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![Page 1: 1 Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee afl2@york.ac.uk Department of Chemistry.](https://reader036.fdocuments.net/reader036/viewer/2022062620/551b341f550346d41a8b5021/html5/thumbnails/1.jpg)
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Electrochemistry & Solutions
1. Solutions and Mixtures
Year 1 – Module 38 Lectures
Dr Adam [email protected]
Department of Chemistry
![Page 2: 1 Electrochemistry & Solutions 1. Solutions and Mixtures Year 1 – Module 3 8 Lectures Dr Adam Lee afl2@york.ac.uk Department of Chemistry.](https://reader036.fdocuments.net/reader036/viewer/2022062620/551b341f550346d41a8b5021/html5/thumbnails/2.jpg)
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Aims
To: • Understand physical chemistry of
solutions and their thermodynamic properties
predict/control physical behaviour
improve chemical reactions
• Link electrochemical properties to chemical thermodynamics
rationalise reactivity.
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Synopsis• Phase rule
• Clapeyron & Clausius-Clapeyron Equations
• Chemical potential
• Phase diagrams
• Raoults law (Henry’s law)
• Lever rule
• Distillation and Azeotropes
• Osmosis
• Structure of liquids
• Interactions in ionic solutions
• Ion-ion interactions
• Debye-Huckel theory
• Electrodes
• Electrochemical cells
• Electrode potentials
• Nernst Equation
• Electrode types
Recommended ReadingR.G. Compton and G.H.W. Sanders, Electrode Potentials Oxford Chemistry Primers No 41.
P. W. Atkins, The Elements of Physical Chemistry, OUP, 3rd Edition, Chapters 5, 6 & 9.
P. W. Atkins, Physical Chemistry, OUP, 7th Edition, Chapters 7, 8 & 10 OR 8th Edition, Chapters 4, 5, 6 & 7.
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Signifi cance of Areas, Lines and Points
Areas: single phase exists (specif y p and T state of matter)
Lines: 2 phases coexist in ium (specif y p or T need to know what 2 phases present)
P
T
S
L
G
P
T
S L
G
P
T
S
L
L
G
S
G
Phase Diagrams
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Signifi cance of Areas, Lines and Points
Areas: single phase exists (specif y p and T state of matter)
Lines: 2 phases coexist in ium (specif y p or T need to know what 2 phases present)
P
T
S
L
G
P
T
S L
G
P
T
S
L
L
G
S
G
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S-L line change in m.pt. with pressure S-G line " in sub.pt. with pressure L-G line " in b.pt. with pressure (isoteniscope explores how pressure aff ects boiling of liquid)
Triple Point: all 3 phases coexist in ium (unique value of p and T)
Critical Point: conditions (T and p) above which liquid state matter ceases to exist
VacuumPump
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Slope of Solid/ Liquid Line
I ce- water system: Slopes Backward: (rare) i.e. dT/ dP = - ve m.pt. pressure - ice-skating also note ice floats on water (ice< water)
CO2 system:
Slopes Forward: (usual) i.e. dT/ dP = + ve m.pt. pressure solid CO2 sinks in liquid CO2
L
S G
P
T
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These phenomena can be understood by simple (!) thermodynamics BUT FIRST
The Phase Rule obtained f rom experiment derived/ proven by TD allows rigorous discussion of phase behavior
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The Phase Rule
I t states: (only at ium)
F = c - p + 2
Components - smallest no. of substances needed to describe system e.g. H2O = 1 (ice, water and steam all H2O) EtOH + H2O = 2 H2O + NaCl = 2 (despite NaCl Na++Cl-)
No. of degrees of freedom
No. of components
No. of phases
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Phases - no. of unif orm, homogeneous regions within the phase diagram e.g. ice = 1 water + ice = 2 water + ice + steam = 3 Degrees of Freedom - smallest no. of independent State Variables needed to describe ium system
(e.g. p, V, T) A state function (or f unction of state) depends only on the current state of the system, and not on the route by which it was reached.
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Consider Water Diagram
At X P = 1 (solid) F = C - P + 2 = 2
At Y P = 2 (L + G) F = 1
At Z P = 3 (S + L + G) F = 0
S
L
G
P
T
X
Y
Z
must specif y 2 variables to locate position on diagram
position determined by only 1 variable
can only co-exist f or fi xed p and T
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Thermodynamics of Two-Phase Equilibria of a Single Component:
The Clapeyron Equation Thermodynamic eqns. needed 1. G = H - TS
2. G = 0 at ium (S=H/ T) 3. dG = Vdp - SdT (f undamental eqn. f or G)
concerned with equilibria along lines S
L
G
P
T
phase
phase
e.g. ice/water at m.pt.
1839 -1903
Josiah Willard Gibbs
Gibbs Free Energy
American mathematical physicist developed theory of chemical thermodynamics. First US engineering PhD…later Professor at Yale.
