Chemistry
2nd Midterm Exam
2010/12/10
Periodic Table of Elements
Formulae
E = h
 = (1/n12 – 1/n22),  = 3.29x1015 Hz
En = n2h2/8mL2
1/2mev2 = h - 
Constants
R = 8.314 J / mol K
= 0.0821 L atm / K mol
= 8.314 L kPa / K mol
Mass of e , me = 9.109  10 -31 kg
Nature Exponential Base e = 2.71828
 2 2

  V ( x, y , z )  ( x , y , z )  E  ( x , y , z )

 2m

H  E
1 atm = 760 Torr = 1.01x105 Pa
c = 2.99108 m/s
h = 6.6310-34 Js
Pa
=
Torr
=
bar
=
psi (pound/inch2)
1.
(4%) 1 atm =
2.
Using (a) the ideal gas equation (3%) and (b) the van der Waals equation (3%),
calculate the pressure exerted by 1.00 mol CO2(g) when confined in a volume of
2
0.500 L at T=298K. (c) Do attractive or repulsive forces dominate? a = 3.640 L
-2
-1
atm mol , b = 0.04267 L mol . (4%)
3.
(a) Calculate the root mean square speeds of N2 at 300 and 1000 K, respectively.
(4%) (b) Roughly mark them in the graph of Maxwell distribution of speeds of N2
at these temperatures. (4%) (c) What happens to the fraction of molecules having
the most probable speed as T increases? (2%)
4.
Consider the Haber process for the synthesis of ammonia. (a) What volume of
hydrogen at 15.00 atm and 350 ℃ must be supplied to produce 1000 kg of NH3?
(6%) (b) What volume of H2 is needed in part (a) if it is supplied at 376 atm and
250 ℃? (3%)
5.
(a) The ionic radii of S2- and Zn2+ are 184.7 pm and 73.2 pm respectively. Which
crystal structure drawn below may be the ionic compound of S2- and Zn2+. (2%)
Please explain your answer. (4%)
Crystal A
Crystal B
Crystal C
(b) What are the coordination numbers of Zn2+ and S2- ions, respectively, in this
crystal? (4%)
6.
Ice (H2O), glucose, and benzophenone (C6H5COC6H5) are examples of compounds
that form solids. The structures of glucose and benzophenone are given here. (a)
What types of forces hold these molecules in a solid? (8 %) (b) Place the solids in
order of increasing melting points (3 %).
7.
Which of the following is the strongest intermolecular force between molecules with
similar mass? (2%)
(A) London (B) ion-dipole (C) hydrogen bonding (D) dipole-dipole (stationary)
2
8.
Which of the following has the lowest boiling point? (2%)
(A) HF (B) HI (C) CH4 (D) PH3 (E) H2S
9.
What is the coordination number of cesium in CsCl? The ionic radii of Cs+ and Clare 170 and 181 pm, respectively. (2%)
(A) 12 (B) 8 (C) 2 (D) 4.
10. What is responsible for capillary action, a property of liquids? (2%)
(A) surface tension and viscosity
(B) adhesion forces and viscosity
(C) surface tension only
(D) adhesion forces and cohesive forces
(E) viscosity only
11. What is the simplest formula of a solid containing A, B, and C atoms in a cubic
lattice in which the A atoms occupy the corners, the B atoms the body-center
position, and the C atoms the faces of the unit cell? (2%)
(A) ABC3 (B) A8BC6 (C) ABC6 (D) A4BC3 (E)ABC
12. Find a best match of materials (1-11) and their properties (a-k). Please write your
answer carefully. (1% each, 11% total)
1. Solder
2. Graphene
3. Sapphire
4. Asbestos
5. Carborundum
6. Aerogel
7. Germanium doped with phosphors
8. MgB2
9. Y2O2S
10. Ferrofluid
11. CdSe/ZnS quantum dots
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
A network solid of Al and O that is hard and rigid
A N-type Semiconductors
A superconductor
A viscous, oily liquid contains finely powdered magnetite
A network solid material of Si and C for use in industrial machinery, tools,
abrasives
A fibrous material with formula of Ca2Mg5(Si4O11)2(OH)2
An alloy of tin and lead
A nanomaterial that can emit light with controlled wavelength
A luminescent material (as red phosphor used for TV monitor)
A ceramic with the density close to air and low thermoconductivity
A single atomic layer of graphite
3
13. Iron pyrite is known as Fool’s Gold because of its resemblances to gold metal. The
unit cell of iron pyrite is shown in the figure below. The cell constant a = b = c =
5.42 Å ,  =  =  = 90
Answer the following questions:
(a) From the list of some Bravais lattices below, please select the most correct
crystal system for Iron pyrite. (2%)
Bravais lattices: tetragonal system, cubic system, orthorhombic system,
monoclinc system
(b) How many iron and sulfur atoms are located in a unit cell. (All sulfur atoms
are located inside the unit cell.) (4%)
(c) Write the formula of iron pyrite. (3%)
(d) Assume the oxidation number of S is 2-, what is the oxidation number of Fe
(2%)
(e) If the density of gold is 19.28 g/cm3, can it be distinguished with the density
of iron pyrite? Calculate the density of pyrite to answer this question. (4%)
Fe
Fe
S
Fe
Fe
S Fe
Fe
S
S
S
Fe
Fe
Fe S
Fe
Fe
Fe
S
S
S
S
Fe
Fe
S
Fe
S
Fe
S
Fe
Fe
Fe
Tilted view
S
S
Fe
Fe
S
Fe
Fe
side view
14. A PbSe quantum dot sample is prepared by mixing 6.0 mL of 0.0012 M Pb(NO3)3(aq)
with 6.0 mL of 0.0012 M K2Se(aq). Each quantum dot contains an average of 165
formula units of PbSe.
(a) What is the concentration of the quantum dots in the final suspension? (3%)
(b) What is the average molar mass of the quantum dot? (3%)
(c) The suspension fluoresces at 601 nm. What is the diameter of the quantum dots it
contains? (4%)
4
Ans:
1. 1 atm = 101.325
2.
Pa = 760 Torr =
1.01325
bar =
14.7
psi
(a) (3%) pressures are calculated very simply from the ideal gas law:
P
nRT (1.00 mol)(0.082 06 L  atm  K 1  mol1 )(298 K)

