Download Condensed Matter 2

Condensed Matter 2
Physical Chemistry and
Applications of
Liquids and Solutions
Course Aims

The course aims to apply the principles of
thermodynamics and kinetics to the study of the
physical properties of gases, liquids and solids
on both microscopic and macroscopic scales.
 There will be particular emphasis on the
properties of liquids and solutions.
 This course provides a bridge between the
elementary study of condensed matter phases
and the advanced study of solid-state physics.
Learning Objectives
 After
completion of this course the student
is expected to have a firm understanding
the relationship between pressure,
temperature, and volume on the behaviour
of solids liquids and gasses.
 They should be able to apply these
relationships to understand phenomena
such as vapour pressure, surface tension,
liquid crystals, etc.

The student should understand the properties of
liquids and solutions (both ideal and real) on
microscopic and macroscopic scales, including
electrolyte solutions, solvation, and equilibrium
in solution, and appreciate the application of
these principles to devices such as batteries and
fuel cells.
 The student should have some understanding of
intramolecular (chemical) and intermolecular
bonding; including the bonding and structure of
crystals.
 The student should be able to quantitatively
solve problems involving all of the above.
Course Content

The Relative Stability of Gases Liquids and
Solids (4 lectures):
Including vapour pressure; surface tension;
supercritical fluids; liquid crystals.

Ideal and real solutions (4 lectures):
Binary solution model; Gibbs-Duhem equation;
suppression and elevation of freezing and
boiling temperatures; osmotic pressure; dilute
solutions; equilibrium in solution.


Electrolyte Solutions (4 lectures):
H, S, and G of ion formation; solvation;
calculating activity coefficients by Debye-Huckel
theory; equilibrium of electrolyte solutions.


Electrochemical cells (4 lectures):
Electrode potentials and cell potential; cell EMF
and the equilibrium constant; electrochemical
cells; batteries; fuel cells, atomic-scale
electrochemistry and nano-machining; half-cell
potentials.


Introductory Solid State Physics (4 lectures):
Including intermolecular and intramolecular
bonding; and heat capacities of solids.
Course Structure
 Lectures:
Fridays 0900 and 1100
The course will consist of
approximately 20 lectures.
All lectures are compulsory and an
attendance register will be taken.
Failure to attend lectures at least 75 % of
lectures may result in deregistration from
the course.
8
Weekly Assignments (best 6 count)
(Weeks, 4-12, save Reading Week).
 These sets are compulsory. Failure to
submit at least 75 % of the assignments
may result in deregistration.
 Deadline:1700 on the Wednesday
preceding the problem class.
 The assignments will be available on the
Module’s homepage
(www.ph.qmul.ac.uk/~phy226)
on the Monday of the week preceding the
deadline.
 Problem
Classes:
Thursdays 1000
8 Weekly Problem Classes. (Weeks, 4-12,
save Reading Week)
All problem classes are compulsory and
an attendance register will be taken.
Failure to attend lectures at least 75 % of
lectures may result in deregistration from
the course.
Course Assessment
 One
Examination paper: 2 hours 15
minutes (80%).
 Eight
assessed problem sets, of which the
best six sets contribute to the mark (12%).
 Two
in-class tests (8 % in total) to be run
during the last lecture before reading week
and the last lecture of term.
Suggested Reading
 Physical
Chemistry, P. W. Atkins.
 Properties of Liquids and Solutions, J.N.
Murrell.
 Introduction to Solid-State Physics, C.
Kittel.
Equilibrium
Things in Balance
Mechanical Equilibrium
 Forces
in balance.
 e.g., two equal masses, a rope and a
pulley.
 From Newton II, the net force is zero.
 Note the system is reversible.
Thermal Equilibrium
 Two
objects in thermal contact.
 Heat flows from the warmer to the cooler.
Until they reach the same temperature.
 (metal objects at room temperature initially
feel cold – sensation diminished and
vanishes at equilibrium)
Chemical Equilibrium
 In
a closed system, reactants turn into
products.
 Eventually this process stops
 A chemical reaction is in equilibrium when
there is no tendency for the quantities of
reactants and products to change.
Chemical Equilibrium
+I2 → 2HI (synthesis)
 2HI → H2 + I2 (dissociation)
 H2
The same chemical reaction system, but
the roles are reversed.
Both yield the same mixture when the
change is completed (equilibrium mixture).
Chemical Equilibrium
 composition
of a chemical reaction
system will tend to change in a
direction that brings it closer to its
equilibrium composition.
Chemical Equilibrium
 The
end is the same no mater where you
start.
Complete Reactions
+ B → C + D
 A reacts with B to form C and D.
 Implies that at the end of the reaction
there is only C and D.
 Implies that C + D → A + B does not
occur.
 The reaction is complete (the reactants
turn completely into products).
A
Reversible Reactions

