NAME:____________________________
Spring 2006
Student Number:________________________
Chemistry 2000 In-Class Test #2A
INSTRUCTIONS:
____/ 50 marks
1) Please read over the test carefully before beginning. You should have 6
pages of questions, a sheet of graph paper and a formula/periodic table
sheet (8 pages total).
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 60 minutes to complete this test.
1.
For each reaction, indicate whether or not it is reasonable for it to have only one elementary
step. Briefly, justify your answer.
[4 marks]
(a)
CH3Cl → CH3 + Cl
could be / could not be elementary
unimolecular reaction; only one bond to be broken
(b)
4 Fe + 3 O2 → 2 Fe2O3
could be / could not be elementary
highly unlikely for seven atoms/molecules to collide at once; cannot be a one-step reaction
2.
Define the terms “intermediate” and “transition state”, clearly indicating the difference
between them. (Full marks were given for less detailed answers than that below.) [4 marks]
Intermediates are only formed in multi-step reactions. An intermediate is a product of one
elementary step and a reactant in a later step.
Transition states are formed in all reactions. A transition state is the high-energy state that
forms in the middle of an elementary step while bonds are in the process of breaking and
forming. As such, transition states are often considered to have “partial bonds”.
e.g.
Multi-step reaction making C–A from A and C (using B as a catalyst)
Step 1:
A + B → A–B
Step 2:
C + A–B → C–A + B
A–B is an intermediate.
There would be two transition states: A - - B (for step 1) and C - - A - - B (for step 2)
On an energy diagram, a transition state is higher in energy than the
reactants/products/intermediates on either side of it while an intermediate is lower in energy
than the transition states on either side of it. (see diagram in notes/text)
NAME:____________________________
3.
The reaction below proceeds according to first order kinetics:
X-Y(g)
[7 marks]
X(g) + Y(g)
Temperature
(K)
273.15
298.15
310.15
(a)
Student Number:________________________
Temperature
(˚C)
0.00
25.00
37.00
Half-Life
(hours)
1.250
0.682
0.527
Rate Constant
(hours-1)
0.5545
1.02
1.32
Calculate the activation energy for this reaction.
Step 1: Calculate rate constants for at least two temperatures.
e.g.
ln(2) = k t½
k = ln (2) = ln(2) = 0.5545 h-1
at 273.15 K
t½
1.250 h
Step 2: Use comparative form of Arrhenius equation to calculate activation energy.
ln
k1
k2
= -
Ea
R
R ln
Ea = 1
T1
1
T1
1
T2
k1
k2
-
1
T2
8.3145 J mol-1 K-1 ln
Ea = -
(
1.02 h-1
0.5545 h-1
1
1
298.15 K
273.15 K
)
Ea = 1.64 x 104 J/mol = 16.4 kJ/mol
(b)
Calculate the frequency factor (A) for this reaction.
Use the Arrhenius equation to find A at one temperature
e.g.
k = A e-Ea/RT
A =
k
e-Ea/RT
=
0.5545 h-1
e-(16410 J/mol)/(8.314 J/molK)(273.15 K)
= 762 h-1
(= 12.7 min-1 = 0.212 s-1)
Because A is a constant for the reaction, values calculated for 25.00 ˚C or 37.00 ˚C should
be within ~1 h-1 of this value.
NAME:____________________________
4.
Student Number:________________________
Wayne is studying the decomposition of phosphine at 750 ˚C:
4 PH3(g)
[10 marks]
P4(g) + 6 H2(g)
By monitoring the concentration of phosphine remaining, he obtains the data below:
Time
(s)
0
10
20
30
40
50
(a)
[PH3(g)]
(mol L-1)
1.41 × 10-3
9.94 × 10-4
7.03 × 10-4
4.97 × 10-4
3.51 × 10-4
2.49 × 10-4
[PH3(g)]-1
(L mol-1)
709
1006
1622
2012
2849
4016
ln[PH3(g)]
-6.56
-6.91
-7.26
-7.61
-7.95
-8.30
Write the rate equation for this reaction.
Step 1: Add two columns to the concentration table: ln[PH3] and 1/[PH3]
Step 2: Plot [PH3] vs. time (zero order?), ln[PH3] vs. time (first order?) and 1/[PH3] vs. time (second
order?). You can evaluate the data by inspection, but you should always plot your “looks
most linear” data and, unless you get a perfectly straight line, you should make the other two
graphs as well. The graph you initially thought would be linear may not be as linear as one
of the other two.
