1
Practice Test 1 Spring 2019
Practice Test for Exam 1
Dr Cabirac Lecture
1. Chlorine dioxide reacts in basic water to form chlorite and chlorate according to the following chemical
equation:
2ClO2 (aq) + 2OH- (aq) → ClO2- (aq) + ClO3- (aq) + H2O(l)
Under a certain set of conditions, the initial rate of disappearance of chlorine dioxide was determined
to be 2.30 x 10-1 M/s. What is the initial rate of appearance of chlorite ion under those same conditions?
A. 5.75 x 10-2 M/s
B. 1.15 x 10-1 M/s
C. 2.30 x 10-1 M/s
D. 4.60 x 10-1 M/s
E. 9.20 x 10-1 M/s
2. The units for a rate constant, k, are:
A. s-1
B. M-1 s-1
C. M-1
D. a rate constant is unitless
E. dependent upon the overall order of the reaction
3. The following graphs were constructed using a set of experimental data for a reaction:
What is the order of [A] for the reaction?
A. zero order
B. first order
C. second order
D. the order cannot be determined
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Practice Test 1 Spring 2019
4. The data below were determined for the reaction shown below.
S2 O82- + 3I- (aq) → 2SO42- + I3-
The rate law for this reaction must be:
A. rate = k[S2 O82-][I-]3
B. rate = k[S2 O82- ]
C. rate = k[S2 O82- ]2 [I- ]2
D. rate = k[I-]
E. rate = k[S2 O82- ][I-]
5. The isomerization of cyclopropane to form propene is a first-order reaction.
At 760 K, 15% of a sample of cyclopropane changes to propene in 6.8 min. What is the half-life of
cyclopropane at 760 K?
A. 3.4 x 10-2 min
B. 2.5 min
C. 23 min
D. 29 min
E. 230 min
6. A reaction was experimentally determined to follow the rate law, Rate = k[A]2 where k = 0.456 s-1M-1 .
Starting with [A]0 = 0.500 M, how many seconds will it take for [A]t = 0.250 M?
A. 2.85 x 10-2 s
B. 1.14 x 10-1 s
C. 1.52 s
D. 4.39 s
E. 5.48x10-1 s
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Practice Test 1 Spring 2019
7. The thermal decomposition of acetaldehyde, CH3 CHO → CH4 + CO, is a second-order reaction.
The following data were obtained at 518°C.
Calculate the rate constant for the decomposition of acetaldehyde from the above data.
A. 2.2 x 10-3 /s
B. 0.70 mmHg/s
C. 2.2 x 10-3 /mmHg·s
D. 6.7 x 10-6 /mmHg·s
E. 5.2 x 10-5 /mmHg·s
8. Concerning the decomposition of A, A → products which of the following methods could be
used to determine the order of the reaction with respect to A?
I. Plot [A] vs time, ln[A] vs time, and 1/[A] vs time and identify which plot yields a straight line.
II. Vary the concentration of A and note by what factor the rate changes.
III. Identify if successive half-lives of A double, halve, or stay constant.
A. I only D. I and III
B. II only E. I, II, and III
C. III only
9. The activation energy for the following first-order reaction is 102 kJ/mol.
N2O5 (g) → 2NO2 (g) + ½ O2 (g)
The value of the rate constant (k) is 1.35 x 10-4 s-1 at 35°C. What is the value of k at 0°C?
A. 8.2 x 10-7 s-1
B. 1.9 x 10-5 s-1
C. 4.2 x 10-5 s-1
D. 2.2 x 10-2 s-1
10. The reaction C4H10 → C2 H6 + C2H4 has an activation energy (Ea) of 350 kJ/mol, and the Ea of
the reverse reaction is 260 kJ/mol. Estimate ∆H, in kJ/mol, for the reaction as written above.
A. –90 kJ/mol
B. +90 kJ/mol
C. 350 kJ/mol
D. –610 kJ/mol
E. +610 kJ/mol
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Practice Test 1 Spring 2019
11. An increase in the temperature of the reactants causes an increase in the rate of reaction. The best
explanation for this behavior is that as the temperature increases,
A. the concentration of reactants increases.
B. the activation energy decreases.
C. the collision frequency increases.
D. the fraction of collisions with total kinetic energy greater than Ea increases.
E. the activation energy increases.
12. The rate law for the reaction 2NO2 + O3 → N2O5 + O2 is rate = k[NO2 ][O3 ]. Which one of the
following mechanisms is consistent with this rate law?
