Chapter 3/9 Chemical Reactions and Quantities The Mole Calculations Using Molar Mass.
Chapter 5 Chemical Potential Phase Transitions Mixtures...Duesberg, Chemical Thermodynamics Chapter...
Transcript of Chapter 5 Chemical Potential Phase Transitions Mixtures...Duesberg, Chemical Thermodynamics Chapter...
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Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg
Chapter 1 : Slide 1
Chapter 5
Chemical Potential Phase Transitions
Mixtures
Chapter11111 1 : Slide 1
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Chemical Thermodynamics : Georg Duesberg
Chemical potential
ii
i nPTnG ,,⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=µ
Where µ = chemical potential (kJ/mol) ΔG = free energy (kJ) ni = moles of component (i)
iiii STHG ⋅−=≡µ
For a single component system containing 1 mole of substance:
µ=mG
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Chemical Thermodynamics : Georg Duesberg Chemical Thermodynamics : Georg Duesberg
Properties of the Gibbs energy
G = H - TS
dG = dH –TdS - SdT
dG = dU + pdV + Vdp –TdS - SdT dU = dqrev +dwrev = TdS –pdV (for closed systems and no non-volume work) Explanation: Both are not state functions but the sum dq + dw can be treated to be reversible because they only depend on the state functions
dG = TdS – pdV + pdV + Vdp –TdS - SdT
dG = Vdp - SdT
G = f ( p, T )
dH = dU +pdV + Vdp
H = U + pV
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Chemical Thermodynamics : Georg Duesberg
4
Properties of the Gibbs energy dG = Vdp - SdT
STG
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂V
pG
T
=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
V is positive so G is increasing with increasing p
G
T (constant p)
Slope = -S G
P (constant T)
Slope = V
S is positive (-S is negative) so G is decreasing with increasing T
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Chemical Thermodynamics : Georg Duesberg
5
Dependence of G on p
It would be useful to determine the Gibbs energy at one pressure knowing its value at a different pressure.
dG = Vdp - SdT
We set dT = 0 (we make sure that the temperature is steady) and integrate:
∫+=f
i
)()( if
p
pVdppGpG
∫∫ =f
i
f
i
p
p
G
GVdpGd
Chemical Thermodynamics : Georg Duesberg
∫=−=Δf
i
)()( if
p
pVdppGpGG
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Chemical Thermodynamics : Georg Duesberg
6
Dependence of G on p in Liquids and Solids
∫+=f
i
)()( if
p
pVdppGpG
Only slight changes of volume with pressure mean that we can effectively treat V as a constant, and take it out of the integral.
)()()()( iifif pGppVpGpG ≈−+=
pVpGpG Δ+= )()( if
Often V Δp is very small and may be neglected i.e. G for solids and liquids under normal conditions is independent of p. (For Water 0.02 kJ for going from 1 to 10 bar) Exception: Geography , very high pressures…
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Chemical Thermodynamics : Georg Duesberg
7
Dependence of G on p: Ideal Gases.
∫+=f
i
)()( if
p
pVdppGpG
For gases V cannot be considered a constant with respect to pressure. For a perfect gas we may use:
i
fiif ln)( )()( f
i ppnRTpG
pdpnRTpGpG
p
p+=+= ∫
We can set pi to equal the standard pressure, pθ ( = 1 bar). Then the Gibbs energy at a pressure p is related to its standard Gibbs energy, Gθ, by:
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Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
θθ
ppnRTGpG f
f ln)( +=
Example: Going from 1 to 10 bar: dG ≈ 6 kJ (R = 8.31 JK-1, T = 298 K, ln 10= 2.3) Ø Can derive that, for a gas: Ø ΔGm = RT ln(pf/pi)
Ø For p -> 0 Gm -> - ∞
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Variation of G with pressure
VpG
T
=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
V is positive so G is increasing with increasing p
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Fugacity (lat.) = “urge to flee”
f= same units as pressure
The fugacity is a parameter which enables us to apply the perfect gas expression to real gases…. In the thermo dynamical treatment of the Chemical Potential
Chemical potential of real gases
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Chemical Thermodynamics : Georg Duesberg
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Dependence of G on p of real Gases
θθ
pfRTGpG ln)( mfm +=
For real gases we modify the expression for a perfect gas and replace the true pressure by a new parameter, f, which we call the fugacity.
