Chemistry I -- Final Exam (總分110分)
2013/1/18
Periodic Table of Elements
1
H
1.0
3
Li
6.9
11
Na
23.0
19
K
39.1
37
Rb
85.5
55
Cs
132.9
4
Be
9.0
12
Mg
24.3
20
21
22
23
24
25
26
27
28
29
30
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4
38
39
40
41
42
43
44
45
46
47
48
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4
56
57
72
73
74
75
76
77
78
79
80
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
137.3 138.9 178.5 181.0 183.8 186.2 190.2 192.2 195.1 197.0 200.6
5
B
10.8
13
Al
27.0
31
Ga
69.7
49
In
114.8
81
Tl
204.4
6
C
12.0
14
Si
28.1
32
Ge
72.6
50
Sn
118.7
82
Pb
207.2
7
N
14.0
15
P
31.0
33
As
74.9
51
Sb
121.8
83
Bi
209.0
8
O
16.0
16
S
32.1
34
Se
79.0
52
Te
127.6
84
Po
(209)
Formulae
PV = nRT =
 Z 2 hR
En 
, n  1, 2, 3, 
n2
m e4
R  e3 2  3.29  1015 Hz
8h  0
2
𝑛𝑀𝑣𝑟𝑚𝑠
3
2

n 
 P  a    (V  nb)  nRT
V  

3
2
Mv
 M  2 2  2 RT
f  v   4 
 ve
 2 RT 
Constants
R
= 8.314 J / mol K
1 atm = 760 Torr = 1.01x105 Pa
= 0.0821 L atm / K mol
c = 2.99108 m/s
= 8.314 L kPa / K mol
h = 6.6310-34 Js
Mass of e , me = 9.10939  10 -31 kg
Avagadro number, NA =
6.022 x 1023
k = 1.3810-23 J/K
9
F
19.0
17
Cl
35.5
35
Br
79.9
53
I
126.9
85
At
(210)
2
He
4.0
10
Ne
20.2
18
Ar
40.0
36
Kr
83.8
54
Xe
131.3
86
Rn
(222)
(1 – 7) Choose the correct answer. (14 points, 2 points each)
1. Which one of the following will have the highest boiling point?
(1) CH4 (2) SiH4 (3) Ar (4) H2S (5) NH3
2. Which one of the following will have the highest viscosity?
(1) ethanol (2) water (3) diethyl ether (4) glycerol
(5) benzene
3. Which one of the following will have the highest conductivity in the solid state?
(1) Ta (2) NaCl
(3) fructose
(4) quartz
(5) water
4. Which one of the following solids has the diamond-like crystal structure?
(1) ice (ignore H atoms) (2) NaCl
(3) quartz (SiO2) (4) zinc blend (ZnS)
(5) B
5. At a certain temperature, a metal’s crystal structure changes from face-centered cubic to
hexagonal close-packed. The density of the metal:
(1)
increases largely (2) increases slightly (3) stays about the same
(4)
decreases slightly (5) decreases largely
For 6 – 7, a hypothetical crystal is composed of cubic close-packed element A. All of the octahedral
holes are occupied by element B while the tetrahedral holes are occupied by element C.
6. The formula of this compound is:
(1) ABC
(2) ABC2
(3) AB2C2
(4) AB2C
(5) A2B2C
7. Assuming that all neighboring atoms are in close contact, fraction of the space occupied in the
crystal is:
(1) > 1 (2) < 0.74
(3) < 1 and > 0.74 (4) < 0.4
(5) = 0.74
8. (7 points)
(a) What does Trouton’s rule state? (3 points)
(b) Ethanol’s Svapo is high, 124 J∙K-1∙mol-1 while helium’s Svapo is low, 20 J∙K-1∙mol-1. Consider
intermolecular forces and explain why they are different from the value suggested by Trouton’s
rule. (4 points)
9. (10 points)
The density and the atomic radius of a primitive cubic element are 9.196 g∙cm-3 and 168 pm.
(i) How many atoms are in a unit cell? (2 points)
(ii) Calculate the length of the unit cell. (2 points)
(iii)Calculate the molar mass for this element. (4 points)
(iv) What is this element? (2 points)
10. (3 points) 1 Torr =
atm =
Pa =
bar
11. (8 points, 2 points each) One mole of neon gas effuses through a porous plug in 85.3 s. Calculate
