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Transcript
Unit 3:
Thermochemistry
Chemistry 3202
1
Unit Outline
 Temperature
and Kinetic Energy
 Heat/Enthalpy Calculation
Temperature changes (q = mc∆T)
Phase changes (q = n∆H)
Heating and Cooling Curves
Calorimetry (q = C∆T & above
formulas)
2
Unit Outline
 Chemical
Reactions
PE
Diagrams
Thermochemical Equations
Hess’s Law
Bond Energy
 STSE:
What Fuels You?
3
Temperature and Kinetic Energy
Thermochemistry is the study of energy
changes in chemical and physical changes
eg. dissolving
burning
phase changes
4
Temperature - a measure of the average
kinetic energy of particles in a substance
- a change in temperature means
particles are moving at different speeds
- measured in either Celsius degrees or
degrees Kelvin
Kelvin = Celsius + 273.15
5
The Celsius scale is based on the
freezing and boiling point of water
The Kelvin scale is based on absolute
zero - the temperature at which
particles in a substance have zero
kinetic energy.
6
p. 628
7
K
°C
50.15
48
450.15
-200
8
300 K
# of particles
500 K
Kinetic Energy
9
Heat/Enthalpy Calculations
system - the part of the universe being studied
and observed
surroundings - everything else in the universe
open system - a system that can exchange
matter and energy with the surroundings
eg. an open beaker of water
a candle burning
closed system - allows energy transfer but is
closed to the flow of matter.
10
isolated system – a system completely closed
to the flow of matter and energy
heat - refers to the transfer of kinetic energy
from a system of higher temperature to a
system of lower temperature.
- the symbol for heat is q
WorkSheet: Thermochemistry #1
11
Part A: Thought Lab (p. 631)
12
Part B: Thought Lab (p. 631)
13
Heat/Enthalpy Calculations
specific heat capacity – the energy , in
Joules (J), needed to change the
temperature of one gram (g) of a
substance by one degree Celsius (°C).

The symbol for specific heat capacity is
a lowercase c
14

A substance with a large value of c can
absorb or release more energy than a
substance with a small value of c.
ie. For two substances, the substance
with the larger c will undergo a smaller
temperature change with the same loss
or gain of heat.
15
FORMULA
q = mc∆T
q = heat (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change
= T2 – T1
= Tf – Ti
16
eg. How much heat is needed to raise the
temperature of 500.0 g of water from
20.0 °C to 45.0 °C?
q = m c ∆T
for c, m, ∆T, T2 & T1
 p. 634 #’s 1 – 4
 p. 636 #’s 5 – 8
WorkSheet: Thermochemistry #2
Solve
17
heat capacity - the quantity of energy , in
Joules (J), needed to change the
temperature of a substance by one
degree Celsius (°C)
The symbol for heat capacity is
uppercase C
 The unit is J/ °C or kJ/ °C

18
FORMULA
C = mc
q = C ∆T
Your Turn
C = heat capacity
c = specific heat
capacity
m = mass
∆T = T2 – T1
p.637 #’s 11-14
WorkSheet: Thermochemistry #3
19
Enthalpy Changes
enthalpy change - the difference between the
potential energy of the reactants and the
products during a physical or chemical
change
AKA: Heat of Reaction or ∆H
20
Endothermic Reaction
Products
PE
∆H
Reactants
Reaction Progress
21
Endothermic Reaction
Products
PE
Enthalpy
∆H
∆H
Reactants
Reaction Progress
22
Products
Enthalpy
∆H is +
Reactants
Endothermic
23
reactants
Enthalpy
∆H is products
Exothermic
24
Enthalpy Changes in Reactions
 All
chemical reactions require bond
breaking in reactants followed by
bond making to form products
 Bond breaking requires energy
(endothermic) while bond formation
releases energy (exothermic)
see p. 639
25
26
Enthalpy Changes in Reactions
endothermic reaction - the energy
required to break bonds is greater than
the energy released when bonds form.
ie. energy is absorbed
exothermic reaction - the energy
required to break bonds is less than the
energy released when bonds form.
ie. energy is produced
27
Enthalpy Changes in Reactions
1.
