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‘ ’
Module 1em ca nerget cs
Lecturers: Prof Bob Gilbert [email protected]
Dr Horst Schirra [email protected]
Prof Matt Trau [email protected]
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
em ca nerge cs eac v y
Text: Chemistr . Blackman Bottle Schmid Mocerino and
Wille.
John Wiley. 2008
For the Electronic Course Profile – see Blackboard
(http://blackboard.elearning.uq.edu.au)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
ALWAYS BRING A CALCULATOR + PAPER + PEN
Blackboard (Bb) is thecentre of the CHEM1020
universe!
Every Monday a course Bb
announcement willsummarise the weeks
act v t es ea nes not ces
n ema w not e sent
every time new information is
a e . ou must c ecregularly.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Sign up for….
• Lecture streams
Lecture notes & Lectopia for each stream will be
available on Blackboard (Bb) site. Hard copies available
for purchase from POD at back of UQ Coop Bookshop
• Lecture timetable – check on SI-net
Rooms may change up to Week 3!
• Practicals: 5 x experiments (3 hrs long)
Induction sessions begin in week 1, sign up via SI-net &
manuals will be provided at 1st
session. Consider CASPiE URE as an alternative lab route.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• PASS (attendance very strongly recommended)
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50% MCQ and SAQ exam at end of semester (must gain 40%or higher in final exam to gain a grade of 4 or higher).
20% practical work
12% 2 QUIZZES
18% IS-IT Chemistry Collaborative Research Task (begins inweek 2 – details via Bb)
• See ECP, POD lecture notes AND BLACKBOARD
(Bb) site for complete details of the course includingt e assessment sc e u e.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Course Coordinator: Dr Gwen Lawrie [email protected]
eLearnin for Students
Blackboard Home: http://blackboard.elearning.uq.edu.au
ourse ro e ome: p: www.courses.uq.e u.au
Course Resources
se t e ourse ro e n or a t e eta s assoc ate
with your course.
For student t echnical support and for Blackboard Training
one:
Email: [email protected]
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Web: http://askit.uq.edu.au/
Check Course Site on Bb frequently:
This is where you will find allAnnouncements
the information that you require!
Course Profile (ECP)
Learning Materials
nnouncemen s rem n ers
Schedules and uidelines
External Links
Assessment
M Grades
Communications
Calendar
UQ Library
Course Profile
eTute HELP
Class ResourcesWileyPLUS
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
a s on
–
• These are online tests throughout the semester.
• .• Each QUIZ is worth a maximum of 6% of the total course
mark.
.
• Full details are available under the new ‘QUIZ’ menu link
Student Feedback 2009 & 2010: These QUIZZES are the bestpreparation for the exam!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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–
• You must complete the induction in week 1 or 2.
• Tie up your long hair (both genders) and wear closed-in
shoes or ou will be turned awa .
All lab info and contacts for reschedules are provided
under the Lab Resources Menu Tab on Bb!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Collaborative Group Task (weeks 2-12):
• ou c oose a scenar o – represen g o a sc ence ssues.
• You will be working in Groups of 4
• Groups are assembled by one of three ways (either
choose your team members; opt into a group; or be placed
- .
• Full details & sign on will be available in week 2 through
‘ ’- .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
What is PASS?
PASS provides one hour of mentored group study, whereyou are able to:
•Clarify concepts not fully understood
•Practice questions in exam format
•Revise your own learning
•Identify gaps in your own knowledge•Practice learning in the language of the discipline
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
•Be part of an interactive learning community!
CHEM1020 S2 2009Mean Grades by PASS Attendance
5.5
e
e a n G r a d
4.5 v e r a l l M
4
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
0 (295) 1‐‐4 (251) 5‐‐8 (238) 9+ (287)
PASS Session Attendance #
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CHEM1020 S2 2009Grade Distribution: Effect of PASS Attendance
45%
50%
35%
40%
25%
30%
d e n t s
15%
20% S t
5%
10%
0%
1 2 3 4 5 6 7
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
0 (295) 1‐‐4 (251) 5‐‐8 (238) 9+ (287)
PASS Classes
begin in Week 3:
on ay u us
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Strategies for learning quantitative chemistry
In module 1 …
• Reading before lectures
• In class model Qs
• JITTerbug quizzes (Feedback)
…
• In-class problem-solving
• Homework between lectures
In module 3 …
• In-class problem-solving
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• Homework between lectures
Module 1: 12 Lectures 1. States of Matter . Chapters 6 and 7 (p202 – 281)
- Gases and Gas Laws
- Intermolecular Forces- Phase Changes
2. Thermochemistry . Chapter 8 (p281 – 331)- The First and Second Laws
- Defining Internal Energy and Enthalpy- Quantifying Enthalpy
- Defining Entropy- The Gibbs Function
3. Equilibrium . Chapter 9 (p332 – 372)- Chemical Equilibrium
- Equilibrium and Gibbs Free Energy- How systems at equilibrium respond to change
- Equilibrium calculations
= A Key Concept oo or e con
with page numbers!= A worked example in the text book you should look at
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
ALWAYS BRING A CALCULATOR + PAPER + PEN
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School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
evident in this image?
Think Back!
How can we explain why some substances exist as gases at room
temperature and atmospheric pressure, but others are liquids or
=> Forces that occur between individual atoms,mo ecu es or ons nt ermo ecu ar orces
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
et s eg n w t gases s nce t ey are t e eas est ….
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
(Elements in their pure state are mostly solids, with exceptions)
We must understand gases to understand and solve someimportant problems:
• Greenhouse Effect
• o u on, p o oc em ca smog
• Energy production
• Biochemistry of lungs
•
What are the characteristics of ases?
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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6.2, 203
A gas will expand to occupy all available space it is in and will
.
shape.
Atoms in a gas move independently of each other and move
randomly (Because the atoms in a gas are moving at random,
the gaseous state is the simplest to describe mathematically by
.
Gases exert a ressure . .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Gas Pressure
Gas molecules are in constant motion and collide with each other and the
container walls
they exert a pressure on the surroundings .
= =
The pressure is defined as: p =FORCE (N)
2
A
F
Gas molecules have mass(m) and velocity (v)
Gas molecules have mass(m) and velocity (v)
Opposing forceOpposing force
N
-2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2m
1 Nm-2 = 1 Pascal (Pa) = 1 J m-3
You are going to come across other units:
Atmosphere (atm)
1atm = 1.01325 105 Pa (101.325 kPa)= 760 mm Hg (= 760 Torr)
= 14.696 psi (pound per square inch)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
‘ ’
Standard tem erature = 0 oC = 273 K
Standard pressure: Two definitions:
1 atmosphere = 100 000 Pa (IUPAC + Blackman Chemistry)
-1
SI units:
Temperature = KVolume = m3 (103 dm3 or 103 L)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
and think about units!
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The Ideal Gas E uation
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.2, 205
p = n ea as quat on
One of the most important equations in physical chemistry.
Describes how an ideal gas behaves under given conditions of
p = pressure (Pa or atm)
,
V = volume (m3 or L)
n = molar amount of gas (mol)
11
)(molKJ8.314 SIRR = gas constant
=
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
moatm.
Assum tions in the Ideal Gas Law?
6.2, 211-13
• The molecules are very small compared to the distance between
em = ey ave neg g e vo ume .
• The molecules do not ‘see’ each other (= no intermolecular
interactions.
• The molecules undergo elastic collisions with container walls (= no
energy is lost due to the collision).
• The molecules move in com letel random motion different s eeds
and direction).
=> e n on o an ea as:
If a as meets these conditions then it will obe
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
pV = nRT , and it is ‘Ideal’
6.2, 204 - 205
a oes e ea
Gas Equation tell us?
ow oes e pressure o a gas c ange w en we vary e
volume, temperature, or amount of gas (number of moles)?
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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11V p 11
T n
22FinallyV p
R 22n
2211 V p V p If one variable remains
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2211 , .
Example
A sample of helium gas is held at constant temperature inside a cylinder
with a volume 0.57 L. The external pressure is 2.1 x 105 Pa. If the piston is
withdrawn until the gas volume is 2.55 L. Calculate the final pressure.
the no of moles of He & T are constant in this problem
p 1V 1 = p 2 V 2 and rearranging to: p 2 = p 1V 1
V 2
p 2 = (2.1 x 105 Pa) x (0.57 x 10-3 m3) = 0.47 x 105 Pa
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2.55 x 10-3 m3
Units! 1 L = 1 x 10-3 m3
1. Watch out for units:
If R is 8.314 J K mol-1, p is Pa (Jm-3 or Nm-2) and V is m3
If R is 0.08206 L atm K-1 mol-1, p is atmos, V is L
2. A useful value (by substituting into the ideal gas law):
1 mole of gas occupies 22.4 L @ 273 K & 1 atm
or 24.5 L @ 298 K & 1 atm
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
A sample of chlorine gas occupies a volume of 946 mL at a pressure
of 726 mmHg. Calculate the pressure of the gas if the volume isreduced to 154 mL (assume temperature remains constant)
Solut ion: Since pV = nRT = constant, p1V 1 = p2V 2, hence
mL946
2
112
V
V p p
Hgmm104.46
mL154gmm
3
kPa101.3Hgmm760
Hgmm104.46 3
2 p or to convert to Pa:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
kPa594=
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Module 1: Lecture 2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.2, 204
ases an so ute emperature
o e a e n ercep o eac ne es a - . , nown as a so ute zero .
0 K = -273.15 °CV
1
P2
P3Absolute Zero
T (°C)-273.15 °CAbsolute zero: the lowest temperature that can be. It is theoretical and hasnot been reached (Liquid Helium gets to 4 K). All calculations done using
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
negative values).
