unit 3: thermochemistry - nlesdschool.nlesd.ca/~glennlegge/listing folder/chem 3202/unit 3... ·...
TRANSCRIPT
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Unit 3:
Thermochemistry
Chemistry 3202
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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)
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Unit Outline
Chemical Reactions
PE Diagrams
Thermochemical Equations
Hess’s Law
Bond 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 a solid
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
Kelvins
Kelvin = Celsius + 273.15
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p. 628
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K 50.15 450.15
°C 48 -200
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# o
f part
icle
s
500 K
300 K
Kinetic Energy
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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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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
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specific heat capacity - the amount of
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
amount of heat applied
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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 #3
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Enthalpy Changes
The difference between the potential energy of the reactants and the products during a physical or chemical change is the Enthalpy change or ∆H.
AKA: Heat of Reaction
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Reaction Progress
PE
Reactants
Products
∆H
Endothermic Reaction
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Reaction Progress
PE
Reactants
Products
∆H ∆H Enthalpy
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. 639
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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.
exothermic reaction - the energy
required to break bonds is less than the
energy released when bonds form.
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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)
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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
4. Solution Formation
∆Hsoln – enthalpy of solution
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There are three ways to represent any 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/mol Enthalpy
(H)
enthalpy
diagram
examples: pp. 641-643
questions p. 643 #’s 15-18
WorkSheet: Thermochemistry #4
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Calculating Enthalpy Changes
FORMULA:
q = n∆H
q = heat (kJ)
n = # of moles
∆H = molar enthalpy
(kJ/mol)
n m
M
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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
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eg. How much heat is released when
50.00 g of CH4 undergoes complete
combustion?
n
50.0 g
16.05 g / mol
3.115 mol
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 / mol
1.110 mol
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 / mol
0.4996 mol
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?
ΔHcomb = -1250.9 kJ/mol
q = - 405 kJ
H
q n
q = nΔH
kJ/mol 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. 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
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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)
0 100 80 20
10 80
20 60
30 50
40 50
50 50
60 40
70 30
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Cooling curve for p-dichlorobenzene
Temp.
(°C )
50
80
KE
PE
KE
Time
solid
freezing
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
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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 -40.0 °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 = 2.03 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)(2.03)(0 - -40.0)
= 3248 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
= (2.220 mol)(6.02 kJ/mol)
= 13.364 kJ
boiling water:
q = n∆H
= (2.220 mol)(40.7 kJ/mol)
= 90.354 kJ
n = 40.0 g
18.02 g/mol
= 2.220 mol
moles of water:
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Total Energy
90.354 kJ
13.364 kJ
3248 J
3216 J
16736 J
126.9 kJ
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Practice p. 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.
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Simple
Calorimeter
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 ignored
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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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(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 or 0.02057 mol
Find ΔH
Calorimeter
100 ml or m = 100 g
c = 4.184
T2 = 40.7
T1 = 20.4
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7. System
ΔH = -53.4 kJ/mol
n = CV = (0.0550L)(1.30 mol/L) = 0.0715 mol
Calorimeter
110 ml or m = 110 g
c = 4.184
T1 = 21.4
Find T2
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6.
qMg = -qcal
nΔH = -mcΔT
n = 0.50 g
24.31 g / mol 0.02057 mol
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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)
(∆Hcomb = -3225 kJ/mol), is burned in a
bomb calorimeter. The temperature of the
calorimeter and its contents increases
from 23.62 °C to 27.14 °C. Calculate the
heat capacity of the calorimeter.
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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. Enthalpy of the production of
carbon monoxide
Pathway #1: 2-step mechanism
C(s) + O2(g) → CO2(g) ∆H = -393.5 kJ
CO(g) + ½ O2(g) → CO2(g) ∆H = -283.0 kJ
C(s) + ½ O2(g) → CO(g) ∆H = -110.5 kJ
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eg. Enthalpy of the production of
carbon monoxide
Pathway #2: formation from the elements
C(s) + ½ O2(g) → CO(g) ∆H = -110.5 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
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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
∆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
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Determine the ΔH value for:
C4H10(g) + 6½ O2(g) → 4 CO2(g) + 5 H2O(g)
using the formation equations below.
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. Calculate the enthalpy of reaction for
the complete 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 & 22
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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 = O C C
[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,& 26
p. 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 ΔH
eg. nuclear power, reactions in the sun
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STSE
What fuels you? (Handout)
#’s 1 – 4, 7, 9, 10, 13, 14, & 16
(List only TWO advantages for #2)
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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
mc T = - mc T
(5.20)(c)(22.68 - 525 ) = -(150.0)(4.184)(22.68 – 19.30)
-2612 c = -2121
c = 0.812 J/g.°C
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copper
m = 12.8 g
T2 = ºC
c = 0.385 J/g.°C
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
T1= -202 ºC
calorimeter
C = 1.05 kJ/°C
T1 = 25.00 ºC
T2 = 23.94 ºC
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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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Group 2 Total Mass = 2.05 g
Group 5 Total Mass = 1.86 g