∆H can represent the enthalpy change for
a number of processes
Chemical reactions
∆Hrxn – enthalpy of reaction
∆Hcomb – enthalpy of combustion
(see p. 643)
28
2.
Formation of compounds from elements
∆Hof – standard enthalpy of formation
The standard molar enthalpy of formation is
the energy released or absorbed when one
mole of a compound is formed directly from
the elements in their standard states. (see
p. 642)
eg.
C(s) + ½ O2(g) → CO(g)
ΔHfo = -110.5 kJ/mol
29
Use the equations below to determine the
ΔHfo for CH3OH(l) and CaCO3(s)
2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ
2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)
30
Phase Changes (p.647)
∆Hvap – enthalpy of vaporization
3.
∆Hfus – enthalpy of melting
∆Hcond – enthalpy of condensation
∆Hfre – enthalpy of freezing
eg. H2O(l)  H2O(g)
Hg(l)  Hg(s)
ΔHvap = +40.7 kJ/mol
ΔHfre = -23.4 kJ/mol
31
4.
Solution Formation (p.647, 648)
∆Hsoln – enthalpy of solution
eg.
ΔHsoln, of ammonium nitrate is +25.7 kJ/mol.
NH4NO3(s) + 25.7 kJ → NH4NO3(aq)
ΔHsoln, of calcium chloride is −82.8 kJ/mol.
CaCl2(s) → CaCl2(aq) + 82.8 kJ
32
Three ways to represent an enthalpy change:
1. thermochemical equation - the energy
term written into the equation.
2. enthalpy term is written as a separate
expression beside the equation.
3. enthalpy diagram.
33
eg. the formation of water from the elements
produces 285.8 kJ of energy.
1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ
thermochemical
equation
2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol
34
enthalpy
diagram
H2(g) + ½ O2(g)
3.
Enthalpy
(H)
examples:
questions
∆Hf = -285.8 kJ/mol
H2O(l)
pp. 641-643
p. 643 #’s 15-18
WorkSheet: Thermochemistry #4
35
Calculating Enthalpy Changes
FORMULA:
q = n∆H
q = heat (kJ)
n = # of moles
m
n
M
∆H = molar enthalpy
(kJ/mol)
36
eg. How much heat is released when
50.0 g of CH4 forms from C and H ?
(p. 642)
n 

50.0 g
16.05 g / mol
3.115 mol
q = nΔH
= (3.115 mol)(-74.6 kJ/mol)
= -232 kJ
37
eg. How much heat is released when
50.00 g of CH4 undergoes complete
combustion?
(p. 643)
50.0 g
n 
16.05 g / mol
 3.115 mol
q = nΔH
= (3.115 mol)(-965.1 kJ/mol)
= -3006 kJ
38
eg. How much energy is needed to change
20.0 g of H2O(l) at 100 °C to steam at 100 °C ?
Mwater = 18.02 g/mol
ΔHvap = +40.7 kJ/mol
20.0 g
n 
18.02 g / mol
 1.110 mol
q = nΔH
= (1.110 mol)(+40.7 kJ/mol)
= +45.2 kJ
39
∆Hfre and ∆Hcond have the opposite sign
of the above values.
40
eg. The molar enthalpy of solution for
ammonium nitrate is +25.7 kJ/mol. How
much energy is absorbed when 40.0 g of
ammonium nitrate dissolves?
40.0 g
n
80.06g / mol
 0.4996 mol
q = nΔH
= (0.4996 mol)(+25.7 kJ/mol)
= +12.8 kJ
41
What mass of ethane, C2H6, must be
burned to produce 405 kJ of heat?
ΔH = -1250.9 kJ
- 405 kJ
n
q = - 405 kJ
 1250.9 kJ
q = nΔH
q
n
H
n = 0.3238 mol
m= nxM
= (0.3238 mol)(30.08 g/mol)
= 9.74 g
42
Complete:
p. 643 #’s 15 - 18
p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29
19. (a) -8.468 kJ (b) -7.165 kJ
20. -1.37 x103 kJ
21. (a) -2.896 x 103 kJ
21. (b) -6.81 x104 kJ
21. (c) -1.186 x 106 kJ
22. -0.230 kJ
23. 3.14 x103 g
43
24. 2.74 kJ
25.(a) 33.4 kJ (b) 33.4 kJ
26.(a) absorbed (b) 0.096 kJ
27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq)
(b) 1.69 kJ
(c) cool; heat absorbed from water
28. 819.2 g
29. 3.10 x 104 kJ
44
p. 638 #’ 4 – 8
pp. 649, 650 #’s 3 – 8
p. 657, 658 #’s 9 - 18
WorkSheet: Thermochemistry #5
45
Heating and Cooling Curves
Demo: Cooling of p-dichlorobenzene
Time (s) Temperature (°C) Time (s)
Temperature (°C)
46
Cooling curve for p-dichlorobenzene
Temp. 80
(°C )
50
KE
liquid
PE
freezing
KE
solid
20
Time
47
Heating curve for p-dichlorobenzene
80
Temp.
(°C )
50
KE
PE
20
KE
Time
48
What did we learn from this demo??
 During
a phase change temperature
remains constant and PE changes
 Changes
in temperature during
heating or cooling means the KE of
particles is changing
49
p. 651
50
p. 652
51
p. 656
52
Heating Curve for H20(s) to H2O(g)
1.
2.
A 40.0 g sample of ice at -40 °C is
heated until it changes to steam and
is heated to 140 °C.
Sketch the heating curve for this
change.
Calculate the total energy required for
this transition.
53
q = mc∆T
140
100
q = n∆H
q = mc∆T
Temp.
(°C )
q = n∆H
0
q = mc∆T
-40
Time
54
Data:
cice = 2.01 J/g.°C
cwater = 4.184 J/g.°C
csteam = 2.01 J/g.°C
ΔHfus = +6.02 kJ/mol
ΔHvap = +40.7 kJ/mol
55
warming ice:
q = mc∆T
= (40.0)(2.01)(0 - -40)
= 3216 J
warming steam:
q = mc∆T
= (40.0)(2.01)(140 -100)
= 3216 J
warming water:
q = mc∆T
= (40.0)(4.184)(100 – 0)
= 16736 J
56
moles of water:
n = 40.0 g
18.02 g/mol
= 2.22 mol
melting ice:
q = n∆H
= (2.22 mol)(6.02 kJ/mol)
= 13.364 kJ
boiling water:
q = n∆H
= (2.22 mol)(40.7 kJ/mol)
= 90.354 kJ
57
Total Energy
90.354 kJ
13.364 kJ
3216 J
3216 J
16736 J
127 kJ
58
Practice
p. 655: #’s 30 – 34
WorkSheet:
pp. 656: #’s 1 - 9
Thermochemistry #6
p. 657 #’s 2, 9
p. 658 #’s 10, 16 – 20
30.(b) 3.73 x103 kJ
31.(b) 279 kJ
32.(b) -1.84 x10-3 kJ
33.(b) -19.7 kJ -48.77 kJ
34. -606 kJ
59
Law of Conservation of Energy (p. 627)
The total energy of the universe is constant
∆Euniverse = 0
First Law of Thermodynamics
Universe = system + surroundings
∆Euniverse = ∆Esystem + ∆Esurroundings
∆Euniverse = ∆Esystem + ∆Esurroundings = 0
OR
OR
∆Esystem = -∆Esurroundings
qsystem = -qsurroundings
60
Calorimetry (p. 661)
calorimetry - the measurement of heat
changes during chemical or physical
processes
calorimeter - a device used to measure
changes in energy
2 types of calorimeters
1. constant pressure or simple
calorimeter (coffee-cup calorimeter)
2. constant volume or bomb calorimeter.
61
Simple
Calorimeter
p.661
62
a
simple calorimeter consists of an
insulated container, a thermometer, and
a known amount of water
 simple calorimeters are used to measure
heat changes associated with heating,
cooling, phase changes, solution
formation, and chemical reactions that
occur in aqueous solution
63
 to
calculate heat lost or gained by a
chemical or physical change we apply the
first law of thermodynamics:
qsystem = -qcalorimeter
Assumptions:
- the system is isolated
- c (specific heat capacity) for water is not
affected by solutes
- heat exchange with calorimeter can be
ignored
64
eg.
