Handout Version (Temperature Dependent Term to Kinetic Models
Temperature and Kinetic Energy Heat/Enthalpy...
Transcript of Temperature and Kinetic Energy Heat/Enthalpy...
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Unit Outline
Temperature and Kinetic Energy
Heat/Enthalpy Calculation
Temperature changes (q = mcΔT)
Phase changes (q = nΔH)
Heating and Cooling Curves
Calorimetry (q = CΔT & above formulas)
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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 or physical changes
eg. dissolving
burning
phase changes
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Temperature, T, measures the average kinetic energy of particles.
A change in temperature means particles are moving at different speeds
Temperature is 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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K 321.15 50.15 73.15 450.15
°C 48 -223 -200 177
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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
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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
the specific heat capacity, c, of a substance reflects how well a substance can store energy
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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 heat loss or gain.
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FORMULA
q = mcΔT q = heat (J)
m = mass (g)
c = specific heat capacity
ΔT = 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? (cH2O = 4.184 J/g.oC) 17
What is the temperature change for a 15 g piece of iron that absorbs 26.5 J of heat?
(cFe = 0.444 J/g.oC)
A) 0.25 oC
B) 0.78 oC
C) 1.3 oC
D) 4.0 oC
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If 23.9 g of an unknown metal requires 343 J of energy to change its temperature from
24.5 oC to 85.0 oC, what is the specific heat capacity of the metal?
A) 0.237 J/g.oC
B) 0.568 J/g.oC
C) 4.22 J/g.oC
D) 868 J/g.oC
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Make sure you are comfortable in taking
q = m c ΔT and solving for c, m, ΔT, T2 & T1
p. 634 #’s 1 – 3 p. 632-->c values
p. 636 #’s 5 – 8 p. 659--> answers
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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
Your Turn p.637 #’s 11-14
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31. Which is the amount of energy required to raise the temperature of 1.0 g of a substance by 1.0 oC?
(A) heat capacity
(B) molar enthalpy
(C) molar heat
(D) specific heat capacity (June ’09)
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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 ΔH
Enthalpy of Reaction
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● Endothermic Reaction
PEEnthalpy
Reactants
ΔH
Products
ΔH
Reaction Progress
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Enthalpy Diagram
Reactants
Enthalpy
Products
ΔH is +
Endothermic
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Enthalpy Diagram
Enthalpy
Reactants
Products
ΔH is - Exothermic
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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)
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Enthalpy Changes in Reactions
endothermic reaction - the energy needed to break bonds is greater than the energy released when bonds form.
ie. energy is absorbed
exothermic reaction - the energy needed to break bonds is less than the energy released when bonds form.
ie. energy is produced
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Enthalpy Changes in Reactions
ΔH can represent the enthalpy change for a number of processes
1.Chemical reactions
ΔHrxn – enthalpy of reaction
ΔHcomb – enthalpy of combustion
(see p. 643)
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2.Formation of compounds from elements
ΔHo
f – 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) ΔHo
f = -110.5 kJ/mol
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Use the equation below to determine the ΔHo
f 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
ΔHo
f= -238.6 kJ/mol
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Use the equation below to determine the ΔHo
f 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
ΔHo
f = -1206.9 kJ/mol
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Use the equation below to determine the ΔHo
f 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
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3.Phase Changes (p.647)
ΔHvap – enthalpy of vaporization (l → g)
ΔHfus – enthalpy of melting (fusion: s → l)
ΔHcond – enthalpy of condensation (g → l)
ΔHfre – enthalpy of freezing (l → s)
eg.
