Download Document

Physical Chemistry
•
Physical
Ph
sical chemistry
chemistr is the
branch of chemistry that
establishes and develops the
principles
p
p
of Chemistry
y in
terms of the underlying
concepts of Physics
Physical Chemistry
Main book: Atkins’ Physical
y
Chemistry
y 8th
edition, Peter Atkins and Julio de Paula
Course web site:
http://homes.nano.aau.dk/lg/PhysChem2009.htm
1.
2.
3.
4.
5.
6.
6
7.
Schedule
Physical transformation of pure substances. Nucleation.
Simple mixtures. Activities. Colligative properties.
Two components systems and phase diagrams. Phase
diagrams
g
in the material science.
Chemical Equilibrium. Equilibrium electrochemistry.
Molecular motion in gases and liquids.Diffusion and
Conductivity.
The rates of chemical reactions
Kinetics of complex reactions.
Oxford On-Line Resources
• http://www.oup.com/uk/orc/bin/9780198700722/
• e-book:
Lecture 1
13 10 2008
13-10-2008
• Lecture:
–
–
–
–
–
thermodynamic functions (recap)
single component phase diagrams
phase transition and phase bondaries
surface tension and nucleation
new problems
Thermodynamic Functions
2´
1
H1, S1,
U1 etc
H2´, S2´,
U2´ etc
3
H3, S3,
U3 etc
2
H2, S2,
U2 etc
f
•
•
State functions: properties that
depend on the current state only
and on the way it was prepared
(
(e.g.
energy and
d enthalpy)
th l )
Path functions: relate to the
preparation of the state (e.g. work
and heat produced)
ΔU = U f − U i
ΔU = ∫ dU
i
Exact differential
f
q=
∫
i , path
d
dq
Enthalpies of transition
• For a chemical reaction, if we define ”standard”
conditions for reagents
g
and p
products the change
g in the
thermodynamic (state) function will not depend on the
path or intermediate steps.
p
p
• Hess’s law: Standard enthalpy (entropy, Gibbs free
energy etc.) of an overall reaction is the sum of the
enthalpies (entropies, Gibbs free energies etc.) of the
individual reaction it mayy be divided into.
• Standard state of a substance at a specified
temperature is its pure form at 1 bar
CH 4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2 H 2O(l ) ΔH ø = −890kJ / mol
1mol
2mol
Unmixed reagents in std.form
1mol
2mol
Unmixed products in std.form
Hess’s Law: Example
Calculate standard reaction enthalpy of combustion of propene if
standard reaction enthalpies
p
for combustion of p
propane
p
and
hydrogenetaion of propene into propane are known.
CH 2 = CHCH 3 ( g ) + H 2 → CH 3CH 2CH 3 ( g )
−124kJ / mol
CH 3CH 2CH 3 ( g ) + 5O2 → 3CO2 ( g ) + 4 H 2O (l )
−2220kJ / moll
1
H 2O(l ) → H 2 ( g ) + O2 ( g )
2
9
CH 2 = CHCH 3 ( g ) + O2 → 3CO2 ( g ) + 3H 2O (l )
2
+286kJ / mol
−2058kJ / mol
Irreversible processes.Clausius inequality
• More energy
gy flows as work under reversible
condition:
− dw ≥ − dw, or dw − dw
rev
rev
≥0
• On the other hand as internal energy is a state
function
dU = dwrev + dq
qrev = dw + dq
q
⇒ dqrev − dq = dw − dwrev ≥ 0 ⇒ dqrev / T ≥ dq / T
⇒ dS
d ≥ dq
d /T
• In an isolated system
y
the entropy
py cannot decrease
spontaneously
dS ≥ 0
Helmholtz and Gibbs energy
dq
≥0
T
Clausius inequality:
dS −
At constant volume:
dU
dS −
≥0
T
At constant pressure:
dS −
Let’s define:
•Helmholtz energy
•Gibbs energy
dH
≥0
T
TdS − dU ≥ 0
TdS − dH ≥ 0
A = U − TS
G = H − TS
Helmholtz and Gibbs energy
• Helmholtz
e
o e
energy:
e gy
dAT ,V = 0
– At equilibrium
reaction would favorable if dAT ,V < 0 meaning that
total entropy of the system increases (dS entropy
change of the system
system, -dU/T – of the surrounding)
– The change in Helmholtz energy is equal to the
maximum work accompanying the process
Helmholtz and Gibbs energy
• Gibbs energy
gy (free
(
energy):
gy)
– At constant T and P chemical reactions are
spontaneous
p
in the direction of decreasing
g Gibbs
energy
– Maximum additional ((non-expansion)
p
) work is g
given
by the change in Gibbs energy.
