Chem 402 Lecture 33: L33 4:14:17

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University College - Chemistry
University College - Chemistry Chem 402
Barnes Alexander

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
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