1 CHAPTER 5 - Simple Mixtures I. Thermodynamics of binary mixtures. A. PMQ’s - partial molar quantities. let Q = an extensive quantity (volume, free energy, etc.) PMQ = (∂Q/∂n) T,p,n’ = how property Q of the whole mixture changes as one varies moles of one component, holding T, p and moles of other components fixed. B. Easiest pmq is partial molar volume (PMV). PMV of component α: V α = (∂V/∂n α ) T,p,n β (V = total volume of mixture) e.g. add MgSO 4 to H 2 O. V α varies with concentration. dV = (∂V/∂n α ) T,p,nβdn α + (∂V/∂n β ) T,p,n dn β ↑ ↑ V α V β V(c) = n α V α (c) + n β V β (c) ↑ ↑ ↑ total pmv of pmv of vol at at conc c at conc c conc c Bottom line: Volumes are NOT additive. We have made it additive only by using a conc-dependent volume (pmv) contributed by each component. In other words, V(total) not equal to n α V α (pure α) + n β V β (pure β). Why aren’t volumes additive? Molecules of substances pack with themselves differently than they pack with other substances.
Transcript
1
CHAPTER 5 - Simple Mixtures I. Thermodynamics of binary mixtures.
A. PMQ’s - partial molar quantities.
let Q = an extensive quantity (volume, free energy, etc.)
PMQ = (∂Q/∂n)T,p,n’ = how property Q of the whole mixture changes as one varies moles of one component, holding T, p and moles of other components fixed.
B. Easiest pmq is partial molar volume (PMV).
PMV of component α:
Vα = (∂V/∂nα)T,p,nβ (V = total volume of mixture) e.g. add MgSO4 to H2O.
Vα varies with concentration.
dV = (∂V/∂nα)T,p,nβdnα + (∂V/∂nβ)T,p,ndnβ ↑ ↑
Vα Vβ
V(c) = nαVα(c) + nβVβ(c) ↑ ↑ ↑ total pmv of pmv of vol at at conc c at conc c conc c
Bottom line: Volumes are NOT additive. We have made it additive only by using a conc-dependent volume (pmv) contributed by each component.
In other words, V(total) not equal to nαVα(pure α) + nβVβ(pure β).
Why aren’t volumes additive? Molecules of substances pack with themselves differently than they pack with other substances.
2
C. Most important pmq is µ (chemical potential):
µα = (∂G/∂nα)T,p,nβ
chemical potential of α = partial molar free energy of α
total free energy of mixture is then G = nαµα + nβµβ + …
Now we can expand the fundamental equation of Chapter 3 to include effect of changes in composition of mixtures
Now mix, and bring total pressure back to p: p = pA + pB Gfinal = nA(µA° + RT ln pA/p°) + nB(µB° + RT ln pB/p°) Now find free energy of mixing: ΔGmix = Gfinal – Ginitial, (µα° terms cancel) = nA RT ln pA/p° + nB RT ln pB/p° - nA RT ln p/p° - nB RT ln p/p° Combine terms: ΔGmix = nA RT ln pA/p + nB RT ln pB/p Express in terms of mole fractions remembering:
xα = pα/p (ideal gas)
xα = nα/n (p = total pressure; n = total moles)
ΔGmix = nxART ln xA + nxBRT ln xB
ΔGmix = n RT (xA ln xA + xB ln xB) Note ΔGmix < 0 always, since xA, xB ≤ 1 always.
4
This confirms common sense: remove the partition, gases A + B spontaneously mix. Mixing, apart from intermolecular interactions, always produces a lower free energy.
Also shows ΔGmix is independent of total pressure p. F. Entropy of mixing: Since (∂G/∂T)p,n = -S ΔSmix = - (∂ΔGmix/∂T) p,n ΔSmix = -nR(xA ln xA + xB ln xB) (ideal) Always ΔSmix > 0. Makes sense! This is the driving force behind mixing. Further, since: ΔGmix = ΔHmix - TΔSmix (ΔSmix differs from ΔGmix by only factor -T ) Then: ΔHmix = 0 (p, T const, ideal gas) This is true since perfect gases are noninteracting. Of course: ΔVmix = 0 if p, T are truly held constant and ideal gas. G. Binary Liquid Mixtures - Raoult’s and Henry’s Law.