First-Order Plot for Decomposition of Phosphine
0.0016
0.0014
0.0012
0.001
0.0008
0.0006
0.0004
0.0002
0
-6
-6.5
0
10
20
50
-7.5
y = -0.0347x - 6.5659
-8.5
10
20
30
40
50
-9
Time (s)
Calculate the numerical value of the rate
constant,
clearly
indicating
the
appropriate units.
For a first order reaction, the slope of
ln[PH3] vs. time will equal –k. Therefore,
k = 0.035 s-1 at 750 ˚C.
What is the overall order of this reaction?
Time (s)
Second-Order Plot for Decomposition of Phosphine
1/[PH3] (L/mol)
By comparison of these three graphs, we
can confidently say that this reaction is
first order in phosphine.
(c)
40
-8
0
(b)
30
-7
ln[PH3]
[PH3] (mol/L)
Zero-Order Plot for Decomposition of Phosphine
4500
4000
3500
3000
2500
2000
1500
1000
500
0
0
10
20
30
Time (s)
first order (since there is only one reactant)
40
50
NAME:____________________________
5.
Student Number:________________________
Michelle is studying the deprotonation of H2PO4- in aqueous solution:
-
H2PO4(aq)
+ OH(aq)
She obtains the data below.
Run
1
2
3
4
5
(a)
2-
HPO4(aq) + H2O(l)
[10 marks]
Initial Concentration
(mol L-1)
[H2PO4-]
[OH-]
0.060
0.0040
0.060
0.0080
0.120
0.0080
0.120
0.0120
0.180
0.0040
Write the rate equation for this reaction.
Initial Rate of Reaction
(mol L-1 s-1)
48
192
384
864
144
rate = k [H2PO4-] [OH-]2
Step 1: Write basic rate equation
rate = k [H2PO4-]x [OH-]y
Step 2: Find x by comparing two runs where the only variable is [H2PO4-]
e.g.
Compare runs 1 and 5
rate5 = k [H2PO4-]5x [OH-]5y
rate1 k [H2PO4-]1x [OH-]1y
144 mol L-1 s-1 = k (0.180 mol/L)x (0.040 mol/L)y
48 mol L-1 s-1
k (0.060 mol/L)x (0.040 mol/L)y
3.0 = 3.0x
x=1
Step 3: Find y by comparing two runs where the only variable is [OH-]
same process as in step 2 should give y = 2
(compare runs 1 and 2, or 3 and 4)
(b)
Calculate the numerical value of the rate constant, clearly indicating the appropriate units.
Step 1: Use one run to calculate k
e.g.
Run #1
rate = k [H2PO4-] [OH-]2
k =
rate
=
(48 mol·L-1·s-1)
= 5.0 × 107 L2·mol-2·s-1
- 2
-1 2
[H2PO4 ] [OH ]
(0.060 mol·L ) (0.040 mol·L)
***2 sig. fig.***
(c)
What is the overall order of this reaction?
third order
NAME:____________________________
6.
(a)
(b)
Student Number:________________________
Nitrogen monoxide and bromine combine to make nitrosyl bromide (NOBr). At 20 ˚C, a
250 mL container contains 1.45 g NO(g), 5.63 g Br2(g) and 0.573 g NOBr(g) once equilibrium
has been reached.
[8 marks]
Calculate Kc for this reaction.
Calculate Kp for this reaction.
T = 293.15 K (3 sig. fig.)
V = 0.250 L
2 NO(g)
mass:
molar mass:
moles:
molarity:
pressure:
(a)
Kc =
(b)
Kp =
+
1.45 g
30.0061 g/mol
0.0483 mol
0.193 mol/L
4.65 atm
Br2(g)
5.63 g
159.808 g/mol
0.0352 mol
0.141 mol/L
3.39 atm
[NOBr]2 =
(0.0209)2
= 0.0826
[NO]2[Br2]
(0.193)2(0.141)
(pNOBr)2
=
(0.502)2
2
(pNO) (pBr2)
(4.65)2(3.39)
= 0.00342
2 NOBr(g)
0.573 g
109.9109 g/mol
0.00521 mol
0.0209 mol/L
0.502 atm
***3 sig. fig.***
***3 sig. fig.***
Alternate approach to (b)
Kp = Kc(RT)∆n
∆n = # moles products – # moles reactants = 2 – 3 = -1
Kp = (0.0826)[(0.082057)(293.15)]-1
Kp = 0.00342
***3 sig. fig.***
Remember:
• Use a balanced reaction equation!