H2O(l) H2O(g) ΔHvap = +40.7 kJ/mol
Hg(l) Hg(s) ΔHfre = -23.4 kJ/mol
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4.Solution Formation (p.647, 648)
ΔHsoln – 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
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Three ways to represent an enthalpy change:
1. A thermochemical equation has the energy term written into the equation.
2. The enthalpy term is written separate beside the equation.
3. An 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 (Thermochemical equation)
2. H2(g) + ½ O2(g) → H2O(l) ΔHo
f = -285.8 kJ/mol
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3. H2(g) + ½ O2(g)
ΔHo
f = -285.8 kJ/mol
H2O(l)
Enthalpy (H)
examples: pp. 641-643 questions p. 643 #’s 15-18 WorkSheet: Thermochemistry #4
Enthalpy Diagram
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Calculating Enthalpy Changes
FORMULA:
q = heat (kJ)
q = nΔH n = # of moles = mass (n = m ) Molar mass M
ΔH = molar enthalpy (kJ/mol)
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Compound ΔHo
f (kJ/mol)
CO(g) -110.5CO
2(g) -393.5
CH4(g) -74.6
eg. How much heat is released when 50.0 g of CH4 forms from C and H ?(Pg. 642)
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eg. How much heat is released when 50.00 g of CH4
undergoes complete combustion?
(Pg.643)
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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 ?
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ΔHfreez
and ΔHcond
have the opposite sign of the
above values. ΔH
vap = -ΔH
cond & ΔH
melt = -ΔH
freez
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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 (NH
4NO
3)
dissolves?
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What mass of ethane, C2H6, must be burned to produce 405 kJ of heat?
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FOR YOU TO COMPLETE
p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29
Worksheet: Thermochemistry #5
ANSWERS: Pg. 659
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Heating and Cooling Curves
Cooling of p-dichlorobenzene (aka moth balls)
https://www.youtube.com/watch?v=JKCJG-Jco_8
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Cooling curve for p-dichlorobenzene
● Check your notes for the graph!
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Heating curve for p-dichlorobenzene
● Check your notes for the graph!
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What we learned from those graphs
During a phase change temperature remains constant and PE changes● PE increases in a heating curve and decreases
in a cooling curve
Changes in temperature during heating or cooling means the KE of particles is changing● KE increases during a heating curve and
decreases during a cooling curve
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June 2009 # 38 http://www.ed.gov.nl.ca/edu/k12/evaluation/chem3202/chem3202_jun09.pdf
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Pg. 651
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Pg. 652
q = mcΔT
q = nΔH
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Pg. 656
q = mcΔT
q = nΔH
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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.
*Check your notes for the heating curve
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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: (from -40 ºC to 0 ºC)
q = mcΔT
= (40.0)(2.01)(0 - -40)
= 3216 J
warming water: (from 0 ºC to 100 ºC)
q = mcΔT
= (40.0)(4.184)(100 – 0)
= 16736 J
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warming steam: (from 100 ºC to 140 ºC)
q = mcΔT
= (40.0)(2.01)(140 -100)
= 3216 J
moles of water: n = 40.0g
18.02g/mol
= 2.22 mol
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melting ice: (fusion)
q = nΔH
= (2.22 mol)(6.02 kJ/mol)
= 13.364 kJ
boiling water: (vaporization)
q = nΔH
= (2.22 mol)(40.7 kJ/mol)
= 90.354 kJ
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Total Energy
90.354 kJ
13.364 kJ
3216 J
16736 J
3216 J
90.354 kJ
13.364 kJ
3.216 J
16.736 J
3.216 J
126.9kJ
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PRACTICE
p. 655: #’s 30 – 34
p. 659 – Answers
WorkSheet: Thermochemistry #6
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Law of Conservation of Energy (p. 627) The total energy of the universe is constant
ΔEuniverse = 0
Universe = system + surroundings
ΔEuniverse = ΔEsystem + ΔEsurroundings
ΔEsystem + ΔEsurroundings = 0 → 1st Law of Thermodynamics
OR ΔEsystem = -ΔEsurroundings
OR qsystem = -qsurroundings
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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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Bomb Calorimeter
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a simple calorimeter is 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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Example
A simple calorimeter contains 150.0 g of water. A 5.20g 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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Solutionaluminum alloy water
m = 5.20g m = 150.0 g
T1 = 525 ºC T1 = 19.30 ºC
T2 =22.68 ºC T2 = 22.68 ºC
FIND c for Al c = 4.184 J/g.ºC
qsys
= - qcal
mcΔT = - mcΔT
(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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Example
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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Solutioncopper Calorimeter
m = 12.8 g C = 1.05 kJ/°C
T2 =23.94 ºC T1 = 25.00 ºC
c = 0.385 J/g.°C T2 = 23.94 ºC
FIND T1 for Cu
qCu
= - qcal
mcΔT = - CΔT
(12.8)(0.38s5)(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
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More Book Questions!!
