Unit 3: Thermochemistry
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Transcript of Unit 3: Thermochemistry
Unit Outline
Temperature and Kinetic Energy Heat/Enthalpy Calculation
Temperature changes (q = mc∆T)Phase changes (q = n∆H) Heating and Cooling CurvesCalorimetry (q = C∆T & above
formulas)
2
May 11
Unit Outline
Chemical ReactionsPE DiagramsThermochemical EquationsHess’s LawBond Energy
STSE: What Fuels You?
3
May 11
Temperature and Kinetic Energy
Thermochemistry is the study of energy changes in chemical and physical changes
eg. dissolving
burning
phase changes
4
May 11
Temperature , T, measures 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
May 11
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
May 11
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
May 11
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
May 11
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
Heat/Enthalpy Calculations
14
May 12
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
May 12
FORMULA
q = mc∆T
q = heat (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change
= T2 – T1
= Tf – Ti
16
May 12
eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C?
Solve q = m c ∆T
for c, m, ∆T, T2 & T1
p. 634 #’s 1 – 4 p. 636 #’s 5 – 8
WorkSheet: Thermochemistry #2
17
May 12
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
May 13
FORMULA
C = mc
q = C ∆T
C = heat capacity
c = specific heat capacity
m = mass
∆T = T2 – T1
Your Turn p.637 #’s 11-14
WorkSheet: Thermochemistry #319
May 13
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
May 18
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. 63925
May 18
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 produced27
May 18
Enthalpy Changes in Reactions
∆H can represent the enthalpy change for a number of processes
1. Chemical reactions
∆Hrxn – enthalpy of reaction
∆Hcomb – enthalpy of combustion
(see p. 643)
28
May 18
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
May 18
Use the equation below to determine the ΔHfo
for CH3OH(l)
2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ
1 C(s) + 2 H2(g) + ½ O2(g) → 1 CH3OH(l) + 238.6 kJ
∆H = -238.6 kJ/mol
30
May 18
Use the equation below to determine the ΔHfo
for CaCO3(s)
2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)
2 Ca(s) + 2 C(s) + 3 O2(g) → 2 CaCO3(s) + 2413.8kJ
1 Ca(s) + 1 C(s) + 1.5 O2(g) → 1 CaCO3(s) + 1206.9 kJ
∆H = -1206.9 kJ/mol
31
May 18
Use the equation below to determine the ΔHf
o for PH3(g)
4 PH3(g) → P4(s) + 6 H2(g) + 21.6 kJ
a) +21.6 kJ/mol
b) -21.6 kJ/mol
c) +5.4 kJ/mol
d) -5.4 kJ/mol
32
May 18
3. Phase Changes (p.647)∆Hvap – enthalpy of vaporization (l → g)
∆Hfus – enthalpy of melting (fusion: s → l)
∆Hcond – enthalpy of condensation (g → l)
∆Hfre – enthalpy of freezing (l → s)
eg. H2O(l) H2O(g) ΔHvap =
Hg(l) Hg(s) ΔHfre =33
+40.7 kJ/mol
-23.4 kJ/mol
May 18
34
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
May 18
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.
35
May 18
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
2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol
thermochemical equation
36
May 18
3.
H2O(l)
H2(g) + ½ O2(g)
∆Hf = -285.8 kJ/molEnthalpy (H)
enthalpy diagram
examples: pp. 641-643questions p. 643 #’s 15-18
WorkSheet: Thermochemistry #4 37
May 18
Calculating Enthalpy Changes
FORMULA:
q = n∆H
q = heat (kJ)
n = # of moles
∆H = molar enthalpy
(kJ/mol)
38
M
mn
May 24
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 / m ol
q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ
39
3 .115 m ol
May 24
eg. How much heat is released when 50.00 g of CH4 undergoes complete combustion?
(p. 643)
n 50.0 g
16.05 g / m ol
3 .115 m ol
q = nΔH = (3.115 mol)(-965.1 kJ/mol) = -3006 kJ
40
May 24
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
n 20.0 g
18.02 g / m ol
1 .110 m ol
q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ
41
May 24
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?
n 40.0 g
80.06 g / m ol
0 .4996 m ol
q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ
43
May 24
What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?
