Chemistry 3202 Unit 1 Test 1 2005-06
page 1
CDLI - Chemistry 3202 Unit 1 Test 1
Reaction Kinetics and Equilibrium
December 2006
Part I - Multiple Choice: Using the form on page 4, circle the letter of the best response.
1.
Which states that particles of matter are in constant random motion?
A. collision theory
B. Hess’s law
C. kinetic molecular theory
D. law of chemical equilibrium
2.
Reactants A and B collided with the correct orientation required to form product C; however,
product C did not form. What explains the observation?
A. the activated complex was stable
B. the collision lacked sufficient intensity
C. the kinetic energy of the colliding particles was too high
D. the potential energy of the colliding particles was too low
3.
In the diagram on the right, which label
represents the activation energy of the
reverse reaction?
A. Q
B. R
C. S
D. T
4.
In the diagram on the right, which label
represents the heat of reaction?
A. Q
B. R
C. S
D. T
Q
T
EP
R
S
Progress of Reaction
5.
The activation energy for this reaction, Cu + S → CuS + 53.1kJ , is 50 kJ. What is the
activation energy of the reverse reaction?
A. 3.1 kJ
B. 50 kJ
C. 53.1 kJ
D. 103.1 kJ
6.
Substance X is a highly unstable, undetectable, short lived structure formed upon the collision of
reactants. How should substance X be classified?
A. activated complex
B. catalyst
C. elementary particle
D. reaction intermediate
Chemistry 3202 Unit 1 Test 1 2005-06
7.
page 2
Which reaction should proceed at a faster rate if surface area of a reactant and pressure are
increased?
A. Zn(NO3 )2(aq) + Na2S (aq) → ZnS (s) + 2 NaNO3 (aq)
B.
ZnS(l)
→
Zn(s)
C.
Zn (s)
+
Cl2
D.
Zn(s)
+
H2SO4(aq)
(g)
+
S (l)
→
ZnCl2
(s)
H2(g)
→
ZnSO4
+
8.
What effect does a catalyst have on a chemical reaction?
A. it alters the reaction mechanism
B. it decreases the heat of reaction
C. it increases the intensity of collisions
D. it inhibits the activated complex
9.
Consider this proposed reaction mechanism:
Step 1 :
H2
Step 2 :
N (g)
NO (g)
+
(g)
+
NO (g)
H2O (g)
→
→
(aq)
N2
(g)
N (g)
+
+
O (g)
Step 3 : H2 (g) + O (g) → H2O (g)
______________________________________________
2 H2
+
(g)
2 NO (g)
→
What is the role of O(g) in this reaction mechanism?
A. activated complex
B. catalyst
C. product
D. reaction intermediate
10.
Which is the rate determining step?
1
3
2
4
EP
Progress of Reaction
A.
B.
C.
D.
step 1
step 2
step 3
step 4
2 H2O (g)
+
N2
(g)
Chemistry 3202 Unit 1 Test 1 2005-06
11.
page 3
Which change will result in a faster rate for the overall reaction in this proposed mechanism?
Step 1 :
Cl2
Step 2 :
CH4
(g)
(g)
→
+
2 Cl(g)
Cl (g)
fast
CH3 (g)
→
+
HCl (g)
slow
Step 3 : CH3 (g) + Cl (g) → CH3Cl (g)
fast
______________________________________________
Cl2
A.
B.
C.
D.
(aq)
+
CH4
(g)
→
CH3Cl (g)
+
HCl (g)
increase the concentration of CH3(g)
increase the concentration of CH4(g)
reduce the pressure on the system
reduce the temperature of the system
12.
What is the catalyst for the formation of ground level ozone?
A. carbon monoxide
B. methyl tert-butyl ether
C. nitrogen monoxide
D. tetraethyl lead
13.
What is equal in an equilibrium?
A. activation energy of forward and reverse reactions
B. final concentrations of reactants and products
C. opposing rates of change
D. starting concentrations of reactants
Use this equation A ⇌ B and the graph to the
right to answer the next two items. Labels T1 to
T4 represent four times when concentrations of
species A and B were sampled.
14.
15.
When is the forward reaction proceeding at
the fastest rate?
A. T1
B. T2
C. T3
D. T4
When is equilibrium achieved?
A. T1
B. T2
C. T3
D. T4
T1 T2 T3 T4
Concentration
A
B
Progress of reaction
Chemistry 3202 Unit 1 Test 1
page 4
Name: _____________________________________ Site: ________________________________
Chemistry 3202 Unit 1 Test 1 Answer Sheet
Part 1: For each item, circle the letter corresponding to your choice.
1.
A
B
C
D
6
A
B
C
D
11.
A
B
C
D
2.
A
B
C
D
7.
A
B
C
D
12.
A
B
C
D
3.
A
B
C
D
8.
A
B
C
D
13.
A
B
C
D
4.
A
B
C
D
9.
A
B
C
D
14.
A
B
C
D
5.
A
B
C
D
10.
A
B
C
D
15.
