Free-Response Review Questions - Kinetics & Equilibrium
PSI Chemistry
Name________________________________
1.
2 NO(g) + Br2(g)  2 NOBr(g)
a) For the reaction above, what are 3 ways the rate of reaction could be measured?
b) Is the rate of disappearance of bromine greater than, less than, or equal to the rate of
appearance of NOBr?
2. In the graph below, label the time at which equilibrium is achieved.
P Cl 3
c onc e ntr a tion
Cl 2
P Cl
5
tim e
3.
Two reactions are represented below. The potential-energy diagram for reaction I is
shown below. The potential energy of the reactants in reaction II is also indicated on the
diagram. Reaction II is endothermic, and the activation energy of reaction I is greater than that
of reaction II.
(I)
A2 + B2  2 AB
(II)
X2 + Y2  2 XY
0
Complete the potential-energy diagram for reaction II on the graph above.
4. Answer the following questions regarding the kinetics of chemical reactions.
(a) The diagram below shows the energy pathway for the reaction
O3 + NO  NO2 + O2. Clearly label the following directly on the diagram.
(i) The activation energy (Ea) for the forward reaction
(ii)
H) for the reaction
(b) The reaction 2 N2O5  4 NO2 + O2 is first order with respect to N2O5.
(i) Using the axes below, draw the graph that represents the change in [N2O5] over time
as the reaction proceeds.
In itial
[ N™O£ ]
Time
(ii) Describe how the graph in (i) could be used to find the reaction rate at a given time, t.
5. For the system 2 SO2(g) + O2(g)  2 SO3(g) , H is negative for the production of SO3.
Assume that one has an equilibrium mixture of these substances. Predict the effect of each of
the following changes on (i) the value of the equilibrium constant and (ii) on the number of
moles of SO3 present in the mixture at equilibrium. Briefly account for each of your predictions.
(Assume that in each case all other factors remain constant.)
(a) Decreasing the volume of the system.
(b) Adding oxygen to the equilibrium mixture.
(c)Raising the temperature of the system.
6.
C(s) + H2O(g)  CO(g) + H2(g)
Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at
equilibrium. State whether the number of moles of CO(g) in the container will increase,
decrease, or remain the same after each of the following disturbances is applied to the original
mixture. For each case, assume that all other variables remain constant except for the given
disturbance. Explain each answer with a short statement.
(a) Additional H2(g) is added to the equilibrium mixture at constant volume.
(b) The temperature of the equilibrium mixture is increased at constant volume.
(c)The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
Le Chatelier’s Principle
Le Chatelier’s Principle states that when a system at equilibrium is subjected to a stress, the
system will shift its equilibrium point in order to relieve the stress.
Complete the following chart by writing left, right, or none for equilibrium shift; and decreases,
increases, or remains the same for the concentrations of reactants and products, and for the
value of K.
N2(g) + 3H2(g) ↔ 2NH3(g)
Stress
1. Add N2
ΔH = -22.0kcal
Equilibrium
Shift
[N2]
[H2]
[NH3]
K
Right
-----
Decreases
Increases
Remains
the same
2. Add H2
-----
3. Add NH3
4. Remove N2
5. Remove H2
6. Remove
NH3
7. Increase
Temperature
8. Decrease
Temperature
9. Increase
Pressure
-----
-----
-----
-----
10. Decrease
Pressure
Le Chatelier’s Principle Continued
H2(g) + I2(g) ↔ 2HI(g)
Stress
1. Add H2
ΔH = +12.6kcal
Equilibrium
Shift
[H2]
[I2]
[HI]
K
Right
-----
Decreases
Increases
Remains
the same
2. Add I2
-----
3. Add HI
-----
4. Remove H2
-----
5. Remove I2
-----
6. Remove HI
-----
7. Increase
Temperature
8. Decrease
Temperature
9. Increase
Pressure
10. Decrease
Pressure
NaOH(s) ↔ Na+(aq) + OH-(aq)
affect equilibrium values)
Stress
1. Add
NaOH(s)
2. Add NaCl
(Adds Na+)
3. Add KOH
(Adds OH-)
4. Add H+
(Removes
ΔH= -10.6kcal (remember that pure solids and liquids do not
Equilibrium
Shift
Amount
NaOH(s)
[Na+]
[OH-]
-----------------
K
5.
6.
7.
8.
OH-)
Increase
Temperature
Decrease
Temperature
Increase
Pressure
Decrease
Pressure
Answers:
1) a) By observing the change in concentration of NO being consumed over a definite
amount of time time.