Kelvin: absolute temperature scale.
(Room Temp) 25 °C 298 K
0 K = -273.15 °C
° = .
Note: man of the e uations we
will use require the use of temperature
in Kelvins.
Also note: Temperature differences are the same in Kelvin or °C
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at
pressure of 1 atm. If all this oxygen were converted to ozone (O3) at
same an , w a wou e e vo ume
2O3O
No. moles O3 produced = 32
Omol2Omol0.50
2mo
= 0.33 mol O3
Since V/n is constant,2
2
1
1
n
V
n
V
1
212
n
nV V
L8.1
L12.2mol0.50
mol0.33
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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Air Bags
Sodium azide decomposes rapidly
gas.
a 3 s a s + 2 g
Causes inflation in < 0.05s
How many litres of nitrogen would100 NaN roduce at room tem
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
www.howstuffworks.com
and pressure?
2NaN3 s 2Na s + 3N2 g
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.4, 214/215
Determination of molar mass M
Number of moles (n ) = mass/molecular weight = m/M R
pV = nRT <=> pV = m /M R RT <=> M R = mRT / pV
= m/V
Using this equation and n = m /MR, we can rearrange pV=nRT to give:
gas =
m
V =
p MR
RT
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Worked examples in Blackman 6.4 p214/5
ne c eory o ases
The Ideal Gas Law provides a numerical route for predicting gas
behaviour based on macroscopic observations (changes inpressure and volume).
behaviour on a molecular basis?
Richard Feynman: “It’s all about atoms jiggling about!”
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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6.2, 204 - 205
a oes e ea
Gas Equation tell us?
ow oes e pressure o a gas c ange w en we vary e
volume, temperature, or amount of gas (number of moles)?
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Descri tion of an ideal as from a molecular
viewpoint.
t . e d u /
velocity (v ) – therefore have kinetic
energy: / w w w . p i
E =1
mv2 t o n h t t p :
o
h n N o r
We assume that translational kinetic energy is the only internal ener ossessed b a as molecule. We i nore rotational
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
vibrational etc energy).
6.3, 208
Ekinetic =1
2mv2
Gas molecules are in constantrandom motion. The energy of amolecule is related to its speed.
There is a distribution of speeds, v , amongst a population
o gas mo ecu es.
l e c u l e s
Mean energy
e r o f M
N u
m
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Relationshi between Kinetic Ener & Tem erature
Experiments show that
temperature increases
To find the average kinetic energy we add the individualmolecular kinetic energies and divide by the total number of
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
molecules
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6.3, 211
Ekinetic =3RT
= kT3
(kinetic energy of one gas molecule)A
Multi l b N to et avera e kinetic ener er mole of as
=3 This expression is independent
ne c, mo ar2 of the identity of the gas!
Temperature: A measure of the average kinetic energy of a gas.
e tota net c energy s t e t erma energy.
Im ortant! Tem erature and thermal ener describe different
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
properties!!
–
kin
Molecules are in constant motion and collisions can occur betweenmo ecu es or w e oun ar es o e con a ner.
The number of collisions/unit area = Force/unit area = pressure
Elastic Collisions
x
x
e ore
after
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Derivation (for your information only)
L
F
Area
Force p
2
Elastic Collisionsx
-x
before
x v v 2
x
Lt
Lv
L
mv
L
v v m
t
v m ma F x x
x i
222
a terx v t
Force of one particle hitting one side of the cube
mv mv mv F p i
i 112 2
3
2
2
2
26 sides to a cube => pressure of one particle
mv nN pV mv
nN p AA 1 2
2
This looks already like the ideal gas law
pressure from many particles:
E nN
V A 2 kin kin E mv mv E 21 22
RT nN A 32
3
kin
RT kT E
33
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
N A23 A. . .
Historically, it went the other way round…
mv nN pV V
mv nN p AA
3
1
3
22
This looks already like the ideal gas law
E nN
pV kin A 2
3 kin kin E mv mv E 2
2
1 22
RT n
pV nRT pV
kT RT pV
E kin
333
=> Using the ideal gas law to
define absolute tem erature inAA
terms of kinetic energy of molecules
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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Rates of as movement
6.3, 217
E =3RT
E = 1 mv2and2NA 2
So2
mv2
2NA
=
Solve for v (noting that m NA is the mass of 1 mole of
gas = M R)
R
M
RT v
3 _
=> The heavier a gas molecule, the slower its
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
diffusion and effusion.
•
molecules and hence the number of collisions per unit area
and increases pressure (constant V & n )
• Reducing the amount of gas molecules, n, decreases the
num er o co s ons per un area re uces pressure
(constant V & T ) or volume (constant p & T )
can be explained in terms of kinetic theory!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Module 1: Lecture 3
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
’
6.5, 219
.
‘The total ressure of a mixture of ases is the sum of the ressures that each gas would exert if it were alone in the container.’
T = 1 2 3……. ii =
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
www.kentchemistry.com/links/GasLaws/dalton.htm
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If a sample contains gases A and B, the total pressure is given by:
pt =
nTRT
V
= RT (nA + nB)
= n RT + n RT
V
V V
= pA + pB
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
pA nART/V
pA
pT
=(nA + nB)RT /V
= nA
=pA + pB
(nA + nB)
= xA
xA is the mole fraction of A, xA: 0 1
pA = xApT
xA0 1
pB = xBpT
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
In general for component i: pi = xipT.
ues on:
A SCUBA diver is using EANx32, an enhanced
Nitrox gas mixture. This mixture contains 32%
Oxygen with 68% Nitrogen. If the cylinder is at
a pressure of 250 kPa what is the partial
pressure of Oxygen?
http://www.oceanearth.org/pastinfo.html
Mole fraction of oxygen = 0.32
p O2 = 0.32 x 250 kPa
= 80 kPa
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Deep sea divers use helium instead of nitrogen (Heliox) – why?
What ou should be able to do at this sta e:
1. List the qualities that define an ideal gas.
2. Describe ideal gas behaviour in terms of the relationship, , .
3. Apply the ideal gas law pV = nRT
4. Explain how and why kinetic energy influences p, V and T
(microscopic).
5. Understand the difference between temperature and thermalenergy.
6. Describe how the partial pressures of individual gases in a
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
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What about ‘Real Gases’?
The ideal gas law begins to
rea own a ow empera ures
or pressures.
Why?
Because gas molecules possess
a vo ume an can see eac
other (=intermolecular
interactions)
Solution?
Modify the ideal gas law to incorporate corrections for real gases
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
=> e.g. van der Waals Equation (2nd year)
6.7, 226
a es o a er – n ermo ecu ar orces
increased numbers of collisions (increase in p), intermolecular
forces become more relevant.
Intermolecular forces are the forces that occur between molecules.A chemical bond is not an intermolecular force (intramolecular
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
.
-
Three t es of molecular force de endin on molecule olarit :
1.Dispersion Forces.
2.Dipolar forces.
. .
Relative Strength:
Dispersion forces < Dipolar forces < Hydrogen Bonds
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.8, 231
-
Molecules with dipole moments can line up so that negative
an pos ve en s are c ose – ea s o an e ectrostat c attraction. Generally important over short distances.
A dipole on a molecule arises from the combination of molecular
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
shape and bond polarity (revise).
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Why do non-polar atoms or molecules exist in condensed states
under some conditions?
For example helium exists as a liquid below about 4 K.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
phys.kent.edu
6.8, 229
Transient changes in the charge
distribution around a nucleus or
molecule) can occur, and when this
happens near another atom, the
other atom.
This results in a weak and short-
lived interaction known as London dispersion force .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.8, 231
Intermolecular Forces
Halogens are diatomic (F2) and interact
via DISPERSION forces.
More electrons = more easily
polarisable = stronger bond
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
6.8, 233
Hydrogen bonds are a special form of dipolar forces. Due toig y po ar on s e.g O-H wit an e ectron e icient
hydrogen, and the small size of hydrogen. Usually involve Hattached to O, N or F.
A network of H-bonding
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
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7.1, 243
Surface Tension.
.
Molecules in a liquids experienceattractive forces from other molecules,but molecules at the surface experienceuneven forces surface tension.
The surface area stays as small as.
o es ve orces: a rac mo ecu es nthe liquid to one another.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Intermolecular Forces in Liquids.
Capillary action.
Adhesive forces: attractmolecules in the liquid to thewalls of the container.
Capillary action occurs whenthe adhesive forces are
encarta.msn.com
stronger than cohesive forcesand gravity.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Intermolecular Forces in Liquids.
7.1, 244
Vapour Pressure
Some mo ecu es are a e to escape rom t e sur aceof the liquid. These gaseous molecules exert a pressure.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Intermolecular Forces in Solids.
7.2, 245
In solids, intermolecular forces are so strong that there is noarge sca e movemen o mo ecu es.
Amorphous – no extended order
Solids
(e.g. glass, plastics)
Crystalline – highly ordered(e.g. diamond)
In liquids the boiling point tells us about strength of
intermolecular forces. In solids, it is the melting point.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
See table 7.1, pg 246
7 2 246 7 2 247
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7.2, 246
.
lattice positions (eg ice). Intermolecular forces are relatively much
weaker than the covalent intramolecular forces.
Sugar Dry IceNapthalene
The molecules are typically bound by dispersion, dipolar or H-bonds.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
7.2, 247
But, there are other solids where bonding is very strong:
.
2. Metallic
3. Ionic solids.
Network > Metallic > Ionic
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Bondin in Solids
7.2, 247
Network solids (highest melting point).
Network solids have a array of covalent bonds whichlinks each atom to its neighbouring atoms.