The fugacity is a parameter we have simply invented to enable us to apply the perfect gas expression to real gases.
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Chemical Thermodynamics : Georg Duesberg
Dependence of G on p Real Gases.
We may show that the ratio of fugacity to pressure is called the fugacity coefficient:
γ=pf
Where γ is the fugacity coefficient
γ is related to the compression factor Z and the Viral coefficients:
dppZp
∫−
=0
1lnγ
1
2lnffRTG =ΔWe may then write
..)'' 2+=×= PCpBepfViral coefficients
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Chemical Thermodynamics : Georg Duesberg
A real gas is in its standard state when its fugacity is equal to 1 bar and it is behaving as if it were an ideal gas at some specified temperature (the diagram shown below is exaggerated to make this point):
In the limit of zero pressure real gases behave more and more like ideal gases and the fugacities of real gases approach their partial pressures: lim fi --> Pi
Pi --> 0
ideal gas fi = Pi
real gas
fugacity
pressure
f = 1 bar
P = 1 bar
hypothetical standard state
not the standard state
The standard state of a pure liquid or solid at a fugacity of 1 bar and some specified temperature. Since the molar volumes of solids and liquids are generally small and relatively insensitive to pressure, the activities of solids and liquids at pressures that are not too far removed from 1 bar remain close to unity.
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Pressure region Z f α γ
I (very Low) ≈1 P ≈0 ≈0
II (moderate) <1 <P >0 >0
III (high) >1 >P <0 <0
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Chemical Thermodynamics : Georg Duesberg
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Dependence of G on T
STG
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂
Using the same procedure as for the dependence of G on p we get:
TSdGd ∫∫ −=
To go any further we need S as a function of T
G = H - TS
G depennce stronger in the case for gases than for fluids or solids, because S is much higher in the gaseous state
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Chemical Thermodynamics : Georg Duesberg
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Dependence of G on T
STH
TG
−=
Let G/T = x STHx −=
2TH
Tx
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂
2
)/(TH
TTG
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂
This is the Gibbs-Helmholtz Equation Important when dealing with chemical reaction and phase changes.
G = H - TS
Put back G/T = x
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Chemical Thermodynamics : Georg Duesberg
17
Gibbs-Helmholtz Equation
2
)/(TH
TTG
p
Δ−=⎟
⎠
⎞⎜⎝
⎛∂
Δ∂ STG
p
Δ−=⎟⎠
⎞⎜⎝
⎛∂
Δ∂
Two expressions:
Changes in entropy or, more commonly, changes in enthalpy can be used to show how changes in the Gibbs energy vary with temperature. For a spontaneous (ΔG < 0) exothermic reaction (ΔH < 0) the change in Gibbs energy increases with increasing temperature.
ΔG/T
T (constant p)
Slope = -ΔH/T2 = positive for exothermic reaction
Very negative
Less negative
STG
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂from
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18
Variation of G with temperature
ΔGm = -SmΔT Can help us to understand why transitions occur
The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal.
The two phases are in EQUILIBIRIUM at this temperature
STG
p
−=⎟⎠
⎞⎜⎝
⎛∂
∂
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19
Variation of G with pressure
19
VpG
T
=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
V is positive so G is increasing with increasing p
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Chemical Thermodynamics : Georg Duesberg
µ
T
High P
low P
Tf Tb Tf Tb
Variation of G with temperature and pressure
Freezing and boiling point raise with pressure – much prounced Effect for boiling point – anomaly: water
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21
Phase Equilibria Phase transitions Changes in phase without a change in chemical composition. Gibbs Energy is at the centre of the discussion of transitions
If two phases are in equilibrium as depicted here, along the phase transition line, then both phases have the same chemical potential. If a pressure is applied, which shifts the system out of equilibrium then the temperature will change (as a result of some particles migrating from one phase to the other) until equilibrium is re-established. The slope of the phase boundary is dp/dT.