If at equilibrium there are significant quantities of
all reactants, the reaction is not complete, and
the reaction is reversible.
 A + B = C + D
 In principle, all chemical reactions are reversible,
but this reversibility may not be observable if the
fraction of products in the equilibrium mixture is
very small, or if the reverse reaction is kinetically
inhibited (very slow.)
Law of ‘Mass Action’

The law of mass action states that any chemical
change is a competition between a forward and a
reverse reaction.
 The rate of each of these processes is governed
by the concentrations of the substances reacting;
as the reaction proceeds.
 These rates approach each other, and at
equilibrium they become identical.
Law of ‘Mass Action’
Rate of forward reaction is:
k f A B 

Rate of reverse reaction is:
k r C  D 

Thus, at equilibrium,

a
b
c
d
k f A B  kr C  D
a
b
c
d
How do we know a reaction is at
equilibrium?
 Consider.
 2H2
+ O2 = 2H2O.
 We add the required amounts of H2 and
O2 to a reaction vessel and leave them for
a month.
 Is the mixture in equilibrium?
How do we know a reaction is at
equilibrium?
 [set
of a spark in the vessel to find out].
 The reaction now rapidly proceeds to
equilibrium [provided the reaction vessel
survives].
 This mixture was not at equilibrium.
 Although the reaction is
thermodynamically favoured it was
kinetically inhibited.
How do we know a reaction is at
equilibrium?
 Test
for equilibrium:
 Change P, or T, or an amount of one
reactant (discussed later).
 If there is a change in the system we were
at equilibrium.
 If no change is observed we were not at
equilibrium (the reaction is kinetically
inhibited).
The Le Chatelier Principle
The Le Chatelier Principle
 If
a system at equilibrium is subjected to a
change of pressure, temperature, or the
number of moles of a component, there
will be a tendency for a net reaction in the
direction that reduces the effect of this
change.
The Le Chatelier Principle
 Example
(amount of a reactant change)
 2HI → H2 + I2 (in an arbitrary eq. mix)
 Now inject more H2:
 H2 now exceeds eq. concentration.
 No longer in equilibrium.
 A net reaction ensues until a new
equilibrium is reached.
The Le Chatelier Principle
 This
happens in a direction that reduces
the effect of the added H2.
 This can happen by some of the excess
H2 reacting with some of the I2 to form
more HI.
 Chemists would say "the equilibrium shifts
to the left."
The Le Chatelier Principle
The Le Chatelier Principle
 Example
(pressure change )
 2NH3 → 3H2 + N2 (in an arbitrary eq.
mix, in a closed reaction vessel)
 Now increase the size of the vessel.
 Effect of increasing the size is that the
pressure inside the vessel falls.
 How does the system react?
The Le Chatelier Principle
→ 3H2 + N2
 If only 2NH3 present, the pressure is 2
arbitrary units.
 If only 3H2 + N2 present, the pressure is 4
arbitrary units.
 Increasing volume decreases the
pressure.
 Therefore system reacts to increase
pressure
 2NH3
The Le Chatelier Principle
→ 3H2 + N2
 Pressure in increased by forming more
products (and decomposing more NH3).
 Equilibrium shifts to the right.
 2NH3
The Le Chatelier Principle
 Example
(temperature change)
 heat + N2 + O2 → 2 NO (endothermic)
 Regard heat as a "reactant" or "product" in
an endothermic or exothermic reaction
respectively.
 Apply the Principle.
The Le Chatelier Principle
 Suppose
this reaction is at equilibrium at
some temperature T1 .
 Increase temperature to T2.
 The system reacts in a way that absorbs
heat.
 Equilibrium shifts to the right (product)
 [by analogy, equilibrium shifts to the left for
exothermic reactions].
Equilibrium Quotient

