ANSWERS TO EXERCISES AND REVISION QUESTIONS20B
(i)
E3.00^10(^8) 3.99 10 – 13
700 10 –^9 171 kJ mol–^1
(ii)
E3.00^10
(^8) 3.99 10 – 13
560 10 –^9 214 kJ mol–^1
20C
EhNA, therefore
E
hNA100 kJ mol
– (^1) /3.99 10 – (^13) kJ s mol– 1
2.51 1014 Hz (i.e. s–^1 )
20D
(i)If 10% of the light is absorbed, 90% of the light is
transmitted, i.e.
(
Io
)
(^100) 1.11
I 90
absorbanceAlog(II^0 )log(1.11)0.0453
(ii)
(
Io
)^10 3.001.0^103 .·.(
I
II)0.0010
o
therefore, the percentage of light transmitted is
100 I 100 0.00100.10%
I 0
20E
The higher temperatures mean that many hydrogen
atoms are present in the n2 level.
20F
The energy gap between the n 4 and n 1 level is
1228 kJ mol–^1. Using the equation EhNA
E
hNA1228/3.99^10
– (^13) 3.08 1015 Hz
20G
The rate at which the excited states is produced can be
controlled like any other chemical reaction.
(i)Lowering the temperature of the reaction mixture will
slow down the reaction and decrease the intensity of the
emitted light.
(ii)Increasing temperature increases the emission
intensity.
20H
(i)red
(ii)blue green
(iii)colourless (absorptions are outside the visible
region).
20I
Only absorptions above 400 nm are relevant. The
absorptions between 400 nm and 520 nm peak at about
430 nm, suggesting that the solution would be yellow
(this is the observed colour).
20J
(i)propionic acid and ethanal
(ii)propionic acid and butanol
(iii)all three.
20K
Its simplest formula is C 4 H 11 N; infrared suggests –NH 2
present;m(C 4 H 11 N ) 73 u (as confirmed by mass
spectrum). Possible structures include (CH 3 ) 2 CHCH 2 NH 2
and CH 3 (CH 2 ) 3 NH 2.
20L
(i)3 (protons from CH 3 , CH 2 and CHO)
(ii)2 (protons from CH 3 and CH).
20M
(i)
(ii)1:6:15:20:15:6:1
20N
(i)(b)
(ii)(a)
(iii)(c)
20O
(i)A cb, so that cA / (^) b1.1/(1.1 104
0.00010)1.0 mol m–^3 , i.e. 0.0010 mol dm–^3.
(ii)(a)cA / (^) b0.010/(1000 0.010)0.0010
mol m–^3
(b) 0.00025 mol m–^3.
(iii)c2.0 mol m–^3 ; (^) A /cb0.40/(2 1.0
10 –^2 )20 mol–^1 m^2.
20P
For 2-nitrophenol, A 2 2 c 2 b(where the subscript ‘2’
stands for 2-nitrophenol).
For 4-nitrophenol, A 4 4 c 4 b(where the subscript ‘4’
stands for 4-nitrophenol).
4 is given in Example 20.5 as 9.6 102 mol–^1 m^2.
Therefore, the total absorbance at 313 nm is
2 c 2 b 4 c 4 b(1.2 102 10 –^3 10 –^2 )
(9.6 102 10 –^3 10 –^2 )
0.00120.00960.0108
20Q
(i)The concentrations are converted to mol m–^3 by
multiplying by 1000. The plot is:
(ii)Slope b(0.500.00)/(17.5 10 –^2 0.0)
2.9 mol–^1 m^3. Given that b 10 –^2 m,
(^) 526 nmslope/b2.9 102 mol–^1 m^2.
Revision questions
20.1 (i)c/ 3.00 108 /2 10 –^7
1.5 1015 Hz, (ii)6.0 1014 Hz,
(iii)3.00 1013 Hz.
20.2The missing words are emission (or fluorescence),
absorbance, chemiluminescence and molar absorption
coefficient.
20.3c/ 3.00 108 /1000 10 –^9
3.00 1014 Hz
EhNA3.99 10 –^13 3.00 1014
120 kJ mol–^1.
20.4EhNA;E/hNA30/3.99 10 –^13
7.5 1013 Hz. Figure 20.1 shows that this
corresponds to the infrared region of the
electromagnetic spectrum.
20.5 (i)Percentage transmitted light 100 84
16%, therefore
16%^100 II
0
By rearrangement,
I (^0) (^100) 6.25
I 16
Alog(6.25)0.796
(ii)1.00 10 –^5 mol dm–^3 1.00 10 –^2 mol M–^3
A cb 1000 1.00 10 –^2 10 –^2 0.100
I (^0) 10 0.1001.26
I
II
I 0 0.79,
100
I 0 79%
To give A 0.010, dilute by a factor of 10.
20.6E/hNA494/3.99 10 –^13
1.24 1015 Hz.
c/3.0 108 /1.24 1015
2.4 10 –^7 m (240 nm)
20.7 (i)E1310/50^2 0.524 kJ mol–^1.
(ii)Eforn 3 level is 1310/3^2 145.5 kJ mol–^1.
Therefore,
E0.524() 145.4 144.9 kJ mol–^1
E/hNA144.9/3.99 10 –^13 3.63 1014 Hz
20.8 (i)Dark lines show wavelengths at which elements
in the sun’s cooler outer region absorb light. (Some lines
are also produced by absorbing elements in the Earth’s
atmosphere.)
(ii)
There are three possible transitions in absorption, and
six in emission. (The smaller number in absorption
follows from the fact that all molecules start at Eo.)
Generalizing, an emission spectrum contains more lines
than the corresponding absorption spectrum because
the excited state does not have to fall to the ground state
in ‘one go’.
20.9 (i)It is likely that the absorption spectrum will
contain a single strong peak between 490 and 530 nm.
(ii)The absorptions below 400 nm may be ignored. The
colour cheese allows us to predict that the solution will
absorb light at about 430 nm in wavelength and that it
will therefore appear yellow.
(iii)The mixture will absorb virtually all the colours of
white light. Very little light will be reflected and the paint
will appear black.
20.10The plot is very similar to Fig. 22.2 (see Chapter
22). Concentration of ozone cis
cP 100.0^102
RT8.3145 298 4.04^10
– (^2) mol m– 3
Therefore,
A(254nm) cb 369 4.04 10 –^2 0.1001.49
20.11
37.5% C and 12.5% H leads to the ratio of H to C as 4:1.
Infrared spectrum contains a broad band at about 3400
450
TMS
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