the time required for the same amount of (a) CO2, (b) C2H4, (c) H2 and (d) NO2 to effuse under
the same conditions of pressure and temperature.
12. (7 points) Using (a) the ideal gas equation and (b) the van der Waals equation, calculate the
pressure exerted by 1.00 mol CO2(g) when confined in a volume of 0.500 L at T=298K. (4
points, 2 points each) (c) Do attractive or repulsive forces dominate? (3 points)
2
2
a = 3.640 L atm/mol , b = 0.04267 L/mol.
13. (4 points) The equation of state for real gases that do not obey the ideal gas law is usually written
in the following form
𝑃𝑉 = 𝑛𝑅𝑇 (1 +
𝐵
𝐶
+ 2 +⋯)
𝑉m 𝑉m
What is the name of this equation? (1 point) What is the name of the coefficients 𝐵 and 𝐶? (2
points) Are these coefficients temperature-dependent? (1 point)
14. (3 points) What is the Joule-Thomson effect?
15. (22 points) A certain amount of neon gas (Ne) in a cylinder of volume 2 L at 25℃ and 4 atm
undergoes the changes in two steps sequentially. The temperature and pressure of the
surroundings is 25℃ and 1 atm.
(a) Step 1: The gas expands isothermally and reversibly to 2 atm. Please calculate q, w,
∆𝑈, ∆𝑆 and ∆𝑆𝑠𝑢𝑟𝑟 of this step. (10 points, 2 points each)
(b) Step 2: After step 1, the gas is cooled at constant volume to 1 atm.
i.
Please calculate the final temperature T. (2 points)
ii.
Please calculate q, w, ∆𝑈, ∆𝑆 and ∆𝑆𝑠𝑢𝑟𝑟 of this step. (10 points, 2 points each)
16. (10 points) The sublimation temperature of CO2 is -78. 5 ℃. The enthalpy of sublimation at
that temperature is 26.1 kJ/mol. Please calculate q, w, ∆𝑈, ∆𝐻 and ∆𝑆 for the following change:
(2 points each)
2 mol CO2 (solid, 1 atm, -78. 5 ℃) →
2 mol CO2 (gas, 1 atm, 25 ℃)
Please treat CO2 as an ideal gas. Important hint: the volume of dry ice can be neglected in
comparison to the volume of gas.
17. (22 points) CO gas, used extensively in organic synthesis, can be obtained by reacting CO 2 gas
with graphite. Here are some important thermodynamic data at 298K:
(a) Please write the correct chemical equation to synthesis CO by reacting CO2 with graphite.
(4 points)
(b) Calculate Hro, Sro and Gro at 298K. (6 points, 2 points each)
(c) If Hro and Sro are constant at any temperature, please predict the temperature at which
this reaction will happen spontaneously. (4 points, 2 points each)
(d) In fact, Hro and Sro are dependent on temperature. Here, both CO and CO2 are treated as
ideal gases. Cp,m of graphite is 8.53 J ∙ K −1 ∙ mol−1 . Please calculate Hro and Sro at
500℃. (8 points, 4 points each)
Answer:
(1 – 7) Choose the correct answer. (10 points, 2 points each)
1. Which one of the following will have the highest boiling point?
(1) CH4 (2) SiH4 (3) Ar (4) H2S (5) NH3
2. Which one of the following will have the highest viscosity?
(1) ethanol
(2) water
(3) diethyl ether (4) glycerol
(5) benzene
3. Which one of the following will have the highest conductivity in the solid state?
(1) Ta
(2) NaCl (3) fructose
(4) quartz
(5) water
4. Which one of the following solids has the diamond-like crystal structure?
(1) ice (ignore H atoms)
(2) NaCl (3) quartz (SiO2) (4) zinc blend (ZnS) (5) B