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22
Derivation of Clapeyron equation dGm(1) = dGm(2)
with dGm = Vmdp – SmdT Vm(1)dp – Sm(1)dT = Vm(2)dp – Sm(2)dT {Vm(2) – Vm(1)}dp = {Sm(2) – Sm(1)}dT
ΔtrsV dp = ΔtrsS dT Or
dp/dT = ΔtrsS/ ΔtrsV (Clapeyron equation) with ΔtrsS=ΔtrsH/T
dp/dT = ΔtrsH/(T ΔtrsV)
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Location of phase boundaries
Ø Clapeyron equation
Ø Clausius-Clapeyron
TVTHptrs
trs ΔΔΔ
=Δ
( )
constant11lnln
ln
1212
2
+⎟⎟⎠
⎞⎜⎜⎝
⎛−
Δ−=
ΔΔ
=Δ
TTRH
pp
TRTH
p
vap
vap
Constant is
ΔvapS/R
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24
Phase diagrams
Ø Map showing conditions of T and p at which various phases are thermodynamically stable
Ø At any point on the phase boundaries, the phases are in dynamic equilibrium
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25
Characteristic points
Ø When vapour pressure is equal to external pressure bubbles form: boiling point
Normal bp: 1 atm, Standard bp: 1 bar Ø When a liquid is heated in a closed vessel the liquid density eventually
becomes equal to the vapour density: a supercritical fluid is formed.
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26
Water Ice I structure
Ø Solid-liquid boundary slopes to the left with increasing pressure
Ø volume decreases when ice melts, liquid is denser that solid at 273 K
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27
first-order phase transition. Remember that at the transition point, we are changing the enthalpy of the system but not its temperature. Thus, the heat capacity Cp at the transition point is infinite. In other words, as we heat the system that is at the transition point, no temperature change happens because all the heat is going into the phase transition.
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 28
Chapter 6
Chemical Potential in Mixtures
Partial molar quantities
Thermodynamics of mixing The chemical potentials of liquids
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Concentration Units
• There are three major concentration units in use in thermodynamic descriptions of solutions.
• These are – molarity – molality – mole fraction
• Letting J stand for one component in a solution (the solute), these are represented by
• [J] = nJ/V (V typically in liters) • bJ = nJ/msolvent (msolvent typically in kg) • xJ = nJ/n (n = total number of moles of all species
present in sample)
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 30
PARTIAL MOLAR QUANTITIES
Recall our use of partial pressures:
p = xA p + xB p + …
pA = xA p is the partial pressure
We can define other partial quantities…
For binary mixtures (A+B): xA + xB= 1
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 31
PARTIAL MOLAR QUANTITIES
The contribution of one mole of a substance to the volume of a mixture is called the partial molar volume of that component.
( )...,,, BA nnTpfV =
AnTpAA n
VV≠
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
,,
...+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛
∂
∂= B
BA
A
dnnVdn
nVdV
At constant T and p
In a system that contains at least two substances, the total value of any extensive property of the system is the sum of the contribution of each substance to that property.
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 32
Partial molar volume
Very Large Mixture of A and B
Add nA of A to mixture
When you add nA of A to a large mixture of A and B, the composition remains essentially unchanged. In this case: Va can be considered constant and the volume change of the mixture is nAVA. Likewise for addition of B.
constnVV
AnTpAA ≈⎟⎟
⎠
⎞⎜⎜⎝
⎛
∂
∂=
≠,,
The total change in volume is nAVA + nBVB . (Composition is essentially unchanged).
...++= BBAA nVnVVScoop out of the reservoir a sample containing nA of A and nB of B its volume is nAVA + nBVB . Because V is a state function:
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Duesberg, Chemical Thermodynamics
Volume Vs. Composition ¡ The partial molar volume of a
substance l slope of the variation of the total
sample volume plotted against composition.