P. 664-665 #'s: 1,4
Thermal Equilibrium
- all components in a calorimeter have the
same temperature
In the last two examples, notice that the final temperature of the water/calorimeter is the same as the final temperature of the sample
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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?
(-5700 kJ/mol)
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System (octane) Calorimeter
m = 11.0g C = 28.0 kJ/ºC
Find ΔHcomb T2 = 39.6 ºC
n = 11.0g T1 = 20.0 ºC
114.26g/mol
=0.09627 mol
qsys = - qcal
n ΔH = -CΔT
(0.09627) ΔH = - (28.0)(39.6 – 20.0)
ΔH = -5.70x103 kJ/mol
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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. (ΔHcomb = -3225 kJ/mol)
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n = 1.26 g
122.13 g/mol
= 0.01032 mol
qsys = - qcal
n ΔH = -CΔT
(0.01032)(-3225) = - (C)(27.14 – 23.62)
C = 9.45 kJ/ ºC
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Your Turn!
Pg. 675 #10
WorkSheet: Thermochemistry #7
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Hess’ Law of Heat Summation
● the enthalpy change (ΔH) of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products)
● ΔH is independent of the pathway and/or the number of steps in the process
● ΔH 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) ΔH = -110.5 kJ
CO(g) + ½ O2(g) → CO2(g) ΔH = -283.0 kJ
____________________________________________
C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
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eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
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Using Hess’ 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.
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Ex: reverse? multiply?
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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Determine the ΔH value for:
4 C(s) + 5 H2(g) → C4H10(g)
using the equations below.
ΔH (kJ)
C4H10(g) + 13/2 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) + 13/2 O2(g) +110.5kJ
5(H2(g) + ½ O2(g) → H2O(g) -241.8 -241.8kJ)
4(C(s) + O2(g) → CO2(g) -393.5 -393.5kJ)
4 CO2(g) + 5 H2O(g) → C4H10(g) + 13/2 O2(g) +110.5kJ
5 H2(g) + 5/2 O2(g) → 5 H2O(g) -1209.0 kJ
4C(s) + 4 O2(g) → 4 CO2(g) -1574.0kJ
Ans: -2672.5 kJ
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Practice for you!!
pg. 681 #’s 12 &14
WorkSheet:
Thermochemistry #8
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Recall
ΔHo
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.
ΔHo
f = 0 kJ/mol
for elements in the standard state
The more negative the ΔHo
f, the more
stable the compound
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June ‘08 #39
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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)
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
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Using Hess’s Law and ΔHf
ΔHrxn = ΣΔHf (products) - ΣΔHf (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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ΔHrxn = ΣΔHf (products) - ΣΔHf (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 kJ/mol
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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
−824.2 kJ/mol -110.5kJ/mol −393.5 kJ/mol
0 kJ/mol
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eg.
The combustion of phenol is given below:
C6H5OH(s) + 7 O2(g) → 6 CO2(g) + 3 H2O(g)
If ΔHcomb = -3059 kJ/mol, calculate the heat of
formation for phenol. ΔHo
f
H2O(g) -241.8 kJ/mol
CO2(g) -393.5 kJ/mol
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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 = ΣBE(reactants) - ΣBE (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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C-C = 347kJ/mol; C-H = 338kJ/mol; O=O = 498kJ/mol; C=O = 745kJ/mol
H-O = 460kJ/mol
[2(347) + 2(6)(338) + 7(498)] - [4(2)(745) + 6(2)(460)]
8236 – 11480
= -3244 kJ
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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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#36 June ‘06
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#37 Aug. ‘08