NAME:____________________________
Spring 2007
(b)
(c)
(d)
(e)
____/ 52 marks
Chemistry 2000 Midterm #1A
INSTRUCTIONS:
1.
(a)
Student Number:________________________
1) Please read over the test carefully before beginning. You should have 5
pages of questions, a sheet of graph paper and a data sheet.
2) If your work is not legible, it will be given a mark of zero.
3) Clearly label any graphs with the question number and a brief title. Axes
should be clearly labeled (including units).
4) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
5) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
6) You may use a calculator.
7) You have 90 minutes to complete this test.
[11 marks]
Draw the phase diagram of water, indicating the triple point at 0.0098 C and 0.006 atm, and
the critical point at 374.4 oC and 217.7 atm. (6 marks)
On your phase diagram, draw and label a line between two points over which the melting
process occurs under constant pressure. (1 mark)
On your phase diagram, draw and label a line between two points over which the boiling
process occurs under constant pressure. (1 mark)
On your phase diagram, draw and label between two points over which the sublimation
process occurs under constant pressure. (1 mark)
On your phase diagram, draw and label a line that shows that you can induce melting by
changing the pressure while keeping temperature constant. Explain why this phenomenon is
virtually unique to water.(2 marks)
o
Since the density of ice is less than water as you increase pressure the melting point
increases, hence the negative slope. Therefore as you increase the pressure at a
temperature near the melting point melting can be induced.
Critical Point
(374 oC, 217 atm)
b) melting
e) melting
P (atm)
c) boiling
Solid
Liquid
Gas
d) Sublimation
Triple Point
(0.0098 K, 0.006 atm)
T (oC)
NAME:____________________________
2.
(a)
Student Number:________________________
[9 marks]
Explain why the order of boiling points for the hydrogen halides is HCl < HBr < HI given
that the electronegativity of Cl > Br > I.(3 marks)
The bond dipole increases in the order HI < HBr < HCl which is opposite to the trend in
boiling point. Therefore the dispersion forces, dipole-induced dipole and induce-dipoleinduced-dipole forces, must dominate the behavior. Since the dispersion forces increase
with increasing number of electrons, their contribution to the total energy of interaction
increases as HCl < HBr< HI. As the IMF’s increase it takes more energy for molecules to
escape the solution state; therefore, the boiling points increase in the order HCl HBr and
HI.
(b)
Draw the structures of CH3CH2OH and CH3OCH3, and explain why their boiling points
differ by nearly 80o C.(3 marks)
H
H
C
H
C
H
O
H H
H
H
C
H
H
C
O
H
H
Both CH3CH2OH and CH3OCH3 are molecules that have a bend configuration at the
oxygen, and are therefore both polar. The difference between them is that CH3CH2OH, has
a H attached directly to O, while CH3OCH3 does not, meaning that CH3CH2OH can
undergo hydrogen bonding interactions as O has sufficient size/electronegativity. Only N, O,
F are electronegative enough and small enough to undergo H-bonding interactions. As a
result the intermolecular forces in CH3CH2OH will be greater than CH3OCH3 thus
requiring more energy for its molecules to escape the liquid state leading to a much larger
boiling point than CH3OCH3.
(c)
Predict which has the larger vapour pressure, CHCl3 or CCl4, and give an explanation for
your answer. (3 marks)
CHCl3 has a permanent dipole while CCl4 does not. The dipolar interactions in CHCl3 are
stronger than the induced-dipole-induced-dipole forces of CCl4. Therefore the stronger
intermolecular forces in CHCl3 hold the molecules more tightly in the liquid phase over
CCl4 , hence reducing vapor pressure when compared to CCl4.
NAME:____________________________
3.
(a)
Student Number:________________________
An unknown organic compound was dissolved in water to make a solution with a melting
point of -0.5120 oC. Assume that the unknown organic compound is a non-dissociating and
non-volatile solute.
[13 marks]
Calculate the molality of the solution. (2 marks)
i=1
m = ∆T/Kf = -0.5120oC/(-1.860 oC kg/mol) = 0.2753 m
(b)
Calculate the molecular mass of the compound given that the solution was prepared using
5.000 litres of water (density 0.997 g/mL) and 176.1 g of the compound. (2 marks)
Moles of compound = (0.2753 mol/Kg)*(0.9970 kg/L)*(5.000 L) = 1.372 mol
Molecular mass of compound = (176.1 g)/ (1.372 mol) = 128.3 g/mol
(c)
Calculate the concentration of this solution in wt % and ppm. (2 marks)
Wt solute *100%/ Wt solution = 0.1761 *100%/ (0.1761 + 5.000*0.9970)
= 3.412 %
Wt solute *106/ Wt solution = 0.1761 *106/ (0.1761 + 5.000*0.9970)
= 34,120 ppm
(d)
The solution is heated up to 100 oC under ambient pressure. Calculate the vapour pressure
of water above the solution (in Pa). (6 marks)
XH20 = moles(H2O)/( moles(H2O)+ mol cpd)
moles(H2O)=mass(H2O) /MH2O
= VH2O*density/MH2O
= (5.000 L)*(0.997 kg/L)*(1000 g/kg)/(18.02 g/mol) = 276.6 mol
XH20 =276.6/ (276.6+ 1.372) = 0.9951
P = XH2OPo = 0.9951*101325 = 100800 Pa= 101000 Pa
[1 mark for sig. figs and units]
NAME:____________________________
4.