The extent of covalent bonding influences properties.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Diamond Graphite
sp 3
dispersion3550°C
sp 2
Graphite
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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Buckyballs
7 2 248
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7.2, 248
Bondin in Solids
Metallic solids.
Bonding in metals is characterised by non-directional covalent
bonding.
Electrons (covalent) are arranged around metal atoms in such away that they are mobile (non-directional).
Why are the electrons delocalised?
When two metal atoms interact, two molecular orbitals result. Withmany atoms, a large number of molecular orbitals result – all
.around different atomic molecular orbitals (Band Model).
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Bondin in Solids
Metallic solids.
Metals have unique properties: high electrical and thermal conductivity,
malleability
Electrons are arranged around metal atoms in such a way that they
are mobile.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
“electron sea” model
7.2, 249
Bondin in Solids
Ionic solids.
Ionic solids contain cations or anions that are heldtogether by opposing electrostatic forces.
These are strong interactions so meltingtemperatures are generally high.
NaCl Na+ + Cl- Fp = 801°CNa
NaCl
The stoichiometry of an ionic solid reflects that they
must be electrically charged.
CaF2 Ca2+
+ 2F-
CaF2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Module 1: Lecture 4
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
7 3 250
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7.3, 250
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
a ac ors ec o u y
o u y maximum amount of solute that will dissolve in a given amount of
solvent at specific temperature (heterogeneous equilibrium)
Polar solvent/polar solute or nonpolar solvent/nonpolar solute favoured
eg hexane (C6H14) is better solvent for grease (C20H14) than methanol
(CH3OH)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Revision
Depends on bond polarity and molecular shape …..
nonpolar polar ionic
Blue = ositive char e
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Red = negative chargeChpt 5.5
Atom Electrone ativit
A. C-CH 2.1
B. C-HC. H-Br
C 2.5Br 2.8
D. N-H
E. C-OCl 3.0
F. O-H.
O 3.5
S 2.5
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
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Jud in olarit of molecules!
•The addition of a carbonyl
functional rou -C=Ointroduces a dipole.
•
localised in one region of
the molecule.
• Dipole-dipole interactions
can occur
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Assumed
knowledge from
Senior Chem
Dispersion
Di ole-di oleIncreasing
hydrogen bonding (special dipole-dipole case)
H2O CCl4
Sucrose
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Polar Nonpolar
- -
O
H
O
H
H H
CC
H
O
H
H
H H
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Blackman Chapter 10 p 378
Consider 3 solvents:
Water (H2O)
Hexane (C6H14)
Carbon tetrachloride (CCl4)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://jchemed.chem.wisc.edu/JCESoft/CCA/pirelli/pages/cca2like.html
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Which of these vitamins do you predict to be soluble in water?
H3C CH3
CH3 CH3
CH
OH Hydrophobic
Fat soluble: A, D, E, K
Vitamin A
Like dissolves Like!
OH
Hydrophilic
Water soluble: B C
OOH
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
HO OHVitamin C
Will this molecule dissolve
.
2. Yes
3. I don’t know!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Solute-solvent interactions must overcome
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
p 374-
are attracted to the polar water molecules:
A nonpolar solid will only dissolve in a nonpolar solvent
because the dispersion forces are of comparable strength
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
p 380
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ons n e crys a a ce can
interact with polar solvents
through formation of ion-dipolebonds.
In water, each ion is surrounded
by water molecules which form a
hydration shell and the ion is
solvated (‘aquated’)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
p 379
Enthalpy diagram for a solid dissolving in a liquid
ahead:
enthalpy is a
function!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
What we measure is the change represented by the purple arrowp 380
Chapter 8
e a y o pre c e o e rec on an e ex en
of spontaneous chemical and physical change under
.
‘Thermochemistry’
e s u y o energy c anges nvo ve w c em ca reac ons
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://www.youtube.com/watch?v=KTHiIwxcexI
y are we o ng ermoc em stry
un y ng ramewor or no on y un ers an ng s eam eng nes,but for understanding a diverse range of chemical reactions:
Why some reactions are spontaneousWhy gases expand and solutions mix
Why chemical reactions proceed towards equilibriumWhy proteins fold/unfold, form multimers
Refrigeration…..…..
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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What we will investigate:
The role of ene r gy in chemical reactions
The Fi r s t Law o f Ther m odynamics
The Second Law o f The r modynam ics
How to understand and manipulate thermochemical data to pred ic t how ene r gy w i l l change in chemical reactions.
How energy changes can p r ed i ct spon tane i t y .
m ax m u m w o r reaction.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
What is Energy??
1. The capacity to do work.
E.g. Lifting an object.
Work done = force x distance
. .
Heat = the process of t r ans fe r o f t he r m a l ene r gy between two
o es or sys ems a eren empera ures. e cons er ea asbeing unable to do work.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
=> Heat is a means of energy transfer, not energy itself!!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
nergy anges n em ca eac ons
r ua y a c em ca reac ons a sor or re ease energy.
In order to understand this, we need to focus on a limited, well-
defined part of universe, called the system . Everything else is called
the surroundings .
The Surroundings +
System = Universe.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.2, 284
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Open
System
Matter
Energy
pen sys em:
can exchange matter and energy with the
surroundings.
ClosedMatter Closed system:
SystemEnergy
surroundings.
IsolatedSystem
Matter
Ener
Isolated system:
can exchange neither matter nor energy
with the surroundin s.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
e proper es o a sys em a any one me s s a e
Module 1: Lecture 5
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Chemical Reactions involve
8.2, 284
Energy Transfer
The system is usually the chemical
reactants and products. The system is ourframe of reference.
Energy is transferred to/from a system from/to the surroundings.
or w an ea q are e wo un amen a ways n
which energy is transferred to or from a system.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
nergy os y sys em = energy ga ne y surroun ngs
The First Law of Thermodynamics
Energy cannot be created or destroyed, it
another.
=> Internal energy difference between reactants and products.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.3, 285The Energy that is transferred comes from the system’s internal energy U
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The Energy that is transferred comes from the system s internal energy, U .
n erna nergy an e rs aw oThermodynamics
The internal energy, U is the sum of all the energies – e.g. potential,kinetic – for all particles in the system.
.
The change in internal energy: U = U final – U initial
If U is positive – system gains energy
If U is ne ative – s stem loses ener
All energy must be released or gained from the surroundings:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
U system = -U surroundings
CH + 2O CO + 2H O l
Exam le: combustion of methane:
Internal Energy system
CH4(g), 2O2(g)Uinitial
internal energy than
reactants
q
CO2(g), 2H2O(l)Ufinal
In the combustion of methane the system has lower energy at the end
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
– .
=> Exothermic reaction
Exam le: Combustion of methane ener O2HCOO2CH
CH4
CO2
Combustion of methane releases energy in form
O2
H2O
of heat (energy flows from system to
surroundings). => exothermic reaction.
q
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
reactants
Other reactions cause energy to flow from surroundings to system:
.
)2NO(energy)(O)(N 22 ggg e.g.
NO(g)Uinitial
4
O2
CO2
N2 NO
H2ONO N2(g), O2(g)Ufinal
Endothermic: reactants have lower internal
energy than productcs
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
,
place to another
8.3, 285
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In chemical reactions energy is exchanged with the surroundingsas either heat (q) or work (w).
With respect to chemical reactions, the First Law can be expressedin terms of w and q.
U = q + w
ere q = ea a e o sys em q = ea remove
w = work done on system (w<0 = work by system)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Positive work or heatmp es t at energy o
system increases.
Negative work or heat
implies that energy of
system decreases.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
‘ ’
8.3, 285
There are many different types of work – electrical, mechanicaletc. At this stage we are interested in work associated withcontraction and expansion of a gas (“pV” work).
e.g. If we raise the temperature of a gas the volume increases.As the as ex ands it ushes back the surroundin s i.e. doeswork.
= - ext
= final – initial
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Rationalisation:
Work = F x d = F x h (Change in height)
= p x A x h
= pV
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.3, 287
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xamp e:A gas expands in volume from 0.1 m3 to 0.2 m3 against a
.
V p w ext Solution:
m0.1m0.2mN10100
4
3323
kJ10
.
Gas “does” -10 kJ of work.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
(negative sign= work done by system)
‘ ’
Heat = the process of t r ans fe r o f t he r m a l ene r gy betweentwo bodies or systems at different temperatures. We consider
heat as being unable to do work.
Heat is a m e a n s of energy transfer, not energy itself!!
We cannot measure heat directly, but we can measure the
body to another:q = C T
Where T = Tfinal –Tinitial and C=heat capacity
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
=> Calorimetry: (Blackman p287-293 – your responsibility!)
U = q + w ,
At constant volume V is 0, so w = 0 and
U = q v
=> =
Energy as q
Internal energy changes by q U r can bemeasured using a
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
bomb calorimeter
8.3, 284
But, in the ma ority of cases reactions occur under conditions
.
of constant pressure (atmospheric).
= ,
U = qp – pV <=> qp = U +pV
Need a new thermodynamic property that is equivalent to theheat of reaction at constant pressure – ENTHALPY (H )
H = U + pV General definition
H = U + pVEnthalpy changes (e.g.
during a chemical reaction)at constant pressure
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Enthalpy
E h i E d h i
8.3, 292
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Enthalpy.
U = qp – pV Energy as PV work
qp = U + pV
H = U + VEnergy as q
=
Enthalpy changes by q
p
=> H = heat of reaction at constant pressure
Enthalpy change for a chemical reaction:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
= products - reactants
Exothermic Endothermic
R , H P
, H
P E n t h a l p
RH > 0 (positive)H < 0 (negative)
E n t h a l p
Exothermic: Endothermic:Hproducts < Hreactants, H < 0 Hproducts > Hreactants, H > 0
The ‘coffee cup’ calorimetercan measure H directl .
(specific heat/heat capacity andcalorimetry: Blackman p287-293 –
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
chemlab.truman.edu
your responsibility!)
Internal Ener versus Enthal .
H = U + (pV)
,
(i.e. H ≈ U).
For a reaction invo ving an i ea gas:
H = U + (nRT)
At constant temperature:
H = U + RTn
= –
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Internal Ener vs Enthal .
C2H5OH(l) + O2(g) CH3COOH(l) + H2O(l), H° = -492.7 kJ
What is the standard internal energy changefor this reaction at 298K ?
Solution:At constant temperature we know that:
U = H – RTn
In this case n = 0 mol – 1 mol = -1 mol
U = H – RT -1 mol = -492.7 x 103 J – (8.314 JK-1mol-1 x 298 K x -1 mol)
= -490.2 x 103 J
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The difference = -2.5 kJ = work done by the system.
I t t E th l i t
I t t E th l i t
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Im ortant Enthal oints.
1.Enthapy is an ex tens i ve property.If the amounts of reactants doubles, theenthalpy doubles. Implicit that a quoted H
.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ
2CH4(g) + 4O2(g) 2CO2(g) + 4H2O (g) H = -1604 kJ
‘ ’
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
(it includes the enthalpy change of the reaction)
Im ortant Enthal oints.
2. Enthalpy change of the rever se reac t i on is of s am e m a g n i t u d e but oppos i te s ign .
4 g 2 g 2 g 2 g = -
2 g 2 g 4 g 2 g =
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Im ortant Enthal oints.
3. Enthalpy change for a reaction depends onthe state of reactants and products.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ
CH (g) + 2O (g) CO (g) + 2H O(l) H = -890 kJ
Difference is:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2 g 2 = -
Module 1: Lecture 6
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
State Functions Pro ert of State
8.x, xxx
State Functions Pro ert of State
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State Functions Pro ert of State .
When a system changes state, the State Functionsare onl de endent on initial and final states.
The change of state is independent of the path
.
Work and heat are dependent on the way in which.
Functions.
10°C 10°C
na s a e n a s a en a s a e
heat freeze
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://itl.chem.ufl.edu/2045_s00/lectures/lec_7.html
State Functions Pro ert of State .
The Final State is independent of the path taken
The distance between
remains the same (a
state function) but the
distance travelled to go
between them is
variable (not a state
function).
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Corollary: If you do a round trip, nothing has happened! ( X = 0)
State Functions Pro ert of State .
The Final State is independent of the path taken
Analogy:
between two points on amap remains the same (a
s a e unc on u e
distance travelled to go
between them or the actual
work going up and down are
variable (not a state
function .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Corollary: If you do a round trip, nothing has happened! ( X = 0)
State Functions Pro ert of State .
The Final State is independent of the path taken
Analogy:
between two points on amap remains the same (a
s a e unc on u e
distance travelled to go
between them or the actual
work going up and down are
variable (not a state
function .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Corollary: If you do a round trip, nothing has happened! ( X = 0)
State Functions Pro ert of State
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State Functions Pro ert of State .
The Final State is independent of the path taken
Analogy:
between two points on a
map remains the same (a
state function) but the
distance travelled to go
between them or the actual
work going up and down are
variable (not a state
function .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Corollary: If you do a round trip, nothing has happened! ( X = 0)
important and useful?
The Final State is independent of the path
taken.
‘The Enthalpy change for a reaction is dependent
products and is independent of the path or numberof ste s taken’ Hess’s Law.
Enthalpy change = H(final) – H(initial)
= H(products) – H(reactants)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Germain Henri Hess (1802 – 1850)
Hess’s Law.
8.x, xxx
,
H 3 = -112kJ
H ( k J
)
2NO2(g ) 2NO2(g )H 2 = 180kJ
N2(g ), 2O2(g )
1 2 31
N2(g ), 2O2(g )
–
= H(products) – H(reactants)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Hess’s Law
.
Can use Hess’s Law to calculate H for many reactionsusing tabulated enthalpies.
Important tabulated values are Standard Enthalpies (or heats)of Formation.
En tha lpy o f Fo r ma t ion (Hf ): enthalpy change associatedwith formation of 1 mole of a substance from its elements (C +
2
But: depends on conditions.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Standard Enthal of Formation Standard Enthal of Formation
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Standard Enthal of Formation.
Depends on state (s, l, g), temperature andpressure.
–at 1 atm and temperature of interest (usually 298K).
For compounds: Gases – 1 atmLiquids – pureSolutions: 1M
. . , .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Standard Enthal of Formation.
Standard enthalpy of formation (H°f )
‘Enthalpy change that results when 1 mole of a compound is
formed from its elements, with all substances in standardstates’
6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)
Not:
6CO2(g) + 6H2O(l) C6H12O6 (s) + 6O2(g)
.
H2(g) + O2(g) H2O2(l), will not happen.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Standard Enthal of Formation.
The standard enthalpy of formation for an elementin its standard state is ZERO.
‘Elements are not formed, they just……are!’
O2(g) O2(g) H°f = 0
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Note: H°f is in kJ/mol
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Blackman, Appendix A, pageA-1 to A-3
How do we use it? This means we can mani ulate
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How do we use it?
r calorimetry.
u : e ous, me consum ng an no very exc ng o a waysfeasible.
We can use thermodynamic tables to determine H rxn,
w ithout entering a laboratory.
‘If a reaction is carried out in a series of steps, the enthalpy
individual steps.’
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
This means we can mani ulatechemical reactions.
1. Take reactants apart to their constituent elements
2. Form products from the elements
Reactants ProductsHrxn
H°f (r))
H°f (p))
Constituent Elements
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
For a general (hypothetical) reaction:
Then:
° ° ° ° °rxn f f - f f
H°rxn = nH°
f (products) - nH°f (reactants)
Where n = coefficients in the balanced equation.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
for combustion of ammonia in air:
3 2 2 2
H°rxn = [cH°
f (C) + dH°f (D)] - [aH°
f (A) + bH°f (B)]
H°rxn = [4H°
f (NO2(g)) + 6H°f (H2O(l))] - [4H°
f (NH3(g)) + 7H°f (O2(g))]
= 4 33.2 + 6 -285.8 – 4 -46.1 – 0
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
= -1396 kJ Is this a sensible answer?
A ain: im ortant thin s to remember
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’ Example 8.9, page 300 of Blackman.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
A ain: im ortant thin s to rememberwhen manipulating reactions.
1. If you reverse a reaction be sure to change the sign on H.
CuO(s) + H2(g) Cu(s) + H2O(l) H = -130.6 kJ
Cu(s) + H2O(l) CuO(s) + H2(g) H = +130.6 kJ
[You will come across this concept again in Module 4 (electrochemistry)]
2. If you multiply by a coefficient, multiply H by the same
coefficient (H is extensive).
CuO(s) + H2(g) Cu(s) + H2O(l) H = -130.6 kJ
-
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2 2 - .
Exam le:
Two forms of carbon are graphite and. ,
calculate H for the conversion of graphite to diamond.
www.encarta.msn.com
C(graphite) + O2(g) CO2(g) H = -394 kJ
C(diamond) +O2(g) CO2(g) H = -396 kJ
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
xamp e:
Solution:
Reverse the second reaction and sum the two
C(graphite) + O2(g) CO2(g) H = -394 kJ
CO2(g) C(diamond) +O2(g) H = +396 kJ
C ra hite C diamond H = +2 kJ
graphite to diamond (endothermic process)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Exam le:
Example:
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Ethanol in petrol
With petrol approaching $2/litre it is cheaper to buy ethanol
. ,effect does it have on fuel efficiency?
Relevant data:
1
188
52
molkJ251)](HC[
molkJ278)](OHHC[
l H
l H
f
f
1
2
1
2
molkJ286)](OH[
molkJ5.393)](CO[
l H
g H
f
f
How much heat is roduced by burnin one ram of ethanol,
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
compared to one gram of octane?
p
The combustion of octane:
Ethanol in petrol
On a mass basis:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Example: Ethanol in petrol
The combustion of ethanol:
On a mass basis:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
RACQ - ‘Blends of up to 10 percent ethanol in petrol (E10) have been declared suitable for use in most –but not all – post-1986 petrol-engine vehicles in Australia. E10 increases fuel consumption between 2.6
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
and 5 per cent relative to unblended petrol’.
So: it’s cheaper and potentially more environmentally friendly but you use more of it………..
Enthal and Phase Chan es.
7.3, 251
What you should be able to do at this stage:
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Heat is also required for a phase change (Module 2)..
dependent on the strength of the intermolecular forces.
solid liquid (heat in) fusH
liquid gas va H
Criticalpoint
solid gas subHSolid Liquid
s s u r e ( a t m )
liquid solid (heat out) solidH = -fusH
= -
Gas P r e
Triple
point
condensation
vap
gas solid depositionH = -subHTemperature (°C)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Note: you may want to go back and review this slide and intermolecular forces, once you have encountered phasechanges.
1. Define and explain the First Law of Thermodynamics and its relation to
chemical reactions.
2. Explain what is meant by the terms System, Surroundings and the Universe
with respect to chemical reactions.
3. Explain how energy is transferred in chemical reactions.
4. Explain what Internal energy (U ) and Enthalpy (H ) are and how they are
re a e .
5. How is ‘work’ defined in chemical reactions (with relation to enthalpy)?
. .
