DOC PREVIEW
U-M CHEM 260 - CHEM 260 PROBLEM SET

This preview shows page 1 out of 4 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 4 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Last Problem Set1. More BoltzmanPart a. There is 1 possibility to be in the 2s state and 3 in the 2p, excluding spins, or 2and 6 respectively if you include spins. Either set will work throughout this problem aslong as you are consistent.Part b. 2s: S= k ln Ω S = 1.38 *10-24 J/K * ln 2 = 9.57*10-24 J/K 2p: S = k ln Ω S = 1.38 *10-24 J/K * ln 6 = 2.47*10-23 J/KPart c.∆G = ∆H - T∆S and we are saying ∆G ≈ ∆A = ∆E - T∆S. (Note these are per atom)∆S = S2p – S2s = 1.51*10-23 J/K∆E = 3*10-20 JSo, ∆G = 3*10-20 J – 298 K * 1.51*10-23 J/K = 2.53 *10-20 JPart d.∆E = 3*10-20 J (Remember we need to consider the relative number of states)3-10*2.04 ====−−−−7.29-298K*J/K10*1.38J10*3kT(2s2pe 3e 3 e 3n n2320Part e. *TJ/K10*1.38J10*3kT(2s2p2320e 3 e 3n n1−−−−===T = 1980 K2. Rotational and Vibration SpectraPart a2B = 20 cm-1 so B = 10 cm-1()()()()()()()( )10m-1.317ER =−====−=−+−−−=+=20m1.734ERBc4RRc4cI4B27Kg1.614E265.81E271.66e265.81e271.66EMMMM222ClHClHµπµππhhhPart bµ = 3.14E-27 KgB = 5.16 cm-1Series of lines equally spaced at 10.31 cm-1Part cFirst you must calculate k (spring constant) for HCl (or DCl)476670kkk5.434E14c2~c2~hchc((2========ππhhnow calculate for DCl12068cm−====~3.8962E14kkDClHClPart dZeroPoint HCl = ½(2885cm-1) = 2.87E-20 JZeroPoint DCl = ½(2068cm-1) = 2.05E-20 J3. NMRThe idea here is to calculate the energies of the states, then the energies of thetransitions, and then calculate the frequency of the radio waves for those transitions.Constantsg= 5.6 h= 6.63E-34 JsUn = 5.05E-27 J/T sa 0.10 TB = 10 T sb 0.20 TJ1.00E-26JProblem 3Part ag Un B (1-sa)mag Un B (1-sb)mbtotalalpha alpha -1.2726E-25 -1.1312E-25 -2.4038E-25 Jalpha beta -1.2726E-25 1.1312E-25 -1.414E-26 Jbeta alpha 1.2726E-25 -1.1312E-25 1.414E-26 Jbeta beta 1.2726E-25 1.1312E-25 2.4038E-25 JPart b Freqalpha alpha to beta alpha 2.5452E-25 J 3.84E+08 hzalpha beta to beta beta 2.5452E-25 J 3.84E+08 hzalpha alpha to alpha beta 2.2624E-25 J 3.41E+08 hzbeta alpha to beta beta 2.2624E-25 J 3.41E+08 hzPart c J ma mb Interactionalpha alpha 2.5E-27 -2.3788E-25 Jalpha beta -2.5E-27 -1.664E-26 Jbeta alpha -2.5E-27 1.164E-26 Jbeta beta 2.5E-27 2.4288E-25 JFreqalpha alpha to beta alpha 2.4952E-25 J 3.77E+08 hzalpha beta to beta beta 2.5952E-25 J 3.92E+08 hzalpha alpha to alpha beta 2.2124E-25 J 3.34E+08 hzbeta alpha to beta beta 2.3124E-25 J 3.49E+08 hzSo, for part b the spectrum looks like 2 lines, and in part c the spectrum looks like 4lines. Each line in part b, splits into 2 lines in part c, one ½ J below the original lineand one ½ J above the original line. So, the fine structure splitting is equal to J.4. Identical NucleiThe idea here is to calculate the energies of the states, then the energies of thetransitions, and then calculate the frequency of the radio waves for those transitions.Constantsg= 5.6 h= 6.63E-34 JsUn = 5.05E-27 J/T sa 0.10 TB = 10 T sb 0.10 TJ1.00E-26JProblem 4Part ag Un B (1-sa) mag Un B (1-sb) mb totalalpha alpha -1.2726E-25 -1.2726E-25 -2.5452E-25 Jab + ba -1.2726E-25 1.2726E-25 0 Jbeta beta 1.2726E-25 1.2726E-25 2.5452E-25 Jab - ba 1.2726E-25 -1.2726E-25 0 JPart b FreqT- to To 2.5452E-25 J 3.84E+08 hzTo to T+ 2.5452E-25 J 3.84E+08 hzPart c J ma mb Interactionalpha alpha 2.5E-27 -2.5202E-25 Jab + ba 2.5E-27 2.5E-27 Jbeta beta 2.5E-27 2.5702E-25 Jab - ba -2.5E-27 -2.5E-27FreqT- to To 2.5452E-25 J 3.84E+08 hzTo to T+ 2.5452E-25 J 3.84E+08 hzIn this case, you always get only one line in the spectrum. One line with nosplitting. Two identical nuclei do not split each


View Full Document

U-M CHEM 260 - CHEM 260 PROBLEM SET

Download CHEM 260 PROBLEM SET
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view CHEM 260 PROBLEM SET and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view CHEM 260 PROBLEM SET 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?