Midterm Exam Guidelines
CHEM 102 Winter 2023
Chemistry 102
Midterm Exam 1 Instructions
Date/Time: Monday, February 6th from 6:30 – 7:30 PM
1 hour writing time
Location: Check your room assignment on eClass
Item Checklist:
 Pen (black or dark blue ink). Do not use red ink or light colours.
 Approved scientific non-programmable calculator
 ONECard or valid photo ID (driver’s licence, passport)
Sample Questions
Please attempt the mock exam to become familiar with the format of questions on your midterm
exam.
*Disclaimer: You are responsible for all material covered in Unit 1 and first part of Unit 2 (up
to and including Section 3).
Do not expect the midterm to have the same topics or weight of topics as these practice questions.
These questions were chosen simply to provide more experience. Midterm 1 will cover all of
UNIT 1 (with emphasis on topics covered that were not covered on Quiz #1), as well introductory
Topics from Equilibrium (10% of the exam).
When attempting questions for practice:
DO NOT EXPECT THAT YOU WILL GET THE SAME QUESTIONS -- REVIEW YOUR
CLASS NOTES, PROBLEM SETS, AND QUIZZES.
If you require additional practice, problem sets and textbook problems can also be utilized
(suggested problems are provided in the lecture notes).
I recommend that you try the sample exam questions in a “closed-book format”. By simulating
exam conditions, you get a better sense of your confidence and preparedness.
Please note that the actual midterm exam will be longer. It should take you no longer than 40
minutes to finish the sample exam questions.
Department of Chemistry, University of Alberta (2023)
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Sample Exam Questions
CHEM 102 Winter 2023
1. Multiple Choice
(a) (2 marks) The reaction between A and B has a rate constant of 9 x 10 -3 M-1/2s-1. Which of the
following is a possible rate law for this reaction? Circle one.
(i)
(ii)
(iii)
(iv)
(v)
Rate = k [A]2[B]2
Rate = k [A]1/2[B]1/2
Rate = k [A][B]-1/2
Rate = k [A]1/2[B]
None of the above
Make sure units cancel so that the rate
(on the left side of eqn.) is in units of
M/s or mol/L s
(b) (2 marks) For the reaction,
A (g) + B (g)  C (g)
The following data was obtained at constant temperature:
Expt. Initial [A] mol/L Initial [B] mol/L Initial Rate mol/L s
1
0.10
0.20
0.31
2
0.20
0.20
1.24
3
0.20
0.10
0.62
The rate law is Rate = k[A]x[B]y where
(i) x = 1, y = 1
When [A] is doubled reaction rate quadruples  order with
respect to [A] is 2. When [B] is halved, rate is halved  order
(ii) x = 1, y = 2
with respect to [B] is 1. OR solve mathematically:
(iii) x = 2, y = 2
rate2/rate1 = (1.24)/(0.31) = [(0.20)/(0.10)]x
(iv) x = 2, y = 1
4 = (2)x
 x=2
(v) x = 1, y = 0
rate2/rate3 = (1.24)/(0.62) = [(0.20/(0.10)]y
2 = (2)y
 y=1
(d) (2 marks) Consider the formation of ethane, C2H2(g) + 2H2(g)  C2H6(g) H°rxn = –311 kJ
Under which reaction conditions would you expect to have the greatest equilibrium yield of ethane?
(i)
(ii)
(iii)
(iv)
(v)
high temperature, high pressure
High pressure will shift eq. position to side with less
low temperature, high pressure
moles of gas (towards ethane)
high temperature, low pressure
Low temperature  exothermic rxn favoured
low temperature, low pressure
none of the above, unless a catalyst is present
2. True/False and Short Answer
(a) (2 marks) Indicate whether the statement is true or false. Justify your choice with a brief explanation.
The units of the rate of reaction depend on the order of the reaction.
True
False
The units of the rate are concentration/time, hence they are not influenced by the order of the reaction.
The units of the rate constant, k however, do change with order.
Department of Chemistry, University of Alberta (2023)
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Sample Exam Questions
CHEM 102 Winter 2023
(b) (3 marks) Excluding enzymes, there are two general types of catalysts. List them below and provide
an example of each.
Type of catalyst
Homogeneous catalyst
Example
I- in H2O2 decomposition (discussed
in class)
Type of catalyst
Heterogeneous catalyst
Example
Metal surface for hydrogenation of
double bonds; metal surface in
catalytic converter;
3. Numerical
The reaction below occurs via a second order process.