¡ partial molar volumes vary with solution composition
In general, partial molar quantities vary with the composition, as shown by the different slopes at the compositions a and b. Note that the partial molar volume at b is negative: the overall volume of the sample decreases as a is added. (e.g. MgSO4 in water)
AnTpAA n
VV≠
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
,,
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 34
Partial molar volume
A different answer is obtained if we add 1 mol of water to a large volume of ethanol.
3
,OHOH cm18
2
2=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
TpnVVThe change in
volume is 18cm3
3
OH)CHCH(,,OHOH cm14
232
2=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
nTpnVVThe change in
volume is 14cm3
Example: What is the change in volume of adding 1 mol of water to a large volume of water?
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35
Introduction to mixtures
Ø Homogeneous mixtures of a solvent (major component) and solute (minor component).
Ø Introduce partial molar property: contribution that a substance makes to overall property.
V = nAVA + nBVB
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 36
VA is not generally a constant: It is a function of composition, because the environment and therefore their interaction of the molecules changes :
PARTIAL MOLAR QUANTITIES
with the composition :
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Exercise Calculate the density of a mixture of 20 g of water and 100g of ethanol.
• First calculate the mole fractions. – 20 g H2O = 1.11 mol; – 100 g EtOH = 2.17 mol – xH2O = 0.34; xEtOH = 0.66
• Then interpolate from the mixing curve – VH2O = 17.1 cm3 mol-1; – VEtOH = 57.4 cm3 mol-1
• Then plug the moles and partial molar volume – (1.11 mol)(17.1 cm3/mol) + (2.17 mol)
(57.4 cm3/mol) =19.0 cm3 + 125 cm3 = 144 cm3
• Finally, the total mass is divided by the total volume: 120 g/144 cm3 = 0.83 g/ cm3
Task: You mix 300 ml alcohol with a bottle of juice (water, 0.7 l) - Will you really end up with for your x-mas party supply (1l)? -Of course not… - How much do you have to mix to get 1 l of the same strength?
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The chemical potential, µ
• We can extend the concept of partial molar properties to state functions, such as Gibbs energy, G.
• This is so important that it is given a special name and symbol, the chemical potential, µ.
G = nAGA + nBGB
G = nAµA + nBµB
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Duesberg, Chemical Thermodynamics
Chapter 6 : Slide 39
The Thermodynamic Criterion of Equilibrium
At equilibrium, the chemical potential of a substance is the same throughout a sample, regardless of how many phases are present.
µ1 µ2
Consider this single substance system:
dG = -µ1 dn dG = +µ2 dn
Total dG = 0 only if µ1 = µ2
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Duesberg, Chemical Thermodynamics
Chapter 6 : Slide 40
Chemical Potential
1,,1
µ=⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
jnpTnG
Partial molar Gibbs energy Chemical potential for component 1
...2,,2
1,,1
+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛
∂
∂= dn
nGdn
nGdG
jj npTnpT
If T and p are kept constant (i.e. dT and dp = 0) then
Or: ...2211 ++= nnG µµ
For a single component system containing 1 mole of substance:
µ=mG
...2211 ++= dndndG µµ
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Chemical Potential
dG is negative
Phase β Phase α Phase γ
dG = 0
Phase β Phase α Phase γ
γβα µµµ iii >>
γβα µµµ iii ==
dniα
dnjα
dniβ
dnjβ
dniγ
dnjγ
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 42
The Partial Molar Gibbs Energy
Ap,T,nAA n
Gµ≠
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
The partial molar Gibbs energy is called the “chemical potential”
...+⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+⎟⎟
⎠
⎞⎜⎜⎝
⎛
∂
∂= B
BA
A
dnnGdn
nGdG
G = nAµA + nBµB
At constant T and p
...++= BBAA dndndG µµ(At equilibrium dG = 0)
The implication of G = f ( p,T, ni ) is that we now write:
∑++−= iidnVdpSdTdG µ
...++++−= AAAA dndnVdpSdTdG µµ
Fundamental Equation of Chemical Thermodynamics.