– Standard Gibbs energy
gy can be defined as:
ΔrG 0 = Δr H 0 − T Δr S 0
Further, standard Gibbs energy of formation can be introduced:
ΔfG =
0
∑
νΔ f G −
Products
0
∑
νΔ f G
Reactants
0
Combining the First and Second Law
dU = dq + dw
For closed system
y
in the absence of additional work
TdS
− pdV
dU = TdS − pdV
Fundamental equation
Can be applied to reversible and irreversible changes
Mathematically:
So,,
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜
⎟ dS + ⎜
⎟ dV
⎝ ∂S ⎠V
⎝ ∂V ⎠ S
⎛ ∂U ⎞
⎜
⎟ =T
⎝ ∂S ⎠V
⎛ ∂U ⎞
⎜
⎟ = −p
⎝ ∂V ⎠ S
Properties of Gibbs energy
G = H − TS
dG = − SdT + Vdp
⎛ ∂G ⎞
⎜
⎟ = −S
⎝ ∂T ⎠ p
⎛ ∂G ⎞
⎜
⎟ =V
⎝ ∂p ⎠V
• G decreases with temperature, proportional to entropy
Gibbs-Helmholtz equation
G−H
⎛ ∂G ⎞
=
⎜
⎟
T
⎝ ∂T ⎠ p
H
⎛ ∂ G⎞
=
−
⎜
⎟
T2
⎝ ∂T T ⎠ p
Phase diagrams
•
•
•
•
Phase – a form of matter that is uniform through in chemical composition and physical state
Phase transition – a spontaneous
p
conversion of one p
phase into another
Transition temperature – a temperature at which two phases are in equilibrium
Metastable phase – thermodynamically unstable phase that persist because the transition is
kinetically hindered
Phase boundaries
Phase boundaries
B ili point
Boiling
i
Boiling temperature – the temperature at which vapour pressure is equal to
external pressure
Phase boundaries
Supercritical liquid
water: 3740C and 218 atm
CO2: 310C and 72.9 atm
Water:
273.16K,, 611 Pa
Heating liquid in a closed vessel
Applications: Supercritical drying; dry cleaning; supercritical fluid chromatography
Phase boundaries
Water: 273
273.16K
16K and 611 Pa
Melting (freezing) temperature – the temperature at which liquid and solid
phases coexist in equilibrium (at given pressure). Melting temperature at 1 atm
called normal,
normal at 1bar – standard
Triple point (T3) - the point at which the three phases coexist.
Phase diagrams: CO2
At atmospheric
pressure “dry-ice”
forms directly
form gas
Phase diagrams: water
Negative slope
Phase diagrams: 4He
Li id li id ttransition
Liquid-liquid
iti
Stays liquid at 0K
Phase transitions
• Chemical potential μ = Gm for a one-component system
• At equilibrium chemical potential is the same through the
system
Ot e se, if chemical
Otherwise,
c e ca pote
potential
t a is
s
different, we can find a place with higher
chemical potential m1 and exchange
place at
material from there with another p
chemical potential m2.
dG = μ2 dn − μ1dn = ( μ 2 − μ1 )dn < 0
Spontaneous process could take place, so
the system is out of equilibrium
Phase transitions
Temperature dependence
⎛ ∂μ ⎞
⎜
⎟ = − Sm
⎝ ∂T ⎠ p
dG = − SdT + Vdp, μ = Gm
Melting point vs. applied pressure
⎛ ∂μ ⎞
⎜
⎟ = Vm
⎝ ∂p ⎠T
Phase transitions
The effect of applied pressure on vapour pressure
pressure applied
pp
to a condensed p
phase its vapour
p
p
pressure rises:
• when p
molecules squeezed and escape as a gas.
p = p* exp(Vm ΔP / RT )
vapour pressure
New vapour pressure in the absense of add.pressure
I d d
Indeed:
d μ (gas) = d μ (liq)
RTdp
dp / p = Vm ((liq)
q)d
dP
p
dp
RT ∫
=
p
p*
p* +ΔP
∫
Vm dpp
p*
⎛ p⎞
RT ln ⎜ * ⎟ = Vm ΔP
⎝p ⎠
p = p* exp(Vm ΔP / RT )
p = p* (1 + Vm ΔP / RT )
Location of phase boundaries
μα ( p, T ) = μ β ( p, T )
d μ = − S m dT + Vm dp
− Sα ,m dT + Vα ,m dp = − S β ,m dT + Vβ ,m dp
Cl
Clapeyron
equation:
ti
dp Sα ,m − S β ,m
=
dT Vα ,m − Vβ ,m
•Exact
E
t equation
ti that
th t applies
li tto any phase
h
ttransition
iti off pure substance
b t
Location of phase boundaries
Solid-liquid boundary
usually>0
dp Δ fus H
=
dT T Δ fusV
Δ fus H
p
usually>0 and small
T
dT
∫* dp = Δ fusV ∫* T
p
T
p= p +
*
p≈ p +
*
Δ fus H
⎛T ⎞
ln ⎜ * ⎟
Δ fusV ⎝ T ⎠
Δ fus H
T −T )
(
V
T Δ fus
*
*
T −T*
when
1
*
T
Location of phase boundaries
Liquid-vapour boundary
usually>0 and large
dp Δ vap H
=
dT T Δ vappV
usually>0 and large
If we assume perfect gas and Vm (liq) Vm (gas)
Δ vap H
dp
=
dT T ( RT / p )
Clausius-Clapeyron equation
d ln p Δ vap H
=
d
dT
RT 2
* −χ
p= pe
Δ vap H ⎛ 1 1 ⎞
χ=
⎜ − *⎟
R ⎝T T ⎠
Location of phase boundaries
solid-vapour boundary
Δ fus H + Δ vap H
dp Δ sub H
=
=
dT T Δ subV
T Δ subV
≈ ΔVvap
Steeper slope
Ehrenfest classification of phase transitions
⎛ ∂μ β ⎞ ⎛ ∂μα ⎞
⎜
⎟ −⎜
⎟ = Vβ ,m − Vα ,m = Δ trsV ,
∂
∂
p
p
⎠T
⎝
⎠T ⎝
•
⎛ ∂μ β ⎞ ⎛ ∂μα ⎞
Δ trs H
−
=
−
+
=
Δ
=
S
S
S
⎜
⎟ ⎜
trs
β ,m
α ,m
⎟
Ttrs
⎝ ∂T ⎠ p ⎝ ∂T ⎠ p
First order transition: first derivatives of the chemical potential with respect to
temperature and pressure are discontinuous.
