Háskóli Íslands/
University of Iceland
Raunvísindadeild
/Science faculty,
Eðlisefnafræði B / Physical chemistry B / EFN403G
Dæma og verkefnaskammtur III /problems and projects III, 12.04.11
(updated 21.04.11)
Tímadæmi:
NB: Students (see names in brackets) will solve the problems/projects during the lecture / problem
class (12.04.11) and hand in solutions in a computerised form for displaying on the web, no later
than one week after the problem class.
ATH: Tilgreindir nemendur leysa viðkomandi dæmi í tímanum ( 12.04.11) og skila úrvinnslum á
tölvutæku formi til birtingar innan viku eftir það .
Tímadæmi A. (Eiríkur Þórir Baldursson & Svanfríður Harpa Magnúsdóttir)
Örbylgju-litrófsgreining / Microwave spectra and spectra analysis
In the microwave oven, microwaves are absorbed by water molecules in liquid phase to cause
rotational excitation. Enclosed is a calculated rotational spectrum of water (gas phase). The
strongest absorption peak appears at 87.072 cm-1. a) What rotational transition does that peak
correspond to and b) what would be the electromagnetic wavelength (mm) which would cause that
transition (absorption) to occur? c) What is the rotational constant (cm-1) for H2O according to
this? d) What is the rotational energy (J mol-1) of the molecules which absorb the 87.072 cm-1
radiation?
50
100
150
-1
 [cm ]
200
250
300
Fig. 1 Calculated microwave absorption spectrum for water (g)
Tímadæmi B (Sigríður Rós Einarsdóttir & Hörður Kári Harðarson) IR litrófsgreining:
Á meðfylgjandi mynd sést litrófstoppur í gleypnirófi sem er til kominn vegna v = 0 → v = 1
örvunar 12C16O sameindarinnar. Tvær raðir toppa eru merktar á myndinni, P-röðin og R-röðin. Í
meðfylgjandi töflu eru nokkur tíðnigildi toppa úr þessum röðum gefin í cm-1. Gerið ráð fyrir að
titringsorkan sé á forminu Ev = e (v+½) – exe(v+½)2 og að snúningsorkan sé á forminu EJ =
BvJ(J+1) innan hvers titringsþreps v. Notið öll orkugildi í einingunni cm-1.
a) Ritið upp jöfnuna fyrir staðsetningar toppanna í R-röðinni og P-röðinni á myndinni sem fall af
skammtatölunni J. Notið B0 og B1 fyrir snúningsfasta ástandanna v=0 og v=1.
b) Notið tíðnigildi toppanna sem eru gefin í töflunni og jöfnurnar úr a)-lið til að meta B0 og B1
fyrir 12C16O sameindina. Reiknið svo r0 og r1, þar sem rv er meðafjarlægð milli kjarna í
titringsþrepi v.
c) Hvaða snúningsþrep 12C16O sameindarinnar hefur stærstu hýsingu í v = 0 miðað við þær
aðstæður sem rófið á myndinni var mælt við? Hver er margfeldni þessa snúningsþreps?
Toppur
P(1)
P(2)
P(3)
P(4)
P(5)
Tafla með dæmi B
Tíðni
Toppur
-1
(cm )
2139,43
R(0)
2135,55
R(1)
2131,63
R(2)
2127,68
R(3)
2123,70
R(4)
Tíðni
(cm-1)
2147,08
2150,86
2154,59
2158,31
2161,97
Tímadæmi C (Þorvaldur Snæbjörnsson & Stefanía Magnúsdóttir) NMR.
a) What is the difference in spin frequency (Hz) of a 1H nucleus attached to an aromatic ring ( =
7) and a 1H nucleus in a TMS (tetra methyl silan) ( = 0) in a fixed magnetic field of a 60 MHz
NMR equipment?
b) Kjarnsýrusameind getur myndað tvenns konar spíral lögun (helical structure), A- og B- form.
Mismunurinn felst meðal annars í mismunandi lögun sykursameinda hluta keðjanna. Ribosa
sameindahluti RNA keðju, þar sem skipt hefur verið á H og D frumeindum að hluta, er sýndur hér
fyrir neðan:
Hliðrunargildi Hi´ róteinda beggja formana eru þau sömu, en viðkomandi kúplingsfastar eru hins
vegar frábrugðnir. Í meðfylgjandi töflu eru gefin upp gildi fyrir A formið:
Kjarnar
H1´
H2´
H3´
H4´
Hliðranir
(ppm)
6.0
5.0
4.5
4.0
kúplingsfastar (A-form)
Jij
Hz
J1´2´
0
J2´3´
5
J3´4´
9
Rissið upp 1H-NMR róf A-forms RNA og auðkennið klofnanir með því að tilgreina Jij. Tiltakið
hvort NMR merki einstakra róteinda séu “singlettar”, “doublettar” eða “triplettar” o.s.frv.