( )...,,, BA nnTpfG =
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 43
The chemical potential of a substance is the slope of the total Gibbs energy of a mixture with respect to the amount of substance of interest. In general, the chemical potential varies with composition, as shown for the two values at a and b. In this case, both chemical potentials are positive.
∑= iidndG µ
In case of constant temperature and pressure reduces to:
∑++−= iidnVdpSdTdG µ
Chemical Potential = Partial Molar Gibbs Energy
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 44
The Wider Significance of the Chemical Potential
∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂++−= i
npTi
dnnGVdpSdTdG
j,,
∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂++= i
npSi
dnnHVdpTdSdH
j,,
∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+−−= i
nVTi
dnnApdVSdTdA
j,,
∑ ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+−= i
nVSi
dnnUpdVTdSdU
j,,
µi
jnpTinG
,,⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
jnpSinH
,,⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
jnVTinA
,,⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
jnVSinU
,,⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂
All extensive thermodynamic properties change with composition!
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chem "chemical work"j jj
dw dnµ− =∑Example: E-chem cell:
Chemical Potential
∑== iiadd dndGdw µmax,In case of constant temperature and pressure also:
j jj
j jj
j jj
j jj
dU TdS PdV dn
dH TdS VdP dn
dA SdT PdV dn
dG SdT VdP dn
µ
µ
µ
µ
= − +
= + +
= − − +
= − + +
∑
∑
∑
∑
, , , ,
, , , ,
j i j i
j i j i
ii iS V n S P n
i iT V n T P n
U Hn n
A Gn n
µ≠ ≠
≠ ≠
⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞∂ ∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 46
2211 nnG µµ +=
22112211 µµµµ dndndndndG +++=
The Gibbs-Duhem Equation
by differentiating:
At equilibrium: dG = 0 0 = µ1dn1 + µ2dn2
22110 µµ dndn +=∴
21
21or µµ d
nnd −=
The Gibbs-Duhem Equation. A similar expression may be deduced for all partial molar quantities (If one is changed the other changes as well…)
...2211 ++= dndndG µµ
(For binary systems)
∑= iidndG µand recall
iidnor µ∑=0
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Duesberg, Chemical Thermodynamics
Chapter 7 : Slide 47
The Gibbs-Duhem Equation
0j jj
SdT VdP n dµ− + =∑
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The chemical potential of perfect gases in a mixture
Recall that Gm(pf) = Gm(pi) + RT ln (pf/pi)
At standard pressure Gm(p) = Gm° + RT ln (p/p°)
Therefore, for a mixture of gases
µJ = µJ° + RT ln (pJ/p°) More simply (at p° = 1 bar)
µJ = µJ° + RT ln pJ
System is at equilibrium when µ for each substance has the same value in every phase
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Gas mixtures
Compare ΔGmix = nRT {xAln xA+ xB ln xB}
ΔG = ΔH – TΔS
Therefore ΔH = 0
ΔSmix = − nR {xAln xA+ xB ln xB}
Perfect gases mix spontaneously in all proportions
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Ideal Liquid Solutions
pJ = xJpJ*
Due to effect of solute on entropy of solution
Raoult’s Law
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Real Solutions
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Chemical potential of a solvent
Ø At equilibrium µA(g) = µA(l) µA(l)= µA°(g) + RT ln pA
µA(l)= µA°(g) + RT ln xApA* µA(l)= µA°(g) + RT ln pA* + RT
ln xA └────────────────┘
µA*
Ø µA(l)= µA*+ RT ln xA
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Is solution formation spontaneous?
G = nAµA + nBµB Can show that ΔGmix = nRT {xAln xA+ xB ln xB} and
ΔH = 0
ΔSmix = −nR {xAln xA+ xB ln xB}
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Ideal-dilute solutions Ø Raoult’s law generally describes well solvent vapour
pressure when solution is dilute, but not the solute vapour pressure
Ø Experimentally found (by Henry) that vp of solute is proportional to its mole fraction, but proportionality constant is not the vp of pure solute.
