SLIDE 1
1
2
1 Thermochemistry 2 Table of Contents The Nature of Energy State - - PDF document
1 Thermochemistry 2 Table of Contents The Nature of Energy State - - PDF document
1 Thermochemistry 2 Table of Contents The Nature of Energy State Functions** Click on the topic to go to that section Enthalpy Measuring Enthalpy Changes: Calorimetry Energy Associated with Changes of State Enthalpies of
SLIDE 2
SLIDE 3
3
Table of Contents
- The Nature of Energy
Click on the topic to go to that section
- State Functions**
- Enthalpy
- Enthalpies of Formation
- Energy in Foods and Fuels
- Hess's Law
- Measuring Enthalpy Changes: Calorimetry
- Energy Associated with Changes of State
- Enthalpies of Reaction
SLIDE 4
4
The Nature of Energy
Return to Table of Contents
SLIDE 5
5
Thermochemistry
Pr es en at We know chemical and physical processes release and absorb
- energy. We use these thermochemical principles to design air
conditioners and refrigerators as well as foot warmers that allow us to stay comfortable as we "go big" on the hill!
SLIDE 6
6
Potential Energy is the energy that objects have energy due to their position. Gravitational Potential Energy GPE = mgh Elastic Potential Energy EPE = 1/2 kx2 Electric Potential Energy UE = kQ1Q2 r2
A Review of Energy from Physics
SLIDE 7
7
Kinetic Energy is the energy that an object has by virtue of its motion: KE = 1/2 mv2 Work is defined by the formula W = Fdparallel
A Review of Energy from Physics
SLIDE 8
8
Algebraically, these two statements combine to become: E0 + W = Ef Since Ef Eo = ∆E, this can also be written as ∆E = W
A Review of Energy from Physics
work The total energy of an isolated system is constant. An outside force can change the energy of a system by doing work on it.
SLIDE 9
9
Another unit of energy is the calorie (cal).
1 cal = 4.184 J
The energy of food is measured in Calories (C). [note the capital "C"]
1 Calorie = 1000 calories = 4184 Joules
Units of Energy
The SI unit of energy is the Joule (J).
SLIDE 10
10
1 A reaction produces 3.8 cal of energy. How many joules
- f energy is produced?
Answer
15.9 J
SLIDE 11
11
2 A reaction uses 235 J of energy. How many calories have been burned?
Answer
56 cal
SLIDE 12
12
3 A 20 ounce coke contains 240 Calories. How many kilojoules of energy are present in a 20 ounce Coke?
Answer
1.0 x 103 kJ
SLIDE 13
13
From last year, we know that ∆E = W. This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q). When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object.
∆E = w + q
*Note, we use a lower case "w" in chemistry.
Energy & Heat A B
heat flow T = 20 T = 10
SLIDE 14
14
The First Law of Thermodynamics
∆E = w + q
Energy is neither created nor destroyed. In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.
Initial state Final state E of system decreases Internal energy, E E < E0
∆E < 0 ()
E E0 Energy lost to surroundings Initial state Final state E of system increases Internal energy, E Energy gained from surroundings E > E0
∆E > 0 (+)
E E0
SLIDE 15
15
System and Surroundings
The system includes the reactants and products (here, the hydrogen, oxygen and water molecules). The surroundings are everything else (here, the cylinder and piston). Surroundings system When considering energy changes, we need to focus on a well defined, limited part of the universe. The portion we focus on is called the system and everything else is called the surroundings. Consider the following reaction occurring within a metal cylinder.
2H2(g) + O2(g) > 2H2O(g)
SLIDE 16
16
Changes in Internal Energy
If ∆E > 0, Efinal > Einitial The system absorbed energy from the surroundings. If ∆E < 0, Efinal < Einitial The system released energy from the surroundings. H2(g) + O2 (g)
Internal energy, E
∆E < 0
(negative)
∆E >0
(positive)
H2O(l)
SLIDE 17
17
4 Ten grams of table salt in dissolved in water in a 250 mL
- beaker. Which of the following is a component of the
system? A NaCl B water C Na+ D beaker E A and B F A, B, and C G A, B, and D
Answer
F
SLIDE 18
18
5 When a strong acid is added to a flask containing water the flask becomes warm to the touch. This is because... A the reaction performed work on the flask B the system absorbed heat from the surroundings C the system released heat to the surroundings D the surroundings released heat to the system
Answer
C
SLIDE 19
19
6 When a strong acid is added to a flask containing water the flask becomes warm to the touch. Which correctly describes the change in energy? A ∆Esys is positive and ∆Esur is negative B ∆Esys is positive and ∆Esur is positive C ∆Esys is negative and ∆Esur is positive D ∆Esys is negative and ∆Esur is negative
Answer
C
SLIDE 20
20
Changes in Internal Energy
System
∆E>0
Heat q > 0 Work w > 0 Surroundings
When energy is exchanged between the system and the surroundings, it is either exchanged as either heat (q)
- r work (w).