7. Explain what a State Function is and how Hess’s law exploits it.
‘ ’. .9. Use thermochemical data to predict the enthalpy change for a reaction.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
. xp a n w a an xo erm c reac on an an n o erm c reac on mean n
relation to enthalpy changes.
Module 1: Lecture 7
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
What we will investi ate:
.
The role of energy in chemical reactions
First Law of Thermodynamics
How to understand and manipulatethermochemical data to redict how enerwill change in chemical reactions.
How energy changes can predict spontaneity.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
What determines whether or not a
A S ontaneous Process:
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process will be spontaneous?
We observe that many exothermic reactions aren ee spon aneous.
+ - -2 - .
u : so are some en o erm c reac ons.
= °2
2
.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Lets look at some spontaneous processes
A hot object cooling
Not spontaneous
Hot Coldspontaneous
Hotter
o o ec a ways coo s, a o o ec w never spon aneous y
absorb heat from colder surroundings to heat up.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
A S ontaneous Process:A hot object cooling
Hot ColdA natural change
What is different between the hot cube and the cold cube?
ENERGY
– the hot cube is characterised by molecules having
temperature).
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Think: where has this energy gone with cooling?
“SPONTANEOUS”
occurs w ou“outside intervention”
Thermodynamics tell us about the directiono a process – o e spee o e process
e.g. n pr nc p e a amon s ouchange spontaneously to graphite- owever: ery s ow process
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
– .
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feeling or native tendency without external
.
There is nothing in the definition of ‘spontaneous’ that describes how FAST the process is.
ne cs e s us ow as a reac on s o u e .Thermodynamics tells us if it will proceed at all.
Remember this!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
e.g.
• A Ball rolls down a hill never spontaneously up a hill
• A gas fills a container uniformly
• Heat flow always from hot to cold object – never the reverse
• o ,
> 0 oC ice spontaneously melts
WHY?
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
No: The meltin of ice occurs s ontaneouslabove 0oC – Endothermic process
Looking ahead: The driving force forspontaneity is an “increase” in the
“entro ” of the universe
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.4, 304
o w a s n ropy
n ropy s a measure o nergy s r u on
um er o arrangemen savailable to a system
Nature spontaneouslyproceeds towards statesthat have the highest
probability of existing.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Entro is a measure of Ener distribution
8.4, 305
Entro is a measure of Ener distribution
8.4, 305
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High-energy molecules Low-energy moleculesLet’s bring 3 high-energy molecules in contact with 3 low-
energy molecules => What is going to happen??
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
High-energy molecules Low-energy moleculesLet’s bring 3 high-energy molecules in contact with 3 low-
energy molecules => What is going to happen??
Number of arrangements available to a system
0 energy units transferred 1 energy unit transferred 2 energy units transferred 3 energy units transferred
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Nature spontaneously proceeds towards states that have the highest probability of existing.
Entro is a measure of Ener distribution
8.4, 305
High-energy molecules Low-energy moleculesLet’s bring 3 high-energy molecules in contact with 3 low-
energy molecules => What is going to happen??
Macrostates
Number of arrangements available to a system
0 energy units transferred 1 energy unit transferred 2 energy units transferred 3 energy units transferred
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Nature spontaneously proceeds towards states that have the highest probability of existing.
Entro is a measure of Ener distribution
8.4, 305
High-energy molecules Low-energy moleculesLet’s bring 3 high-energy molecules in contact with 3 low-
energy molecules => What is going to happen??
Macrostates
Microstates = Number of arrangements available to a system
0 energy units transferred 1 energy unit transferred 2 energy units transferred 3 energy units transferred
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Nature spontaneously proceeds towards macrostates that have the highest probability ofexisting = the highest number of microstates.
Lets think in terms of ‘Probabilit ’
8.4, 305
This is im ortant:
8.4, 306
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p=5% p=45% p=45% p=5%
Which do you think is the most ‘probable’ state?
Spontaneous processes proceed from macrostates of low probability
to macrostates of higher probability
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
‘Spon t aneous p rocesses tend t o d i sperse energy ’ .
‘Spon t aneous p rocesses tend t o d i sperse energy ’ .
We can now introduce a quantity that is a measure for how
–
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
8.4, 307
c e s s i b
l e
e l s
b e r o f a c
e n e r g y l e
N u m
The distribution of an identical amount of energy is greater in the larger volume.
This is important:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
,factor when considering entropy!!
Second Law of Thermod namics8.4, 309
increase in the Entropy of the Universe.
Suniverse = Ssys + Ssurr
If Suniv = positive (>0) – Entropy of the universe increases.
If Suniv = negative (<0) – process is spontaneous in the
opposite direction
If Suniv = 0 – equilibrium has been reached.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.4, 309
Our two Law s of Thermodynamics
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The Second Law of Thermod namics.
Nature spontaneously proceeds
diffusion of energy and thus have
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Our two Law s of Thermodynamics.
Ener is conserved.
Entropy in maximised.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
We are interested in Entro chan es.
Entropy is a state function, so
S = Sfinal – Sinitial
Similarly to Enthalpy, we have to take into account the system.
Where the surroundings + system = universe.
Suniv = entropy change in the universe
=sys
Ssurr = entropy change in the surroundings
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Interplay of S and S determines thesign of Suniv
Suniverse = Ssystem + Ssurr > 0
Zumdahl, 2006
Note: one of the entropy changes may be negative, but sum must be positive.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
If Suniv < 0, process will not be spontaneous as written, but is spontaneous in thereverse direction.
Entropy changes and real life.
A thermod namic caveat.
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How do we end up withcomplex systems that have
no a arent entro y – i.e.how do we end up with asystem having localised
Suniverse = Ssystem + Ssurr > 0
Large and positive
Negative
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Transferrin the Second Law to ‘Life’ is not a ood idea!Remember these laws were essentially developed for chemicalreactions (steam trains and reactions in test tubes) where the
= ……..
i.e. You can ask yourself if life violates the Second Law but’
D. R. Brooks and E. O. Wiley.. , .
Thorne K. S. et al.–
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
– .QB843,B55.B59
8.4, 309
To determine spontaneity (=Suniv), we need to know
Ssys and Ssurr .
How does the entropy of the system and the surroundingsc ange ur ng a process
The only way we can exchange energy between system andthe surroundings is via work (w) or heat (q).
=> We need to find a t h e r m o d y m a n i c definition of entropy,
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
factor.
ENTROPY is:
q r S = J K-1
(So heat flow is the important factor in determining entropy.)
It all started with steam engines…
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Optional reading: Carnot Heat Engines
The Carnot C cleIn the days of steam engines
The Carnot C cle
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In the days of steam engines…
http://commons.wikimedia.org/wiki/File:Tower_bridge_steam_engine.jpg
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://cache-media.britannica.com/eb-media/11/71511-004-321E1907.jpg http://www.todayinsci.com/H/Hornblower_Jonathan/HornblowerJonathanPatent1298.htm
Cyclic process with 4 steps:1. Isothermal expansion at T1 (hot)2. Adiabatic expansion (T1 → T2
3. Isothermal compression at T2 (cold)
4. Adiabatic compression (T2 → T1)
We can look at Work , Heat and efficienc for this circular rocess:
= Model for any cyclic process!!
q1=-wtot + -q2 -wtot = q1+q2
=> = -w = +
T1 t t
Also: = (T1-T2)/T1 (derivation: RED BOX CHALLENGE!)
system
q1
-wtot
=(q1+q2)/q1 = (T1-T2)/T1 1+ q2 /q1 = 1 – T2 /T1
=> q1 /T1 + q2 /T2 = 0 for this processT2
-q2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
=> q/ T = state function!! => Entropy!!!
We can define a standard entropy (S°)
Notice enthalpy as H
But entropy as just S
y
To understand that we need to focus onthermodynamic entropy = Entropy as it relates to
the occupancy and probability of energy levels with
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
re ation to t erma energy – i.e. eat.
°definition of Entropy:
The thermodynamic definition of ENTROPY is:
q r -
T
At absolute zero T=0 so heat cannot be transferredso = eore ca y
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
increases until we arrive at S° (298 K, 1 atm)
8.4, 308
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School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Consider a perfect crystal of HCl at 0 K:
Entropy of such a crystal is+- +- +- +- +-
,way to achieve perfect
order
+- +- +- +- +-
+-+- +- +- +-
crystal at 0 K is zero.
Disorder appears as crystal is warmed, leading to increasein entropy. Values of S° are absolute entropies (cf H , U ,
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
etc)
Module 1: Lecture 8
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Example: Calculate S° for the synthesis of ammonia from
2 an 2:
N2(g) + 3H2(g) 2NH3(g)
S°rxn = 2 S°(NH3) – [S°(N2) + 3 S°(H2)]
= 2 * 192.5 – [191.6 + 3 * 130.7]= -198.7 JK-1
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Note: S° for elements is not equal to zero
Srxn < 0 for the Haber process: N2(g) + 3H2(g) 2NH3(g) ‘Rules’ for Entro Chan es
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rea er n ropy Less Entropy
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Reduction of ways in which energy can be distributed
(1) S° for elements is not equal to zero
(2) Reactions that consume or generate gases can have
.Could predict that S° for Haber reaction < 0, since thereare 4 moles of gas on the left and 2 on the right.
N2(g) + 3H2(g) 2NH3(g)
I2(g) 2I(g) S0 > 04Fe(s) + 3O2(g) 2Fe2O3(s) S0 < 0
Example: Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
1. Prediction: the roblem of rust
+
4 mol Fe (s)
3 mol O2 (g)
2 mol Fe2O3 (s)
S < 0
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2. Prediction: Com lex molecules
Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
As 3 mol hydrogen forms 3 mol water vapour one
would assume S to be zero (no change in number of gas molecules).