Henry’s Law
pB = xBKB
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Gas solubility
KH/(kPa m3 mol−1)
Ammonia, NH3 5.69
Carbon dioxide, CO2 2.937
Helium, He 282.7
Hydrogen, H2 121.2
Methane, CH4 67.4
Nitrogen, N2 155
Oxygen, O2 74.68
Henry’s law constants for gases dissolved in water at 25°C
Concentration of 4 mg/L of oxygen is required to support aquatic life, what partial pressure of oxygen can achieve this?
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Application-diving
Table 1 Increasing severity of nitrogen narcosis symptoms with depth in feet and
pressures in atmospheres.
Depth P Total P N2 Symptoms
100 4.0 3.0 Reasoning measurably slowed.
150 5.5 4.3 Joviality; reflexes slowed; idea fixation.
200 7.1 5.5 Euphoria; impaired concentration; drowsiness.
250 8.3 6.4 Mental confusion; inaccurate observations.
300 10. 7.9 Stupefaction; loss of perceptual faculties.
Gas narcosis caused by nitrogen in normal air dissolving into nervous tissue during dives of more than 120 feet [35 m]
Pain due to expanding or contracting trapped gases, potentially leading to Barotrauma. Can occur either during ascent or descent, but are potentially most severe when gases are expanding. Decompression sickness due to evolution of inert gas bubbles.
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Real Solutions-Activities µJ = µJ° + RT ln aJ
Substance Standard state Activity, a
Solid Pure solid, 1 bar 1
Liquid Pure liquid, 1 bar 1
Gas Pure gas, 1 bar p/po
Solute Molar concentration of
1 mol dm−3 [J]/co
po = 1 bar (= 105 Pa), co = 1 mol dm−3.
* Activities are for perfect gases and ideal-dilute solutions; all activities are dimensionless.
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Colligative properties Properties of solutions that are a result of changes in the disorder of the solvent, and rely only on the number of solute particles present
Lowering of vp of pure liquid is one colligative property
Freezing point depression
Boiling point elevation
Osmotic pressure
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Colligative properties Ø Chemical potential of a solution
(but not vapour or solid) decreases by a factor (RTlnxA) in the presence of solute
Ø Molecular interpretation is based on an enhanced molecular randomness of the solution
Ø Get empirical relationship for FP and BP (related to enthalpies of transition)
mKTmKT
bb
ff
=Δ
=Δ
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Cryoscopic and ebullioscopic constants
Solvent Kf/(K kg mol−1) Kb/(K kg mol−1)
Acetic acid 3.90 3.07
Benzene 5.12 2.53
Camphor 40
Carbon disulfide 3.8 2.37
Naphthalene 6.94 5.8
Phenol 7.27 3.04
Tetrachloromethane 30 4.95
Water 1.86 0.51
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Osmotic pressure
Van’t Hoff equation
MRT=Π
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Phase diagrams of mixtures
We will focus on two-component systems (F = 4 ─ P), at constant pressure of 1 atm (F’ = 3 ─ P), depicted as temperature-composition diagrams.
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Fractional Distillation-volatile liquids
Important in oil refining
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Exceptions-azeotropes Azeotrope: boiling without changing High-boiling and Low-boiling
Favourable interactions between components reduce vp of mixture Trichloromethane/propanone HCl/water (max at 80% water, 108.6°C)
Unfavourable interactions between components increase vp of mixture Ethanol/water (min at 4% water, 78°C)
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Liquid-Liquid (partially miscible)
Ø Hexane/nitrobenzene as example
Ø Relative abundances in 2 phases given by Lever Rule n’l’ = n’’l’’
Ø Upper critical Temperature is limit at which phase separation occurs. In thermodynamic terms the Gibbs energy of mixing becomes negative above this temperature
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Other examples
Water/triethylamine Weak complex at low temperature disrupted at higher T.
Nicotine/water Weak complex at low temperature disrupted at higher T. Thermal motion homogenizes mixture again at higher T.
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Liquid-solid phase diagrams