1st consider pure(*) liquid A, w/vapor pressure pA
* & chem. pot. µA*(l):
µA*(l) = µA° + RT ln pA
*/p° Now add component B: µA(l) = µA° + RT ln pA/p° with partial pressure pA
5
Subtract pure case: µA(l) = µA
*(l) + RT ln pA/pA*
↑ ↑ ↑ chem pot pure A important of A liquid ratio in mix Raoult observed that, for the solvent, i.e., component present in gross excess:
pA = xApA* (Raoult’s Law)
Here pA is vapor pressure A above mix; xA is mole fraction in liquid, pA
* is vapor press pure A If this is true at all compositions, then the solution is ideal. (e.g. two
chemically and physically similar liquids: benzene, toluene.) Ideal solutions obeys Raoult’s law everywhere, so
µA(l) = µA*(l) + RT ln xA (xA ≈ pA/pA*)
This is definition of ideal solution. Look at ideal p vs. composition diagram:
6
Real solution diagram (w/positive deviation from Raoult’s law)
Henry’s Law for solute A:
pA = xAKA
where KA is Henry’s Law constant for solute A
Solutions dilute enough to obey Henry’s Law are called “ideal dilute solutions”.
They are not ideal, but their behavior is at least linearly predictable.
Conclusion: A nearly pure solvent obeys Raoult’s Law while a dilute solute obeys Henry’s Law.
7
Solvent molecules are surrounded
mostly by other solvent molecules, so their environment differs little from the pure solvent. They obey Raoult’s Law then.
Solute molecules are in an
environment completely different than pure solute, being surrounded completely by solvent molecules. Thus solute molecules obey Henry’s Law.
Henry’s Law can also be viewed as an equation for the solubility of a gas
in a liquid.
Gas pressure above the liquid = pA = xAKA
xA represents solubility in the liquid.
KA represents a solubility constant.
Reformulate in terms of molality of solute mA.
XA=nA
nA+n
B
≈nA
nB
=nA
mass B MB
=mA∗M
B
where MB is molar mass of B solvent in kg/mol So now:
pA=m
AMBKA( )
=mA!KA
!KA = Henry’s constant in units (pressure*kg/mol)
Negative deviation from Raoult’s Law !
Figure 5A.15
Figure 5-16
8
H. ΔGmix for 2 liquids forming “ideal” solution:
Already did for 2 perfect gases. Works the same way for liquids.
Separated components:
Ginitial = nAµA*(l) + nBµB
*(l) * = pure liquids
Mixed:
µα = µα*(l) + RT ln pα/pα
* ↑ ↑
A, B if ideal solution, pα/p
α* = xα (Raoult’s law)
µα = µα*(l) + RT ln xα
Gfinal = nA(µA* (l) + RT ln xA) + nB(µB
* (l) + RT ln xB)
ΔGmix = Gfinal – Ginitial = n RT(xA ln xA + xB ln xB)
This is for 2 liquids which form ideal solution.
Note: same as 2 gases.
Clarification: Ideal gas means no interaction between molecules. Ideal liquid mixture cannot mean no interactions. Simply means A - B interactions are same as A - A or B - B. ΔSmix equation same as before also.
When A - A and B - B and A - B interactions are all of different strengths.
Side note: If they are too different, will get only partial miscibility.
(weak) (strong) benzene-benzene H2O-H2O
(weak) benzene-H2O
9
Thermodynamic properties of real solutions are expressed in terms of excess functions.
SE = ΔSmix - (-nR(xA ln xA + xB ln xB)) excess real ideal entropy value value Since ΔHmix = 0 for ideal solution, any enthalpy change (heat liberated or
absorbed) would be the excess enthalpy.