A simple calorimeter contains 150.0 g of
water. A 5.20 g piece of aluminum alloy at
525 °C is dropped into the calorimeter
causing the temperature of the calorimeter
water to increase from 19.30°C to
22.68°C.
Calculate the specific heat capacity of the
alloy.
65
eg.
The temperature in a simple calorimeter
with a heat capacity of 1.05 kJ/°C changes
from 25.0 °C to 23.94 °C when a very cold
12.8 g piece of copper was added to it.
Calculate the initial temperature of the
copper. (c for Cu = 0.385 J/g.°C)
66
Homework
p. 664, 665 #’s 1b), 2b), 3 & 4
 p. 667, #’s 5 - 7

67
p. 665 # 4.b)
(60.4)(0.444)(T2 – 98.0) = -(125.2)(4.184)(T2 – 22.3)
26.818(T2 – 98.0) = -523.84(T2 – 22.3)
26.818T2 - 2628.2 = -523.84T2 + 11681
550.66T2 = 14309.2
T2 = 26.0 °C
68
6. System (Mg)
m = 0.50 g
= 0.02057 mol
Find ΔH
Calorimeter
v = 100 ml
so m = 100 g
c = 4.184
T2 = 40.7
qMg = -qcal
T1 = 20.4
nΔH = -mcΔT
7. System
ΔH = -53.4 kJ/mol
n = CV
= (0.0550L)(1.30 mol/L)
= 0.0715 mol
Calorimeter
v = 110 ml
so m = 110 g
c = 4.184
T1 = 21.4
Find T2
69
Bomb
Calorimeter
70
Bomb Calorimeter
used to accurately measure enthalpy
changes in combustion reactions
 the inner metal chamber or bomb contains
the sample and pure oxygen
 an electric coil ignites the sample
 temperature changes in the water
surrounding the inner “bomb” are used to
calculate ΔH

71
 to
accurately measure ΔH you need to
know the heat capacity (kJ/°C) of the
calorimeter.
 must account for all parts of the
calorimeter that absorb heat
Ctotal = Cwater + Cthermom.+ Cstirrer + Ccontainer
NOTE: C is provided for all bomb
calorimetry calculations
72
eg. A technician burned 11.0 g of octane in a
steel bomb calorimeter. The heat capacity of
the calorimeter was calibrated at 28.0 kJ/°C.
During the experiment, the temperature of the
calorimeter rose from 20.0 °C to 39.6 °C.
What is the enthalpy of combustion for octane?
73
eg.
1.26 g of benzoic acid, C6H5COOH(s), is burned in
a bomb calorimeter. The temperature of the
calorimeter and contents increases from 23.62 °C
to 27.14 °C. Calculate the heat capacity of the
calorimeter. (∆Hcomb = -3225 kJ/mol)
Homework
p. 675 #’s 8 – 10
WorkSheet: Thermochemistry #7
74
Hess’s Law of Heat Summation
 the
enthalpy change (∆H) of a physical or
chemical process depends only on the
beginning conditions (reactants) and the
end conditions (products)
 ∆H is independent of the pathway and/or
the number of steps in the process
 ∆H is the sum of the enthalpy changes of
all the steps in the process
75
eg. production of carbon dioxide
Pathway #1:
2-step mechanism
C(s) + ½ O2(g) → CO(g)
∆H = -110.5 kJ
CO(g) + ½ O2(g) → CO2(g)
∆H = -283.0 kJ
C(s) + O2(g) → CO2(g)
∆H = -393.5 kJ
76
eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g)
∆H = -393.5 kJ
77
Using Hess’s Law
 We
can manipulate equations with
known ΔH to determine the enthalpy
change for other reactions.
NOTE:
 Reversing an equation changes the sign
of ΔH.
 If we multiply the coefficients we must
also multiply the ΔH value.
78
eg.
Determine the ΔH value for:
H2O(g) + C(s) → CO(g) + H2(g)
using the equations below.
C(s) + ½ O2(g) → CO(g)
H2(g) + ½ O2(g) → H2O(g)
ΔH = -110.5 kJ
ΔH = -241.8 kJ
79
eg.