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Derivation of Clapeyron Eqn.:
convert 1 mol f rom state to state at ium (i.e. @ transition)
G = G - G = 0
G = G
dG = dG
Since, dG = Vdp - SdT
dG = Vdp - SdT = Vdp - SdT = dG
(V - V)dp = (S - S)dT
G.F.E = G G
1799 - 1864
Benoit Paul Emile Clapeyron
Parisian engineer and mathematician. Derived differential equation for determining heat of melting of a solid
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Application of Clapeyron Eqn. to Solid/ Liquid Phase
Process: f usion (melting)
T = melting point, Tm
H = enthalpy (fH)
V = volume (fV)
= slope of line =
I nverting: = change in Tm = with p
mdTdp
VTHfm
f
dpdTm
HVT
f
fm
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What is fV?
V = 1/ density = 1/ = 1/ gcm-3 = cm3g-1
fV = 1/ l - 1/ s
Sign of slope: Water dp/ dT = - ve = and fH = positive fV = - ve
s < l hence ice floats
VTHfm
f
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Modified eqn. for Liquid/Gas and Solid/Gas Lines:
Clausius-Clapeyron Eqn. Maths dx/x = dlnx dp/p = dlnp
= = =
=
=
dxxn
1nx 1n
2x
dx
12x 12
x1
2T
dT
T1
dTdp
VTH Clapeyron
Equation
Rudolf Julius Emmanuel Clausius
1822 -1888
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Derivation
e.g. liquid/ gas system
V = Vg - Vl
Vg >> Vl
V Vg
For an I deal Gas,
pVg = RT (per mol)
Vg = RT/ p = V
substituting into Clapeyron eqn
= =
dTdp
RTTpH
pdp
2RT
dTH
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Hence,
dlnp = =
I ntegrating,
lnp = - + constant
I soteniscope expt.
NB - normal boiling point is T at 1 atmosphere - T is always in K not C
2TR
dTHdT
plnd2RT
H
T1
RH
y = m x + c
lnp
1/ T
slope = -H/ R X
X X
X
X
Clausius-ClapeyronEquation
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Example Problems The densities of a substance in solid & liquid states at the melting point (23 C) are 0.875 & 0.901 gcm3 respectively. Heat of f usion is 276 J g-1. At what temperature will the substance melt at an applied pressure of 100 atmos.?
Use, =
fV = 1/ l - 1/ s = 1/ 0.901 - 1/ 0.875 = -0.033 cm3g-1 = (23+273) x (-0.033)
convert cm-3m-3..... x 10-6
= -296 x 0.033x10-6 KPa-1
= -0.35 K
mdTdp
VTHfm
f
dpdTm T V
(cm3g-1)
H 276 (J g-1)
dpdTm
276
99
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The vapour pressure of liquid napthelene is 15.5 mmHg at 95 C and 505.7 mmHg at 200.5 C. Calculate the heat of vapourisation and the normal boiling point of napthalene.
Use, ln =
ln =
H = 47,900 J mol = 47.9 kJ mol-1
Normal b.pt. (i.e. Tb f or p = 1atm = 760 mmHg)
ln =
Tb = 490 K (217C)
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v
T1
T1
RH
2
1
p
p
5.157.505
338
15.473
1314.8
Hv
5.15760
338
1T1
314.8900,47
b
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Consider ideal gas at constant temperature:
dG = Vdp – SdT = Vdp
Since pV = nRT,
2
1
2
11
2lnp
pnRT
p
nrTdpVdpG
22 lnpnRTGG If initial state 1 = STP (1 atm)
In general for a mixture AB:
GA = GoA + nART ln pA, GB = Go
B + nBRT ln pB
since GA = nAA
A = Ao + RT ln pA
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Phase Diagrams in Binary
Systems
Composition:
mol f raction of A = nA / nA+nB
f or liquids - xA, xB
xA =
f or vapour phase - yA, yB
nA pA, nB pB (pA,pB - partial pressures)
yA = pA / (pA + pB) = pA / p
BBAA
AA
M/wM/wM/w
xA + xB = 1
yA + yB = 1
p = pA + pB
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yA yA yB yB
nA nB nA+ nBGinitial = nAA + nBB
= nA[ + RTlnp] + nB[B + RTlnp]
Gfinal = nA[ + RTlnyAp] + nB[B + RTlnyBp]
G = Gfinal - Ginitial = RT[nAlnyAp – nAlnp
+ nBlnyBp - + nBlnp]
ya
yb
ya ya yb yb
Initial Final
A B
Gmixing
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yA yA yB yB
ya
yb
yA yA yB yB
Gmixing Smixing
ya
yb
001
1
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Chemical Potential (in English!)
