PS 17.1 Review: Chapter 17 Equilibrium
(p 1 of 2)
1. The collision model successfully explains why reactions
go faster when the concentrations of reactants
_________ and temperatures ______________.
a. decrease; decrease
b. increase; increase
c. increase; decrease
d. decrease; increase
2. Catalysts speed up the rates of chemical reactions by
lowering the ________________.
a. temperature
b. activation energy
c. collisional frequency
d. kinetic energy
3. Chlorofluorocarbons (CFCs) are notorious as major
contributors to ozone layer damage. They break down in the
presence of UV radiation and release atomic Cl into the
stratosphere. Atomic Cl acts as a __________ in the
breakdown of ozone, and as such is not consumed in the
reaction.
a. reactant
b. modifier
c. catalyst
d. enzyme
4. Chemical equilibrium is a dynamic process in which the
______ of the forward reaction equals that of the reverse
reaction.
a. rate
b. reactivity
c. heat
d. enthalpy
5. In the previous chapter we learned that the partial
dissociation of weak acids such as acetic acid could be
written thus:
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
Since we now know that the double arrow indicates that
this reaction is in equilibrium, write the equilibrium
constant expression for the dissociation of acetic acid.


a.
b.
[H  ][C 2 H 3O 2 ]
[C H O ]

K
c.
K
K  [H ] 
[HC 2 H 3O 2 ]
[HC 2 H3O2 ]

[H  ][C2 H3O2 ]
d.
K
2
3
2
[HC 2 H 3O 2 ]
[HC 2 H 3O 2 ]
[H  ]
6. The For the reaction below, calculate the value of the
equilibrium constant, given the equilibrium concentrations.
N2O4(g)  2NO2(g)
[N2O4] = 0. 055 M
[NO2] = 0.060 M
a.
K = 15
b.
K = 1.5
c.
K = 1.1
d.
K = 0.065
7. Write out the equilibrium expression for the reaction:
CaF2(s)  Ca+2(aq) + 2F-(aq)
[Ca 2 ][F - ]
[CaF2 ]
a.
K
c.
K  [Ca 2 ] 
[ F- ]
[CaF2 ]
[Ca 2 ][ F- ]
[CaF2 ]
b.
K
d.
K  [Ca 2 ][ F- ]2
8. In the equilibrium reaction:
N2(g) + 3H2(g)  2NH3(g)
If the system is at equilibrium, and the total pressure is
increased by reducing the volume, the position of
equilibrium ______________.
a.
stays constant
b.
shifts to the right
c.
shifts to the left
d.
cannot be predicted
9. When solid sodium hydroxide dissolves in water, the
process is exothermic: NaOH(s)  Na+(aq) + OH-(aq) +
heat
If you add NaOH to water and it doesn’t all dissolve, would
you heat it or cool it to get more to dissolve?
10. NO is formed at high temperatures from the
combustion of atmospheric N2 in the internal combustion
engine:
N2(g) + O2(g) + heat  2NO(g)
In order to cut down on the production of NO, a
greenhouse gas and a precursor of NO2, which is an
ingredient of smog, would you increase or decrease the
combustion temperature of the engine?
11. At a given temperature, K = 50 for the reaction:
H2(g) + I2(g)  2HI(g)
Calculate the equilibrium concentration of H2 given:
[I2] = 1.5 x 10-2 M and [HI] = 5.0 x 10-1 M
a.
1.5 x 10-2 M
b.
3.0 x 10-2 M
c.
5.0 x 10-1 M
d.
3.3 x 10-1 M
12. Write out the Ksp expression for the reaction:
Ag2SO4(s)  2Ag+(aq) + SO42-(aq)
[Ag  ]2
a. .
K sp 
b.
2
[SO 4 ]
+ 2
Ksp = [Ag ] [SO42-]
2
c.
d.
K sp 
[Ag  ]2 [SO 4 ]
[Ag 2SO 4 ]
Ksp = [Ag+]2
PS 17.1
Review: Chapter 17 Equilibrium
(p 2 of 2)
13. If a saturated solution of PbCl2 is prepared by dissolving some of the salt in distilled water and the concentration of Pb 2+ is
determined to be 1.6 x 10-2 M, what is the value of Ksp?
a.
2.6 x 10-4
b.
2.0 x 10-4
c.
3.2 x 10-2
d.
1.6 x 10-5
ANSWERS
1. Choice B is consistent with collision theory, which states that molecules must collide in order to react. Higher concentrations
lead to more frequent collisions. Higher temperatures provide more energy for successful collisions, and also increase the
number of collisions because the kinetic energy increases (KE = ½ mv2; mass is constant so velocity goes up, leading to more
collisions).
2. Choice B provides a reasonable answer. Catalysts provide an alternate pathway of lower activation energy; thus, more
molecules have sufficient energy to react upon collision.
3. C provides the correct response. Catalysts such as Cl take part in one step of a reaction and are regenerated in another step.
4. A correctly gives the answer. When forward and reverse reaction rates are equal, there are no longer any changes in
concentration of reactants or products, and equilibrium has been reached.
5. A correctly expresses the equilibrium constant expression.
2
6. D is correct. Putting concentrations into the equilibrium expression yields: K  (0.060)
(0.055)
7. D correctly expresses the equilibrium expression. CaF2 does not appear in the expression because it is a solid.
8. B reflects the change in the equilibrium position. If pressure is increased, Le Châtelier’s Principle tells us that the
equilibrium will shift to accomplish a reduction in pressure (relieve the stress). Since there are 4 moles of gas on the left and
only 2 moles on the right, the system will shift towards the smaller number of moles, toward the right, so that the pressure will
be reduced.
9. Cool it to make it more soluble. Think of heat as a product that, if removed, will make more product form, i.e. more NaOH
will dissolve as the position of equilibrium shifts to the right.
10. Decrease the temperature to shift the position of equilibrium to the left. If we lower the temperature we remove heat,
and so the equilibrium shifts to produce more heat. By doing so, less NO will be produced.
11. D 4 gives the numerically correct answer, which is arrived at by substituting into the equilibrium expression and solving for
[H2].
Rearranging :
[HI] 2
(5.0x10 -1 ) 2
(5.0x10 -1 ) 2
K  50 

[
H
]

 3.3x10 -1 M
-2
2
[H 2 ][ I 2 ] [H 2 ](1.5x10 )
(50)(1.5x10 -2 )
12. Choice B is correct. Remember that because Ag2SO4 is a solid, it does not appear in the equilibrium expression.
13. D should be selected. The equilibrium for the dissolution of the salt is written as: PbCl 2(s)  Pb2+(aq) + 2Cl-(aq)
We are given a value for the concentration of lead(II) ion. The concentration of chloride ion must be twice that of the lead.
Next we put these values into the Ksp expression and solve:
Ksp = [Pb2+] [Cl-]2 = (1.6 x 10-2) (3.2 x 10-2)2 = 1.6 x 10-5