110c-lecture16

# 110c-lecture16 - Activity of Solvent(a1 and Solute(a2 vs...

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Activity of Solvent (a 1 ) and Solute (a 2 ) vs. Standard State Solvent (1) + Solute(2) Solution Initial Final P * 1 P 1 n 1 Raoult’s Law Standard State (solvent): 1 , 1 * 1 1 1 ln 1 1 ln ln x RT P P RT G x liq s γ µ = = = 1 , 1 * 1 1 , 1 x P P a x x = = Initial state: pure solvent Initial Final P * 2 P 2 n 2 Henry’s Law Standard State (volatile solute): x H liq s k P RT P P RT G , 2 * 2 2 2 ln 2 2 ln ln = = = 2 , 2 , 2 , 2 x k P a x x H x = = Initial state: pure solute Initial Final P º 1 P 1 n 2 1m ?m Standard State of non-volatile solute (m 2 º =1molal): ( ) 2 , 2 2 2 2 2 2 ln 2 2 ln ln ln m RT m m RT P P RT G m s = = = = o o o ) 1 ( 1 1 , 2 2 , 2 , 2 o P P x a x x x = = Initial state: 1 molal solution

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Colligative Properties of Solutions and Osmotic Pressure Colligative properties (e.g. melting temperature and osmotic pressure): Depend only on number of dissolved species regardless of size, charge or mass. Provides a way to count molecules dissolved in solution. For dilute solution, the solvent obeys Raoult's law (P 1 =x 1 P 1 *): 1 1 1 ln x RT liq sln + = µ liq sln 1 1 < because 0 ln 1 < x for x 1 < 1 (i.e. solvent in solution has lower free energy due to higher entropy--less order) For an ideal solution, 1 H = 0 1 1 ln x R S = and 0 ln 1 1 ln 1 1 < = = x RT G liq s Dissolving a solute causes x 1 < 1 and liq sln 1 1 < 1 1 1 S T T G P P = = ∂µ for the solvent (component 1) in solution. sol S > liq S >> vap S Dissolving a solid solute: 1. Lowers melting temp of solvent. 2. Raises boiling temp of solvent. 3. Increases osmotic pressure. 1(sln) + 2(sln) 1(liq) Solvent(1) + Solute(2) Solution
Melting Temperature Depression At the melting point (T fus ), solid and liquid solvent (component 1) are in equilibrium: () () fus sln fus solid T T 1 1 µ = 1 1 1 1 ln a RT T T liq fus sln fus solid + = = where liq s P P a 1 ln 1 1 = 1 1 1 ln a RT liq solid + = RT a liq solid 1 1 1 ln = Apply the Gibbs-Helmholtz eqn, p903 2 1 1 , 1 / T H T T x P = ∂µ 2 2 1 1 2 1 1 1 1 1 , 1 ln RT H RT H H RT H H RT T T a fus s l l s l s x P = = = = 2 1 1 2 1 1 2 RT H H RT H H RT H l s s l fus = = (replace with H fus and get sign change) dT RT H a d fus 2 1 ln = Fusion (melting): solid liquid Freezing: liquid solid 1(sln) + 2(sln) 1(s) ) ( ln 1 fus s T ) ( 1 fus solid T

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dT RT H a d fus 2 1 ln = Integrate both sides from pure liquid (a 1 =1, T= * fus T ) to a final solution with activity 1 a = = = = * * 2 * 2 1 1 1 1 1 1 1 ln ln fus fus fus fus T fus T fus fus T fus T fus a T T R H T dT R H dT RT H a a d = fus fus fus T T R H a 1 1 ln * 1 assuming H fus is constant.
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110c-lecture16 - Activity of Solvent(a1 and Solute(a2 vs...

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