GibbsFreeEnergy
Pressure
Molecules acquire more spare energy
Greater “chemical potential”
Low Pressure
Constant Temperature
High Pressure
G ln(pressure)
Gmolar = G molar + RT lnp
Energy free for moleculesto “do stuff”at STP
Effect of environmenton this free energy
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For single component e.g. pure H2O
Why do we use Chemical Potential?
No real need to use
Gibbs Free Energy (G) is total energy in entire system available to “do stuff”
- includes all molecules, of all substances, in all phases
G = nAA + nBB
For mixturese.g. H2O/C2H5OH
G = nH2OH2O
Free energy only comes from H2O
Free energy from 2 sources
G = nH2OH2O+nEtOHEtOH
tells us how much fromH2O versus C2H5OH
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Chemical Potential :1. A measure of "escaping tendency" of components in a solution
2. A measure of the reactivity of a component in a solution
Why do we different molecules have different Chemical Potentials?
Ethanol can soak up much more energy in extra vibrational modes and chemical bonds
- will respond differently to pressure/temperature increases
FreeEnergy(G)
&
Gas-phase molecule
Liquid-phaseSolid-state
Pressure
VolatileInvolatile
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Raoult's law
I deal solns. obey Raoult's Law:
Partial pressures pA and pB vary linearly with
the mol f ractions xA and xB in a liquid.
i.e.
pA = poA xA
pB = poB xB
p = pA + pB = poA xA + po
B xB
y = m . x + c (=0)
Raoult's laweqns. of straight linespassing thru origin
Total pressure above boiling liquid
Volatility
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xA runs between 0 (none present) to 1 (pure solution)
Mixing (diluting a substance) always lowers xA
this means mixing ALWAYS gives a negative lnxA
Pure A
0
lnxA
Pure B
xA
10
Dilution
Mixing alwayslowers
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Vapour-Pressure Diagrams
Plots, on the same diagram, of vapour
pressure against composition of (i) the liquid
phase (ii) the vapour phase.
(a) Vapour pressure vs. liquid composition
(the liquid line)
p = pA + pB = poA xA + po
B xB
= poA xA + po
B (1 - xA)
= poA xA + po
B - poB xA
= poB + (po
A - poB) xA
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BUT
p depends on liquid composition (x)
A + B
P I f applied pressure > p
vapour liquid
Liquid poA
p
1 0 xoA
0 1 xoB
Pure A
Pure B
Volatile Involatile
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however
(b) Vapour pressure vs. vapour composition
p is not a linear f unction of y
poA
poB p
1 0 yoA
0 1 yoB
Pure A
Pure B
Vapour
Volatile Involatile
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The Vapour-Pressure Diagram
po
A
poB p
mol f raction A
liquid line
liquid linevapour line
L + V
B
liquid only
vapour only
liquid+ vapour
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Deviations f rom I deality
Associated with intermolecular attractions
(e.g. Van der Waals, H-bonding)
Strength of interaction heat of mixing
i.e. mH f or A-B system
3 cases
A B
A
B Case 1: Ideal
A equally attracted by B
as by itself + vice-versa
(e.g. alkanes) mixH = 0 Ideal mix
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A B
A
BCase 2: -ve deviation
A more attracted by B(e.g. CHCl3 + acetone)
mixH = < 0
A B
A
BCase 3: +ve deviation
A less attracted by B(e.g. EtOH + water)
mixH = > 0
b.pt. > ideal
b.pt. < ideal
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Summary: Raoult’s Law for Solvents
pA = xA . pAΘ
Liquid po
A
1 0 x o
A
p Partial pressure of A
Total pressure
0 1 xoB
Partial pressure of B
pB = xB . pBΘ
pBΘ
Proportionality constant
Involatilelow vapour pressure
Volatilehigh vapour pressure
A B
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p0ө = vapour pressure
= tendancy of system to increase S
High order: low S Less order: higher S
Strong desire to S
Boiling of Afavoured
Less need to S
A happier in liquid
A
A
AA
High p0ө(A) Low p0
ө(A)
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Proportionality constant
Amount in solution
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For dissolution of oxygen in water, O2(g) O2(aq), enthalpy change under standard conditions is -11.7 kJ/mole.
Dissolution isEXOTHERMIC
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Consider O2 dissolution in water:
Important in Green Chemistry for selective oxidation
Cinnamyl AlcoholCinnamic Acid
Cinnamaldehyde
Solvent: H2O Solute: O2
pH2O
pO2
Aspects of Allylic Alcohol OxidationAdam F. Lee et al, Green Chemistry 2000, 6, 279
Henry's law accurate for gases dissolving in liquids when concentrations and partial pressures are low.
As conc. and partial pressures increase, deviations from Henry's law become noticeable
Similar to behavior of gases- deviate from the ideal gas law at high P and low T.
Solutions obeying Henry's law are therefore often called ideal dilute solutions.