ΔH = -1250.9 kJ
q = - 405 kJ
m = ?
H
q n
q = nΔH
kJ 1250.9
kJ 405- n
n = 0.3238 mol
m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g
44
May 24
Complete: p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29
p. 638 #’ 4 – 8
pp. 649, 650 #’s 3 – 8
p. 657, 658 #’s 9 - 18
45
WorkSheet: Thermochemistry #5
May 24
19. (a) -8.468 kJ (b) -7.165 kJ 20. -1.37 x103 kJ
21. (a) -2.896 x 103 kJ (b) -6.81 x104 kJ
21. (c) -1.186 x 106 kJ 22. -0.230 kJ
23. 3.14 x103 g 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 kJ46
Heating and Cooling Curves
Demo: Cooling of p-dichlorobenzene Time (s) Temperature
(°C)Time (s) Temperature (°C)
47
May 25
Cooling curve for p-dichlorobenzene
KE
PE
KE
solidfreezing
liquidTemp. (°C )
50
80
Time
20
48
May 25
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
50
May 25
Heating Curve for H20(s) to H2O(g)
A 40.0 g sample of ice at -40 °C is heated until it changes to steam and is heated to 140 °C.
1. Sketch the heating curve for this change.
2. Calculate the total energy required for this transition.
55
May 25
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
57
May 25
warming ice: (from -40 ºC to 0 ºC)
q = mc∆T
= (40.0)(2.01)(0 - -40)
= 3216 J
warming water: (from 0 ºC to 100 ºC)
q = mc∆T
= (40.0)(4.184)(100 – 0)
= 16736 J58
May 26
warming steam: (from 100 ºC to 140 ºC)
q = mc∆T
= (40.0)(2.01)(140 -100)
= 3216 J
59
n = 40.0 g 18.02 g/mol
= 2.22 mol
moles of water:
May 26
melting ice: (fusion)
q = n∆H
= (2.22 mol)(6.02 kJ/mol)
= 13.364 kJ
boiling water: (vaporization)
q = n∆H
= (2.22 mol)(40.7 kJ/mol)
= 90.354 kJ
60
May 26
Practicep. 655: #’s 30 – 34
pp. 656: #’s 1 - 9
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
WorkSheet: Thermochemistry #6
62
May 27
Law of Conservation of Energy (p. 627)
The total energy of the universe is constant
∆Euniverse = 0
Universe = system + surroundings
∆Euniverse = ∆Esystem + ∆Esurroundings
∆Euniverse = ∆Esystem + ∆Esurroundings = 0
OR ∆Esystem = -∆Esurroundings
OR qsystem = -qsurroundings
First Law of Thermodynamics
63
May 30
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. 64
May 30
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
66
May 30
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
ignored67
May 30
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.
68
May 30
aluminum alloy water
m = 5.20 g m = 150.0 g
T1 = 525 ºC T1 = 19.30 ºC
T2 = ºC T2 = 22.68 ºC
FIND c for Al c = 4.184 J/g.ºC
qsys = - qcal
mcΔT = - mc ΔT
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
c = 0.812 J/g.°C69
May 31
22.68
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)
70
May 31
copper
m = 12.8 g
T2 = ºC
c = 0.385 J/g.°C
FIND T1 for Cu
qsys = - qcal
mcT = - CT(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
T1= -202 ºC
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
71
May 31
23.94
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
73
6. System (Mg)m = 0.50 g = 0.02057 molFind ΔH
74
Calorimeter v = 100 ml
so m = 100 gc = 4.184T2 = 40.7T1 = 20.4
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 gc = 4.184T1 = 21.4Find T2
qMg = -qcal
nΔH = -mcΔT
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
76
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
77
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?
