Kinetic Energy notes and slides from MUN uni

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)

Chemical Reactions
PE Diagrams
Thermochemical Equations
Hess’s Law
Bond Energy
STSE: What Fuels You?

Temperature and Kinetic Energy
Thermochemistry is the study of energy
changes in chemical or physical changes
eg. dissolving
burning
phase changes

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

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.

K 321.15 50.15 73.15 450.15
°C 48 -223 -200 177

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.

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

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

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.

FORMULA
q = mcΔT
q = heat (J)
m = mass (g)
c = specific heat capacity
ΔT = temperature change
= T2 – T1
= Tf – Ti

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.o
C)
A) 0.25 o
C
B) 0.78 o
C
C) 1.3 o
C
D) 4.0 o
C

If 23.9 g of an unknown metal requires 343 J of
energy to change its temperature from
24.5 oC to 85.0 o
C, what is the specific heat capacity of
the metal?
A) 0.237 J/g.o
C
B) 0.568 J/g.o
C
C) 4.22 J/g.o
C
D) 868 J/g.o
C

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

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

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)

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

● Endothermic Reaction
PE
Enthalpy
Reactants
ΔH
Products
ΔH
Reaction Progress

Enthalpy Diagram
Reactants
Enthalpy
Products
ΔH is +
Endothermic

Enthalpy Diagram
Enthalpy
Reactants
Products
ΔH is - Exothermic

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)

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

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)

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

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

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

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

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

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

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.

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

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

Calculating Enthalpy Changes
FORMULA:
q = heat (kJ)
q = nΔH n = # of moles = mass (n = m )
Molar mass M
ΔH = molar enthalpy (kJ/mol)

Compound ΔHo
f
(kJ/mol)
CO(g) -110.5
CO2
(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)

eg. How much heat is released when 50.00 g of CH4
undergoes complete combustion?
(Pg.643)

eg. How much energy is needed to change 20.0 g of H2O(l)
at 100 °C to steam at 100 °C ?

ΔHfreez
and ΔHcond
have the opposite sign of the
above values.
ΔHvap
= -ΔHcond
& ΔHmelt
= -ΔHfreez

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 (NH4
NO3
)
dissolves?

What mass of ethane, C2H6, must be burned to produce
405 kJ of heat?

FOR YOU TO COMPLETE
p. 645; #’s 19 – 23
pp. 648 – 649; #’s 24 – 29
Worksheet: Thermochemistry #5
ANSWERS: Pg. 659

Heating and Cooling Curves
Cooling of p-dichlorobenzene (aka moth balls)
https://www.youtube.com/watch?v=JKCJG-Jco_8

Cooling curve for p-dichlorobenzene
● Check your notes for the graph!

Heating curve for p-dichlorobenzene
● Check your notes for the graph!

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

June 2009 # 38
http://www.ed.gov.nl.ca/edu/k12/evaluation/chem

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

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

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

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

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

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

PRACTICE
p. 655: #’s 30 – 34
p. 659 – Answers

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

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.

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

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

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.

Solution
aluminum 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
q
sys
= - q
cal
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

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)

Solution
copper 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

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

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

● 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

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)

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

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)

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

Your Turn!
Pg. 675 #10

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

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

eg. production of carbon dioxide
Pathway #2: formation from the elements
C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ

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.

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

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

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

Practice for you!!
pg. 681 #’s 12 &14
WorkSheet:
Thermochemistry #8

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

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)
ΔH
f
(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
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)

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

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

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

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.

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.

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

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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