CHEMISTRY 213 MID-TERM EXAM B FALL 2007 ANSWERS Answer questions on the green computer sheet provided with a PENCIL YOU WILL DRAW THE STRUCTURES OF MOLECULES BA AND BB ON THE LAST PAGE OF THIS EXAM - REMOVE IT NOW AND FILL IN YOUR NAME AND REGISTRATION NUMBER. ALSO DO THIS ON THE GREEN COMPUTER SHEET NOW (Where it says identification). Ignore course, section, etc., BUT CODE BOTH YOUR NAME & NUMBER. THERE SHOULD BE 7 PAGES OF QUESTIONS IN THIS EXAM, PLUS THIS PAGE, PLUS THE STRUCTURE PAGE TO BE HANDED IN. There are 31 (+Q#1) QUESTIONS WORTH 31 MARKS AND 15 MARKS FOR THE TWO STRUCTURE PROBLEMS (BA and BB) and 50 minutes available, so ration your time accordingly. Questions 2-32 have equal marks. NO BOOKS, DATA SHEETS OR OTHER MATERIALS ALLOWED EXCEPT CALCULATORS ------------------------------------------------------------------------------------------------------- DATA YOU MAY NEED DURING THIS EXAM: N = 6 x 10 23 h = 6.6 x 10 -34 J s c = 3 x 10 8 m s -1 k = 1.383 x 10 -23 J K -1 R = 8.3 J K -1 mole -1
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CHEMISTRY 213
MID-TERM EXAM B
FALL 2007
ANSWERS
Answer questions on the green computer sheet provided with a PENCIL
YOU WILL DRAW THE STRUCTURES OF MOLECULES BA AND BB
ON THE LAST PAGE OF THIS EXAM - REMOVE IT NOW AND
FILL IN YOUR NAME AND REGISTRATION NUMBER. ALSO DO THIS
ON THE GREEN COMPUTER SHEET NOW (Where it says identification).
Ignore course, section, etc., BUT CODE BOTH YOUR NAME & NUMBER.
THERE SHOULD BE 7 PAGES OF QUESTIONS IN THIS EXAM, PLUS
THIS PAGE, PLUS THE STRUCTURE PAGE TO BE HANDED IN.
There are 31 (+Q#1) QUESTIONS WORTH 31 MARKS AND 15 MARKS FOR
THE TWO STRUCTURE PROBLEMS (BA and BB) and 50 minutes available,
so ration your time accordingly. Questions 2-32 have equal marks.
6. Which two functional group(s) are present in BA
DETERMINE TWO POSSIBLE STRUCTURES FOR COMPOUND BA AND ENTER THEMON THE LAST PAGE OF THIS PAPER IN THE SPACE PROVIDED
contains a conjugated aldehyde C=C-CHO = C3Oso we have another 2C to find942 is between a trans double bond (970) and a gem double bond (900)so is either H2C=C(CH2CH3)CHO or CH3CH2-CH=CH-CHOcannot be the cis compound (700) or mono substituted (not then a conj aldehyde)Could maybe be CH3CH=C(CH3)CHO
Atmospheric S16O2, has IR active fundamental vibrations at A:1362 B: 1358 C: 1151 D: 1147 E: 518 F: 515 cm-1
[there are two isotopes of sulfur, 32S and 34S ]
7. Which would be the symmetric stretching vibration of 34SO2
asym>sym and light mass> heavy massso sym/heavy is lowest of stretching freqencies = 1147 = D
8. Since these vibrations were ir active, is the molecule linear =G or bent = Hfor symmetric stretch to be active, molecule must be bent = H
9. What is the total number of fundamentals for 34S16O2
10. Chlorine (Cl2) is a pale green gas at room temperature (300K). Basing your answer onBoltzmann, how many molecules are in the first excited electronic state at roomtemperature - assume 105 molecules are in the ground state. [hint: you will need toestimate the energy gap using data we discussed in the first week of term!]
green gas means absorbs in visible, where ΔE = ~ 200-300kJ/moleNU/NL = e-(DE/RT) is then 0 ( cannot have a fraction of a molecule in Nu)
11. If you had a mixture of PH3 and 13CF4 in an NMR spectrometer operating at fixedfrequency but which could vary the magnetic field, and you scanned from low field tohigh field, in which order would the nuclei achieve resonance.
A: C-F-H-P B: H-P-F-C C: H-F-P-C D: H-C-F-PE: P-C-F-H F: C-P-F-H G: C-P-H-F H: H-F-C-PI: P-C-H-F J: C-F-P-Hthe relative γ values are H=1, F=0.94, P=0.4 and C=0.25 (you don’t need to know values, butmust know order!) so for 2.3 Tesla, H = 100MHz but C = 25 MHz, to get C at 100MHz wouldtake 4x the fieldso order as you scan from low to high field is H<F<P<C = C
Of the choices:A: Avogadro’s number (N)B: nuclear spin quantum number (I)C: induced fieldD: applied fieldE: spin-spin couplingF: rotational quantum number (J)G: magnetogyric ratio (()
Which is chiefly responsible for12. the different frequencies needed to observe P, F ,13C and H in the same magnetic fieldG: magnetogyric ratio (()
Chem 213 Mid-Term Exam 2007 B 4
13. the difference in Hz between CHCl3 and CHI3 in a 60MHz and 300MHz spectrumD: applied field14. the fact that the 13C spectra of CH2Cl2 and CDCl3 both show three linesB: nuclear spin quantum number (I) = 1 for D (3 lines) and 1/2 for H (3 lines because 2 protons)
Of the choices:A: singlet B: doublet C: quartet D: sextetE: septet F: doublet of quartets G: quartet of quartetsH: doublet of doublet of triplets I: quartet of doublets J: doublet of doublet of quartets
What pattern would you observe for the 15. F in PF6 d B16. P in PF6 septet E17. H in 13CH3CF3 dq F18. H in 13CF3-PH2 ddq J
19. What pattern of intensities would you expect for the 13C of CF313CH3
λ = 4μm = 4 x 10-6m E=(Nhc/λ) = (6 x 1023 x 6.6 x 10-34 x 3 x 108)/4 x 10-6
= 29.7 x 103 J mole-1 =H
24. If a sample of water was placed near to the source in Q#23, what is the most likelychange that would occur in the water - hint: think where water absorbs in cm-1.
A: nothing B: sample boils C: spin change D: bond rotationE: electronic transition F: bond stretching/bending G: bond dissociation
4μm = 2500 cm-1 and water does NOT absorb at 2500 cm-1, so nothing, A
25. What wave number would radiation of frequency 500 MHz have
26. What is the energy of the fundamental frequency [also = E1-E0]E = Nhν = Nhc/λ but wn = 1/λ so E = Nhc.wn= 6 x 1023 x 6.6 x 10-34 x 3 x 1010 x 1715 {c in cm.sec-1} = 20.4 kJ/mole = B
27. What is the zero point energyZPE = hν/2 = 20.4/2 = 10.2 kJ/mole = A
28. What is the energy of E1
E1 = hν x 3/2 = 30.6 kJ/mole = D
29. What is the probable energy of E2
a bit less than 5/2 x hν (51) so choose 50.8 = G
Chem 213 Mid-Term Exam 2007 B 9
30. What is the energy of the first overtoneE2-E0 = 50.8-10.2 = 40.6 kJ/mole = E
31. What is the dissociation energy of the bond (kJ mole-1)