H2O O2
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The Tie-Line
Horizontal line thru a point on V-P diagram
defi ning p and composition of the system
join compositions of co-existing
phases at equilibrium
poA
poB p
mol f raction (liq, vap)
L
V
A B
A
xAyA
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The Lever Rule
tie-lines composition of phases But how much total material is in each phase Defi ne no. of mols in liquid & vapour:
nl = (nA)l + (nB)l
nv = (nA)v + (nB)v
Lever Rule:
mol f raction
poA
poB p
A B
l v a
allengthavlength
n
n
v
l
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A A
AB
A
A
BA
B B
BA
B
B
AB
Tie-line
LIQUID
GAS
LIQUID
GAS
Lever Rule
A-B Composition Liquid-Gas Distribution
B A
Volatile Involatile
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Boiling Point Diagrams
1. liquids boil when p = prevailing pressure
(f or normal b.p. , p = 1atm.)
2. b.p. is 1/ vapour pressure (po)
B.P. diagrams are of opposite slope to V-P
Thus
Fixed T Fixed p
0 xA 1
liquid
vapour
0 xA 1
vapour
liquid TB,A
TB,B L + V L + V
T
poB
poA
p
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Example Problem
The following temperature/composition data were obtained for a mixture of octane (O) and toluene (T) at 760 Torr, where x is the mol fraction in the liquid and y the mol fraction in the vapour at equilibrium
The boiling points are 110.6 C for toluene and 125.6 C for octane. Plot the temperature/composition diagram of the mixture. What is the composition of vapour in equilibrium with the liquid of composition:1. x(T) = 0.25 2. x(O) = 0.25
x(Toluene) y(Toluene) Temperature / C0.908 0.923 110.90.795 0.836 1120.615 0.698 1140.527 0.624 115.80.408 0.527 117.3
0.3 0.41 1190.203 0.297 1200.097 0.164 123
P.W. Atkins, Elements of Phys.Chem. page 141
Boiling/Condensation Temperature
Liquid Vapour
x(T) = 1, T = 110.6 C
x(O) = 1, x(T) = 0, T = 125.6 C
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Principle of Distillation Take liquid x1 & heat to bubble line
vapour composition y
Cool vapour liquid composition x2 REPEAT!!
T
mol fraction 1 0
x1
y2
y1
boiling
x2
cool
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Extreme Deviations f rom I deality
Result in:
maxima/ minima in B.Pt + V.P. diagrams
AZEOTROPES
a - without zeo - boil
tropos - change
means composition does not change during boiling
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A B
A
BCase 2: -ve deviation
A more attracted by B(e.g. CHCl3 + acetone)
mixH = < 0
A B
A
BCase 3: +ve deviation
A less attracted by B(e.g. EtOH + water)
mixH = > 0
b.pt. > ideal
b.pt. < ideal
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Vapour-Pressure diagrams
Boiling-Point diagrams
P
A composition B
Vapour
Liquid
L+V L+V
L+V L+V
A composition B
Vapour
Liquid
+- ve - ve
T
A composition B
Vapour
Liquid
L+V L+V
L+V L+V
A composition B
Vapour
Liquid
azeotropic composition
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Distillation of Azeotropes
T
A composition B
V
L
L+V
A B
P
T
A composition B
V
L
L+V
A
B
Distillate always ends up at azeotropic composition Residue is pure A or B
Distillate always ends up as pure A or B Residue is mixture with azeotropic composition
-EtOH cannot be dried by distillation P = 96% EtOH, 4% H2O
A
B
A B
Residue Distlllate
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Topics Covered (lectures 2-4)
Chemical Potential - A(l) =
A(l) + RTlnxA
- A(g) = A(g) + RTlnpA
Mol fractions - A = nA / nA+nB
Raoult’s Law - pA = po
A xA pB = poB xB
- ideal solutions
- +ve/-ve deviations
Vapour-pressure diagrams - Tie-lines
- Lever Rule
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Solutions EquationsPhase Rule
F = c - p + 2 c = componentsdegrees of freedom p = no. of phases
Clapeyron EquationH = enthalpy of phase change
V = volume change associated with phase change
Clausius-Clapeyron Equation
or
Mol fractionsxA = nA / nA+nB ni = mols of i
yA = pA / pA + pB pi = partial pressure of i
Raoults LawpA = po
A xA and pB = poB xB
Lever Rule (for tie-line joining phases via point a)
nl =no. moles in liquid phasenv =no. moles in liquid phase
dTdp
VTH
dTplnd
2RTH
21
v
2
1
T1
T1
RH
pp
ln
allengthavlength
nn
v
l
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Contains solute (e.g. NaCl, glucose)
WHY?!!!
Solvent AA(l) =
A(l) + RTlnxA
A
A(solution) < A(solvent)
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