AP Chemistry Practice Exam 2 ❮ 357- A—The higher boiling isomer is more polar than
the other isomer because the two very electroneg-
ative chlorine atoms are on one side, which leads
to their polar bonds working together. When the
chlorine atoms are on opposite sides, their polar
bonds work against each other. - D—Add the equations together and cancel any
species that appear on both sides (intermediates). - A—The first step in the mechanism is the slow
(rate-determining) step. It is the slowest because it
has the highest activation energy. For the first step to
be the slow step, the first peak must be the highest. - C—The rate law always considers the slowest
step in a mechanism. There is one molecule of
N 2 O 5 as the reactant in the slow step; therefore,
the rate law will only use the concentration
of this reactant raised to a power equal to the
number of molecules in the slow step. - C—Use Graham’s law; a molecule with one-
fourth the molar mass will diffuse at double the
rate. Neon is the nearest to one-fourth the molar
mass of krypton. - A—This is a consequence of kinetic molecular
theory. The average kinetic energy depends only
on the absolute temperature. - D—Heavier nonpolar species exhibit greater
London dispersion forces, and stronger attractive
forces lead to greater deviation from ideal behav-
ior under a given set of conditions. - A—An increase in volume will cause an equilib-
rium to shift toward the side with more gas. There
are four gas molecules on the left side and five gas
molecules on the right side; therefore, an increase
in volume will result in a shift to the right, which
increases the amount of HCl (and O 2 ). B will
have the opposite effect. Cooling an endothermic
equilibrium will cause a shift to the left, which will
decrease the amount of HCl. D will yield no change
because helium is not part of the equilibrium. - D—The amount of time necessary is a kinetics
problem. There is not kinetic data presented to
make the determination possible.
33. A—The ICE table for this equilibrium is:
Cl 2 H 2 O HCl O 2
1.0 atm 1.0 atm 0 0- 2 x - 2 x + 4 x +x
1.0 - 2 x 1.0 - 2 x + 4 x +x
From the equilibrium line on the table, the equi-
librium pressure should be:
(1.0 - 2 x) + (1.0 - 2 x) + (+ 4 x) + (+x) = 2.0 + x
Therefore, the equilibrium pressure will be
greater than 2.0 atm by an amount equal to x.- D—To determine the amount, it would be neces-
sary to know the value of the equilibrium con-
stant. A cannot be correct because it is not possible
for the amount of any of the materials to be zero
at equilibrium. B and C cannot be correct because
at least some of the O 2 would be converted to Cl 2
and H 2 O leading to a decrease in the amount. - B—The pH of a 1.0 M methylamine solution
is the highest; therefore, it is the strongest of the
bases. For this reason, the pH at the equivalence
point of the methylamine titration will be the
highest. - A—As in ammonia, all these compounds behave
as Brønsted–Lowry bases by accepting a hydro-
gen ion. The reaction involves the hydrogen ion
attaching to the lone pair on the nitrogen atoms. - C—These are all bases because the nitrogen atom
is capable of reacting with a hydrogen ion by
donating its lone pair to the hydrogen ion. The
oxygen atom pulls electron density away from
the nitrogen atom causing the nitrogen atom to
attract the lone pair more strongly making it less
able to donate the pair to a hydrogen ion. - A—Only A can undergo oxidation, as there are
higher oxidation states of nitrogen. For example,
HNO 3 is the most likely oxidation product and,
unlike HNO 2 , it is a strong acid. - D—Hydrogen bonding is possible when hydrogen
is attached to N, O, and F. D is the only compound
in the diagram where this is true. The simple pres-
ence of hydrogen and N, O, or F is insufficient.
21-Moore_PE02_p341-370.indd 357 31/05/18 1:54 pm