A. NO2 + NO2 → N2O4 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + 5/2 O2 (slow)
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)
13. With respect to the figure below, which choice correctly identifies all the numbered positions?
A. A D. D
B. B E. E
C. C
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Practice Test 1 Spring 2019
14. Aspirin, C9H8O4 , slowly decomposes at room temperature by reacting with water in the atmosphere to
produce acetic acid, HC2H3O2 , and 2-hydroxybenzoic acid, C7H6O3 (this is why old bottles of aspirin
often smell like vinegar):
C9H8O4 + H2O → HC2H3O2+ C7H6O3
Concentration and rate data for this reaction are given below.
Write the rate law for this reaction and calculate k (be sure to include the correct units).
A. rate = k[C9H8O4][ H2O]2 , and the rate constant is 1.2 x 10-9 M-2 s-1 or 1.2 x 10-9 1/(M2s)
B. rate = k[C9H8O4], and the rate constant is 1.2 x 10-9 or 1.2 x 10-9 1/ s
C. rate = k[C9H8O4]2 [H2O], and the rate constant is 1.2 x 10-9 M-2 s-1 or 1.2 x 10-9 1/(M2s)
D. rate = k[C9H8O4][ H2O], and the rate constant is 1.2 x 10-9 M-1 s-1 or 1.2 x 10-9 1/(Ms)
E. None of the above
15. For the following exothermic reaction, the rate law at 298 K is rate = k [H2][I2].
H2 (g) + I2 (g) → 2HI(g)
Which of the following, if any, would cause the rate of the reaction to increase?
a. Addition of hydrogen gas at constant temperature and volume
b. Increase in volume of the reaction vessel at constant temperature
c. Addition of a catalyst
d. Increase in temperature
A. Only a and b would increase the rate.
B. Only a, c, and d would increase the rate.
C. Only b and d would increase the rate.
D. All of the above would increase the rate.
E. None of the above would increase the rate.
16. Which is the correct equilibrium constant expression for the following reaction?
Fe2O3 (s) + 3H2 (g) ⇌ 2Fe (s) + 3H2O (g)
A. Kc = [Fe2O3] [H2]3 / [Fe]2 [H2O]3
B. Kc = [H2] / [H2O]
C. Kc = [H2O]3 / [H2]3
D. Kc = [Fe]2 [H2O]3 / [Fe2O3] [H2]3
E. Kc = [Fe] [H2O] / [Fe2O3] [H2]
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Practice Test 1 Spring 2019
17. Based on the graph below, what is the equilibrium expression for the reaction?
A. 𝐾 = [𝐴][𝐵]
[𝐶]
B. 𝑲 = [𝑪]𝟐
[𝑨][𝑩]𝟐
C. 𝐾 = [𝐶]
[𝐴][𝐵]
D. 𝐾 = [𝐶]3
[𝐴]2[𝐵]3
E. Not enough information
18. A reaction with an equilibrium constant Kc = 1.5 x 10-2 would consist of which of the
following at equilibrium:
A. approximately equal reactants and products
B. some reactants and products with reactants slightly favored
C. some reactants and products with products slightly favored
D. essentially all reactants
E. essentially all products
19. Given the following information:
2A (g) + B (g) ⇌ A2B (g) Kp1
2A (g) + C2 (g) ⇌ 2AC (g) Kp2
3/2 A2 (g) + B(g) + C(g) ⇌ AC(g) + A2B(g) Kp3
Which relationship represents the equilibrium constant for the reaction:
3A2 (g) + 3B (g) + 2C (g) ⇌ 3A2B (g) + C2 (g) Knet = ?
A. Knet = Kp1 x Kp2x Kp3
B. Knet = Kp1 x 2 Kp3 / Kp2
C. Knet = Kp1 – Kp2 + 2 Kp3
D. Knet = Kp1 x Kp32 / Kp2
E. Knet = Kp1 x Kp2 x Kp32
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Practice Test 1 Spring 2019
20. Phosgene, COCl2, a poisonous gas, decomposes according to the equation
COCl2 (g) ⇌ CO (g) + Cl2 (g). Calculate Kp for this reaction if Kc = 0.083 at 900ºC.
A. 0.125
B. 8.0
C. 6.1
D. 0.16
E. 0.083
21. 2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it
was found that 28% of the NOCl had dissociated according to the equation
2NOCl (g) ⇌ 2NO (g) + Cl2 (g).
Calculate the equilibrium constant, Kc, for the reaction.