5. At a certain temperature, a metal’s crystal structure changes from face-centered cubic to
hexagonal close-packed. The density of the metal:
(1) increases largely (2) increases slightly
(3) stays about the same
(4) decreases slightly (5) decreases largely
For 6 – 7, a hypothetical crystal is composed of cubic close-packed element A. All of the octahedral
holes are occupied by element B while the tetrahedral holes are occupied by element C.
6. The formula of this compound is:
(1) ABC (2) ABC2
(3) AB2C2
(4) AB2C (5) A2B2C
7. Assuming that all neighboring atoms are in close contact, fraction of the space occupied in the
crystal is:
(1) > 1
(2) < 0.74
(3) < 1 and > 0.74
(4) < 0.4
(5) = 0.74
8. (7 points)
(a) What does Trouton’s rule state? (3 points)
(b) Ethanol’s Svapo is high, 124 J∙K-1∙mol-1 while Helium’s Svapo is low, 20 J∙K-1∙mol-1. Consider
intermolecular forces and explain why they are different from the value suggested by Trouton’s
rule. (4 points)
Ans:
(a) Svapo of various liquids are close to 85 J∙K-1∙mol-1. (3 points)
(b) Ethanol’s Svapo is higher than 85 J∙K-1∙mol-1 because ethanol has hydrogen bonding which
provides more orderness in the liquid state. (2 points)
Helium’s Svapo is lower than 85 J∙K-1∙mol-1 because only weak London force (dispersion
force) holds He atoms together in a less ordered way in its liquid state. (2 points)
9. (10 points)
The density and the atomic radius of a primitive cubic element are 9.196 g∙cm-3 and 168 pm.
(v) How many atoms are in a unit cell? (2 points)
(vi) Calculate the length of the unit cell. (2 points)
(vii) Calculate the molar mass for this element. (4 points)
(viii) What is this element? (2 points)
Ans:
(a) Primitive cubic, 1 (2 points)
(b) The length of the unit cell is twice of the sphere radius (atomic radius), 336 pm. (2 points)
(c) The molar mass can be calculated from M = d x (8 r3 x NA)
(correct formula, 2 points)
M = 9.196 g∙cm-3 x 8 x (168 pm)3 x 6.02 x 1023 mol-1
= 210 g∙mol-1
(correct answer, 2 points)
(d) Bi, Po, or At (只要答對任一元素即可得 2 points)
d = m / a3
= M / (8 r3 x NA )
50
Solution (10) (3 points, 1 point each)
1 Torr = 1.32x10-2 atm = 1.33x102 Pa = 1.33 x10-2 bar
Solution (11) (8 points, 2 points each)
The rate of effusion is inversely proportional to the square root of the molar mass.
rate1
1
=
rate2
M1
1
M2
=
M2
M1
The rate will be equal to the number of molecules N that effuse in a given interval and will be the
same for given condition to the selected gas. Atomic mass of neon at 1mol is 20.2 g/mol.
N
-1
Time = 85.3 = 20.2g × mol
N
Time
M1
85.35
\Time = 85.3´
M1
20.2g × mol-1
Thus, if molar mass is known, effusion time can be calculated.
(a) CO2, molar mass: 44.0 g/mol
TimeCO = 85.3´
2
44.0g × mol-1
= 125.9s (2 points)
20.2g × mol-1
(or 126 s)
(b) C2H4, molar mass: 28.0 g/mol
TimeC H
2
4
28.0g × mol-1
= 85.3´
= 100.4s (2 points)
20.2g × mol-1
(or 100 s)
(c) H2, molar mass: 2.00 g/mol
TimeH = 85.3´
2
2.00g × mol-1
= 26.8s
20.2g × mol-1
(2 points)
(or 27 s)
(d) NO2, molar mass: 46.0 g/mol
TimeNO = 85.3´
2
46.0g × mol-1
= 128.7s
20.2g × mol-1
(2 points)
(or 129 s)
Solution (12) (7 points)
(a) pressures are calculated very simply from the ideal gas law:
nRT (1.00 mol)(0.082 06 L  atm  K 1  mol1 )(298 K)
P

