AP Chemistry Practice Exam 1 ❮ 325
- C—It is necessary to use the half-life relationship
for first-order kinetics. This relationship is t1/2 =
0.693/k, and t1/2 = (0.693/86 h-^1 )(3600 s/h) =
29 s. To save time on the exam you can approxi-
mate this equation as t1/2 = (0.7/90)(3600).
Dividing 3600 by 90 gives 40, and 40 times 0.7
is equal to 28. If the half-life is ≈ 28 s, then the
time (58 s) is equivalent to about two half-lives,
so one-fourth of the sample should remain. - D—The loss of 0.20 mol of CO means that
0.40 mol of H 2 reacted (leaving 0.20 mol) and
0.20 mol of CH 3 OH formed. Dividing all the
moles by the volume gives the molarity, and:
Kc = (0.20)/(0.40)(0.20)^2 = (0.10)/(0.20)(0.20)^2
= 0.10/(0.20 × 0.04) = 0.10/0.008 = 12 - A—Based upon the stoichiometry of the reac-
tion, H 2 (g) will form twice as fast as CH 3 OH(g)
disappears. The numerical values are the same;
however, the rates have opposite signs. - B—At equilibrium, there is no net change
because the forward and reverse reactions are
going at the same rate. - B—As the reaction approaches equilibrium,
there is a net decrease in the number of moles of
gas present. A decrease in the number of moles
of gas will lead to a decrease in pressure - B—The loss of 1.5 atm of CH 3 OH(g), based on
stoichiometry leads to the formation of 1.5 atm
of CO(g) and 3.0 atm of H 2 (g); therefore, the
net change is (-1.5 + 1.5 + 3.0) atm = 3.0 atm. - A—Oxygen (odorless) evolves at the anode (posi-
tive) and ammonia (distinctive odor) evolves at
the cathode (negative). From the choices given,
the only gas with a distinctive odor that could
form is ammonia, NH 3. For this reason, the
nitrite ion half-reaction must be the cathode reac-
tion. The anode half-reaction must be the reverse
of one of the reactions in the table. The only one
of the half-reactions given that generates an odor-
less gas is the one involving oxygen, O 2. - A—The lead electrode is the cathode (-0.13 V),
and the chromium electrode is the anode (reverse
sign, +0.91 V). Cell voltage = (-0.13 + 0.91) V =
+0.78 V. The standard voltage of a galvanic cell
is always positive.
34. D—It is necessary to consider all possible half-
reactions that might occur during electrolysis.
Of the half-reactions listed, only the reduction
of water and the reduction of aluminum ions are
applicable to this experiment. For any electroly-
sis, the half-reaction requiring the least amount
of energy will take place. So, while it is possible
to reduce both water and the aluminum ion, the
reduction potential for water is lower (requiring
less energy), so water will reduce in preference to
aluminum ion.
35. B—In order for substance A to reduce substance
B, substance A must be a stronger reducing
agent than substance B. Aluminum is a strong
reducing agent as indicated by the large nega-
tive potential in the table (-1.66 V). The only
substance in the table that has a more negative
potential is rubidium, Rb (-2.93 V).
36. C—Silver appears twice in the list of half-
reactions, once in the simple reduction of silver
ions and once in the reduction of silver chloride,
AgCl. The AgCl half-reaction clearly shows that
this compound is a solid. This solid will form as
the silver ion in the silver nitrate solution reacts
with chloride ion from the potassium chloride
in the salt bridge. The formation of solid silver
chloride will alter the concentration of the silver
ion, which leads to a change in the cell potential.
37. B—The weakest acid (smallest Ka) will have the
highest pH at the endpoint.
38. D—The buffer capacity only depends on the
number of moles present. All three solutions
have the same number of moles. - C—Less acid would require less than the ideal
amount of base to reach the endpoint. Therefore,
the endpoint would occur too soon. - D—An acid contaminant was present and it
would be necessary to add additional base to
titrate this acid in addition to the phenol. - B—Since only the plot of 1/[A] versus time
yielded a straight line, this implies that the reac-
tion is second-order in A. - C—The electron configuration of carbon in
1s^2 2s^2 sp^2. The removal of the last electron (2p)
requires the least amount of energy.
20-Moore_PE01_p307-340.indd 325 31/05/18 1:58 pm