However, S is large and positive, why?
Due to formation of a more complexmolecule.
Remember? The distribution of energy, not of molecules is the important
The H2O molecule can vibrate and rotate in several ways.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
factor when considering entropy!!
‘Rules’ for Entro Chan es
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Liquid H2O has a relatively lowentropy. Why?
Remember intermolecular forces?
Just another manifestation of the.
High BP, Low entropy – H-bonds
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• The entropy for gas is larger than for liquid which is larger thanfor a solid
• The more com lex the molecule the hi her the entro
• Forming a solution from a molecular solid yields an increase in
• Forming a solution from a salt can cause either an increase or
• Pressure and Temperature affect entropy
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Pressure and EntroWhich has the higher entropy?
N as at 1 atm or N as at 1x10-2 atm?
pV = nRT
One mole of N2 gas at 1x10-2 atm has a volume 100 timesthat of one mole of N2 gas at 1 atm. Thus, N2 gas at 1x10-2
atm has the higher entropy (particle in a box)
1x10-2 atm >>
1 atm
Slow pressure > Shigh pressure
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Tem erature and Entro
Why is temperature so important?
l e v e l s
e e n e r g
c c e s s i b l
Absolute zero
(BEC)
Temp 1 Temp 2
A change in temperature can make a huge difference
to accessible energy levels and thus the distribution
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
o energy n ose eve s.
What ou should be able to do at thisstage:
Inter la of S and S determines
the sign of Suniv
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g
‘ ’ . e ne w a s mean y a spon aneous reac on .
2. Define the Second Law of Thermodynamics.
3. Explain what is meant by ‘entropy’.
4. Describe how entropy changes for a spontaneous reaction.
5. Predict entro chan es based on anal sis of athermodynamically balanced equation (e.g. formation of gases,higher complexity molecules).
6. Use thermochemical data to predict the entropy change for asystem.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
7. Describe how temperature affects entropy.
the sign of Suniv
Suniverse = Ssystem + Ssurr > 0
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Zumdahl, 2006
We can determine the system entropy8.4, 309
change, but what about entropy changes inthe surroundings?
To determine spontaneity, we need to know Ssys
+ Ssurr . How does the entropy of the
surroundings change during a process?
The only way we can exchange energy with the.
The thermodymanic definition of entropy is:
q / T
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
Heat flow to the surroundin s - S ?8.4, 309
T e entropy c ange in t e surroun ings is ue toheat transfer from the system (conservation of
.
Ssurr = q /T
Therefore spontaneity is often dependent ontemperature. Eg an en ot ermic reaction may espontaneous at high temperature, but not at low.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Remember Enthal ?
Recall:
. surr
flow.
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Heat flow = H at constant
+ sign: endothermic (into the system)- sign: exothermic (out of the system)
So: At constant temperature andpressure.
Ssurr =-H sys
T
+ sign: exothermic
- sign: endothermic
S = S + S > 0
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2. The magnitude of Ssurr depends on
temperature
Driving forceprovided by =
Magnitude ofthe entro =
Quantity of heat (J)
energy flow(heat)
change of thesurroundings
Temperature (K)
=Quantity of heat (J)+SsurrExothermic
=Quantity of heat (J)-SEndothermic
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Temperature (K)
=
=H
- -ve si n because enthal with res ect to s stemT
surr
1. The sign of Ssurr depends on direction of heat flow.
2. The magnitude of Ssurr depends on temperature and
magnitude of enthalpy
As T increases, Ssurr decreases
“The impact of the transfer of a given quantity of energy as heat to
or from the surroundings will be greater at lower temperatures”
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The roblem of rust
4Fe(s) + 3O2( g) 2Fe2O3(s) S0 < 0 (prediction)
4 mol Fe
But experimentally, rusting is spontaneous
-H sys
+ surrT
3 mol O2
- . ,
Ssurr = -(-1648.4 x 103J)/298K = +5529 J/K
Ssys = -549.4 J/K
2 mol Fe2O3
Suniv = -549.4 + 5529= 4980 J/K
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The Gibbs Free Energy
8.2, 283
S = S + S
Converting the ‘general’ equation to one involving only
system variables.
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Suniv = Ssys + Ssurr.
But it’s inconvenient as a measure of spontaneity, since requiredto know about system and surroundings.
We want a simple thermodynamic function related tospontaneity which only requires knowledge of system.
Thanks to J. Williard Gibbs we do:
The Gibbs fun c t i on ( G)
In 1876 Gibbs published the first part of the work for which he is
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
,
publishing the second part of this work in 1878.
S = S + S
HsysS
universe
= S s stem
-
TS universe = Hsystem - TS system
G = H - TThis is the 2nd equation
G is the Gibbs Function of a process at const ant pressure and
o remem er
temperature
Just a more convenient wa of determinin if a rocess is
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
spontaneous……but it is much more useful than that.
Wh is this useful?8.2, 284
If the process is spontaneous:
S universe = S system + S surr > 0
Hsys
TS universe = S system - > 0
TS universe = Hsystem – TS system < 0
G = H – TS < 0
G w ill be negative for a spontaneous process!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
H - T S < 0
i.e. if G < 0 process is spontaneous
if G > 0 reverse process is spontaneous
if G = 0 system is at equilibrium
Two functions that can predict spontaneity:
+,
ΔSuniv: all processes
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Exam le: Wh does ice melt at 10°C but not
at -10°C?
kJ03.6)(OH)(OH0
22 H ls
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H2O(s) H2O(l) H° = 6.01 kJ
º
t < 0ºC
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
T T H° S° S S G°surr un v
C K kJ J K-1 J K-1 J K-1 J mol-1
-10 263 6.03 22.1 -22.9 -0.8 +220
-. . .
+10 283 6.03 22.1 -21.3 +0.8 -220
us a , ce me s spon aneous y, u a - ,water freezes spontaneously.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.5, 314
ummary o e e ec s o e s gns o
and S on spontaneity
Only when the negative
va ue o s arger an
the negative value of TS
H2O(l)→ H2O(s)
This is an exothermic change
accompan e y a ecrease
in entropy so is spontaneouso
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.5, 314
Various combinations are possible:
.
H S G Example
____________________________________________________________
+ + Spont. at high temp. CaCO3(s) everse spon . a ow emp. a s 2 g
Spont. at low temp, reverse CaO(s) + CO2(g) spont. at high temp. CaCO3(s)
+ G always +. 3O2(g) 2O3 (g)
Reverse rxn spont. at all temps.
+ G alwa s . 2O 3O
Spont. at all temps.______________________________________________________________________________________________________________
= –
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
At what temperature would processes with the
Solution:
a)G = 10 000 J + 220 J K-1 x T = 0
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p p
a) H = +10 kJ; S = -220 J K-1
b) H = +10 kJ; S = +184 J K-1
At equilibrium we know G = 0, so given the data we
G = H - T S
for the temperature T.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
- s canno e zero a any empera ure
This process has G +ve at all temperatures.
- -
- this equals zero at T = 54.3 K.
Below 54.3 K, G is > 0 (not spontaneous)
. ,
i.e. we can determine the temperature at which a
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
non-spontaneous reaction becomes spontaneous.
Module 1: Lecture 9
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Standard Gibbs Free Ener Chan es8.5, 314
As for H °, we define G° (standard Gibbs Free Energy
change) as the change for a reaction carried out under
standard state conditions, with all reactants and products in
.
Also use standard free energies of formation (G° f ).
For aA + bB cC + dD,
G°rxn = [cG°
f (C) + dG°f (D)] - [aG°
f (A) + bG°f (B)]
Exactly analogous to use of H
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The Standard Free Energy of Formation
Gf °
Gf
°
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Gf ° is the change in the Gibbs Free Energy when one mole of a
substance is created from its constituent elements, with all species
.
Like Hf °, Gf
° has units of kJ mol-1.
Like Hf ° for an element in its standard state, Gf
°
is defined to be zero.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• G° cannot be measured directly bymeans of any instrumentation
• G° can be calculated from othermeasure quantities
• H can e etermine y measuringheat flow in a calorimeter
• S° can be determined experimentally
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
8.5, 314 - 315
Three ways to determine G°:
1. G°rxn = npG°
f,prod – nrG°f,react
2. G°
= H°
– TS°
3. Reaction Manipulation (calculating from other rns.)
earn ese ways
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Methods for Calculatin G° 1
2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)
G f º -163 kJ/mol 0 kJ/mol -394 kJ/mol -229 kJ/mol
G° = 2Gf °(CO2(g)) + 4Gf
°(H2O(g)) – (2Gf °(CH3OH(g)) - 3Gf
°(O2(g))
= 2mol(-394 kJ/mol) + 4mol(-229 kJ/mol)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
– 2mol(-163 kJ/mol) – 3mol(0 kJ/mol) = -1378 kJ
Methods for Calculatin G° 2
G° = H° - T S° 22))((Of))((SOf(g))(SOf
ggH H H H
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Consider the reaction
2SO2(g ) + O2(g ) 2SO3(g )
carried out at 25C and 1 atm. Calculate H , S , and G using the
following data:
Substance H (kJ/mol) S (J/K mol)
f
SO2(g) -297 248
SO3(g) -396 257
O2(g) 0 205
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
))((Of ))((SOf (g))3
(SOf 22 gg
kJ594kJ792
0kJ/mol)297(mol2kJ/mol)396(mol2
kJ198
22 )(O)(SO)(SO 223 ggg SSSS
J/K205-J/K496-J/K514
mol)J/Kmol(2051mol)(248J/Kmol2mol)J/K(257mol2
J/K187-
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
ST H G
K187)K298(J000,198 J700,55J000,198
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Methods for Calculatin G° 3
The Free Energy is a State Function so we can manipulate any’
Cdiamond(s) Cgraphite(s)
Cdiamond(s) + O2(g) CO2(g) Gº = -397kJ
CO2(g) Cgraphite(s) + O2(g) Gº = 394kJ
Cdiamond(s) Cgraphite(s) Gº = -3kJ
This reaction is spontaneous but VERY slow
º
° ’
.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
Why is G so useful?The Gibbs Free Ener and Chemical
8.5, 318
Equilibrium
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1.Its relationship to equilibrium
. ts re at ons p to wor .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The real power of G lies in its relationship to equilibrium.
