Chem 402 Lecture 6: L6 1:30:17

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Washington University in St. Louis
University College - Chemistry
University College - Chemistry Chem 402
Barnes Alexander

1 January 2017 L6: Entropy of Mixing and Electronic Partition Functions I. Entropy and Partition Functions A. Entropy of Mixing 1. kT at 298K a. (8.314 J/mol•K)  2.48 kJ/mol -23 -11 b. (1.38 x 10 J/molecule•K)  4.11 x 10 J/molecule -1 -1 c. (0.695 cm /K)  203 cm d. Since the translational energy varies withxn , by plotting 2og (energy), it allows us to get evenly spaced energy levels on the plot 2. Setup a. Mixing happens because you can access more microstates, and thus entropy increases (allows spontaneity) b. Want the solution in terms of mole fractions X = N /N = V /V (if constant B b B pressure) 𝑁 c. Entropy (S) = 𝑘𝑙𝑛Ω = 𝑄 = 𝑞 𝑁! 3. For an ideal gas mixture: a. Assume same initial (p, T) for A and B, and thus same (p, T) for mixture • Assume equal molecular volumes and lattice cell sizes b. ∆S = S 1 S 2 c. After mixing, count how many ways to distribute N Aolecules of A and N B molecules of B among the (V/v) lattice states d. There are (V/v) ways to distribute N molecules among (V/v) sites • Correct for indistinguishability, divide by N! (for respective parts) 4. Derivation 𝑉𝐴𝑁 𝑉𝐵 𝑁 𝑉𝐴 𝑁𝐴 (𝑣 ) 𝐴( 𝑣) 𝐵 ( 𝑣) a. 𝑆1= 𝑘𝑙𝑛Ω + 𝐴𝑙𝑛Ω = 𝑘𝑙𝐵Ω Ω = 𝑘𝑙𝑛𝐴 𝐵 = 𝑘(𝑙𝑛 + 𝑁 𝑁 𝐴 𝑁𝐵! 𝑁𝐴! (𝑉𝐵) 𝐵 𝑣 𝑙𝑛 𝑁𝐵! ) 𝑁 𝑞𝑁 ( ) b. 𝑆 = 𝑘𝑙𝑛Ω = 𝑘𝑙𝑛 = 𝑘𝑙𝑛 𝑣 2 𝐹 𝑁𝐴!𝑁𝐵! 𝑁𝐴!𝑁𝐵! 𝑁 𝑉 𝑁 𝑉 𝑁 (𝑁𝐴+𝑁 𝐵 ( ) ( 𝐴) ( 𝐵) ( ) c. ∆𝑆 = 𝑆 2 𝑆 =1𝑘𝑙𝑛 𝑣 − 𝑘𝑙𝑛 𝑣 𝑣 = 𝑘𝑙𝑛 𝑣𝑁 𝑁 = 𝑁𝐴!𝑁𝐵! 𝑁𝐴! 𝑁 𝐵 (𝑉𝐴) 𝐴 (𝐵 ) 𝐵 𝑣 𝑣 𝑉 𝑁𝐴 𝑉 𝑁𝐵 𝑣 𝑁 𝐴 𝑣 𝑁𝐵 𝑉 𝑁𝐴𝑉𝑁𝐵 𝑘𝑙𝑛[( ) ( ) ( ) ( ) ] = 𝑘𝑙𝑛 𝑁 𝑁 𝑣 𝑣 𝑉𝐴 𝑉𝐵 𝑉𝐴 𝐴𝑉𝐵𝐵 𝑁 𝑁 𝑉 𝐴 𝑉 𝐵 𝑉 𝑉 d. = 𝑘𝑙𝑛( )𝑉 (𝑉) = 𝑘(𝑁 l𝐴( ) 𝑉 𝑁 ln( 𝐵) 𝑉 𝐴 𝐵 𝐴 𝐵 e. = −𝑘(𝑁 ln( ) + 𝑁 ln( )) = (−𝑘 𝑁 ln 𝑋 ( ( ) + 𝑁 ln 𝑋 ))) 𝐴 𝑉 𝐵 𝑉 𝑁 𝐴 𝐴 𝐵 𝐵 • Since we have constant pressure, VA/V is equal to the mole fraction of A • Then we multiply through by N/N (=1) to get to the final expression f. ∆𝑆 = −𝑘𝑁(𝑋 𝑙𝑛𝐴 + 𝑋𝐴𝑙𝑛𝑋 )𝐵 𝐵 • The mole fraction is always <1, the ln of a number <1 is always negative, since both mole fractions are negative, and ln𝐴 and ln 𝑋𝐵< 0, the negative corrects for S > 0 (so that S increases for the mixing of ideal gases) B. Quantitative Treatment of Combining Molecular Partition Functions 1. Configurational PF a. Since q conf= gconf= 2 =∑ 𝑖𝑒 −𝜀𝑖/𝑘𝑇= 𝑒 + 𝑒 = 2 • Works when we set these two degenerate configurational states at 0 energy 2. If the energies are independent (and can just add together) the molecular partition functions just multiply a. Ex: molecular energy 𝜀 = 𝜀 𝑡𝑟𝑎𝑛𝑠 + 𝜀𝑟𝑜𝑡+ 𝜀 𝑣𝑖𝑏+ 𝜀 𝑒𝑙𝑒𝑐+ ⋯ b. Molecular partition function: 𝑞 = 𝑞 𝑡𝑟𝑎𝑛𝑠 𝑟𝑜𝑡 𝑣𝑖𝑏 𝑒𝑙𝑒𝑐+ ⋯ ℎ𝑣 2𝜋𝑚𝑘𝑇 3/2 2𝐼𝑘𝑇 𝑒−2𝑘𝑇 𝐷 /𝑘𝑇 c. Ultimately deriving: 𝑞 𝑉,𝑇 = ( ℎ2 ) 𝑉 • 𝜎ℏ 2• −ℎ𝑣 • 𝑔𝑒1 𝑒 1−𝑒 𝑘𝑇 • Translational, then rotational, then vibrational, then electronic
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