Excess enthalpy is often written as follows: HE = nβRTxAxB Where β is a parameter that measures the strength of the AB interactions in units of RT β < 0 for exothermic mixing β > 0 for endothermic mixing Adding this term to the entropy of mixing term we get the free energy of
mixing for real solutions: ΔGmix = n RT(xA ln xA + xB ln xB +βxAxB) Variations of ΔGmix for different values of β −−> Note: this predicts two phases here for β > 2
Figure 5B.3a
Figure 5B.5
10
II. Colligative Properties: A. Generalities (examples): The addition of non-volatile solute to a solvent lowers its freezing point,
raises its boiling point, and produces an osmotic pressure. Effect does not depend on identity of solute, only on moles present in
solution. Dissolved solute is to be considered non-volatile. (i.e., no contrib to vapor
pressure above liquid) Also does not dissolve in any solid phase forming (no solid solution
formed) i.e., solute stays in the liquid phase. Collig. effect caused by reduction of µA(l) of solvent A when adding
solute B. µA(l) = µA*(l) + RT ln xB ↑ negative µA(g) of vapor above liquid mixture is unchanged, because solute is non-
volatile. Similar for solid phase “below” liquid. Therefore, since only the liquid phase has its chemical potential lowered,
adding solute improves “stability” of liquid phase relative to gas or solid.
11
Flux Argument:
Primary effect is entropic, not enthalpic (i.e. its not due to interactions
between solute and solvent). Entropy Argument:
Added solute in liquid phase increases randomness of liquid without
affecting randomness of solid or gas. B. Boiling Point Elevation: A = solvent; B = non-volatile solute
Can derive equation for boiling point elevation using the Gibbs-Helmholtz
equation (see text) using xB<<1 (i.e. pretty dilute solution of B):
Entropy change less positive, spontaneity decreases. Harder to vaporize, boiling temp ↑ Entropy change more negative, spontaneity decreases. Harder to solidify, freezing temp ↓
12
Boiling point elevation given by
ΔTb = (RT*2/ΔHvap)xB ↑ identity of solute irrelevant, only mole fraction. T* = bp of pure solvent ΔHvap = heat of vap of pure solvent.
Most times we re-express ΔTb in terms of molal conc of solute mB added and not mole fraction.
Can show that:
xB ≈ mBMA when B is very dilute (MA is molar mass of solvent in kg/mol)
ΔTb = (RT*2 MA/ΔHvap)mB ↑Kb bp elev constant for that particular solvent (ebullioscopic const)
ΔTb = Kb mB Synoptic Table 5B.1* Freezing-point and boiling-point constants. C. Freezing Point Depression: Just like previous description but vapor phase replace by solid phase. Freezing pt. depression ΔTfp = (RT*2/ΔHfus)xB ↑mole frac solute T* = fp of pure solvent ΔHfus = heat of fusion of solvent
Similarly: ΔTfp = KfmB ↑fp depression const for solvent (cryoscopic constant) Good for molecular weight determination of the solute.
13
D. Osmosis:
Solution “draws” pure solvent A to the right across membrane as A
attempts to equalize concentration of itself on both sides of membrane.
π = osmotic pressure = excess pressure which must be applied to solution side to exactly balance and affect tendency of A to flow into solution side.
Mathematically, equilibrium is achieved when:
µA*(p) = µA(p + π) = µA
*(p + π) + RT ln x A ↑ ↑ of pure of A in solution A on left on right side at side at pressure p + π pressure p
Remaining problem is to see how pressure affects µ.
