Outline
Material and Energy Balances
CHEN 2120
Class Meeting #22
March12th, 2007
“Immiscible and partially miscible solutions”
• Thanks for the evaluations!
• Revisit 6.42
• Wed: Example problem working day
(combine W/F lectures into just F)
• Miscibility/extraction
• Ternary phase diagrams
• Homework Hints
Example: 6.42 revisited
Charlie
Chapters
5 and 6
*How is this different than Exam 1 material?
- Humidity
- Gases (partial pressures, ideality, equilibrium)
*Equations of state: a tool to calculate moles from info
that you are given (volume, pressure, temp.)
*Raoult’s Law/Henry’s Law: a tool to determine comp.
of equilibrium processes.
Miscibility
Miscibility/Extraction
• Chloroform is not miscible with water:
water
chloroform
• A ternary (3 component) system of chloroform, acetone, and water
will separate into two phases; acetone will preferentially go into
phase where it is most miscible.
*2 phases!
“Oil and water”
• Acetone is miscible with either chloroform or water:
chloroform/acetone
water/acetone
*1 phase
*1 phase
water/acetone
chloroform/acetone
• Distribution coefficient (K):
K = [(x)c phase]/[(x)w phase]
where x denotes mass fraction of acetone:
(x)c phase denotes mass fraction of acetone in the chloroform phase
Thus, if you are given K, then you can calculate how much acetone
goes to the chloroform phase vs. the water phase.
1
Miscibility/Extraction
• Extraction – a solute moves from a phase
in which it is less soluble to a phase in
which it is more soluble:
Two-phase, three-component
systems
65 wt% acetone
20 wt% MIBK
15 wt% water
(single phase)
K = [(x)c phase]/[(x)w phase]
Two-phase
region
water/solute
chloroform/solute
If solute is more soluble in c-phase than in w-phase,
will K be greater than 1 or less than 1?
Two-phase, three-component
systems
Point M:
15 wt% acetone
30 wt% MIBK
55 wt% water
At equilibrium:
Separates into 2 phases
1st phase (L): 85% water
12% acetone
3% MIBK
2nd phase (N): 4% water
20% acetone
76% MIBK
Now you try… (Clicker Prob. 22.1)
• A mixture of 20 wt% water, 33 wt% acetone, and
the remainder methyl isobutyl ketone (MIBK) is
brought to equilibrium at 25°C. This results in
two phases. Use Fig. 6.6-1.
What is the mass fraction of MIBK in:
the MIBK-rich phase?
A) 0.35
B) 0.07
C) 0.43
D) 0.58
the water-rich phase?
A) 0.04
B) 0.23
C) 0.43
D) 0.58
Now you try… (Clicker Prob. 22.1)
Water-rich phase
xW = 0.71
xA = 0.25
xM = 0.04
MIBK-rich phase
xW = 0.07
xA = 0.35
xM = 0.58
Original
mixture
• A mixture of 20 wt% water, 33 wt% acetone, and
the remainder methyl isobutyl ketone (MIBK) is
brought to equilibrium at 25°C. This results in
two phases.
If the total mass of the system is 1.2 kg initially,
what is the combined masses of phases 1 and
2?
1.2 kg mixture
m1
Separation
m2
Phase 1
Phase 2
2
Now you try… (Clicker Prob. 22.1)
• A mixture of 20 wt% water, 33 wt% acetone, and
the remainder methyl isobutyl ketone (MIBK) is
brought to equilibrium at 25°C. If there is initially
1.2 kg of the mixture, what are the masses of each
of the two resulting phases?
A) 0.24 kg and 0.96 kg
C) 0.34 kg and 0.86 kg
B) 0.12 kg and 1.08 kg
D) 0.53 kg and 0.67 kg
1.2 kg mixture
xW0 = 0.20
xA0 = 0.33
xM0 = 0.47
m1
Separation
m2
xW1 = 0.07
xA1 = 0.35
xM1 = 0.58
Phase 1
Phase 2
xW2 = 0.71
xA2 = 0.25
xM2 = 0.04
Mass balance: m1 + m2 = 1.2 kg
H2O balance: (0.20)(1.2 kg) = (0.07)(m1) + (0.71)(m2)
Solve simultaneously Æ m1 = 0.96 kg, m2 = 0.24
Homework Hints
• 6.68 – Very similar to problem 6.55 from last week.
Ans: (b) xA = 0.47, yA = 0.66; (cii) ~76 mole% vapor;
(ciii) liq. vol. = ~63 cm3
• 6.78 – nearly identical to example from Friday’s lecture
(3/9/07). Ans: sol/liq mass ratio = ~0.43
• 6.94 – Similar to today’s example (ternary phase
diagram). Ans: ~4200 g in MIBK-rich phase
• 7.6 - easy
• 7.7 - easy
3