∆E = q + w
SLIDE 21
21
q, w, ∆E, and Their Signs
q
+
system gains heat
system loses heat w
+
work done on system
work done by system ∆E
+
net gain of energy by system
net loss of energy by system
Sign Conventions for q, w and ∆E
SLIDE 22
22
7 The ∆E of a system that gains 50 kJ of heat and performs 24 kJ of work on the surroundings is ________ kJ.
A 74 B 26 C D
+26
E +74
Answer
D
SLIDE 23
23
8 The ∆E of a system that releases 120 J of heat and does 40 J of work on the surroundings is ________ J.
A 80 B
160
C D +80 E +160
Answer
B
SLIDE 24
24
9 The ∆E of a system that absorbs 120 J of heat and does 120 J of work on the surroundings is ________ J. A
240
B
120
C
D
+120
E +240
Answer
C
SLIDE 25
25
10 The ∆E of a system that absorbs 12,000 J of heat and the surrounding does 12,000 J of work on the system is _______ J.
A 24000 B
12000
C D
+12000
E +24000
Answer
E
SLIDE 26
26
Exchange of Heat between System and Surroundings
Recall, when heat is absorbed by the system from the surroundings, the process is endothermic . System Surroundings Heat q System Surroundings Heat +q When heat is released by the system into the surroundings, the process is exothermic.
SLIDE 27
27
11 The reaction that occurs inside the foot warmer packet is endothermic? True False
Answer
SLIDE 28
28
12 What will happen when a hot rock is put into cold water? A the water and rock will both gain energy B the water and rock will both lose energy C the rock will gain energy and the water will lose energy D the rock will lose energy and the water will gain energy
Answer
D
SLIDE 29
29
13 If you put a hot rock in cold water, and your system is the rock, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures
Answer
A
SLIDE 30
30
14 If you put a hot rock in cold water, and your system is the water, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures
Answer
B
SLIDE 31
31
15 If you put an ice cube in water, and your system is the ice, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures
Answer
B
SLIDE 32
32
16 If you put an ice cube in water, and your system is the water, the process is _______. A exothermic B endothermic C neither, there is no net change of energy D it depends on the exact temperatures
Answer
A
SLIDE 33
33
17 When NaOH dissolves in water, the temperature of solution increases. This reaction is________. A exothermic B endothermic
Answer
A
SLIDE 34
34
18 When CaCl 2 dissolves in water the temperature of water drops. This reaction is _____. A endothermic B exothermic
Answer
A
SLIDE 35
35
19 Water droplets evaporating from the skin surface will make you feel cold. This process is _____. A exothermic for water B endothermic for skin C exothermic for skin D endothermic for water E A and C F C and D
Answer
F
SLIDE 36
36
State Functions**
Return to Table of Contents
SLIDE 37
37
State Functions
The internal energy of a system is independent of the path by which the system achieved that state. Below, the water could have reached room temperature from either direction...it makes no difference to the final energy of the system if it reached its final temperature by heating up or cooling down. 50g 100C 50g 25C 50g 0C
Cooling Heating
∗ ∗
∗
SLIDE 38
38
State Functions
Internal energy is a state function. It depends only on the present state of the system, not
- n the path by which the system arrived at that state.
D∆E depends only on Einitial and Efinal
∗ ∗
SLIDE 39
39
These multiple paths explains how engines, air conditioners, batteries, heaters, etc. operate.
State Functions
They move between energy states while performing some task.
∗ ∗
SLIDE 40
40
State Functions
Understanding thermodynamics enables us to harness energy flow for useful purposes.
∗ ∗
Charged battery Heat Work
Energy lost by battery
Discharged battery
∆E
Heat
A B
For instance, whether the battery is shorted out or is discharged by running the fan, its ∆E is the same. But q and w are different in the two cases. If the battery shorts
- ut, all of the energy is lost as
heat, whereas if it is used to run the fan, some energy is used to do work. Q and w are NOT state functions.
A B
SLIDE 41
41
Work Done by a Gas
When a gas expands it does work on its surroundings: W = Fd. In this case: F = PA and d = ∆h. W = Fd W = (PA)∆h W = P∆V A = cross sectional area P= F/A P= F/A
∆h ∆V
h Initial state Final state
∗ ∗
SLIDE 42
42
Since the gas expands, and does work on its surroundings, the energy
- f the gas decreases, this is
considered negative work. W = P∆V
Work Done By a Gas
Initial state Final state A = cross sectional area P= F/A P= F/A
∆h
h
∆V ∗ ∗
() work is done by system
SLIDE 43
43
If the surroundings compress the gas, decreasing its volume, this increases the energy of the gas, it is considered positive work. W = + P∆V
Work Done By a Gas
P= F/A P= F/A h
∆V
Initial state Final state A = cross sectional area
∆h ∗ ∗
(+) work is done on system
SLIDE 44
44
We can measure the work done by the gas if the reaction is done in a vessel fitted with a piston.
Work Done by a Gas
∗ ∗
HCl Solution HCl Solution + Zinc
SLIDE 45
45
Enthalpy
Return to Table of Contents
SLIDE 46
46
Enthalpy
The word "enthalpy" is derived from the Greek noun "enthalpos" which means heating. The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance to occupy. It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.
H = E + W ∆H = ∆E + W
SLIDE 47
47
At constant pressure, the change in enthalpy is the heat gained or lost by the system.
Enthalpy
Note: This is only true at constant pressure. See ** for more information (W = P∆V)
∆H = (q+W) W ∆H = q
SLIDE 48
48
Enthalpy
The enthalpy of a system (H) is a combination of the internal energy of a system (E) plus the work that needs to be done to create the space for the substance (PV) to occupy. It is impossible to measure enthalpy, H, directly. Only the change (∆H) can be measured.
H = E + PV ∆H = ∆E + PV
**
SLIDE 49
49
If a process takes place at constant pressure (as most processes we study do), we can account for heat flow during the process by measuring the enthalpy of the system. At constant pressure, the change in enthalpy is the heat gained or lost by the system.
Enthalpy
∆H = ∆E + ∆(PV) ∆H = ∆E + P∆V ∆H = (q+w) w ∆H = qp (at constant pressure)
**
SLIDE 50
50
Enthalpy is an extensive property; its value depends on the quantity of the substance present. Combustion of 16 grams of CH4 ∆H = 891 kJ Combustion of 32 grams of CH4 ∆H = 1782 kJ
∆H for a reaction in the forward direction is equal in size,
but opposite in sign, to ∆H for the reverse reaction. CH4(g) + 2O2(g) > CO2(g) + 2H2O(g) ∆H = 891 kJ 2H2O(g) + CO2(g) > CH4(g) + 2O2(g) ∆H = +891 kJ
∆H for a reaction depends on the state of the products and
the state of the reactants. CH4(g) + 2O2(g) > CO2(g) + 2H2O(g) ∆H = 891 kJ CH4(g) + 2O2(g) > CO2(g) + 2H2O(l) ∆H = 979 kJ
Enthalpy
SLIDE 51
51
20 When 114 grams of gasoline combust, the ∆H is equal to 5,330 kJ. What is the ∆H for combustion of 57 grams of gasoline?
Answer
2665 kJ
SLIDE 52
52
Endothermic and Exothermic Processes
A process is endothermic when ∆H is positive. ∆H>0 System Endothermic Surroundings Heat +q ∆H<0 System Exothermic Surroundings Heat q A process is exothermic when ∆H is negative.