V
V
Calculating for the volumes requested, we obtain P = 48.90 atm
(b)Using the van der Waals equation:

an 2 
 P  2 (V  nb)  nRT
V 

 nRT   an2 
P
 2 
 V  nb   V 
 (1.00 mol)(0.082 06 L  atm  K 1  mol1 )(298 K) 


V  (1.00 mol)(0.04267 L  mol1 )


2
2
2
 (3.640 L . atm . mol )(1.00 ) 


V2


= 53.47 atm – 14.56 atm = 38.91 atm.(3%)
(c)(4%) 以下 2 種答案任一答法皆正確
(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.
5
3.
(a) 517 and 944 m/s.(4%)
(b)
(4%)
(c) The percentage of molecules having the most probable speed decreases as the
temperature is raised (the distribution spreads out).(2%)
4. (a) The number of moles of H 2 needed will be 1.5 times the amount of NH3
produced, as seen from the balanced equation (3%)
1
2
N 2 (g)  23 H 2 (g)  NH 3 (g)
or
N 2 (g)  3 H 2 (g)  2 NH 3 (g)
Once the number of moles is known, the volume can be obtained from the ideal
gas law.(3%)
  3 mol H 2  (1.0  103 kg)(103 g  kg 1 ) 
 
 RT

1
3
2
mol
NH
nH2 RT
17.03
g

mol
2 nNH3 RT
3 


V 


P
P
P
6

10 g
 3 
(0.082 06 L  atm  K 1  mol1 )(623 K)
 
1 
 2   17.03 g  mol 

15.00 atm


 3.0  105 L
6
(b) (3%) The ideal gas equation
PV
PV
PV
PV
1 1
 2 2 simplifies to 1 1  2 2
n1 RT1 n2 RT2
T1
T2
because R and n are constant for this problem.
(15.00 atm)(3.0  105 L) (376 atm)V2

623 K
(523 K)
V2  1.0  104 L
5.
(a) Crystal B (2%);
r(Zn2+)/r(S2-)=73.2/184.7 =0.396 <0.4
Zn2+ should be in the tetrahedral hole of S2- structure
The chemical formula of ionic compound of Zn2+ and S2-) should be ZnS.
So the crystal B fulfill the two above conditions (4%)
(b) 4 (Zn2+) (2%)
4 (S2-) (2%)
6. (a) Ice: dipole-dipole, London forces, and hydrogen bonding (3%)
Glucose: dipole-dipole, London forces, and hydrogen bonding (3%)
Benzophenone: dipole-dipole, London forces (2 points)
(b) Ice (0 oC) < benzophenone (48 oC) < glucose (150 oC) (3%)
7. C
12.
題號
答案
8. C
1
g
9.B
2
k
3
a
10.D
4
f
11. A
5
e
6
j
7
b
8
c
9
i
10
d
11
h
13. Answer
(a) (2%) cubic system
(b) (4%) Fe: 4 atoms; S: 8 atoms in an unit cell
(c) (3%) FeS2
(d) (2%) Fe4+
(e) (4%) Calculated density = 5.01 g/cm3.
[(55.84*4+32.06*8)/6.02x1023]/(5.42x10-8)3
The difference of density between iron pyrite and gold is significant, which can
be distinguished easily.
7
14. Answer
Part (c) needs a formula derived from the particle in a box model to calculate the
diameter
(a) (3%) [PbSe] = 0.006 L*0.0012M / 0.012L = 0.0006 M
[QD(165 PbSe)] = 0.0006/165 = 3.64×10-6 M
(b) (3%) Molar mass of QD(165 PbSe) = 165*(207+79) = 47190 g/mol
(c) (4%) hv=ΔE = E2-E1 = En = 22h2/8meD2 - 12h2/8meD2 =3h2/8meD2
hc/λ= 3h2/8meD2
D = (3hλ/8mec)0.5= [(36.6310-3460110-9)/(89.10910-312.99108)]0.5
=7.410-10 m = 0.74 nm
8