3502_34_dece07_LG

3502_34_dece07_LG - Chem 3502/5502 Physical Chemistry II...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 34, December 07, 2009 (Some material in this lecture has been adapted from Cramer, C. J. Essentials of Computational Chemistry , Wiley, Chichester: 2002; pp. 301-303; 314-315.) Solved Homework From K i / j = A [ ] B [ ] = % E A ( ) % E B ( ) = e " E A / k B T e " E B / k B T = e " E A " E B ( ) / k B T Let us assume K = 1, in which case E A E B is zero. An error of a factor of 10 means K = 10 (or 0.1, the only difference will be in the sign of E A E B and not its magnitude). So, we simply need solve 10 = e " E A " E B ( ) / k B T for Δ E = E A E B , which is, at 25 °C (298 K) " E = # k B T ln10 = # 1.3806 $ 10 # 23 J K -1 ( ) 298 K ( ) 2.3026 ( ) = 9.4733 $ 10 # 21 J At 500 °C (773 K), we have " E = # k B T ln10 = # 1.3806 $ 10 # 23 J K -1 ( ) 773 K ( ) 2.3026 ( ) = 2.4573 $ 10 # 20 J The numbers above correspond to the errors in the predicted energy difference that would lead to an error of an order of magnitude in the equilibrium constant (the numbers are, of course, independent of what the equilibrium constant really is). A quick glance at the form of the equation for the rate constant
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34-2 k = k B T h e " G TS " G A ( ) / k B T should make clear that the energy errors will be the same for the rate constant as for the equilibrium constant (prove this to yourself if you don’t see it! Note that the prefactor of k B T/h makes no difference because an error that is a “factor” of ten implies that the ratio of the wrong and right rate constants is 10, so the prefactor disappears as it is present in both the numerator and denominator of the ratio). These are, of course, tiny energy differences, but that is because we really in this case need to think in terms of large collections of molecules, not single molecules (how can there be an equilibrium constant for one molecule?) So, if we multiply by Avogadro’s number (6.02 x 10 23 mol –1 ) and divide by 1,000 (so we have kJ instead of J), we get the more typical molar quantities 5.7 kJ mol –1 (298 K) and 14.8 kJ mol –1 (773 K) [or, 1.4 kcal mol –1 (298 K) and 3.5 kcal mol –1 (773 K)]. So, not much room for error, particularly at low temperatures. .. (drat interesting chemical quantities always depending exponentially on energy differences!) The final question was what fraction is this energy difference at 298 K of the total energy of HF/STO-3G water. Noting 1 a.u. of energy (1 E h ) is 4.36 x 10 –18 J, the total energy of water, at –74.9659012170 a.u. is 3.27 x 10 –16 J. The energy error 9.47 x 10 –21 J is a mere 0.003% of that energy. So, an error of 0.003% in a calculation of water could throw off an equilibrium constant or a rate constant by an order of magnitude. Imagine how tight the requirements are for a really big molecule! Vibrational Spectroscopy Revisited Last lecture, we considered the solution of the 1-dimensional vibrational Schrödinger equation for an arbitrary potential to compute accurate vibrational wave functions and transition energies. While this approach is quite accurate for select instances, it is too demanding for a molecule with many degrees of freedom. Given the importance of vibrational (infrared) spectroscopy for identification of molecular
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3502_34_dece07_LG - Chem 3502/5502 Physical Chemistry II...

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