Earlier condition for the reaction
aA + bB → cC + dD
can be expressed as a quotient
kf
kr

C  D 

a
b
A B 
c
d
 Qc
Equilibrium Constant
 If
expressed in terms of the equilibrium
concentrations, this becomes:
kf
kr

C  D 

a
b
A B
c
d
 Kc
Equilibrium Constant
 Kc
is the value of Qc when the reaction is
at equilibrium.
 The ratio Kc/Qc serves as an index how
the composition of the reaction system
compares to that of the equilibrium state.
 Thus, it indicates the direction in which
any net reaction must proceed.
Equilibrium Constant
Qc/Kc > 1
 Product concentration too high for
equilibrium;
 net reaction proceeds to left.

Equilibrium Constant
Kc/Qc = 1
 System is at equilibrium
 No net change will occur. < 1Product
concentration too low for equilibrium;
net reaction proceeds to right.

Equilibrium Constant
Kc/Qc < 1
 Product concentration too low for
equilibrium.
 Net reaction proceeds to right.

Relating Q and K to Reactions
 Each
tiny dot on the graph represents a
possible combination of NO2 and N2O4
concentrations that produce a certain
value of Qc for N2O4 → 2 NO2.
Relating Q and K to Reactions
 Only
those dots that fall on the red line
correspond to equilibrium states of this
system (those for which Qc = Kc ).
Relating Q and K to Reactions
 If
the system is initially in a non-equilibrium
state, its composition will change in a
direction that moves it to one on the line.
Relating Q and K to Reactions
 Solid/vapour
equilibria.
 E.g.,: Sublimation of iodine I2(s) = I2(g)
 The possible equilibrium states of the
system are limited to those in which at
least some solid is present (shaded)
Relating Q and K to Reactions
 However,
within this region, the quantity of
iodine vapor (red line) is constant [as long
as the temperature is unchanged].
Relating Q and K to Reactions
 The
arrow shows the succession of states
the system passes through when 0.28
mole of solid iodine is placed in a 1-L
sealed container.
Relating Q and K to Reactions
 The
unit slope of this line reflects the fact
that each mole of I2 removed from the
solid ends up in the vapor.
Relating Q and K to Reactions
 The
decomposition of ammonium chloride
NH4Cl(s) = NH3(g) + HCl(g) is another
example of a solid-gas equilibrium.
Relating Q and K to Reactions
 Arrow
1 traces the states the system
passes through when solid NH4Cl is
placed in a closed container.
Relating Q and K to Reactions
 Arrow
2 represents the addition of
ammonia to the equilibrium mixture;
Relating Q and K to Reactions
 the
system responds by following the path
3 back to a new equilibrium state.
Relating Q and K to Reactions
 This
state contains a smaller quantity of
ammonia than was added (Consistent with
the Le Châtelier principle).
Does the reaction stop at
equilibrium?
 At
equilibrium Qc = Kc.
 There is no change in the concentrations
of the reactants/products.
 The absence of any net change does not
mean that nothing is happening.
 Why?
Does the reaction stop at
equilibrium?





As any reaction is reversible, it can be
expressed at the sum of the forward and the
reverse reactions.
A = B
A → B rate = kf[A].
B → A rate = kr[B].
This implies

B k f
Kc 

A kr
Does the reaction stop at
equilibrium?