• Kc and Kp have no units!
• Concentrations must be in units of mol/L to calculate Kc
• Concentrations must be in units of atm to calculate Kp
• To convert molarity to pressure, recall that PV = nRT therefore P = (n/V)RT (or P = MRT)
• If using alternate approach to part (b), ∆n refers to balanced equation (not actual # moles) and
R must be 0.082057 L·atm·mol-1·K-1 so that you are converting molarities to pressures in atm
NAME:____________________________
7.
(a)
(b)
(c)
Student Number:________________________
The reaction between methanol (CH3OH) and hydrobromic acid to produce bromomethane
(CH3Br) and water is thought to proceed according to one of the three mechanisms below.
The observed rate equation for this reaction is rate = k[CH3OH][HBr][Br-]-1
[7 marks]
Circle the mechanism which is consistent with this information.
Suggest one further test that could be performed to corroborate your chosen mechanism.
Give at least one good reason for eliminating each of the other two mechanisms.
Mechanism A
Step 1:
CH3OH(aq) + HBr(aq)
CH3Br(aq)
+
H2O(aq)
Mechanism B
Step 1:
CH3OH(aq) + HBr(aq)
Step 2:
+
CH3OH2(aq)
+
Br(aq)
+
CH3OH2(aq)
+
Br(aq)
fast
CH3Br(aq)
+ H2O(l)
slow
+
CH3OH2(aq)
+
Br(aq)
fast
Mechanism C
Step 1:
CH3OH(aq) + HBr(aq)
Step 2:
+
CH3OH2(aq)
Step 3:
+
CH3(aq)
+ Br(aq)
+
CH3(aq)
+ H2O(l)
CH3Br(aq)
slow
fast
(b)
To corroborate proposed mechanism C, look for CH3+ and/or CH3OH2+ as they are both
intermediates. (Looking for Br- is less helpful since the observed rate equation already tells
us that it is involved.)
(c)
“Mechanism A” cannot be correct as the rate equation corresponding to that mechanism is
rate = k[CH3OH][HBr] This is not the observed rate equation.
“Mechanism B” cannot be correct as the rate equation corresponding to that mechanism is
rate = k[CH3OH][HBr] This is not the observed rate equation. For “Mechanism B”:
Rate determining step is step 2 therefore:
rate = k2[CH3OH2+][Br-]
Step 1 is a fast equilibrium therefore:
K1 = [CH3OH2+][Br-]
[CH3OH][HBr]
or
[CH3OH2+][Br-] = K1[CH3OH][HBr]
Therefore:
rate = k2K1[CH3OH][HBr] = k[CH3OH][HBr]
where k = k2K1
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u)
1.6605 × 10-24 g
Avogadro's number
6.02214 × 1023 mol–1
Bohr radius
5.29177 × 10-11 m
Charge of electron (e)
1.6022 × 10-19 C
Coulomb constant
8.998 × 109 N·m2·C-2
Ideal gas constant (R)
8.3145 J·mol-1·K-1
Ideal gas constant (R)
0.082057 L·atm·mol-1·K-1
5.4688 × 10-4 u
1.0086649 u
1.0072765 u
6.626 × 10-34 J·s
1.097 x 107 m-1
2.179 x 10-18 J
2.9979 x 108 m·s-1
Mass of electron
Mass of neutron
Mass of proton
Planck's constant
Rydberg Constant (R)
Rydberg unit (Ry)
Speed of light in vacuum
Formulae
n
Psolvent = Xsolvent Posolvent
∆Psolvent = - Xsolutes Posolvent
M =
∆Tbp = Kbp msolute i
∆Tfp = Kfp msolute i
Π = MRT
m
d =
m
U =
V
[R]t - [R]o = - kt
ln
-Ea/RT
ln
k = A e
Kp = Kc (RT)∆n
Xi =
n
[R]t
k1
k2
= -
ni
ntotal
1
1
= kt
[R]t
[R]0
= - kt
[R]0
V
Ea
R
1
T1
1
T2
ln(2) = k t1/2
PV = nRT
1
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
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)
107
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