SLIDE 53
53
21 The reaction A + B > C is endothermic. The DH for this reaction is +50 J. What is the DH for the reaction C > A + B?
A Cannot be determined. B
0.02 J
C
50 J
D 100 J
Answer
C
SLIDE 54
54
22 The reaction A + B > C is exothermic. The ∆H for this reaction is 150 J. What is the ∆H for the reaction C > A + B?
A
+150
B
zero C 150 D this reaction will not happen
Answer
A
SLIDE 55
55
23 Dissolving NaOH in water will increase the temperature of the solution. This reaction is _____
A
exothermic
B
endothermic
C
adiabatic
D
isothermal
Answer
A
SLIDE 56
56
24 NH3NO3 + H2O > NH4
+ + NO3 + OH
∆H = +25.69 kJ/mol. This reaction is exothermic.
Yes
No
Answer
No
SLIDE 57
57
Measuring Enthalpy Changes: Calorimetry
Return to Table of Contents
SLIDE 58
58
Since we cannot know the exact enthalpy of the reactants and products, we measure ∆H through calorimetry, the measurement of heat flow by making use of the expression: ∆H = qp
Measuring Enthalpy Changes: Calorimetry
The subscript p on q means the process is occurring at constant pressure. No work is being done only heat is being
- exchanged. You will not always see the subscript but it is
implied when we are speaking of Enthalpy.
SLIDE 59
59
Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1°C) is its heat capacity.
The amount of energy required to raise the temperature of one gram of a substance by 1 K (1°C) is its specific heat (c).
SLIDE 60
60
25 Heat capacity is an example of an A Intensive property B Extensive property
Answer
B
SLIDE 61
61
26 Specific heat is an example of an A Intensive property B Extensive property
Answer
A
SLIDE 62
62
Heat Capacity and Specific Heat
Specific heat = heat transferred mass x temperature change
m∆T c = q
SLIDE 63
63
27 The specific heat of marble is 0.858 J/gK. How much heat (in J) is required to raise the temperature of 20g of marble from 22 °C to 45 °C?
q = mc∆T
Answer
395J
SLIDE 64
64
28 An 26 g sample of wood (c = 1.674 J/(gK)) absorbs 200 J of heat, upon which the temperature of the sample increases from 20.0 °C to ______°C.
Answer
24.6 C
q = mc∆T
SLIDE 65
65
29 A sample of silver (c = 0.236 J/gK) absorbs 800 J of heat, upon which the temperature of the sample increases from 50.0 °C to 80°C. What is the mass of the sample?
Answer
113 g
q = mc∆T
SLIDE 66
66
sparknotes.com
thermometer styrofoam cup stirrer insulated cover
Constant Pressure Calorimetry
Many chemical reactions happen in aqueous solutions. The apparatus to the left is a calorimeter. How could you use it to measure the heat change for a chemical reaction in an aqueous solution?
Answer
By carrying out a reaction in aqueous solution in a simple calorimeter like this one, the heat change for a system can be indirectly measured by measuring the heat change for the water in the calorimeter.
SLIDE 67
67
Because the specific heat for water is well known (4.184 J/gK), we can measure ∆H for the reaction with this equation: D∆H = q = mc∆T (at constant pressure)
Constant Pressure Calorimetry
sparknotes.com
thermometer styrofoam cup stirrer insulated cover
SLIDE 68
68
Constant Pressure Calorimetry
Example: A student wishes to determine the enthalpy change when ammonium chloride (NH4Cl) dissolves in water. The student masses
- ut 20.00 grams of ammonium chloride and adds it to 500 grams of
water in a styrofoam cup at a temperature of 16.1 Celsius. The student
- bserves the temperature to decrease to 13.2 Celsius. What is the
enthalpy change for the dissolution of ammonium chloride?
- 1. Find enthalpy change of solution using q = mc∆T
= 520 g x 2.9 C x 4.2 J = 6,334 J
- 2. Since heat released by surroundings (solution) is equal
to the heat gained by system (ammonium chloride). ∆H for dissolving of NH4Cl = +6334 J g℃
SLIDE 69
69
Bomb Calorimetry
Reactions can be carried
- ut in a sealed “bomb”
such as this one. The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Thermometer Ignition wires Oxygen atmosphere Sample Water
SLIDE 70
70
Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, ∆E, not ∆H. For most reactions, the difference is very small.
Bomb Calorimetry (Constant Volume)
Thermometer Ignition wires Oxygen atmosphere Sample Water
SLIDE 71
71
30 The reaction below takes place in a bomb calorimeter. If the student found that the temperature of the water was 12.2 Celsius to start and the heat capacity of the calorimeter was 34 J/C, what must be the enthalpy change of the reaction if the temperature of the water increased to 15.6 Celsius? 4Fe(s) + 3O2(g) > 2Fe2O3(s)
Answer
115.6 J
SLIDE 72
72
31 A mysterious meteorite is discovered in your backyard. To determine its identity, you determine its specific heat. The 164.6 gram sample of metal is heated to 90 C and then dumped in 300 grams of water in a styrofoam cup at an initial temperature of 10 C. After the metal is added, the temperature rises to 11.3 C. Identify the metal. A B C Metal Specific Heat (J/gC) Cu 0.386 Au 0.126 Al 0.900
Answer
B
SLIDE 73
73
Energy Changes Associated with Changes of State
Return to Table of Contents
SLIDE 74
74
Energy Changes Associated with Changes of State
Chemical and physical changes are usually accompanied by changes in energy. Recall the following terms: When energy is put into the system, the process is called _____________. When energy is released by the system, the process is called _____________.
Answer
endothermic exothermic
SLIDE 75
75
Endothermic processes Exothermic processes
Energy is taken into the system from the surroundings (∆H > 0) Energy is released from the system to surroundings (∆H < 0)
Energy Changes Associated with Changes of State
Answer
boiling or evaporating a liquid sublimation of a solid L R melting a solid boiling or evaporating a liquid sublimation of a solid freezing a liquid condensing a gas deposition of a gas Fill in the blanks
SLIDE 76
76
Solid Gas Liquid Vaporization Condensation Melting Sublimation Freezing Deposition Energy of system
Phase Changes
SLIDE 77
77
32 Which of the following is/are exothermic?
- I. boiling
- II. melting
- III. freezing
A I only B I and II only C III only D I, II and III
Answer
C
SLIDE 78
78
Energy Changes Associated with Changes of State
The heat of fusion (∆Hfus) is the energy required to change a solid at its melting point to a liquid. The heat of vaporization (∆Hvap) is defined as the energy required to change a liquid at its boiling point to a gas. Heat of fusion(H2O) = 6.0 kJ/mol Heat of vaporization(H2O) = 41.0 kJ/mol Class Question: Why is the heat of vaporization much higher than the heat of fusion for a substance? In order to change phase from a solid to liquid, the particle attractions need only be strained somewhat. When a material changes from a liquid to a gas, the particle attractions must be essentially broken. move for answer
SLIDE 79
79
10 40 80 Heat of vaporization Heat of fusion 24 5 29 41 58 23 6 7
Energy Changes Associated with Changes of State
Note that these quantities are usually per 1.00 mole, whereas q = mc∆T involves mass in grams. Butane Diethyl ether Water Mercury Heat of fusion and vaporization kJ/mol
SLIDE 80
80
33 The heat of vaporization for butane is 24 kJ/mol. How much energy is required to vaporize 2 mol of butane? A 2kJ B 12 kJ C 22 kJ D 48 kJ
Answer
SLIDE 81
81
34 The heat of vaporization for butane is 24 kJ/
- mol. How much energy is required to vaporize
0.33 mol of butane? A 8kJ B 12 kJ C 16 kJ D 72 kJ
Answer
SLIDE 82
82
35 How much energy is required to melt 0.5 mol of water? A 2kJ B 3 kJ C 6 kJ D 12 kJ Heat of fusion for H2O (s) Heat of vaporization for H2O (l) 6 kJ/mol 41 kJ/mol
Answer
SLIDE 83
83
36 How much energy is released when 3.0 mol of water freezes? A 2kJ B 3 kJ C 18 kJ D 123 kJ Heat of fusion for H2O (s) Heat of vaporization for H2O (l) 6 kJ/mol 41 kJ/mol
Answer
SLIDE 84
84
37 How much energy is needed to melt 10.0 mol solid Hg? A 2.3 kJ B 5.8 kJ C 23 kJ D 230 kJ Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kJ/mol 58 kJ/mol E 580 kJ
Answer
SLIDE 85
85
38 How much energy is needed to vaporize 2.0 mol Hg (l)? Heat of fusion for Hg (s) Heat of vaporization for Hg (l) 23 kJ/mol 58 kJ/mol
Answer
116 kJ
SLIDE 86
86
Energy Changes Associated with Changes of State
This graph is called a heating curve. It illustrates how temperature changes over time as constant heat is applied to a pure solid substance. Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 87
87
From A to B, ice is heating up from 25oC to 0oC.
Energy Changes Associated with Changes of State
Since Q = mc∆T ∆T = Q So the slope is (mc) 1 (mc) Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 88
88
From B to C, ice is melting. The added heat is breaking the hydrogen bonds of the solid, so the temperature is constant. That's why the slope is zero.
Energy Changes Associated with Changes of State
Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 89
89
From C to D, liquid water is heating up from 0 C to 100 C. Once again, the slope is 1/(mc). But "c" is different for all the phases
- f a substance, so
the slope is different for solid, liquid and gaseous H2 O.
Energy Changes Associated with Changes of State
Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 90
90
From D to E, liquid water is boiling into vapor. The added heat is breaking the IM forces
- f the liquid, so the
temperature is constant. That's why the slope is zero.
Energy Changes Associated with Changes of State
Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liqui d water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 91
91
From E to F water vapor, steam, is heating up from 100 C to 125 C. Once again, the slope is 1/(mc). But "c" is different for all the phases
- f a substance,
so the slope is different for for solid, liquid and gaseous H2 O.
Energy Changes Associated with Changes of State
Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 92
92
Recall that slope = where m = mass and C = specific heat of the substance
Energy Changes Associated with Changes of State
1 mC (q) Energy Added (Joules) (DT) Temperature (C) A B C D This graph shows heat transfer versus change in temperature for 1 gram of 4 different substances.
SLIDE 93
93
39 Which substance has the lowest specific heat? A A B B C C D D
(DT) Temperature (C) A B C D (q) Energy Added (Joules)
Answer
A
SLIDE 94
94
40 Which substance requires the highest amount of heat added to raise the temperature? A A B B C C D D
(DT) Temperature (C) A B C D (q) Energy Added (Joules)
Answer
D
SLIDE 95
95
41 Which segment(s) contain solid sodium chloride? A AB only B AB and BC C AB, BC and CD D BC, CD and DE
Answer
B
SLIDE 96
96
42 Which segment(s) contain liquid sodium chloride? A AB only B AB and BC C AB, BC and CD D BC, CD and DE
Answer
D
SLIDE 97
97
43 What is the melting point (in oC) of sodium chloride?
Answer
801 C
SLIDE 98
98
44 What is the freezing point (in oC) of sodium chloride?
Answer
801 C
SLIDE 99
99
45 Which is greater: the specific heat of solid NaCl,
- r the specific heat of molten (liquid) NaCl?
A solid B liquid C Cannot be determined. D They are equal.
Answer
B
SLIDE 100
100
Segments AB, CD, EF slope = nonzero T increasing KE increasing PE constant apply q = mc∆T Segments BC & DE slope = 0 ∆T = 0 KE constant PE increasing apply ∆Hfus or ∆Hvap
Features of a Heating Curve
Temperature (
0C)
Heat added (each division corresponds to 4 kJ) Liquid water Ice and liquid water (melting) Liquid water and vapor (vaporization) Water vapor
25 25 75 125 50 100
Ice A B C D E F
SLIDE 101
101
Recall that any given substance has a different value for specific heat as a solid, as a liquid and as a gas. Additionally, melting one mole of a substance (∆Hfus) and vaporizing one mole of the same substance (∆Hvap) require different amounts of energy.
Energy Changes Associated with Changes of State
SLIDE 102
102
Specific heat of ice 2.09 J/g℃ Specific heat of water 4.18 J/g℃ Specific heat of steam 1.84 J/g℃ Heat of fusion (∆Hfus) of water at 0℃ 6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g
Specifics about Water
SLIDE 103
103
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
It is a good idea to sketch out the segments first on a graph.
SLIDE 104
104
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Segment 1 Warming the ice from 15.0°C up to the substance's MP which is 0°C. 14 4 10 20 6 12 24
Temperature (C) Time (s)
SLIDE 105
105
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Segment 2 Melting the ice at 0°C.
14 4 10 20 6 12 24
Temperature (C) Time (s)
SLIDE 106
106
Calculating Energy Changes from a Heating Curve
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Segment 3 Warming the liquid from 0°C to 25°C.