Second order phase transition: first derivative of the chemical potential is continuous
but the second derivative is discontinuous
discontinuous. Though the symmetry changes
abruptly at the point of transition, the state of the phase changes continuously
Lambda-transition – first derivative continuous but heat capacity is infinite at Ttrs.
•
•
I
II
Infinite at
transition
Phase Transitions
First Order
Melting; Vaporization; most of
solid state transitions (like “tin
pest” b (white) ->a (grey))
Second Order
Ferromagnetism (H=0);
Superconductivity; Superfluidic
Liquid crystals
• Phase transition = change
g in symmetry
y
y
Abrupt change, two states coexist
Smooth transition
Example: XY-ferromagnet
Order parameter: local magnetization
Overcooling/overheating is possible
Surface Tension
•
Physical boundary between phases can posses extra energy, therefore
work is required do change the area:
dw = γ dσ
Surface tension
•
Minimization of surface energy often results in formation of a bubble or
y Let’s consider force balance in a spherical
p
cavity:
y
cavity.
dσ = 4π (r + dr ) 2 − 4π r 2 = 8π rdr
dw = 8πγ rdr
F = 8πγ r
F = 4π r 2 p
4π r 2 pin = 4π r 2 pout + 8πγ r
pin = pout + 2γ / r
Vapour pressure of a droplet (from Kelvin equation)
For water:
p = p* exp(2γ Vm / rRT )
p (r = 0.1μ m) = 1.011 p* p (r = 1nm) = 2.95 p*
Surface tension
• Nucleation
Seed of the new phase will grow only if Gibbs free energy of the system decreases
Significant surface energy has to be created to separate phases at the onset of
phase transition
ΔG = −
A
4π r
Δμ + 4π r 2γ
3Vm
3
∂
4π r 2
ΔG = −
Δμ + 8π rγ
∂r
Vm
B
2γ Vm
r =
Δμ
*
As a result at the phase transition only the
nucleation center with infinite radius will
grow. All finite nucleation center require
finite thermodynamic force: Δμ > 0
Problems
To solve in class:
•
•
•
4.7a: An open vessel containing (a) water, (b) benzene, (c) mercury stands in a
laboratoryy measuring
g 5.0 m × 5.0 m × 3.0 m at 25°C. What mass of each
substance will be found in the air if there is no ventilation? (The vapour pressures
are (a) 3.2 kPa, (b) 13.1 kPa, (c) 0.23 Pa.)
4.9a Calculate the melting point of ice under a pressure of 50 bar. Assume that the
density of ice under these conditions is approximately 0
0.92
92 g cm–3 and that of liquid
water is 1.00 g cm–3.
4.10a What fraction of the enthalpy of vaporization of water is spent on expanding
water vapour
To solve at home:
• P4.5: Calculate the difference in slope of the chemical potential against pressure
for water on either side of (a) normal freezing point (b) normal boiling point
point. The
-3
-3
densities of ice and water at 0ºC are 0.917 g cm and 1.0 g cm and those of
water and water vapour at 100ºC are 0.958 g cm-3 and 0.598 g cm-3, respectively.
By how much the chemical potential of water vapour exceed that of liquid water at
1 2 atm and 100ºC
1.2
100ºC.
• P4.7: 50 dm3 of dry air is slowly bubbled through a thermally insulated beaker
containing 250g of water initially at 25ºC. Calculate the final temperature assuming
the vapour pressure of water constant at 3.17kPa, its heat capacityy 75.5 JK-1mol-1
and it behaves as a perfect gas. Assume that the temperature of air is constant.
If you need any extra data to solve the problems, search for them in the Data section in
the end of the Atkins book!