Skiladæmi
ATH: Skilið eigi síðar en í fyrirlestrartímanum, fimmtudaginn 14. apríl, 2011 /
Problems/projects to be handed in no later than 14.04.11.
1) Litrófsgreiningar almennt: Styrkákvarðanir efna með gleypnimælingum. (sbr.
fyrirlestur 17.03.11)
a) 15.8% af 340 nm geislun fer í gegn um 1 cm breiða kúvettu með vökvalausn með NADH.
Gleypnistuðull () NADH fyrir þessa bylgjulengd er 6.22 x 106 cm2 mól-1. Hver er styrkur NADH
í lausninni?
b) Gleypni lausnar af lífhvata (E) og AMP í 1 cm kúvettu mælist 0.46 fyrir 280 nm geislun og
0.58 fyrir 260 nm geislun. Gleypnistuðlar er:
280 / cm2 mól-1
260 / cm2 mól-1
lífhvati (E)
2.96 x 107
1.52 x 107
AMP
2.4 x 106
1.5 x 107
Hverjir eru styrkir efnanna (E og AMP)? (svar: [AMP] = 25.2 mól dm-3)
2) Microwave spectrum for NO and spectral analysis (sbr. fyrirlestur 22.03.11):
* 26.75 cm-1
Microwave spectrum
Molecule = NO(g)
t = 25oC
Bandwidth = 1.5 cm-1
0
20
40
60
80
100
-1
 [cm
]
Enclosed (above) is a microwave spectrum of NO(g) derived at room temperature. The
wavenumber value for the * -marked peak is indicated.
120
a) What transition does the * -marked peak in the spectrum correspond to?
b) Derive the rotational constant (B / cm-1) for the molecule from the data.
(answer: 1.672 cm-1)
c) What is the average spacing (cm-1) between the spectral peaks?
d) What is the average internuclear distance (nm) between the atom nuclei?
e) What is the rotational energy (J mol-1) of the molecules which absorb at 26.75 cm-1 (*) ?
What is the rotational frequency (Hz) of the molecules in (e)?
3) Microwave spectroscopy (sbr. fyrirlestur 22.03.11)
The equilibrium bond length in nitric oxide (14N16O) is 1.15A. Calculate (a) the moment of inertia
(I) of NO and (b) the energy for the J= 0 -> 1 transition. (c) How many times does the molecule
rotate per second at the J = 1 level? (svar: (b): 6.78 x 10-23J).
a. 4) IR spectroscopy (sbr. fyrirlestrar 24-29.03.11)
Enclosed are IR spectra of CO (Fig 1: low resolution, Fig 2: high resolution).
a) Based on the numerical values given in the text of figure 1, derive the vibrational
constants, e and exe for CO and compare with values given in a database. (answer:
calculated e = 2169 cm-1)
b) determine the zero energy for the molecule by using the e and exe constants from (a).
c) What transition do the * -marked peaks in the spectrum (Fig. 2) correspond to?
d) Determine the wavenumber values of the peaks in (c) (the *-marked peaks) by using an
appropriate program to calculate the IR spectrum for CO based on relevant spectroscopic
constants given in a database. (answer: left *: 2139.37 cm-1)
Fig. 1
Fig. 2
5) Titringsróf /IR róf fjölatóma sameinda (sbr. fyrirlestur 29.03.11)
a) Enclosed (fig.1) you will find an IR spectrum for an impure acetylene (C2H2) gas sample.
The spectrum shows two major absorption bands for C2H2 at 1) 3289 cm-1, and 2) 730
cm-1. Figure 2 shows possible vibrational modes for acetylene.
Fig 1.