78
system (octane) calorimeter
m = 11.0 g
T2 = 39.6 ºC
T1 = 20.0 ºC
qsys = - qcal
n ΔH = -CΔT
(0.09627) ΔH = - (28.0)(39.6 – 20.0)
ΔH = -5700 kJ/mol
79
n = 11.0 g 114.26 g/mol = 0.09627 mol
C = 28.0 kJ/ºC
Find ΔHcomb
May 31
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)
80
benzoic acid calorimeter
T1 = 23.62 ºC
T2 = 27.14 ºC
Find C
m = 1.26 g
ΔHcomb = -3225 kJ/mol
May 31
Homework
p. 675 #’s 8 – 10
WorkSheet: Thermochemistry #7
81
qsys = - qcal
n ΔH = -CΔT
(0.01032) ΔH = - (C)(27.14 – 23.62)
C = 9.45 kJ/ ºC
n = 1.26 g 122.13 g/mol
= 0.01032 mol
May 31
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
82
June 2
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 kJ83
June 2
eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ
84
June 2
Using Hess’s Law We can manipulate equations with
known ΔH to determine an unknown enthalpy change.
NOTE: Reversing an equation changes the
sign of ΔH. If we multiply the coefficients we must
also multiply the ΔH value.
85
June 2
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) ΔH = -110.5 kJ
H2(g) + ½ O2(g) → H2O(g) ΔH = -241.8 kJ
86
reverse ? multiply ?June 2
eg.
Determine the ΔH value for:
4 C(s) + 5 H2(g) → C4H10(g)
using the equations below.
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) -393.5
Switch
Multiply by 5
Multiply by 4 87
June 2
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) +110.5
5(H2(g) + ½ O2(g) → H2O(g) -241.8)
4(C(s) + O2(g) → CO2(g) -393.5)
Ans: -2672.5 kJ
4 CO2(g) + 5 H2O(g) → C4H10(g) + 6½ O2(g) +110.5
5 H2(g) + 2½ O2(g) → 5 H2O(g) -1209.0
4C(s) + 4 O2(g) → 4 CO2(g) -1574.0
88
June 2
Review∆Ho
f (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 compound90
June 3
Use the formation equations below to determine the ΔH value for:
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
ΔHf (kJ/mol)
4 C(s) + 5 H2(g) → C4H10(g) -2672.5
H2(g) + ½ O2(g) → H2O(g) -241.8
C(s) + O2(g) → CO2(g) -393.5
Using Hess’s Law and ΔHf
91
June 3
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)
92
June 3
ΔHrxn = ∑ΔHf (products) - ∑ΔHf (reactants)
ΔHf
CO2(g) -393.5 kJ/mol
H2O(g) -241.8 kJ/mol
C6H12O6(s) -1274.5 kJ/mol
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)
ΔHrxn = [6(-393.5) + 6(-241.8)] – [1(-1274.5)+ 6(0)]
= [-2361 + -1450.8] - [-1274.5 + 0]
= - 2537.3 kJ93
June 3
Use the molar enthalpy of formation to calculate ΔH for this reaction:
Fe2O3(s) + 3 CO(g) → 3 CO2(g) + 2 Fe(s)
ΔHrxn = [3(-393.5) + 2(0) ] – [3(-110.5)+ 1(-824.2)]
= [-1180.5 + 0] - [-331.5 + -824.20]
= - 24.8 kJ
p. 688 #’s 21 & 2294
−824.2 kJ/mol−110.5 kJ/mol −393.5 kJ/mol
June 3
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.
95
June 7
−393.5 kJ/mol
−241.8 kJ/mol
ΔHcomb = -27.4 kJ/mol
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.
96
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.
97
C-C = 347C-H = 338O=O = 498
C=O = 745H-O = 460
p. 690 #’s 23,24,& 26p. 691 #’s 3, 4, 5, & 7
= -3244 kJ
98
[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]
+ 7 O = OCC → 4 O=C=O + 6 H-O-H2
8236 - 11480
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 ΔHeg. nuclear power, reactions in the sun
99
aluminum alloy water
m = 5.20 g m = 150.0 g
T1 = 525 ºC T1 = 19.30 ºC
T2 = ºC T2 = 22.68 ºC
FIND c for Al c = 4.184 J/g.ºC
qsys = - qcal
mcT = - mc T
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
c = 0.812 J/g.°C101
copper
m = 12.8 g
T2 = ºC
c = 0.385 J/g.°C
FIND T1 for Cu
qsys = - qcal
mcT = - CT(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
T1= -202 ºC
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
102