A. 0.021
B. 0.039
C. 0.169
D. 26
E. 47
22. If one starts with pure NO2 (g) at a pressure of 0.500 atm, the total pressure inside the
reaction vessel when 2NO2 (g) ⇌ 2NO (g) + O2 (g) reaches equilibrium is 0.674 atm.
Calculate the equilibrium partial pressure of NO2 .
A. 0.152 atm
B. 0.174 atm
C. 0.200 atm
D. 0.326 atm
E. The total pressure cannot be calculated because Kp is not given
23. Consider the reaction N2 (g) + O2 (g) ⇌ 2NO (g) for which Kc = 0.10 at 2,000ºC.
Starting with initial concentrations of 0.040 M of N2 and 0.040 M of O2, determine the
equilibrium concentration of NO.
A. 5.4 x 10-3 M
B. 0.0096 M
C. 0.011 M
D. 0.080 M
E. 0.10 M
24. For the reaction PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) at a particular temperature, Kc = 24.3.
Suppose a system at that temperature is prepared with [PCl3 ] = 0.10 M, [Cl2 ] = 0.15 M,
and [PCl5 ] = 0.60 M. Which of the following is true based on the above?
A. Qc > Kc , the reaction proceeds from left to right to reach equilibrium
B. Qc > Kc, the reaction proceeds from right to left to reach equilibrium
C. Qc < Kc, the reaction proceeds from left to right to reach equilibrium
D. Qc < Kc, the reaction proceeds from right to left to reach equilibrium
E. Qc = Kc, the reaction is currently at equilibrium
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Practice Test 1 Spring 2019
25. Sodium carbonate, Na2CO3 (s), can be prepared by heating sodium bicarbonate, NaHCO3 (s) as shown
below. 2NaHCO3 (s) ⇌ Na2CO3 (s) + CO2 (g) + H2O (g) Kp = 0.23 at 100ºC
If a sample of NaHCO3 is placed in an evacuated flask and allowed to achieve equilibrium at 100ºC,
what will the total gas pressure be?
A. 0.46 atm
B. 0.96 atm
C. 0.23 atm
D. 0.48 atm
E. 0.11 atm
26.For the following reaction at equilibrium in a reaction vessel, which change will cause the Br2
concentration to decrease?
2NOBr(g) ⇌ 2NO(g) + Br2 (g) ∆Hºrxn= 30 kJ/mol
A. Increase the temperature.
B. Remove some NO.
C. Add more NOBr.
D. Compress the gas mixture into a smaller volume.
E. Add a catalyst
27. For the reaction at equilibrium 2SO3 ⇌ 2SO2 + O2 (∆Hºrxn = 198 kJ/mol), increasing the
reaction temperature would:
A. Shift the equilibrium to the right and increase the value of the equilibrium constant K
B. Shift the equilibrium to the left and increase the value of the equilibrium constant K
C. Shift the equilibrium to the right and decrease the value of the equilibrium constant K
D. Shift the equilibrium to the left and decrease the value of the equilibrium constant K
E. Cause no change
28. The equilibrium constants for the chemical reaction N2 (g) + O2 (g) ⇌ 2NO (g)
are Kp = 1.1 x 10-3 and 3.6 x 10-3 at 2,200 K and 2,500 K, respectively. Which one of these statements
is true?
A. The reaction is exothermic, ∆Hº < 0.
B. The partial pressure of NO(g) is less at 2,200 K than at 2,500 K.
C. Kp is less than Kc by a factor of (RT).
D. The total pressure at 2,200 K is the same as at 2,500 K.
E. Higher total pressure shifts the equilibrium to the left.
EQUATIONS
YOU
MAY
(OR
MAY
NOT)
NEED
1
AIL
_
4
IB]
_
1
AIC]
_
4
AID]
aA
+
bB
—
cC
+
dD
rate=-
Sr
Sor
Poa
ga
A
Zero-Order:
[A],
=—kt
+
[A],
ty
=
el
A
0.693
A
First-Order:
In[A],
=—kt+
Inf[A],
In|
(Al
=—kt
12
=
[4]=
[alo
[A],
iv
2
111
|
:
——=——
+kt
typ
=
Second-Order:
[Ai
(Alo
1/2
KIAlo
k.
E,{
1
1
R
=8.314
J/mol-K
Ink=-H21
ing
in|
1|-Fe(
2-2)
RT
ky
R\T
TT
Law
of
Mass
Action:
aA
+
bB
@cD
+
dD
K,
—
lel[p}"
,
°~
[Ay*{B]?
Kp
=
Ke(RT)4°
—»
R
=
0.0821
L-atm/mol-K
(this
equation
only)
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Practice Test 1 Spring 2019
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