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Lecture 7

# Chem 402 Lecture 7: L7 2:1:17 Premium

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School
Washington University in St. Louis
Department
University College - Chemistry
Course
University College - Chemistry Chem 402
Professor
Barnes Alexander
Semester
Spring

Description
1 February 2017 L7: Vibrational and Rotational Statistical Mechanics I. Vibrational Partition Function A. Vibrational Energy 1 1. ๐ ๐ฃ๐๐= โ๐ฃ( 2 v) for v = 0, 1, 2, โฆ (๏ฎ is vibrational constant, v is quantum #) ๐ 1/2 a. ๐ฃ = ( ) /2๐, where k is the force constant of the molecule and ๏ญ is its reduced ๏ญ mass B. Populations of State 1. Qualitatively, most molecules are in the ground vibrational state at room temperature 2. All vibrational energy levels are non-degenerate (g = 1) 3. โฉ๐ธ = ๐ = โ ๐๐๐ ๐ C. Derivation 1 1. ๐ (๐ = โ ๐โ๐๐ฃ/๐๐ = โ ๐โโ๐ฃ(2+v)โg๐ฃ/๐๐= ๐ โโ๏ฎ/2๐๐โ โ (๐โโ๏ฎ/๐๐)๏ฎ ๐ฃ๐๐ ๐ฃ 0 0 a. Substitute ๐๐ฃ๐๐into the partition function b. g vs the degeneracy, and gv= 1 since vibrational states are non-degenerate c. Then distribute and separate out the constant 2. Approximate the sum: โ 0(๐โโ๐ฃ/๐๐ ) as โ 0(๐ฅ)๐ 2 3 ๐ a. ๐ ๐ 1 + ๐ฅ + ๐ฅ + ๐ฅ + โฏ+ ๐ฅ b. ๐ฅ๐ ๐ ๐ฅ + ๐ฅ + ๐ฅ + โฏ+ ๐ฅ ๐+1 ๐+1 ๐+1 c. ๐ ๐ ๐ฅ๐ = ๐ โ ๐ฅ ยป ๐๐(1 โ ๐ฅ) = 1 โ ๐ฅ 1โ๐ฅ๐+1 1 d. ๐ ๐ (1โ ๐ฅ)= ๐ ๐ (1โ ๐ฅ) 1 โข Consider x n+1= โโ๐ฃ/๐๐ which is always less than 1 for large N ๐ โข Since there are many levels (v ~40-100, lots of levels!), this will go to zero โ โโ๐ฃ/๐๐ ๏ฎ 1 e. โ 0(๐ ) = (1โ ๐โ๐ฃ/๐) โโ๏ฎ/2๐๐ ๐ โโ๏ฎ/2๐๐ ( ๐ ) 3. ๐ (๐ = ๐ and therefore ๐ (๐ = (1โ ๐๐ฃ/๐) ๐ฃ๐๐ (1โ ๐โ๏ฎ๐ฃ/๐) ๐ฃ๐๐ ๐! โ๐ 4. Simplify using vibrational temperature,๐ฃ๐๐ = ๐ โฮ๐ฃ๐๐2๐ a. ๐ (๐ = ๐ ๐ฃ๐๐ (1โ ๐ฮ ๐ฃ๐๐) D. Derivation of ๐ธ โช = ๐ ๐ฃ๐๐ ๐ฃ๐๐ ๐ 1. โฉ๐ธโช = ๐ = ๐๐ ( ๐ (๐๐๐)) = ๐๐ ( ๐ (๐๐๐ )) ๐ฃ๐๐ ๐ฃ๐๐ ๐๐ ๐,๐ ๐๐ ๐! ๐,๐ ๐ ๐ = ๐๐ ( (๐๐๐ โ ๐๐๐!)) ๐,๐ = ๐๐๐ (2 (๐๐๐ โ ๐๐๐!)) ๐,๐ ๐๐ ๐๐ a. Substitute in Q, then separate the ln terms and bring N down since constant b. Again, โ-lnN!โ drops out since it is constant, 0 w.r.t. T โฮ /2๐ 2 ๐ ๐ ๐ฃ๐๐ 2 ๐ โฮ ๐ฃ๐๐2๐ โฮ๐ฃ๐๐๐ 2. = ๐๐๐ ( ๐๐(ln((1โ ๐ฮ ๐ฃ๐๐))))๐,๐ = ๐๐๐ ( ๐๐(ln๐ โ ๐๐(1 โ ๐ )))๐,๐ a. Sub in q vibnd separate out the terms ฮ 2 ฮ๐ฃ๐๐ ๐ โ ๐ฃ๐๐ 2 ฮ ๐ฃ๐๐ 1 ๐ 3. = ๐๐๐ ( 2 (โ ๐๐(1 โ ๐ ๐ )))๐,๐ = ๐๐๐ ( 2 โ ฮ๐ฃ๐๐ (1 โ 2๐ ๐๐ 2๐ 1โ ๐โ ๐ ๐๐ ฮ โ ๐ฃ๐๐ 2 ฮ๐ฃ๐๐ 1 ฮ ๐ฃ๐๐ 1 ๐ ๐ ))๐,๐ = ๐๐๐ ( 2โ ฮ ๐ฃ๐๐ 2 (โ ฮ ๐ฃ๐๐)๐,๐ 2๐ 1โ ๐ ๐ ๐ ๐ ๐ ฮ๐ฃ๐๐ a. Bring down the term from the ln since it is a constant, and lne = 0 2๐ b. Then do chain rule differentiation w.r.t. T (first the outside, then the inside) -(x) x c. In the last term inside the parenthesis, it is an term, make this 1/e 2 ฮ๐ฃ๐๐ 1 ฮ๐ฃ๐๐ 1 ฮ๐ฃ๐๐ ฮ๐ฃ๐๐ 1 4. = ๐๐๐ ( 2๐2 โ ฮ๐ฃ๐๐ ๐2 ( ฮ๐ฃ๐๐) ๐,๐= ๐๐( 2 โ ฮ๐ฃ๐๐ ( ฮ๐ฃ๐๐))๐,๐ = ๐โ ๐ โ1 ๐ ๐ ๐โ ๐ โ1 ๐ ๐ ฮ ๐ฃ๐๐ ฮ ๐ฃ๐๐ ฮ๐ฃ๐๐ ฮ๐ฃ๐๐ ฮ๐ฃ๐๐ ฮ๐ฃ๐๐ ๐๐( โ ฮ ฮ ฮ )๐,๐ = ๐๐( โ ฮ )๐,๐ = ๐๐( + ฮ )๐,๐ 2 ๐โ ๐๐+ ๐๐โ๐ ๐๐๐ 2 1โ๐ ๐๐๐ 2 ๐ ๐๐โ1 2 1 a. T will all cancel out, then distribute the (ฮ๐ฃ๐๐ term to the adjacent term ๐ ๐ ฮ ๐ฃ๐๐ฮ๐ฃ๐๐ b. The ๐ โ ๐ + ๐ = ๐ = 1 and then we can rearrange the term to get to the final expression, in Joules E. Derivation of C v, Vib ๐ ๐ธ๐ฃ๐๐ ๐ ฮ๐ฃ๐๐ ฮ๐ฃ๐๐ ๐ ฮ๐ฃ๐๐ ฮ ๐ฃ๐๐ 1. ๐ถ ๐ฃ,๐ฃ๐๐ = (๐๐( + ฮ ))๐,๐ = ๐๐พ ( + ฮ )๐,๐ ๐๐ ๐๐ 2 ๐ ๐๐โ1 ๐๐ 2 ๐ ๐๐โ1 a. C iv always energy over temp b. Sub in U vibrom above, then pull out constants NK โฮ ๐ฃ๐๐ ๐ 1 1 ๐ 1 ๐ ๐ 2. ๐๐พฮ ๐ฃ๐๐ ๐๐ 2+ ฮ๐ฃ๐๐ )๐,๐ = ๐๐พฮ ๐ฃ๐๐๐๐( ฮ๐ฃ๐๐ ( โฮ ๐ฃ๐๐)๐,๐ = ๐ ๐ โ1 ๐ ๐ โ1 ๐ ๐ โฮ๐ฃ๐๐ โฮ๐ฃ๐๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐๐พฮ ( ) = ๐๐พฮ ( ) ๐ฃ๐๐ ๐๐ ฮ๐ฃ๐๐ฮ๐ฃ๐๐ โฮ๐ฃ๐๐ ๐,๐ ๐ฃ๐๐๐๐ โฮ๐ฃ๐๐ ๐,๐ ๐ ๐ ๐ โ๐ ๐ 1โ๐ ๐ a. Pull out ฮ ๐ฃ๐๐, a constant, then ยฝ drops out since it equals zero w.r.t. T โฮ ๐ฃ๐๐ ๐ b. Then multiply by the identity ๐ (over itself, = 1) โฮ๐ฃ๐๐ โฮ๐ฃ๐๐ โ1 โฮ ๐ฃ๐๐ 3. = ๐๐พฮ ๐ ((๐ ๐ )(1 โ ๐ ๐ ) ) = ๐๐พฮ (๐ ๐ ๐ (1 โ ๐ฃ๐๐๐๐ ๐,๐ ๐ฃ๐๐ ๐๐ โฮ โ1 โฮ โ1 โฮ ๐ฃ๐๐
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