dµ = Vmdp - SmdT let dT=0 ↑ molar volume
µ(p2) = µ(p1) + Vmp1
p2
∫ dp
µA*(p + π) = µA
*(p) + Vmp
p+π
∫ dp
↑~ Vmπ
14
Now:
µA*(p) = [µA
*(p) +Vmπ] + RT ln xA ↑ equal ↑
So:
Vmπ = - RT ln xA ↑ ln (1-xB) ≈ -xB ≈ -nB/nA (for dilute B)
Vmπ = RT nB/nA
nAVm = V
πV = nB RT Van’t Hoff equation
Reminiscent of pV = nRT or
π = nB/V RT
π = cB RT ↑ = molar conc of B
III. Thermodynamics of Solubility.
A. Mathematically similar to colligative properties
µB(l) = µB*(l) + RT ln xB
B in pure B solution
So: µB
*(s) = µB*(l) + RT ln xB
Now rearrange:
ln xB = (µB*(s) - µB
*(l))/RT
15
= - ΔGfus (B)/RT = - ΔHfus(B)/RT + ΔSfus(B)/R
Since ΔGfus = 0 at melting point T* of solute B, we may add ΔGfus/RT* to right side.
This expresses the effect of temperature on solubility.
solub (ln xB)↓ as T↓ from T*.
IV. Thermodynamic Description of Real Solutions (activity coefficients)
A. First recall general expression: (real or ideal)
µA(l) = µA*(l) + RT ln pA/pA* (general)
chem pot pure of A in A mixture liquid with B
B. If solution is ideal, Raoult’s Law is obeyed:
pA = xApA* or: pA/pA* = xA
And so:
µA(l) = µA*(l) + RT ln xA (ideal)
C. Let’s define aA as the “activity” of A, where aA is that quantity which could replace xA and preserve the form of the ideal equation but be true for real solution.
µA(l) = µA*(l) + RT ln aA ↑ “activity”
activity = “effective” mole fraction of A in the solution. if solution is ideal, aA = xA; if solution is non-ideal, aA ≠ xA
16
but we could write: aA = γAxA ↑
activity coefficient
so when aA ! xA γA ! 1 γA = 1 is ideal behavior.
D. Since pA/pA* is exact in general equation:
aA = pA/pA* exact
E. Chemical potential of component A in real solution is thus:
µA(l) = µA*(l) + RT ln xA + RT ln γA
↑ non-ideality effect
F. Above equation is especially good for treating the solvent, since it approximately obeys Raoult’s Law:
aA = pA/pA* ≈ xA γA near 1.0
G. For treating the solute, exploit Henry’s Law behavior:
µB+(l) ≡ new standard chem. pot. for an ideal dilute solution
= chem. pot. of solute when solute is obeying Henry’s Law.
µB+(l) = µB
*(l) + RT ln KB/pB*
so µB(l) ≡ µB
+(l) + RT ln xB
Deviations from Henry’s Law characterized by γB ≠ 1, so above equation is modified as follows:
µB(l) = µB+(l) + RT ln γBxB
or
µB(l) = µB+(l) + RT ln aB
17
H. Standard state definitions involving solutions:
1. For the solvent, standard state is pure solvent.
2. Up to now we have defined activities of species in solution in relation to mole fraction
aB = γBxB ↑
activity coefficient
It is a more common practice to define them in relation to molality.
aB = γBmB/m° ↑
activity coefficient
where m° = 1 molal (1 mol/kg)
Then, the final expression for chemical potential of any real solute at any molality is
µB = µB° + RT ln aB
where µB° ≡ chem. pot. of the solute in a hypothetical ideal solution at 1 molal.
Therefore, for solutes, the standard state is the hypothetical state in which concentrations are 1 molal (mol/kg) but the solute is behaving ideally.
3. This is not practical for biological situations, because if [H+] = 1 molal, then pH=0, which is too far removed from any realistic biological solution. Instead pH=7 is defined to be the biological standard state, all other solutes being 1 molal.