Determine the ΔH value for:
4 C(s) + 5 H2(g) → C4H10(g)
using the equations below.
Switch
ΔH (kJ)
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
-110.5
H2(g) + ½ O2(g) → H2O(g)
-241.8
C(s) + O2(g) → CO2(g)
Multiply
by 4
Multiply
by 5
-393.5
80
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g)
+110.5
5(H2(g) + ½ O2(g) → H2O(g)
4(C(s) + O2(g) → CO2(g)
-241.8)
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g)
5 H2(g) + 2½ O2(g) → 5 H2O(g)
4C(s) + 4 O2(g) → 4 CO2(g)
+110.5
-1209.0
-1574.0
-393.5)
Ans: -2672.5 kJ
81
Practice
pg. 681 #’s 11-14
WorkSheet:
Thermochemistry #8
82
Review
∆Hof (p. 642, 684, & 848)
The standard molar enthalpy of formation is
the energy released or absorbed when one
mole of a substance is formed directly from
the elements in their standard states.
∆Hof = 0 kJ/mol
for elements in the standard state
The more negative the ∆Hof, the more
stable the compound
83
Using Hess’s Law and ΔHf
Use the formation equations below to determine
the ΔH value for:
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
4 C(s) + 5 H2(g) → C4H10(g)
H2(g) + ½ O2(g) → H2O(g)
C(s) + O2(g) → CO2(g)
ΔHf (kJ/mol)
-2672.5
-241.8
-393.5
84
Using Hess’s Law and ΔHf
ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants)
eg. Use ΔHf , to calculate the enthalpy of
reaction for the combustion of glucose.
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
85
Use the molar enthalpy’s of formation to
calculate ΔH for the reaction below
Fe2O3(s) + 3 CO(g) → 3 CO2(g) + 2 Fe(s)
p. 688 #’s 21 & 22
86
Eg.
The combustion of phenol is represented by
the equation below:
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(g)
If ΔHcomb = -3059 kJ/mol, calculate the heat of
formation for phenol.
87
Bond Energy Calculations (p. 688)
 The
energy required to break a bond
is known as the bond energy.
 Each type of bond has a specific
bond energy (BE).
(table p. 847)
 Bond
Energies may be used to
estimate the enthalpy of a reaction.
88
Bond Energy Calculations (p. 688)
ΔHrxn = ∑BE(reactants) - ∑BE (products)
eg. Estimate the enthalpy of reaction for the
combustion of ethane using BE.
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)
Hint: Drawing the structural formulas for all
reactants and products will be useful here.
89
2
C
C
+ 7O=O
→ 4 O=C=O + 6 H-O-H
[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]
8236 - 11480
= -3244 kJ
p. 690 #’s 23,24,& 26
p. 691 #’s 3, 4, 5, & 7
90
Energy Comparisons
 Phase
changes involve the least amount
of energy with vaporization usually
requiring more energy than melting.
 Chemical changes involve more energy
than phase changes but much less than
nuclear changes.
 Nuclear reactions produce the largest ΔH
eg.
nuclear power, reactions in the sun
91
STSE

What fuels you? (Handout)
92
aluminum alloy
m = 5.20 g
T1 = 525 ºC
T2 =
ºC
FIND c for Al
water
m = 150.0 g
T1 = 19.30 ºC
T2 = 22.68 ºC
c = 4.184 J/g.ºC
qsys = - qcal
mcT = - mc T
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
BACK
c = 0.812 J/g.°C
93
copper
m = 12.8 g
T2 =
ºC
c = 0.385 J/g.°C
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
FIND T1 for Cu
qsys = - qcal
mcT = - CT
(12.8)(0.385)(23.94 – T1) = -(1050)(23.94 – 25.0)
4.928 (23.94 – T1) = 1113
23.94 – T1= 1113/4.928
23.94 – T1= 225.9
BACK
T1= -202 ºC
94
q
heat
J or kJ
c
Specific heat
capacity
Heat capacity
J/g.ºC
C
ΔH
Molar heat or
molar enthalpy
kJ/ ºC or
J/ ºC
kJ/mol
95