AB (g)  A (g) + B (g)
(a) (3 marks) At a temperature of 25C, a 2.00 M sample of AB has a second half-life of 5.00 s.
Determine how long it would take this sample to undergo 60.0 % decomposition.
Use t1/2 for 2nd order reaction to find k:
t
/
=
[ ]
At the beginning of the reaction, [AB] was 2.00M. At the end of the first half life, [AB] had dropped to
1.00M. So at the beginning of half life #2, [AB] 0 = 1.00M
1
5. 00 s =
k = 0.200 M-1s-1
k(1.00M)
Use second order integrated rate law to determine time necessary for 60.0% decomposition. If the sample
has decomposed by 60.0%, the amount of AB remaining is 0.400(2.00 M) = 0.800 M
1
1
= (0.200 M s )t +
(0.800 M)
2.00 M
t = 3.75s
(b) (3 marks) The activation energy of the reaction is 70 kJ mol -1. Calculate the rate constant of the
reaction at 150C.
=−
ln
Ea
1
R
T2
–
1
where we can use the k and T from (a) of this problem for
T1
k1 = 0.200 M-1s-1 and T = 298K
ln
.
=−
70
8.314
1
∙
(
)
–
1
(298K)
Solve for k2 = 8.45102 M-1s-1
Department of Chemistry, University of Alberta (2023)
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Sample Exam Questions
CHEM 102 Winter 2023
4. Written Response
For the reduction reaction of nitric oxide by H2, the following three-step reaction mechanism has been
proposed:
2 NO (g)  N2O2 (g)
N2O2 (g) + H2(g)  N2O(g) + H2O(g)
N2O (g) + H2(g)  N2(g) + H2O(g)
STEP 1:
STEP 2:
STEP 3:
Fast
Fast
Slow
(a) (1 mark) List the intermediate(s) in the proposed mechanism.
N2O and N2O2 are intermediates
(b) (5 marks) What is the rate law for the overall reaction?
Rate is determined from the rate limiting (rate determining) step:
1 mark
Rate3 = k3 [N2O][H2]
Intermediates cannot appear in the overall rate law. Hence, [N2O] must be substituted:
In STEP 2, equilibrium is established. Therefore, Forward rate = Reverse rate
k2 [N2O2] [H2]= k-2 [N2 O] [H2 O]
Solving for [N2O]: [N2O] =
1.5 marks
k2 [N2O2] [H2]
k-2 [H2 O]
Using STEP 1 to substitute the other intermediate, [N 2O2]:
k1 [N2O2] = k-1 [NO]2
k1 [NO]2
Solving for [N2O2]: [N2O2] =
k-1
Substituting into the rate law given by the rate determining step:
Rate =
1.5 marks
1 mark
k3 k2 k1 [NO]2 [H2 ]2
k-2 k-1 [H2 O]
[Kinetics, Case II mechanism problem - very similar to PS#2 Q4]
concentration
(c) (3 marks) Sketch on the plot below, relative to each other, a reactant’s concentration versus time for a zero
order and for a first order reaction that have the same k and starting concentration. Label the two sketches to
distinguish between them.
Zero Order
Order
First Order
Order
Both curves should be
decreasing. For zero order,
[r] versus time is linear. For
first order, it is nonlinear.
time
Department of Chemistry, University of Alberta (2023)
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Sample Exam Questions
CHEM 102 Winter 2023
Formulas
[A]t = – kt + [A]o
ln [A]t = – kt + ln [A]0
1
[A]t
= kt +
Rate =
or
ln
[A]t
[A]o
= – kt
1
±1 ∆[X]
= k [A] [B] …
x ∆t
Math
/ =
𝑡
/
𝑡
[A]0
k=A𝑒
𝑡
ln (k) =
- Ea
RT
[A]0
2k
k
1
=
/
ln(2)
=
k [A]0
(X = reactant or product)
+ ln (A)
ln
k1
k2
=
Ea
1
R
T2
–
1
T1
Miscellaneous Constants
ln (x) = y  x = ey
ln (xy) = ln(x) + ln(y)
ln
= ln (x) – ln (y)
ln (xa) = a ln (x)
Department of Chemistry, University of Alberta (2023)
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