Fig. 2
i) What is the number of vibrational modes for acetylene and how does it compare with
the possible vibrational modes shown on figure 2? Explain.
ii) Explain what transitions the two acetylene bands correspond to.
b) The fundamental frequencies (wavenumbers) of H2O(g) are ~1  3651.1 cm-1, ~2 
1594.7 cm-1 and ~3  3755.9 cm-1. Predict the values of the following combination
frequencies (wavenumbers) neglecting the anharmonicity factor: (0,0,0) -> (0,1,1),
(0,2,1), (1, 0,1), (1,1,1), (2,0,1) and (2,1,1) (svar: wavenumber for (0,0,0) -> (0,2,1) is
6945.3 cm-1)
6) UV/Vis luminescence (sbr. fyrirlestur 05.04.11)
íslenskt:
Ljómun (I) frá lífrænu litarefni í kjölfar ljósörvunar með LASER geisla blossa, var mæld sem fall
af tíma (t / sek). Í ljós kom að ljómunin (I) minnkaði með tíma frá upphafsgildi (I0) skv.
I = I0 exp(-0,2 t)
a) hver er líftími ljómunarinnar og
b) hvort er um að ræða flúrljómun eða fosfórljómun? Rökstyðjið.
English:
Luminescence from an organic dye following light excitation by use of a laser pulse was measured
as a function of time (t / sec). The luminescence intensity (I) was found to decrease with time from
the initial intensity (I0) as,
I = I0 exp(-0,2 t)
a) what is the luminescence lifetime and
b) argue whether the luminescence is fluorescence or phosphorescence.
7) Ozon shielding (sbr. fyrirlestur 05.04.11)
Enclosed (see below) are absorption spectra of ozon in the near-UV and Visible regions plotted as
cross sections (/cm2) as a function of wavelength. Cross sections () are closely related to molar
absorptivity (molar extinction coefficients; see p: 711 in RC) () and defined by the relationship:
I/I0 = exp(- *l*c) = 10-A
where I0 and I are the intensities before and after passing through a sample,  is the cross section
in cm2, l is the length of the absorbing sample in cm and c is the concentration of the sample in
molecules per cm3. A is the absorbance. In reality ozon is to be found everywhere in the
atmosphere to some extend, but its maximum concentration is close to 20 km above sea level (see
for example HERE and HERE). Therefore, it is customary to define the ozon layer close to that
altitude. The ozon concentration, in the atmosphere, on the other hand, is so little that if it was all
collected together around the earth at 1 bar pressure and room temperature, it would only form
about 3 mm thick gas layer! Judging from the near-UV absorption spectrum of ozon, the cross
section is about 1.15 x 10-17 cm2 for 255 nm but about 0.04x10-17 cm2 for 210 nm.
Determine a) the absorbance (A) and b) % transmission for i) 255 nm, ii) 210 nm and iii) 600 nm
sun radiations by assuming that he ozon layer equals 3 mm thick gas layer of ozon at 25oC and 1
bar pressure. (svar: a, ii) : A = 1.27)
8) Greenhouse effects (sbr. fyrirlestur 05.04.11)
Enclosed is a figure showing total absorption of the solar radiation in the upper atmosphere (top)
as well as absorption spectra of (O2 + O3), CO2 and H2O (below). The two major absorption peaks
of CO2 in the infrared region are marked and indicated. Determine its wavenumber values (cm-1)
and search for relevant data (for example in a relevant database) to determine what energy
transitions these absorption peaks correspond to: specify the type of transitions involved as well
as relevant quantum numbers and quantum number changes.
9) NMR (sbr. fyrirlestrar 05-07.04.11)
a) Describe and explain how electrons in atoms and chemical bonds, close to nuclei, can affect the
energy splitting of different nuclear spin states for chemical samples in a magnetic field.
b) What magnetic field strength (Tesla) is needed to cause absorption of 19F nuclei in a 60 MHz
NMR equipment?
c) What is the population ratio (N(-1/2)/N(+1/2)) for the two spin states of 19F nuclei in (b) at room
temperature (above).
10) Röntgengreining kristalla / X-ray diffraction:
a) The layers of atoms in a crystal are separated by 325 pm. At what angle in a diffractometer will
diffraction occur using:
i)
– molybdenum K X rays ( = 70.8 pm)
ii)
– copper K X rays ( = 154 pm) (answer:  = 13.7o)
b) The smallest observed diffraction angle of silver taken with copper K X rays ( = 154 pm) is
19.076o. This angle is associated with the (111) plane in the cubic close-packed structure of silver.
i)
Determine the value of the unit cell length (answer: 408.6 pm)
ii)
If density,  = 10.500 g cm-3, calculate the number of atoms in the unit cell and assign the
cubic cell (sc, bcc or fcc).
11) Dæmi 14.9 bls. 779 í LMS* (sbr. fyrirlestrar 05-07.04.11)
* LMS: "Physical Chemistry" by K.J.Laidler, J.H. Meiser and B.C.Sanctury, 4th ed., 2003