18
V. Ionic Solutions
A. Ion Activities.
1. Because of strong interactions between ions (Coulombic), deviations from ideality are important.
Interaction energy from Coulomb’s law:
V =qiqj
4πεoεrij
=cZ
iZj
εrij
Zi = charge in units of proton charge (e) (e.g., Na+ has Z = +1) ε = dielectric constant of medium (=78.3 for H2O at 25°C) r = separation distance (use Å) c = 1.39 x 103 Å-kJ/mol
For example Na+ and Cl- separated by 10 Å in water
V = -1.74 kJ/mol ↑ (-) means attractive
Calc: estimate the average ionic separation Na+---Cl- in 1.0 M NaCl?
2 moles ions/1000 cm3
or (1000 cm3)/(2 moles) x (1 mol)/(6 x 1023 ions) x (108 Å/cm)3
≈ 1.66 x 103 Å3/ion
Length of cubic side = (1.66 x 103)1/3 = 11.8 Å
Average V ~ -1.5 kJ of attractions at 11.8 Å
2. Thus, solutions of ions must be very dilute (≤ 10-3M) to ignore deviations from ideality.
19
3. In general, activity a = γm/mθ for the solute.
γ depends on concentrations and temperature.
Extremely dilute, γ → 1 a ≈ m/mθ
4. γ of ions. Consider M+X- electrolyte. (1:1 electrolyte)
Total molar free energy of ionic solution:
Gm = µ+ + µ-
= (µ+ideal + RT ln γ+) +(µ-
ideal + RT ln γ-)
5. Experimentally, cannot separately resolve γ+ and γ-, i.e. the separate contributions to non-ideality.
Rewrite: Gm
= µ+ideal + µ-
ideal + RT ln γ+γ-
6. Define mean activity coefficient γ± (a geometric mean, not an arithmetic one).
γ± = (γ+γ-)1/2 for 1:1
Then:
Gm = µ+ideal + µ-
ideal + RT ln γ±2
= (µ+ideal + RT ln γ± ) + (µ-
ideal + RT ln γ±)
i.e. both ions equally share responsibility for non-ideality.
7. For electrolyte MpXq, general case:
γ± = (γ+p γ-q)1/s s = p + q
Still each ion α has µα = µ α* + RT ln γ±
Total Gm = pµ+ + qµ-
8. Theory for γ± , the Debye-Huckel theory.
Basic physical picture:
20
Then, the average interaction energy between two given ions i and j will not be:
V =cZ
iZj
εrij
(simple Coulomb’s Law with solvent dielectric ε)
But modified:
V =cZ
iZj
εrij
e−κrij
exponential is Debye-Huckel screening term (damps out i-j interaction rapidly vs. distance rij)
On average, a charge’s interaction is “screened” from another charge by the intervening “ion atmosphere.”
κ−1 = Debye screening length = rD
κ = (εRT/2ρF2I)-1/2 = 0.328 I1/2 (in Å-1)
ε = dielectric of solvent ρ = density F = Faraday’s # I = ionic strength
Ionic strength:
I = 12
Zi2 mimo
!
"##
$
%&&
i
species
∑
Sum over different ion types i, Zi = charge of ion i, mi = molality
Example: Na+Cl- soln. at 0.1 molal.
I = (1/2) {(+1)20.1 + (-1)20.1} = (1/2) (0.1 + 0.1) I = 0.1 (simple for 1:1)
21
Example: try MgCl2 at 0.1 molal. I = (1/2) {(+2)2(0.1) + (-1)2(0.2)} = (1/2) (0.4 + 0.2) I = 0.3 molal I = “effective concentration of ions” Mg+2 gets double-weighted because +2.
Problem: How large is D-H screening term for 0.1m NaCl for ions separated by 10Å?
I = 0.1; screening length rD = κ−1 = 1/(0.328 I1/2)= 9.58 Å
DH term = e−r/rD = e−κr = e−10Å/9.58Å = 0.35
DH screening cuts ionic interactions to 35% of full Coulomb.
9. DH limiting law for γ± .
log10 γ± = −A Z+Z− I1/2
A = 0.509 for aqueous solution at 25° C
Valid to only 0.01 molal.
10. Extended DH law.
No longer assume ions are points, but occupy volume. Ion atmosphere cannot extend all the way to the ion core, but is excluded a bit.