We can predict in which direction a system under standard
conditions will a roach e uilibrium G = 0 .
if G < 0, process is spontaneous (products favoured)
if G > 0, reverse process is spontaneous (reactants)
if G = 0, system is at equilibrium
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The Gibbs Free Energy and ChemicalEquilibrium
G < 0
In terms of energy: atequilibrium G = 0 andthe system no longer has
the capacity to do work.
productsreactants
G = 0
G ° = - RT lnK This is the 3rd equationto remember!!!
where K is the equilibrium constant
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
(This concept will be covered in more depth in module 2)
2. The Gibbs Free Ener and Work.8.5, 316
A reversible rocess is done in a s ecific wa so thatwhen undone there is no observable change in thesystem or surroundings.
A Breversible
Remember the First Law?
U = q + w
For the perfect reversible process we would be able toextract the maximum work from the internal energy –
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.relate to?
This is the entropy.
The Gibbs Free Ener is w
U
,
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T=
U
q r w r,max
For a perfectly reversible process, qr , thus the gain inentropy, is minimized.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
r,max
q r w r,max
G = H - T S
G = wr,max
The thermodynamic limit to efficiency.
But: perfectly reversible processes are unattainable,
yet we can determine G for any reaction usingthermodynamic tables??
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
G, H and S are State Functions.
G = w
When G is negative, the magnitude of G is equal to
,
.
When G is ositive the ma nitude of G is e ual to
the minimum work that must be expended to make the
process spontaneous.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Wh the ‘Gibbs Free Ener ’ ?
‘ ’ –work, not pV work done on/by the system, butelectrical, mechanical etc work.
’ n s as s you a so see en ropy e ne asthe ‘energy that can’t do useful work’ or ‘low
’ .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The other extreme – total irreversibilit
8.3, 287
U
–
8.5, 317
U
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q
For a perfectly irreversible process, q is maximizedand work is minimized.
e.g. irreversible battery discharge.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
q r w
In a real process there will always be losses of heat to.
Extra losses will occur due to friction etc.
‘As long as a body retains the ability to transfer heat to another body – a s, a any empera ure a ove a so u e zero – row n an mo on s no
only possible but also i nev i tab le .’ Einstein, 1905.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Another statement of the second law:
In an real c clic rocess work is chan ed toheat in the surroundings, and the entropy of the
universe increases.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Example: What is the maximum work obtainable from
2CH OH l + 3O 2CO + 4H O l
the combustion of methanol
G f º -166.4 kJ/mol 0 kJ/mol -394.4 kJ/mol -237.2 kJ/mol
Gº = (2 x -394.4 + 4 x -237.2) – (2 x -166.3) kJ
= -1404.8 kJ
Ho = (2 x -393.5 + 4 x -285.8) – (2 x -238.7) kJ
= - .
We can expect no better than 96.7 % conversion of the
energy the reaction can provide.
i.e. even done reversibly we still lose 3.3% as heat to the
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
surroundings – entropy has increased.
Example:
How much work can be gained from a reactionwith the following parameters?
Example: The combustion of glucose provides energy for
nervous an muscu ar ac v y. a s e max mum wor
available from combustion of 1 mole of glucose at 37oC (310K)o = -1
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g p
H = +10 kJ; S = +184 J K-1; T = 50K
G = H – TS
= 10 000 – 184 50
=
.
in order for reaction to be at equilibrium (e.g. raise thetem .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
. .
C6H12O6 s + 6O2 6CO2 + 6H2O l
Know: wmax = G, and G = H-TS
So: Hrxn = [6xH(H2O(l)) + 6xH(CO2(g))] – [6xH(O2(g)) + H(C6H12O6)]
= -2802.04 kJ/mol
G0 = H0 - TS0
= -2802.04 x103 J - (310 K x 182.4 J/K)
- .
Thus: Wmax = 2858.6 kJ/mol glucose burnt.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
MW glucose = 180g/mol, then 2858.6/180 = 15.9 kJ/gram glucose.
. ,
need: (70 x 9.8) x 3 = 2.06kJ of energy.
We have 15.9 kJ/gram of glucose, so 70kg person would
need to burn: 2.06/15.9 = 0.13 grams glucose.
In reality, much more than this would be required due to.
Other examples: See Page 317, Blackman.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
4Fe(s) + 3O2( g) 2Fe2O3(s) Suniv = 4980J/K
4 mol Fe
+
G = -TSuniv = H – TS
3 mol O2 = -298 K 4980 J/K
= - 1484.0 kJ
2 mol Fe2O3
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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Affected by:
tem erature Links to
Equilibriumpressure,
concentrations
Thermodynamics Note: In the next few lectures we use specific examples to teach key ideas
(concepts) – you are expected to be able to apply these ideas to any
equilibrium reaction.
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Homogeneous
Module 1
Heterogeneous
Module 2
ChemicalReactions
Acid/Base:Module 4
PhaseChanges
SolubilityElectro-
chemistry:Module 2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Underpinning CHEM2002, CHEM2056, BIOC2000
Examples of equilibrium processes in real life tend to be complex systems so we will be in with sim le exam les. .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://www.rsc.org/education/teachers/learnnet/cfb/transport.htm
Explore your existing understanding of equilibrium …
Write the balanced equation for the decomposition of dinitrogen tetroxide
to nitro en dioxide:
2NO
Write the balanced equation for the dimerisation reaction of nitrogen
ox e o orm n rogen e rox e:
42 g2 g
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
42 2 g422
Blackman Chapt er 9
Consider . . .
dinitrogen
tetroxidenitrogen
dioxide
•N2O4(g ) is in a sealed tube – closed system - begins to dissociate
o orm 2 g , w c n urn s mu aneous y eg ns o recom ne o
form N2O4(g )
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
•Equilibrium is a dynamic process (it can respond to change!)p 334
We can observe the progress of a reaction by monitoring concentration as
a function of time.)(ON 42 g )(2NO2 g
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• The initial concentration of N2O4 decreases with time• Simultaneously the concentration of NO2 increases with time
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• qu r um s reac e w en ere s no c ange
The same equilibrium will be reached by starting with either N2O4(g) or
with NO2(g)
This is an example of a Homogeneous equilibrium
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The position of chemical equilibrium can be expressed in terms of
the equilibrium constant:
)(ON 42 g )(2NO2 g
After concentrations have stopped changing, find that:
C25at1061.4
NO o3
2
2 K ON 42
c
regar ess o s ar ng concen ra ons. s ra o s nown as e
equilibrium expression for the reaction and results in the
equilibrium const ant , K c .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Generic re resentation of the ex ression for the
Equilibrium constant:
p 335
Note:
1. [A] refers to the equilibrium concentration of A in mol L-1 (orM).
2. K c is dimensionless in the sub-discipline of physical chemistry.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The value of Kc is dependent on temperature
)(ON 42 g )(2NO2 g
K ... What does the ma nitude tell us?
Compare:
2H (g) + O (g) 2H O(g)
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]NO[ 2
2
.]ON[ 42c
C25at1061.4]ON[
]NO[ 3
42
2 cK
Kc ... What does the magnitude tell us?
At equilibrium, concentrations of reactants and products will be of thesame order of magnitude at 100 oC.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf
2H2(g) + O2(g) 2H2O(g)
298Kat101.9][O]H[
80
2
2
2
2 cK
2 2
298Kat108.4]NO[ 31
2K
][O]N[ 22
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
equilibrium
K >> 1 : products favoured
K << 1 : reactants avoure
=
will be comparable
When is an equilibrium positioned in the middle?
Rule of thumb …
When K is in the ran e of 0.01-100 10-2 – 102
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The equilibrium constant for the following reaction is
= 5c
CO(g) + Cl2(g) COCl2(g)
What is the position of this equilibrium
at this temperature?
1. Far to the left
.
3. Close to the middle
e uivalent concentrations
of reactants and products)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
An equi l ibr i um expression can be wr i t ten in more than one way … What happens t o value of c f or a r ever se reaction?
If th it ti i id d (i th ti i itt i
Using the Equilibrium Constant to predict composition
The equilibrium constant is useful because it enables us to calculate
e uilibrium concentrations of reactants and roducts:
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If the opposite reaction is considered (i.e the equation is written inreverse) then K c is inverted.
e.g. )(ON 42 g )(2NO2 g
C25at1061.4]ON[
]NO[ 3
42
2 cK
inverted
)(2NO2 g )(ON 42 g
C25at217]ON[2
42 cK
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2
)(ON 42 g )(2NO2 g
The equilibrium expression is:
2
C25at10x4.61]ON[
-
42
2 cK
If, at equilibrium M01.0NO2 , then
M0.0217NOON
2
242
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
c
Using the Equilibrium Constant to predict composition
The equilibrium constant also enables us to calculate equilibrium
concentrations, more complex:
C25at10x4.61]ON[
3-
42
2 cK )(ON42
g )(2NO2
g
If the initial concentration of N2O4 is 0.50 M, what are the equilibrium
To solve this t e of roblem we can use an ‘ICE Table’ where werepresent the concentrations of reactant and product species as a
variable and use algebra!!