V
V
Calculating for the volumes requested, we obtain P = 48.90 atm
(b)Using the van der Waals equation:

an2 
P


(V  nb)  nRT
V2 

 nRT   an2 
P
 2 
 V  nb   V 
(2 points)
 (1.00 mol)(0.082 06 L  atm  K 1  mol 1 )(298 K) 


V  (1.00 mol)(0.04267 L  mol 1 )


 (3.640 L2 . atm . mol 2 )(1.00 2 ) 


V2


= 53.47 atm – 14.56 atm = 38.91 atm.
(2 points)
(c) (i)The effect of the b term in the van der Waals equation was to increase P from 48.9 to 53.47
atm; the effect of a was to decrease P by 14.56 atm. The a term in this case is more influential
than the b term. That is, attractive forces dominate.
(ii) Because P(ideal gas) > P(real gas), attractive forces dominates.
答案(i)或(ii)皆對，只要寫出一種答案即可得 3 points
Solution (13) (4 points)
(i) Virial equation;
(1 point)
(ii) 𝐵 is the second virial coefficient and 𝐶 is the third virial coefficient;
(iii)Yes, 𝐵 and 𝐶 are temperature-dependent.
(2 points)
(1 point)
Solution (14) (3 points)
The Joule-Thomson describes the fact that real gases cool down when they expand. This effect can
be used to liquefy gases.
Solution of (15) (22 points)
(a) Because the process is isothermal, U  0
(2 points)
For a reversible process,
w  nRT ln
V2
V1
n is obtained from the ideal gas law:
n=
𝑃𝑉
4×2
=
= 0.327 𝑚𝑜𝑙
𝑅𝑇 0.082 × 298
4
𝑤 = −0.327 ∗ 8.314 ∗ 298 ∗ ln 2 = −562.2 J (2 points)
Because U  0 ,
∆𝑆 =
𝑞𝑟𝑒𝑣
𝑇
=
q = -w = +562.2 J (2 points)
562.2
298
= 1.89 J ∙ K −1 or
4
∆𝑆 = 0.327 ∗ 8.314 ∗ ln = 1.89 J ∙ K −1
2
(2 points) 兩種算法皆正
確！
−𝑞𝑟𝑒𝑣
∆𝑆surr =
𝑇
=
−562.2
298
= −1.89 J ∙ K −1 or
Because it is a reversible process, ∆𝑆surr = −∆𝑆 = −1.89 J ∙ K −1
(2 points) 兩種算法皆正確！
(b) According to the ideal gas law, the final temperature of Path 2 is
T=
𝑃𝑉
𝑛𝑅
1×4
=
0.327×0.082
= 149 𝐾 (2 points)
Because the volume does not change in Path 2, w=0
(2 points)
3
∆U = n𝐶v,m ∆𝑇 = 0.327 × 2 × 8.314 × (149 − 298) = −607.6 J (2 points)
q = ∆U − w = ∆U = −607.6 J (2 points)
149
3
149
∆S = n𝐶v,m ln 298 = 0.327 × 2 × 8.314 × ln 298 = −2.83 J ∙ K
∆𝑆surr =
−𝑞𝑟𝑒𝑣
𝑇
=
607.6
298
= −2.04 J ∙ K −1
−1
(2 points)
Solution of (16) (10 points)
𝑇𝑖 = 273 − 78.5 = 194.5 𝐾
𝑇𝑓 = 273 + 25 = 298 𝐾
𝑉𝑖 ~ 0 L
𝑉𝑓 =
𝑛𝑅𝑇
𝑃
=
2×0.082×298
1
= 48.9 L
∆𝐻 = ∆𝐻𝑠𝑢𝑏𝑙𝑖𝑚𝑎𝑡𝑖𝑜𝑛 + 𝑛𝐶p,m ∆𝑇
7
= 26100 + 2 × 2 × 8.314 × (298 − 194.5) = 32123.5 𝐽
q = ∆𝐻 = 32123.5 J
w
(2 points)
= −𝑃𝑒𝑥 × ∆𝑉 = −1 × (48.9 − 0) = −48.9 L ∙ atm
= −48.9 × 101.3 = −4953.6 J
(2 points)
∆U = q + w = 32123.5 − 4953.6 = 27170 J
(2 points)
(2 points)
(2 points)
𝑇𝑓
∆𝑆 = ∆𝑆𝑠𝑢𝑏𝑙𝑖𝑚𝑎𝑡𝑖𝑜𝑛 + 𝑛𝐶p,m × ln ( )
𝑇𝑖
=
26100
194.5
7
2
298
= 159 𝐽 ∙ 𝐾 −1
)
194.5
+ 2 × × 8.314 × ln (
(2 points)
Solution of (17) (22 points)
(a) CO2(g) + C(graphite)  2CO(g)
(4 points)
(b) Hro = -110.53*2-(-393.51) = +172.45 kJ/mol
(2 points)
o
Sr = 197.66*2 – 213.79-5.74 = +175.79 J/mol/K
(2 points)
o
o
o
Gr = Hr - T*Sr = 172.45 – 298*175.79/1000 = +120.06 kJ/mol
(2 points)
(c) Gro = Hro - T*Sro = 172.45 – T *175.79/1000 <=0
T >= 172.45/175.79*1000
T >= 981 K (4 points)
(d) Hro(500℃) = Hro(298K) + (3.5R*2 – 8.53 – 3.5R)*(500+273-298)/1000
= +172.45 + (3.5*8.314-8.53) *(500+273-298)/1000
= +182.22 kJ/mol
(4 points)
Sro(500℃)
= Sro(298K) + (3.5R*2 – 8.53 – 3.5R)*ln((500+273)/298)
= +175.79 + (3.5*8.314-8.53) *ln((500+273)/298)
= +195.40 J/mol
(4 points)