false

Unlock Document

University College - Chemistry

University College - Chemistry Chem 402

Barnes Alexander

Spring

Description

14 April 2017
L33: Colligative Properties
I. Colligative Properties
A. Basics
1. Colligative properties are properties of solutions in the dilute limit
a. Where there is a solvent “A” and a solute “B” where n A> n b
• For X A> X B
b. They are changed properties of a solvent by adding solute
2. Colligative properties are a direct result of 𝑥(𝑙,𝑇,𝑝 < 𝑝𝑢𝑟𝑒(𝑙,𝑇,𝑝 )
𝐴 𝐴
3. Previously we defined these properties entropically, but now we will look at them
from a free energy standpoint
a. ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 < 0 for spontaneous processes
b. The “-T∆S” term represents the entropy of the system
c. “∆H” is the entropy of surroundings
d. It really doesn’t matter what happens as long as there is a net uni> 0
• Therefore it is necessary to have ∆H so that there is an increase in entropy of
the surroundings for low temperature limit
4. Using concentration measures
𝑛𝐵 𝑛𝐵
a. Mole fraction: 𝑋 𝐵 𝑛 𝐴𝑛 𝐵 ~ 𝑛𝐴
• This is because in the conditions thatAn >> nb, we can take B out of the
bottom
𝑀𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛𝐵
b. Molality: 𝑚 =𝐵 𝑘𝑔 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑛 𝑀
𝐴 𝐴
• Where M isAthe molar mass of A
c. Molarity: 𝐵 =] 𝑛𝑏
𝑉𝐴
B. Vapor Pressure Lowering
1. Just use Raoult’s Law
a. 𝑝 𝐴 𝑋 𝑝 𝐴 𝐴 − 𝑋 𝑝 𝐵) 𝐴
b. ∆𝑝 = 𝑝 − 𝑝 = 𝑋 𝑝 − 𝑝 = 𝑝 𝑋 − 1 = −𝑋 𝑝 ) ∗
𝐴 𝐴 𝐴 𝐴 𝐴 𝐴 𝐴 𝐵 𝐴
2. The change in vapor pressure due to adding a solute will always be negative
a. ∆𝑝 = −𝑋 𝑝 𝐵 𝐴∗
b. Always a decrease in vapor pressure, thus change is negative
C. Boiling Point Elevation
1. Assumptions
a. A = solvent, B = solute
b. B is a non-volatile solute, thus only A is present in the vapor
2. Derivation
𝑚𝑖𝑥 ∗
a. µ 𝐴 (𝑙,𝑇,𝑝 = µ 𝑔,𝐴,𝑝 )
b. µ 𝑙,𝑇,𝑝 + 𝑅𝑇𝑙𝑛𝑋 = µ 𝑔,𝑇,𝑝 ∗ ( )
𝐴 𝐴 𝐴
c. 𝐺 𝑙,𝑇,𝑝 + 𝑅𝑇𝑙𝑛𝑋 = 𝐺(𝐴,𝑇,𝑝) 𝑅𝑇𝑙𝑛𝑋 = 𝐺 𝑔,𝑇,𝑝𝐴− 𝐺 𝑙,𝑇,𝑝 = ∆𝐺 ( ) 𝑣𝑎𝑝
• Since we are going from liquid to gas, final – initial = gas – liquid
d. 𝑅𝑇𝑙𝑛 1 − 𝑋 𝐵) = ∆𝐺 𝑣𝑎𝑝
1 2 1 3
• If we do a Taylor series expansion of ln 1 − 𝑋𝐵 )= −𝑋 −𝐵𝑋 −2𝑋 𝐵 3 𝐵
• However, for small B, we only take the first term since the others go to 0 ∆𝐺𝑣𝑎𝑝 𝑛𝐵 𝑛𝐵
e. –𝑅𝑇𝑋 = 𝐵𝐺 𝑣𝑎𝑝 𝑋 𝐵 − 𝑅𝑇 = 𝑛 +𝑛 ≈ 𝑛
1 𝐴 𝐵 𝐴
𝑚 𝑛 ∆𝐺𝑣𝑎𝑝
f. Multiply by “1”: ( 1 ) 𝐵 = −
𝑚 𝑛𝐴 𝑅𝑇
𝐴
• Since nA/m A 1/M (tAe molar mass of A)
• Since nB/m A m (mBlality of B)
𝑚 𝑏 ∆𝐺𝑣𝑎𝑝 ∆𝐺̅ 1
g. 1 = − 𝑅𝑇 𝑚 =𝐵 𝑇 (− 𝑀 𝑅 )
𝑀 𝐴 𝐴
• But at this point, we need to get a ∆T in there, so take the partial with respect
to T of both sides keeping a constant pressure
∆𝐺𝑣𝑎𝑝
𝜕𝑚 𝐵 1 𝜕( 𝑇 ) ∆𝐻𝑣𝑎𝑝
h. ( ) = − ( ) = 2
𝜕𝑇 𝑝 𝑀 𝐴 𝜕𝑇 𝑀𝐴𝑅𝑇
𝑝
𝜕 ∆𝐺𝑣𝑎𝑝) ∆𝐻𝑣𝑎𝑝
• We get this result from the Gibb-Helmholtz equation: = − 2
̅ 𝜕𝑇 𝑇
i. ( 𝜕𝑚 𝐵) = 1 (∆𝐻 𝑣𝑎)
𝜕𝑇 𝑝 𝑀𝐴𝑅 𝑇𝐴∗
• Since the T doesn’t change much around boiling point, we can substitute in
the pure boiling point of state A
• Also, we can further approximate the partials as finite changes (“∆”)
∆𝐻 ∗2
j. ( ∆𝑚 𝐵) = 1 𝑣𝑎𝑝 ∆𝑇 = 𝑀 𝐴𝑇 𝐴 (∆𝑚 )
∆𝑇 𝑝 𝑀𝐴𝑅 𝑇𝐴2 ∆𝐻𝑣𝑎𝑝 𝐵
• But, let’s thinking about the change in molality of B: ∆B = m B 0 (mixed –
pure)
• Thus, we can just say that this change is equal toBm
3. Boiling Point Elevation
2 2
∗ 𝑀 𝐴𝑇 𝐴∗ 𝑀 𝐴𝑇 𝐴
a. ∆𝑇 = 𝑇 − 𝑏 = 𝑏 ∆𝐻̅ (𝑚 𝐵) = 𝐾 𝑏 ,𝐵where 𝐾 = 𝑏 ∆𝐻
𝑣𝑎𝑝 𝑣𝑎𝑝
• All of the terms in b are constants and thus can be pulled out, usually this
value of Kbcan be found tabulated in books
b. This expression will always be positive for going from a liquid to a gas
c. ∆T ib always positive since all the terms are positive and everything except m Bs
constant
D. Freezing Point Depression
1. A

More
Less
Related notes for University College - Chemistry Chem 402

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.