- A—When dealing with gaseous equilibria,
volume changes are important when there is a
difference in the total number of moles of gas on
opposite sides of the equilibrium arrow. All the
answers, except A, have differing numbers of
moles of gas on opposite sides of the equilibrium
arrow. - C—Hydrolysis of any ion begins with the inter-
action of that ion with water. Thus, both the ion
and water must be on the left side of the equilib-
rium arrow, and hence in the denominator of the
equilibrium-constant expression (water, as with all
solvents, will be left out of the expression). The
oxalate ion is the conjugate base of a weak acid. As
a base it will produce OH−in solution along with
the conjugate acid (HC 2 O 4 −) of the base. The equi-
librium reaction is: C 2 O 42 −(aq) + H 2 O(l)
OH−(aq) +HC 2 O 4 −(aq). - D—The low value for the equilibrium constant
means that the equilibrium lies to the left. For
this to be true, the weaker acid and the weaker
base must be on the left side. - E—The equilibrium given is actually the sum of
the following three equilibria:
ZnS(s) Zn^2 +(aq) +S^2 −(aq)
Ksp=1.6 × 10 −^24S^2 −(aq) +H+(aq) HS−(aq)
K=1/Ka2+1/1 × 10 –9HS−(aq) +H+(aq) H 2 S(aq)
K′=1/Ka1+1/9.5 × 10 –8
Summing these equations means you need to
multiply the equilibrium constants:Ksum=KspKK′=Ksp/Ka2Ka1=1.6 × 10 −^24 /[(1 × 10 −^19 ) (9.5 × 10 −^8 )]- D—As the reaction moves to the left, the H 2 O
behaves as a base (accepts H+). When the reaction
moves to the right, HPO 42 −behaves as a base. - D—The equilibrium constant for the two succes-
sive ionizations will be the product of the two
equilibrium constants given. Thus, K=Ka1Ka2=
(1.5 × 10 −^3 )(2.0 × 10 −^6 ). - B—The addition or removal of some solid, as
long as some remains some present, will not
change the equilibrium. An increase in volume
will cause the equilibrium to shift towards the
side with more moles of gas (right). Raising the
temperature of an endothermic process will shift
the equilibrium to the right. Any shift to the
right will increase the amounts of the products.- E—Assuming 1 mol of each reactant is used, the
equilibrium quantities would be: [C 2 H 4 ] = 1 −x,
[O 2 ] = 1 − 3 x, [CO 2 ] = 2 x, and [H 2 ] = 2 x. Unless
the value of x is known, it is not possible to relate
the actual concentrations of any reactant to any
product. - B—Using the following table:
The presence of 0.20 mol of CO (0.20 M) at
equilibrium means that 2x=0.20 and that x=
0.10. Using this value for x, the bottom line of
the table becomes:The equilibrium expression is: K =
[CO]^2 [H 2 ]^2 /[CH 4 ][CO 2 ]. Entering the equilib-
rium values into the equilibrium expression gives:
K=(0.20)^2 (0.20)^2 /(0.20)(0.30)- A—The addition of a product will cause the
equilibrium to shift to the left. The amounts of
all the reactants will increase, and the amounts of
all the products will decrease (the O 2 will not go
below its earlier equilibrium value since excess
was added). The value of Kis constant, unless the
temperature is changed. The rates of the forward
and reverse reactions are equal at equilibrium. - C—The only way to change the value of Kis to
change the temperature. For an exothermic
process (ΔH <0), Kis increased by a decrease in
temperature.
[CH 4 ] [CO 2 ] [CO] [H 2 ]
Equilibrium 0.20 0.30 0.20 0.20[CH 4 ] [CO 2 ] [CO] [H 2 ]
Initial 0.30 0.40 0 0
Change −x −x + 2 x + 2 xEquilibrium 0.30 −x 0.40 −x 2 x 2 x236 Step 4. Review the Knowledge You Need to Score High