This implies that the rates of the forward and
backward reactions are equal: not that they are
zero.
 I.e., at equilibrium As are turning into Bs as fast
as Bs turn into As.
 Equilibrium is a dynamic process.
Example
 The
commercial production of hydrogen is
carried out by treating natural gas (mainly
methane) with steam at high temperatures
and in the presence of a catalyst (“steam
reforming of methane”):
CH4 + H2O  CH3OH + H2
Example






Given the following boiling points:
CH4 (methane) = –161°C
H2O = 100°C
CH3OH = 65°
H2 = –253°C.
predict the effects of an increase in the total
pressure on this equilibrium at 50°, 75° and
120°C.
Example






Given the following boiling points:
CH4 (methane) = –161°C
H2O = 100°C
CH3OH = 65°
H2 = –253°C.
predict the effects of an increase in the total
pressure on this equilibrium at 50°, 75° and
120°C.

Solution: Calculate the change in the moles of
gas for each process:
Example
 50°C
CH4 (g) + H2O (l) CH3OH (l) + H2(g)
 Solution:
Calculate the change in the
moles of gas for each process:
One mole gas  one mole of gas
NO CHANGE
Example

75°C
CH4 (g) + H2O (l) CH3OH (g) + H2(g)

Solution: Calculate the change in the moles of
gas for each process:
One mole gas  two mole of gas
To counter the increase in pressure, equilibrium
shifts to the left.
Example
 120°C
CH4 (g) + H2O (g) CH3OH (g) + H2(g)
 Solution:
Calculate the change in the
moles of gas for each process:
two mole gas  two mole of gas
NO CHANGE
How to find the equilibrium
constant for a series of reactions
 Many
chemical changes can be regarded
as the sum or difference of two or more
other reactions.
 If we know the equilibrium constants of the
individual processes, we can easily
calculate that for the overall reaction
according to the following rule:
How to find the equilibrium
constant for a series of reactions
 The
equilibrium constant for the sum of
two or more reactions is the product of the
equilibrium constants for each of the
steps.
Example
Given the following equilibrium constants:
CaCO3 (s) → Ca2+ (aq) + CO32– (aq)
K1 = 10–6.3
HCO3– (aq) → H+(aq) + CO32–(aq)
K2 = 10–10.3
Calculate the value of K for the reaction
CaCO3 (s) + H+(aq) → Ca2+ (aq) + HCO3– (aq)
Example
The net reaction
CaCO3 (s) + H+(aq) → Ca2+ (aq) + HCO3– (aq)
is the sum of Reaction 1
CaCO3 (s) → Ca2+ (aq) + CO32– (aq)
K1 = 10–6.3
And the reverse of Reaction 2
HCO3– (aq) ← H+(aq) + CO3 2– (aq)
K-2 = 10-(-10.3)
Example
Therefore,
K = K1 K-2 = 10(-8.4+10.3) = 10+1.9
The Harber process
 The
Haber process for the synthesis of
ammonia is based on the exothermic
reaction
N2(g) + 3 H2(g) = 2 NH3(g)
Apply the le Châtelier principle in order to
maximize the amount of product in the
reaction mixture.
The Harber process
N2(g) + 3 H2(g) = 2 NH3(g)
Would prefer N2(g) + 3 H2(g) = 2 NH3(l)
The Harber process
N2(g) + 3 H2(g) = 2 NH3(g)
Would prefer N2(g) + 3 H2(g) = 2 NH3(l)
it should be carried out at high pressure
and low temperature.
The Harber process
N2(g) + 3 H2(g) = 2 NH3(g)
But lower temperature slows the reaction.
A choice has to be made.
But Haber solved the first problem by developing
a catalyst that would greatly speed up the
reaction at lower temperatures.
The Harber process
N2(g) + 3 H2(g) = 2 NH3(g)
A catalyst is a substance that lowers the
energy barrier for a reaction.
A catalyst plays no role in the net reaction.
It is regenerated at the end of the reaction.
Thus, they are only needed in trace
quantities.