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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 CurvesCalorimetry (q = C∆T & above
formulas)
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Unit Outline
Chemical ReactionsPE DiagramsThermochemical EquationsHess’s LawBond Energy
STSE: What Fuels You?
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Temperature and Kinetic Energy
Thermochemistry is the study of energy changes in chemical and physical changes
eg. dissolving
burning
phase changes
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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
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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.
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p. 628
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K 50.15 450.15
°C 48 -200
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# of
par
ticle
s
500 K
300 K
Kinetic Energy
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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.
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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
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Part A: Thought Lab (p. 631)
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Part B: Thought Lab (p. 631)
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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
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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.
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FORMULA
q = mc∆T
q = heat (J)
m = mass (g)
c = specific heat capacity
∆T = temperature change
= T2 – T1
= Tf – Ti
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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
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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
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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
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
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Reaction Progress
PE
Reactants
Products
∆H
Endothermic Reaction
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Reaction Progress
PE
Reactants
Products
∆H ∆HEnthalpy
Endothermic Reaction
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∆H is +Enthalpy
Reactants
Products
Endothermic
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Enthalpy
products∆H is -
Exothermic
reactants
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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
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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
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)
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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
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Use the equations below to determine the ΔHf
o 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)
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3. Phase Changes (p.647)∆Hvap – enthalpy of vaporization
∆Hfus – enthalpy of melting
∆Hcond – enthalpy of condensation
∆Hfre – enthalpy of freezing
eg. H2O(l) H2O(g) ΔHvap =
Hg(l) Hg(s) ΔHfre =31
+40.7 kJ/mol
-23.4 kJ/mol
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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
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.
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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
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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 35
Calculating Enthalpy Changes
FORMULA:
q = n∆H
q = heat (kJ)
n = # of moles
∆H = molar enthalpy
(kJ/mol)
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M
mn
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
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3 .115 m ol
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
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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
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∆Hfre and ∆Hcond have the opposite sign of the above values.
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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
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What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?
ΔH = -1250.9 kJ
q = - 405 kJ
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
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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 g43
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
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p. 638 #’ 4 – 8
pp. 649, 650 #’s 3 – 8
p. 657, 658 #’s 9 - 18
WorkSheet: Thermochemistry #5
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Heating and Cooling Curves
Demo: Cooling of p-dichlorobenzene Time (s) Temperature
(°C)Time (s) Temperature (°C)
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Cooling curve for p-dichlorobenzene
Temp. (°C )
50
80
KE
PE
KE
Time
solidfreezing
liquid
20
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Heating curve for p-dichlorobenzene
Temp. (°C )
50
20
80
KE
KE
PE
Time
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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
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p. 651
50
p. 652
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p. 656
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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.
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Time
Temp. (°C )
-40
0
100
140
q = mc∆T
q = n∆H
q = mc∆T
q = mc∆T
q = n∆H
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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
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warming ice:
q = mc∆T
= (40.0)(2.01)(0 - -40)
= 3216 J
warming water:
q = mc∆T
= (40.0)(4.184)(100 – 0)
= 16736 J
warming steam:
q = mc∆T
= (40.0)(2.01)(140 -100)
= 3216 J
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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
n = 40.0 g 18.02 g/mol
= 2.22 mol
moles of water:
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Total Energy
90.354 kJ
13.364 kJ
3216 J
3216 J
16736 J
127 kJ
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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
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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
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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
SimpleCalorimeter
p.661
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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
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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
ignored64
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.
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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)
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Homework
p. 664, 665 #’s 1b), 2b), 3 & 4 p. 667, #’s 5 - 7
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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
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6. System (Mg)m = 0.50 g = 0.02057 molFind ΔH
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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
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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
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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
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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?
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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
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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
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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 kJ76
eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ
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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.
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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
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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 80
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
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Practice
pg. 681 #’s 11-14
WorkSheet: Thermochemistry #8
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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 compound83
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
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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)
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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 & 2286
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.
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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.
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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.
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+ 7 O = OCC
[2(347) + 2(6)(338) + 7(498)]
→ 4 O=C=O + 6 H-O-H
- [4(2)(745) + 6(2)(460)]
p. 690 #’s 23,24,& 26p. 691 #’s 3, 4, 5, & 7
= -3244 kJ
2
8236 - 11480
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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
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STSE
What fuels you? (Handout)
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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.°C93BACK
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
94BACK
q heat J or kJ
c Specific heat capacity
J/g.ºC
C Heat capacity kJ/ ºC or J/ ºC
ΔH Molar heat or molar enthalpy
kJ/mol
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