14 4 10 20 6 12 24
Temperature (C) Time (s)
SLIDE 107
107
q1 = mc∆T q1 = (10.0g) (2.09 J/goC) (15.0oC) q1 = 313.5 J q1 = 0.314 kJ or 313.5J
We are now ready to apply either the calorimetry equation or ∆Hfus or ∆Hvap.
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
Segment 1 Warming the ice from 15.0°C up to the substance's MP which is 0°C.
SLIDE 108
108
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
q2 = (∆Hfus ) (# moles) q2 = (6.01 kJ/mol) [(10.0 g)/ 18.0 g/mol)] q2 = 3.34 kJ
- r
6.01/18 J/g x 10g = 3.34 kJ = 3340J
Segment 2 Melting the ice at 0 °C.
SLIDE 109
109
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
q3 = mc∆T q3 = (10.0g) (4.18 J/goC) (25.0oC) q3 = 1045 J q3 = 1.05 kJ
Segment 3 Warming the water from 0 °C to 25.0 °C.
SLIDE 110
110
Sample Problem: Calculate the enthalpy change (in kJ) for converting 10.0 g of ice at 15.0°C to water at 25.0 °C.
Calculating Energy Changes from a Heating Curve
Now, we add the heat changes for all 3 segments. First, make sure that all quantities are in kilojoules.
Total ∆H = (q1 + q2 + q3) kJ ∆H = (0.314 + 3.34 + 1.05) kJ ∆H = 4.6 kJ or 4698.5J
SLIDE 111
111
46 Calculate the enthalpy change in J of converting 75 g of ice at 11 ℃ to liquid water at 22℃. A 98,654 B 33,371 J C 8,621 D 26,474 E 35,096 J
Specific heat of ice 2.09 J/g℃ Specific heat of water 4.18 J/g℃ Specific heat of steam 1.84 J/g℃ Heat of fusion (∆Hfus)
- f water at 0℃
6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g Answer
SLIDE 112
112
47 Calculate the enthalpy change of converting 25 g of water vapor at 110 ℃ to liquid water at 50℃. A 69,765 B 69,765 C 59,315 D 70,685 J E 13,935
Specific heat of ice 2.09 J/g℃ Specific heat of water 4.18 J/g℃ Specific heat of steam 1.84 J/g℃ Heat of fusion (∆Hfus)
- f water at 0℃
6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g Answer
SLIDE 113
113
48 Calculate the enthalpy change in kJ of converting 31.8 g
- f ice at 0℃ to water vapor at 140℃.
A 109 kJ B 96 kJ C 104 kJ D 88 kJ E 109 kJ
Specific heat of ice 2.09 J/g℃ Specific heat of water 4.18 J/g℃ Specific heat of steam 1.84 J/g℃ Heat of fusion (∆Hfus)
- f water at 0℃
6.01 kJ/mol or 330 J/g Heat of vaporization (∆Hvap) of water at 100℃ 40.7 kJ/mol or 2600 J/g
Answer
SLIDE 114
114
Enthalpies of Reaction
Return to Table of Contents
SLIDE 115
115
Enthalpy of Reaction
The change in enthalpy, ∆H, is the enthalpy of the products minus the enthalpy of the reactants:
D∆H = Hproducts − Hreactants
CH4(g) + 2 O2(g) CO2(g) + 2H2O(l) ∆H1 = 890 kJ Enthalpy ∆H2 = 890 kJ
SLIDE 116
116
This quantity, ∆H, is called the enthalpy of reaction or the heat of
- reaction. The combustion of hydrogen in the
balloon below is an exothermic reaction and the energy released is equal to the heat of reaction.
Enthalpy of Reaction
2H2(g) + O2(g) 2H2O(g) ∆H<0 exothermic Enthalpy
SLIDE 117
117
How much energy is released when 4.0 mol of HBr is formed in this reaction? H2(g) + Br2(g) > 2HBr(g) ∆H = 72 kJ
Enthalpy of Reaction
Example Problem #1 X 4.0 mol HBr(g) 72kJ 2.0 mol HBr(g) X = 72kJ (4/2) X = 144kJ 144kJ of energy are released =
SLIDE 118
118
How much energy is released when 4.0 mol of HBr is formed in this reaction? H2(g) + Br2(g) > 2HBr(g) ∆H = 72 kJ