A
B
C
D
Part II – Constructed Responses: Complete each item in the space provided. State final answers in
sentence form and follow the rules for rounding/significant digits.
16.
Magnesium metal reacts with ethanoic acid.
Mg (s)
a)
(2)
(2)
(2)
+
CH3COOH (aq)
→
H2
(g)
+
Mg ( CH3COO )2
(aq)
Predict the effect of each change on the overall rate of reaction, and explain your prediction in
terms of the collision theory.
i) diluting the ethanoic acid with water
The change should cause a slower rate of reaction. Diluting the ethanoic acid reduces its
molar concentration. This in turn results in fewer overall collisions per unit time and
consequently fewer successful collisions. As a result the rate of reaction should decrease.
ii) increasing the temperature of the ethanoic acid from 20°C to 30°C.
The change should result in a faster rate of reaction. Increasing the temperature
increases the average kinetic energy of the atoms and molecules. Collisions between them
are more intense and result in the formation of more activated complexes per unit of
time, hence the faster rate of reaction.
b)
Describe an experiment you could carry out to verify the prediction you made in item ii above.
Obtain a flask and add some ethanoic acid. Warm or cool the flask so the contents are at
exactly 20°C. Attach a hose and stopper to the flask and run the hose to a water filled cylinder.
Add the magnesium metal to the flask and measure the time it takes to displace the water with
hydrogen gas. Repeat the procedure with ethanoic acid at 30°C.
Chemistry 3202 Unit 1 Test 1
page 5
Name: _____________________________________ Site: ________________________________
(3) 17.
a)
Sketch a potential energy diagram for this uncatalysed reaction
2 O3
+
300 kJ
→
3 O2
assuming an activation energy for the forward reaction of 500 kJ. Label the reactants,
products, activated complex, activation energy of forward reaction, and heat of reaction.
(Leave room for item c below.)
Activated Complex
3 O2
Activation Energy = 500 kJ
Heat of Reaction, /\H = 300 kJ
Ep
2 O3
Progress of Reaction
b)
Consider this proposed mechanism for the catalyzed reaction:
Step 1 :
O3
+
OH
→
O2
+
HO2
fast
Step 2 : HO2 + O3 → OH + 2 O2
slow
____________________________________________
(1)
i)
Identify the reaction intermediate and briefly explain your choice.
HO2 is the reaction intermediate. It is formed in step 1 and consumed again in
step 2.
(1)
ii)
Identify the catalyst and briefly explain your choice.
OH is the catalyst. It reacts in step 1 and is regenerated in step 2.
(1)
c)
On the diagram you produced above, indicate the reaction pathway for the catalyzed reaction
using a dotted line.
See green line on sketch above
Chemistry 3202 Unit 1 Test 1
page 6
Name: _____________________________________ Site: ________________________________
(2)
18. a)
The modern catalytic converter is a ceramic structure coated with metals like platinum,
rhodium and palladium. The structure consists of numerous honeycomb shaped channels. In
terms of rates of reaction, what is the purpose of the honeycomb design? Explain.
The honeycomb design of the catalytic converter affords a very high metal surface area.
This maximizes the contact points for the adsorption step in the reaction mechanism for
conversion of CO and No to CO2 and N2. Higher surface area means more collisions per
unit of time which in turn leads to more successful collisions per unit of time. This results
in a faster rate of reaction – more of the available reactants become product when the
surface area of the catalyst is maximized.
(1)
b)
Give two reasons why tetraethyl lead is no longer added to gasoline.
Lead is a deadly cumulative poison associated with reduced intelligence in children, anemia,
anorexia, and kidney disease. Its toxicity has resulted in a switch from leaded gasoline which
contained tetraethyl lead to unleaded gasoline.
Lead also “poisons” catalytic converters. It undergoes metallic bonding with the metals
typically used in catalytic converter surfaces and blocks the adsorption sites for CO and NO.
This reduced the efficiency of converters. Rather than redesign converters, gasoline was
redesigned instead.
Removal of a lead additive from gasoline solved two serious and costly problems.
19.
Equal amounts of hydrogen and iodine gases are added to an empty round bottomed flask. The gases
react to produce hydrogen iodide and eventually this equilibrium is established.
H2
(3)
+
I2
⇌
2 HI
Describe the changes in the concentrations of the reactants and products as this equilibrium is
established and list the conditions that must be met in order for the equilibrium to be maintained.
Initially, the concentrations of hydrogen and iodine are high. As they react to form HI, their
concentrations decrease. The rate of the reaction between H2 and I2 progressively decreases
as their concentrations decrease.
Initially, the concentration of HI is zero, but it increases as hydrogen and iodine react. Some
of the HI decomposes back into hydrogen and iodine. The rate of this reaction increases as
the amount of HI increases.
Eventually, the rate of the forward reaction (H2 + I2) equals the rate of the reverse reaction
(breakdown of HI) and equilibrium is achieved. The equilibrium is only established and
maintained as long as pressure and temperature are constant and as long as the system is
closed.