By observing the change in concentration of Br2 being consumed over a definite amount
of time.
By observing the change in concentration of NOBr being produced over a definite
amount of time.
b) The rate of disappearance of Br2 is only one half of the rate of appearance of NOBr.
2)
3)
4)
5)
(a) As volume decreases, pressure increases and the reaction shifts in the direction of fewer
molecules (less volume; more SO3) to relieve the stress. Value of Keq does not change.
(b) Additional O2 disturbs the equilibrium and SO3 is formed to relieve the stress. Value of Keq
does not change.
(c) Increase in temperature shifts the reaction to the left to “use up” some of the added heat.
Less SO3 remains. Value of Keq decreases due to the relative greater increase in the rate
of the endothermic reaction (reaction to the left).
6)
(a) CO will decrease. An increase of hydrogen gas molecule will increase the rate of the
reverse reaction which consumes CO. A LeChatelier Principle shift to the left.
(b) CO will increase. Since the forward reaction is endothermic (a H > 0) an increase in
temperature will cause the forward reaction to increase its rate and produce more CO. A
LeChatelier Principle shift to the right.
(c) CO will decrease. A decrease in volume will result in an increase in pressure, the
equilibrium will shift to the side with fewer gas molecules to decrease the pressure, , a
shift to the left.
(d) CO will remain the same. Once at equilibrium, the size of the solid will affect neither the
reaction rates nor the equilibrium nor the concentrations of reactants or products.
7)
Stress
Equilibrium
Shift
[N2]
[H2]
[NH3]
K
1. Add N2
Right
-----
Decreases
Increases
Remains
the same
2. Add H2
Right
Decreases
-----
Increases
Remains
the same
3. Add NH3
Left
Increases
Increases
-----
Remains
the same
4. Remove N2
Left
-----
Increases
Decreases
Remains
the same
5. Remove H2
Left
Increases
-----
Decreases
Remains
the same
Right
Decreases
Decreases
-----
Remains
the same
7. Increase
Temperature
Left
Increases
Increases
Decreases
Decreases
8. Decrease
Temperature
Right
Decreases
Decreases
Increases
Increases
9. Increase
Pressure
Right
Decreases
Decreases
Increases
Remains
the same
10. Decrease
Pressure
Left
Increases
Increases
Decreases
Remains
the same
Stress
Equilibrium
Shift
[H2]
[I2]
[HI]
K
1. Add H2
Right
-----
Decreases
Increases
Remains
the same
2. Add I2
Right
Decreases
-----
Increases
Remains
the same
3. Add HI
Left
Increases
Increases
-----
Remains
the same
6. Remove
NH3
8)
4. Remove H2
Left
-----
Increases
Decreases
Remains
the same
5. Remove I2
Left
Increases
-----
Decreases
Remains
the same
6. Remove HI
Right
Decreases
Decreases
-----
Remains
the same
7. Increase
Temperature
8. Decrease
Temperature
9. Increase
Pressure
Right
Decreases
Decreases
Increases
Increases
Left
Increases
Increases
Decreases
Decreases
None
Remains
the same
Remains
the same
Remains
the same
Remains
the same
10. Decrease
Pressure
None
Remains
the same
Remains
the same
Remains
the same
Remains
the same
Stress
Equilibrium
Shift
Amount
NaOH(s)
[Na+]
[OH-]
K
1. Add
NaOH(s)
None
-----
Remains
the same
Remains
the same
Remains
the same
2. Add NaCl
(Adds Na+)
Left
Increases
-----
Decreases
Remains
the same
3. Add KOH
(Adds OH-)
Left
Increases
Decreases
-----
Remains
the same
4. Add H+
(Removes
OH-)
5. Increase
Temperature
6. Decrease
Temperature
7. Increase
Pressure
Right
Decreases
Increases
-----
Remains
the same
Left
Increases
Decreases
Decreases
Decreases
Right
Decreases
Increases
Increases
Increases
Left
Increases
Decreases
Decreases
Remains
the same
8. Decrease
Pressure
Right
Decreases
Increases
Increases
Remains
the same
9)
10)
1. e
2. c
3. a
4. b
5. f
6. endothermic
7. d
8. f
9. exothermic
10 a. d,c,b
b. decrease
c. unchanged
11) a. The activation energy of photosynthesis is greater than that of oxidation.
b. Photosynthesis
c. Energy change is only based on the initial and final energies of the reaction.