Term
€
e−κr ⇒e−κ(r−B)
(1 + κB)
B ≈ ionic size parameter (diameter) (dist of closest approach of 2 ions)
€
log10 γ± = −AZ+Z− I
1/ 2
1 + BI1/ 2+ CI (Extended DH Law)
Extends range of validity of DH theory to ≈ 0.1 molal.
22
CHAPTER 5 - Part B: Phase Diagrams (Phase Transitions in Complicated Mixtures)
I. Phases, Components, and Degrees of Freedom
The Phase Rule - a general way of predicting phase behavior of arbitrarily complicated mixtures. A. Define phase P = states of matter that are uniform throughout in
chemical composition and physical state. Example: How many phases here?
Complication: Dispersion = seems uniform on macroscale but consists of
granules or droplets when look closely. Macroscale Microscale B. Components C = minimum # of independent species necessary to fully
define the composition of all phases present in the system.
Easy when no reactions.
More complicated when reactions present and species are distributed between phases.
General: C = S - R Components = species - relationships
23
Examples: (figuring out how many components) acetone-chloroform C = 2 NaCl in H2O C = 2 5 species – Na+ (aq) Cl- (aq) H2O H3O+ OH- but: [Na+] = [Cl-] [H3O+] = [OH-] H2O ⇔ H+ + OH- R = 3 C = S - R = 5 - 3 = 2 Common sense! Saturated NaCl in H2O. 6 species - NaCl(s) Na+ (aq) Cl- (aq) H2O H3O+ OH- Add another relation NaCl(s) ⇔ Na+ + Cl- C = 2 still. Buffer solution: HAc and NaAc in H2O. species: H+ + Ac- + Na+ + H2O + OH- + HAc S = 6 HAc ⇔ H+ + Ac- H2O ⇔ H+ + OH- [H+] + [Na+] = [OH-] charge neutrality R = 3 C = 6 - 3 = 3 C. Now, the Gibbs phase rule: F = C - P + 2 F = variance , or # of “degrees of freedom”, or number of intensive
variables that can be changed independently without disturbing the number of phases present.
The “2” in the equation can be thought of as the T and p of system. Now let’s explore the consequences.
24
D. One-Component System example: Think of an experiment of the cooling curve of pure napthalene. Start with liquid napthalene: C = 1, P = 1 phase(liq) F = 1 - 1 + 2 F = 2 degrees of freedom Can still vary T and p and remain a single liquid phase. Cooling Curve:
When solid began to form, variance went from F = 2 to 1. Lost one degree of freedom. However, since pressure of room is also beyond our control, that 1
degree is really superfluous also. So effectively we have 0 freedoms. Temperature must halt until we have 1 phase again.
24
D. One-Component System example: Think of an experiment of the cooling curve of pure napthalene. Start with liquid napthalene: C = 1, P = 1 phase(liq) F = 1 - 1 + 2 F = 2 degrees of freedom Can still vary T and p and remain a single liquid phase. Cooling Curve:
When solid began to form, variance went from F = 2 to 1. Lost one degree of freedom. However, since pressure of room is also beyond our control, that 1
degree is really superfluous also. So effectively we have 0 freedoms. Temperature must halt until we have 1 phase again.
25
E. P-T Diagrams of Pure Substances, revisited
1. Understanding the Triple Point of H2O: Pure H2O: C = 1, P = 3 solid liq gas F = 1 - 3 + 2 = 0 A single unique point in T and p space!
Cannot have a quadruple point!! 2. Crossing the coexistence lines:
25
E. P-T Diagrams of Pure Substances, revisited
1. Understanding the Triple Point of H2O: Pure H2O: C = 1, P = 3 solid liq gas F = 1 - 3 + 2 = 0 A single unique point in T and p space!
Cannot have a quadruple point!! 2. Crossing the coexistence lines:
26
F. Two-Component System example: e.g., napthalene and diphenylamine. Start with liquid solution: C = 2, P = 1 (liq) F = C - P + 2 F = 2 - 1 + 2 = 3 to start “3” means we can vary T, p and one or the other composition and still
retain a 1-phase system. Cooling Curve:
II. Binary Volatile Liquid Mixture. (Acetone-Chloroform. e.g.)
A. Vapor Pressure-Composition Diagrams.
1. Ideal solution of two volatile liquids: Raoult says: pA = xApA* pB = xBpB* total p = pA + pB = xApA* + xBpB* p = pB* + (pA*- pB*)xA shows p is linear function of xA ---!
26
F. Two-Component System example: e.g., napthalene and diphenylamine. Start with liquid solution: C = 2, P = 1 (liq) F = C - P + 2 F = 2 - 1 + 2 = 3 to start “3” means we can vary T, p and one or the other composition and still
retain a 1-phase system. Cooling Curve:
II. Binary Volatile Liquid Mixture. (Acetone-Chloroform. e.g.)
A. Vapor Pressure-Composition Diagrams.
1. Ideal solution of two volatile liquids: Raoult says: pA = xApA* pB = xBpB* total p = pA + pB = xApA* + xBpB* p = pB* + (pA*- pB*)xA shows p is linear function of xA ---
27
2. Calculate composition yi of vapor phase: yA = pA/p yB = pB/p
yA=
xApA*
pB* + (p
A* −p
B*)x
A
yB= 1− y
A
Different curves are different values of the ratio p
A* / p
B* .
3. Then total pressure in terms of vapor compositions is:
p =pA*pB*
pA* − (p
A* −p
B*)y
A
Different curves are different values of the ratio p
A* / p
B* .
€
pA* /pB
*
€
pA* /pB
*
28
B. Temperature-Composition (T-x) Diagrams. For distillations, it is more useful to put both liquid and vapor composition
on the same diagram, vs T. Consider ideal mix first. Here, component A is less volatile than B.
Vapor richer in the more volatile component. Successive vaporization and condensation is fractional distillation.
Number of theoretical plates on Right is 3 and 5. Plates = number of successive vaporizations and condensations to achieve a condensate of a given composition from a given distillate.
28
B. Temperature-Composition (T-x) Diagrams. For distillations, it is more useful to put both liquid and vapor composition
on the same diagram, vs T. Consider ideal mix first. Here, component A is less volatile than B.
Vapor richer in the more volatile component. Successive vaporization and condensation is fractional distillation.
Number of theoretical plates on Right is 3 and 5. Plates = number of successive vaporizations and condensations to achieve a condensate of a given composition from a given distillate.
28
B. Temperature-Composition (T-x) Diagrams. For distillations, it is more useful to put both liquid and vapor composition
on the same diagram, vs T. Consider ideal mix first. Here, component A is less volatile than B.
Vapor richer in the more volatile component. Successive vaporization and condensation is fractional distillation.
Number of theoretical plates on Right is 3 and 5. Plates = number of successive vaporizations and condensations to achieve a condensate of a given composition from a given distillate.
29
Chemical engineering schematic of Continuous Fractional Distillation tower separating one feed stream into four distillate and one bottoms fractions.
30
C. Azeotropic mixtures. azeotrope = liquid mixture which boils to produce a vapor with same
composition as the liquid. No further separation can be achieved with continued distillation. Examples: acetone-chloroform; 95% ethanol-water at 78 °C High-boiling azeotrope low-boiling azeotrope Favorable A-B interactions Unfavorable A-B interactions
31
III. Liquid-Liquid Phase Diagrams - Partially Miscible Liquids.
A. T-x Phase diagrams - examples:
Use of the lever rule ----!
Suppose mixture is 0.59 mol hexane And 0.41 mol nitrobenzene. T=290K So overall XN = 0.41 So there must be two liquid phases. What are their compositions: α) XN = 0.35 β) XN = 0.83 How much of each do we have? lever rule says:
nαα= n
ββ
nα= relative amount of phase α
nα
nβ
=β
α
=0.83 − 0.410.41− 0.35
= 7
32
B. Example 2: another example having a lower crit temp.
C. Example 3: Nicotine-H2O. Has an upper AND lower critical temperature D. Example 4: showing higher temperature portion where vaporization can
occur.
32
B. Example 2: another example having a lower crit temp.
C. Example 3: Nicotine-H2O. Has an upper AND lower critical temperature D. Example 4: showing higher temperature portion where vaporization can
occur.
32
B. Example 2: another example having a lower crit temp.
C. Example 3: Nicotine-H2O. Has an upper AND lower critical temperature D. Example 4: showing higher temperature portion where vaporization can
occur.
33
E. Example 5: like above, but boiling occurs before full liquid miscibility temperature is achieved.
Xα XV Xβ IV. Solid-Liquid Phase Diagrams. (2 components)
A. Simplest - where no solid solutions form. Example: napthalene and diphenylamine.
eu =eutectic mixture = Liq. solution + solid N + solid D P = 3 C = 2 So: F = 2 - 3 + 2 = 1 (which is pressure)
B. Here, the solids are slightly miscible (i.e. solid solutions form):
33
E. Example 5: like above, but boiling occurs before full liquid miscibility temperature is acheived.
IV. Solid-Liquid Phase Diagrams. (2 components)
A. Simplest - where no solid solutions form. Example: napthalene and diphenylamine.
eu =eutectic mixture = Liq. solution + solid N + solid D P = 3 C = 2 So: F = 2 - 3 + 2 = 1 (which is pressure)
B. Here, the solids are slightly miscible (i.e. solid solutions form):
33
E. Example 5: like above, but boiling occurs before full liquid miscibility temperature is acheived.
IV. Solid-Liquid Phase Diagrams. (2 components)
A. Simplest - where no solid solutions form. Example: napthalene and diphenylamine.
eu =eutectic mixture = Liq. solution + solid N + solid D P = 3 C = 2 So: F = 2 - 3 + 2 = 1 (which is pressure)
B. Here, the solids are slightly miscible (i.e. solid solutions form):
33
E. Example 5: like above, but boiling occurs before full liquid miscibility temperature is acheived.
IV. Solid-Liquid Phase Diagrams. (2 components)
A. Simplest - where no solid solutions form. Example: napthalene and diphenylamine.
eu =eutectic mixture = Liq. solution + solid N + solid D P = 3 C = 2 So: F = 2 - 3 + 2 = 1 (which is pressure)
B. Here, the solids are slightly miscible (i.e. solid solutions form):
34
C. Reacting Systems: Compound Formation Case. A + 2B ! C
D. Partial miscibility in a 3-component system . H2O – acetone – phenol.
34
C. Reacting Systems: Compound Formation Case. A + 2B C
D. Partial miscibility in a 3-component system . H2O – acetone – phenol.
The three-component miscibility gap. The H20-acetone-phenol system. The three-component miscibility gap. The H20-acetone-phenol system.
35
E. The Fe-C system.
Consider a system of composition (2) corresponding to about 2.5% C. On
cooling of a molten iron-carbon solution, austenite phase of composition c begins to form, and with further cooling the liquid and solid solutions move in composition toward e and d, respectively, the proportion of the latter type of phase increasing. At 1125°C the eutectic is reached, and the system is now in equilibrium with Fe3C (cementite); during the halt austenite phase of composition d and Fe3C crystallize out together. Below 1125°C the system consists of Fe3C and austenite phase of composition moving along the da line. At 700°C the α-iron-austenite-Fe3C three-phase line is reached, and the further changes are as described earlier. An iron of this composition, unlike a steel, does not become a single phase until the melting point is reached. Such iron is called cast iron - it is not malleable when hot, but has valuable corrosion-resistant properties. If of composition close to the eutectic, its melting point is low enough to allow a fairly easy casting procedure.
