w = -P∆V, it is expressed as.
Qp = ∆H = ∆E + P∆V
Therefore, the heat of a reaction is siad to be equal to
the enthalphy of the reaction. If the value of work is too
small.
∆H = ∆E

64. As mentioned in the prssure, the changes in enthalphy (ΔH)
is equal to q
p
The change in enthalphy for a given chemical reaction can be
taken as
ΔH = ΔH - ΔH
reaction products reactants
if ΔH is positive, then the reaction is endothermic at
constant pressure
reaction

65. Sample: A system recieve 450 kJ of heat from its
surroundings and the surroundings do 50.0 kJ of work
on the system. What is the change internal energy?

66. Sample: A system recieve 450 kJ of heat from its
surroundings and the surroundings do 50.0 kJ of work
on the system. What is the change internal energy?
SOLUTION
Because heat is received by the system (Q is positive)
and work is done on the system (w is positive), so
∆U= Q + W
= 450 kj + 50 Kj

67. Sample: A system recieve 450 kJ of heat from its
surroundings and the surroundings do 50.0 kJ of work
on the system. What is the change internal energy?
SOLUTION
Because heat is received by the system (Q is positive)
and work is done on the system (w is positive), so
∆U= Q + W
= 450 kj + 50 Kj
=500 Kj

68. BOARD WORK
A student must use 225 mL of hot waterin a lab
procedure. Calculate the amount of heat required to
raise the temperature of 225 mL of water from 20.0 /C to
100.0 /C.

69. BOARD WORK
A student must use 225 mL of hot waterin a lab
procedure. Calculate the amount of heat required to
raise the temperature of 225 mL of water from 20.0 /C to
100.0 /C.
Solution:
- since the density of water is 1.00 g /mL, the
mass of 225 mL of water is 225 g
- c for water is 4.184 J/g C °C

70. BOARD WORK
A student must use 225 mL of hot waterin a lab
procedure. Calculate the amount of heat required to
raise the temperature of 225 mL of water from 20.0 /C to
100.0 /C.
Solution:
- since the density of water is 1.00 g /mL, the
mass of 225 mL of water is 225 g
- c for water is 4.184 J/g C °C
q = mcΔT
q = (225 g)(4.184 J/gC °C)(100.0°C - 20.0 °C)

71. BOARD WORK
A student must use 225 mL of hot waterin a lab
procedure. Calculate the amount of heat required to
raise the temperature of 225 mL of water from 20.0 /C
to 100.0 /C.
Solution:
- since the density of water is 1.00 g /mL, the
mass of 225 mL of water is 225 g
- c for water is 4.184 J/g C °C
q = mc) T
q = (225 g)(4.184 J/gC °C)(100.0°C - 20.0 °C)
= 755312 J
= 75.5 kJ

72. BOARD WORK
3000 J of heat is added to a system and 2500 J of work
is done by the system. What is the change in internal
energy of the system?

73. BOARD WORK
3000 J of heat is added to a system and 2500 J of work
is done by the system. What is the change in internal
energy of the system?
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule

74. BOARD WORK
3000 J of heat is added to a system and 2500 J of work
is done by the system. What is the change in internal
energy of the system?
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule
ΔU = Q-W

75. BOARD WORK
3000 J of heat is added to a system and 2500 J of work
is done by the system. What is the change in internal
energy of the system?
Heat (Q) = +3000 Joule
Work (W) = +2500 Joule
ΔU = Q-W
ΔU = 3000-2500
ΔU = 500 Joule

76. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C

77. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C

78. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
ΔT = Tfinal - Tinitial = 85°C - 15°C
= 70.°C

79. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
ΔT = Tfinal - Tinitial = 85°C - 15°C
= 70.°C
Q = mcΔT
= (450 g)•(4.18 J/g/°C)•(70.°C)

80. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
ΔT = Tfinal - Tinitial = 85°C - 15°C
= 70.°C
Q = mcΔT
= (450 g)•(4.18 J/g/°C)•(70.°C)
Q = 131, 670 J
= 130 kJ

81. BOARD WORK
What quantity of heat is required to raise the
temperature of 450 grams of water from 15°C to 85°C?
The specific heat capacity of water is 4.18 J/g/°C
m = 450 g
C = 4.18 J/g/°C
Tinitial = 15°C
Tfinal = 85°C
ΔT = Tfinal - Tinitial = 85°C - 15°C
= 70.°C
Q = mcΔT
= (450 g)•(4.18 J/g/°C)•(70.°C)
Q = 131, 670 J
= 130 kJ

82. BOARD WORK
Calculate the change in the internal changes of a piston
expanding against a pressure 0.75 atm from 10.3 L to
22.4 L in the process 1, 030J of heat is absorbed

83. BOARD WORK
Calculate the change in the internal changes of a piston
expanding against a pressure 0.75 atm from 10.3 L to
22.4 L in the process 1, 030J of heat is absorbed
Solution: Given is the following information:
P= 0.75 atm V2=22.4 L
V1= 10.3 L q= 1,030 J

84. BOARD WORK
Calculate the change in the internal changes of a piston
expanding against a pressure 0.75 atm from 10.3 L to
22.4 L in the process 1, 030J of heat is absorbed
Solution: Given is the following information:
P= 0.75 atm V2=22.4 L
V1= 10.3 L q= 1,030 J
Calculate the change internal energy ΔE
w= -PΔV

85. BOARD WORK
ΔV = 22.4 L - 10.3 L = 12.1 L
w = -(0.75 atm) (12.1 L) = 9.2 L -atm

86. BOARD WORK
ΔV = 22.4 L - 10.3 L = 12.1 L
w = -(0.75 atm) (12.1 L) = 9.2 L -atm
Convert = (-9.1 L-atm)(101.3 J/L-atm) = -9.2 x 10 J
2

87. BOARD WORK
ΔV = 22.4 L - 10.3 L = 12.1 L
w = -(0.75 atm) (12.1 L) = 9.2 L -atm
Convert = (-9.1 L-atm)(101.3 J/L-atm) = -9.2 x 10 J
2
ΔE= q + w

88. BOARD WORK
ΔV = 22.4 L - 10.3 L = 12.1 L
w = -(0.75 atm) (12.1 L) = 9.2 L -atm
Convert = (-9.1 L-atm)(101.3 J/L-atm) = -9.2 x 10 J
2
ΔE= q + w
= 1 030 J + (-9.2 x 10 J)
2

89. BOARD WORK
ΔV = 22.4 L - 10.3 L = 12.1 L
w = -(0.75 atm) (12.1 L) = 9.2 L -atm
Convert = (-9.1 L-atm)(101.3 J/L-atm) = -9.2 x 10 J
2
ΔE= q + w
= 1 030 J + (-9.2 x 10 J)
2
ΔE = 1.1 x 10J
2

91. SEATWORK ASSIGNMENT
(YELLOW PAD PAPER and WHITE
BOND PAPER)