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Strate :
The concentration of the reactant will have decreased by the an amount
,
increased to be (2 x ).
Initial conc: 0.50 0.00
)(ON 42 g )(2NO2 g
Change: - x +2 x
Equilibrium conc: (0.50 – x ) 2 x
Use the stoichiometry of the equation to determine the relative amounts
at equilibrium
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
The equilibrium constant is given by:
2
C25at10x4 61 3-2 K
o e equ r um concen ra ons are:
= – =. . .
[NO ] = 2 x 0 023 M = 0 046 M
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C25at10x4.61]ON[ 42
cK
2
x
x
50.01061.4 3
substituting
Which becomes a quadratic equation and when solved for x:
x = 0.023
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Homework: Write out the steps to solve this equation.
[NO2] = 2 x 0.023 M = 0.046 M
)(ON 42 g )(2NO2 g
. .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
A l in understandin of e uilibrium:
The amount of nitrogen fixed naturally by plantsis too low to meet our demands for food supply -
so the Haber process is one of the most important
commercial rocesses.
Liquid ammonia produced by the Haber process is either added directly to
soil as a fertiliser or is converted into ammonium salts (phosphates &.
)(H)(N gg )(NH3 g
Homework: Balance this equation:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
At 127 ºC the following equilibrium concentrations were found:
[H2] = 3.1x10-3 M
-12 .
[NH3] = 3.1x10-2 M
• Write the equilibrium expression for this reaction
º• c .
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Module 1: Lecture 11
Where we left off …
)(H3)(N 22 gg )(NH2 3 g
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School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
or
)(H)(N 2222 gg )(NH3 g
At 127 ºC the following equilibrium concentrations were
found:
2 = . x -
[N2] = 8.5x10-1 M
NH3
= 3.1x10- M
• Write the equilibrium expression for this reaction
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
• Calculate the value of K c at 127 ºC for this reaction.
Writing and substituting into the equilibrium expression –
there is more than one correct answer!
222
22 3
4
3313
22
3 108.3
101.3105.8
.
HN
cK
What is the value of K* for the reverse reaction?
)(H3)(N 22 gg )(NH2 3 g
5
3
22* 106.211HN
K
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
3108.3NH cK
How will the equilibrium constant change if we alter the
coefficients that we use to balance the equation for the
Haber process?
)(H)(N 223
221 gg )(NH3 g
1
K cWhen would we use
this stoichiometry?
2
3
2
33
23
21
NH NH K
24
2222
109.1108.3 21
21
cK K
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Note that (xa) = xa
For the following reaction, Kc = 7.9 x 1011 at 500 K
(2) g + r 2 g r g
Gas Phase Equilibrium (Homogeneous)
concentration … when we consider gases we can also express K
in terms of the artial ressures of the ases.
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( )
c 4HBr 2H2 + 2Br 2
. .
2. (7.9 x 1011)2
3. 1/[(7.9 x 1011
)2
]
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
)(ON 42 g )(2NO2 g
)(2
2 NOP
)(42
O N p
P
sua y use s an ar s a e o a m, so pressures expresse n
units of atmospheres, and equilibrium constants quoted with no
.
Alternatively: pressure in Pa divided by standard pressure of 5 ‘ ’
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
How are K and K related?
Remember that pressure is related to concentration
Since concentration is Pn
t roug t e ea gas equat on.
RT V
K RT
P
2NO
2
2NO n
RT RT
Pc
4O2N42ON
cP
where n is change in moles of gas as reactants are
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
converted to products
Exam le:)(ON 42 g )(2NO2 g
C25at10x4.61]ON[
]NO[ 3-2
2 cK
What is the value of K ?
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
p337 (9.3)
n
cP RT K K )(
Kc = 4.61 x 10-3
One possible way of removing NO
to react it with CO in the presence of
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K c 4.61 x 10
R = 0.08206 atm L mol-1 K-1
T = 298 K (273 + 25 oC)
n = +1
113 1013.129808206.01061.4 PK
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
to react it with CO in the presence of
suitable catalyst.
2 2
At 300 oC this reaction has K c = 2.2 x 1059.o p
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Caution!!
It is very important that correct units are used in this
equation. This relates to units of the gas constant, R.
- c p
be in atm, respectively, we must use R = 0.08206 atm L mol-1
K-1
Otherwise a conversion must be used.
Note:
‘ ’
very complex due to unit conversions. Refer to p 338-9.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
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Usin Q to redict the res onse of e uilibrium to
change in concentration:
Qualitatively: If we
- 2
2]NO[
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Qualitatively: If we
,
will the system
respond?
)(ON 42 g )(2NO2 g
It will move to restore equilibrium . . .
We can use the reaction quotient, Q, to predict in whichdirection it will move
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2 ]NO[
42ON
By adding more NO2 gas we will increase the concentration
.
So Q > Kc and the reaction will proceed in the reverse
This means that more reactants form and we end up with
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
c
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
…
2SO (g) 2SO (g) + O (g)
= -3 , c .
r te t e express on or c
If we mix together the following reactants:
[SO3] = 2 x 10-3 M
2 = x -
[O2] = 3 x 10-2 M
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
In which direction will the reaction proceed towards equilibrium?
Module 1: Lecture 12
Effect of Chan e in Pressure (or Volume) on Equi l ibrium
Very important if gases are involved . . . pressure can be
c ange n t ree separate ways:
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School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
1. Add or remove a gaseous reactant or product
2. Add an inert gas (eg helium)
3. Change the volume
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
p 352
1. Add or remove a gaseous reactant or product
Changes the concentration, since concentration = n/V , so
treat as e ore.
.
No change in concentrations, so equilibrium position
unaffected.
Concentration of aseous s ecies chan es but ma or
3. Change the volume
may not change equilibrium. This depends on the
stoichiometry of the reaction.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Experimental Observation:
What is the effect of decreasing the volume of this system
at equilibrium at constant temperature?
)(ON 42 g )(2NO2 g
Observe: The concentration of both species increases.
S stem res onds to chan e b shiftin balance to the side
with fewer moles of gases, ie left.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Vol & equilLe Chatelier!
,
2
2NOQ (any conditions)
Effect of decreasin volume b increasin ressure
Volume change:Initial
equilibriumno onger a
equilibriumFinal
equilibrium
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Q (any conditions)
= >
42
, , , .
System adjusts by decreasing [NO2] and increasing [N2O4]
until Q = K once more.
The value of K remains constant.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
q
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Compare the number of molecules at equilibrium in (a) and (c)
H3N NH2
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
reducing the number of molecules.
vessel in which the following equilibrium existed:
2HI(g) H2(g) + I2(g)
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Eff ect of Temperat ure on Equil ibri um
If temperature is changed, the value of K
c anges. rect on o c ange epen s on
the sign of H
OH6CoCl2 Cl4OHCo2
2 + kJ
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g
2OH6CoCl 24 o
62
blue pink
Forward reaction is exothermic (ΔH < 0).
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Cobalt temp final
OH6CoCl 24 Cl4OHCo62 + kJ
blue pink
Heating forces equilibrium to left, cooling to right, ie
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Le Chatelier!
o = + .
What effect would raising the temperature have?
1. No effect2. An increase in the amount of NO2(g)
. 2 4
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
N2O4 effect of T
An increase in the number of NO2 molecules. The
reaction is endothermic and so the input of heatenergy as a reactant shifts the equilibrium to the
right.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
How do we ex lain effect of tem erature uantitativel
The 1st winner of the Nobel
’
p 355
2 RT dT
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’
For his work in thermodynamics
But A so etermine t at car on
forms tetrahedral rather than
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
H is positive H is negativeExothermic
Endothermic
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
ncreas ng ecreases soreactants are favoured
ncreas ng ncreases soproducts are favoured
2
ln H K d
Integrate
212
1ln
T T R K
When we know the standard enthalpy and K at one
temperature, we are able to calculate the K at a different
empera ure.
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Temperature Dependence of K
We can deduce the temperature dependence of the
equi i rium constant using t e o owing:
000
S H 00
S H
R RT 00
1
RT R
,
slope = -H ° / R and y -intercept S° / R
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.p 350
00
RT RK
ln
c xm y
If at temp. T 1 we have K 1 and at temp. T 2, K 2, then
S H K
00
11
ln
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y
l n K
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
1/T
1
S H 00
1 RT R 2
u rac equa on rom :
0
1
2ln T T R
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Subtract SECOND equation from FIRST:
R
S
T R
H K
00
1
1ln
S H
K 00
2
1ln
2
00 11 H H
21
21T R T R
0
1 11H K
212 T T R K o:
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
2 2 3
• By using the equation below, explain the observed values
R
S
RT
H K ln
R
S
T R
H 00
1
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
empera ure epen ence o
N2(g) + 3H2(g) 2NH3(g) - 92.38 kJ
• Exothermic process: H 0 < 0
F th i th l i
The a licabilit of the van’t Hoff e uation at two
0
specified temperatures:
1ln
TTR
T1 = 500 K
K1 = 90
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• For an exothermic rocess the slo e is
positive and therefore as the temperature
increases the equilibrium constant
decreases
00 K
R RT K ln
l
R
S
T R
H 1
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
Slope!1/T
T T R
2
K2 = ?
Ho = - 92 kJ 0 11H
21
21T T R
ln(90)- ln K2 = – – 92 x 103 1 – 1
8.314 500 800
4.50 – lnK2 = 8.30ln K = -3.80
School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium
.
K2 = 0.02