(a)
Student Number:________________________
The initial rates of the reaction of 2 A + B → C were measured at 25 oC, changing the
initial concentrations of A and B as tabulated below.
[11 marks]
Experiment
Initial [A] (M)
Initial [B] (M)
Initial rate (M/s)
1
0.15
0.25
0.152
2
0.15
0.60
0.875
3
0.25
0.25
0.253
4
0.40
0.25
0.405
Determine the order of reaction for A and B. (5 marks)
Rate = k*[A]x * [B]y
Rate1/Rate2 = (k [0.15 M] x [0.25 M] y)/(k [0.15 M] x [0.60 M] y) = 0.152 / 0.875
(0.25/0.60) y = 0.152 / 0.875
(0.42) y = 0.174
yln(0.42)=ln(0.174)
y = ln(0.174) / ln(0.42) = 2
Rate2/Rate3 = (k [0.15 M] x [0.60 M] 2)/(k [0.25 M] x [0.25 M] 2) = 0.875/ 0.253
(0.15/0.25) x = (0.875/ 0.253) [0.25 M] 2 /[0.60 M] 2
(0.60) x = 0.600
xln(0.60)=ln(0.60)
x = ln(0.60 / ln(0.60) = 1
(b)
Write the rate expression for this reaction. (1 mark)
Rate = k*[A] * [B]2
(c)
Calculate the rate constant for this reaction. (2 marks)
k= Rate/[A] * [B]2= (0.152 M/s)/(0.15 M)(0.25 M)2=16.21 L/mol s = 16 L/mol s
(d)
Predict the rate of reaction if [A] = 0.050 M and [B] = 0.030 M at 25 oC. (1 mark)
Rate = k*[A] * [B]2= (16.21L/mol s)(0.050 M)(0.030 M)2 = 7.3*10-4 M/s
[1 mark for sig figs, 1 mark for units]
NAME:____________________________
5.
Student Number:________________________
The decomposition of A2 to A is a first order process, where the loss of A2 is monitored at
25 oC, and tabulated below. Calculate the rate constant. It is not necessary to use graphical
methods.
[3 marks]
Time (s)
[A2] M
0.0
2.50
1.0
1.98
2.0
1.57
3.0
1.25
4.0
0.99
5.0
0.79
6.0
0.63
7.0
0.50
8.0
0.39
9.0
0.31
10.0
0.25
By inspection of the data, one sees that the concentration halves every 3 seconds,
therefore t1/2 = 3.0 s. (1 mark)
k = ln2/ t1/2 = 0.693/(3.0 s) = 0.23 s-1 (1 mark)
[1 mark for sig. figs and units]
Alternately, this question can be solved by plotting ln[A2] vs. time (since it is a first order
reaction) and calculating the slope which is, by definition, equal to –k.
6.
If the rate constant for a reaction doubles when the temperature increases from 0 to 10 oC,
what is the activation energy for that reaction?
[5 marks]
ln(k2/k1) = - Ea/R*(1/T2 – 1/T1)
Ea = (-ln(k2/k1)*R)/(1/T2 – 1/T1)
= -ln(2/1)*(8.314 J/molK)/(1/(283 K)- 1/(273 K))
= 5.76 J/molK /(1.29*10-4 1/K)
= 4.465*104 J/mol = 44.7 kJ/mol (4 marks)
[1 mark for sig. figs and units]
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u) 1.6605 × 10-27 kg
Ideal gas constant (R)
8.3145 J·mol-1·K-1
Avogadro's number
Standard atmospheric pressure 1 atm = 101.325 kPa
6.02214 × 1023 mol–1
Other Useful Data
Kfp of water = -1.860 ˚C kg/mol
Formulae
ni
Xi =
P1
ln
= -
P2
ntotal
∆Hovap
1
R
T1
Psolvent = Xsolvent Posolvent
∆Psolvent = - Xsolutes Posolvent
∆Tbp = Kbp msolute i
∆Tfp = Kfp msolute i
[R]t - [R]o = - kt
ln
-Ea/RT
ln
k = A e
1
[R]t
1
-
T2
msolute =
k1
k2
= -
msolvent
Π = MRT
1
1
= kt
[R]t
[R]0
= - kt
[R]0
nsolute
Ea
R
1
T1
1
T2
ln(2) = k t1/2
Chem 2000 Standard Periodic Table
18
4.0026
1.0079
He
H
2
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
107
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
Np
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré