AP Chemistry Practice Exam 1 ❮ 327- C—There are several ways of solving this prob-
lem. One way is to determine the moles present
in the original containers, which must be the
same as in the final container. In each case,
moles = n = PV/RT. Numbering the containers
from left to right as 1, 2, 3, and 4 gives:
n 4 = n 1 + n 2 + n 3
=
⋅
⋅
+
(1.0 atm)(1L)0.0821Latm
molK(300K)
⋅
⋅
+
(1.0 atm)(2 L)0.0821
Latm
molK(300K)
⋅
⋅
(1.0 atm)(1 L)0.0821
Latm
molK(200 K)
1
(0.0821)(300)
2
(0.0821)(300)
=
+
+
1
(0.0821)(200)
3
(0.0821)(300)
1
(0.0821)(200)
=
+
1
(0.0821)(100)
1
(0.0821)(200)
=
+
()
=
+
1
0.0821
1
100
1
200
molP 4 = n 4 RT/V =()
+
⋅
⋅
=
1
0.0821
1
100
1
200
mol 0.0821
Latm
molK(300K)
4 L
+
=
1
100
1
200
mol(atm) (300)4
(0.010 + 0.0050) atm (75.0) = 1.1 atmOn the exam, it is not necessary to write out all
these steps. Take shortcuts.- A—Stronger intermolecular forces lead to higher
boiling points. Even though COS has dipole-
dipole forces, which are usually stronger than
the London dispersion forces present in CS 2 ,
the greater molar mass of CS 2 leads to a London
dispersion force contribution that is sufficient
to compensate for the general trend of dipole-
dipole forces being stronger than London dis-
persion forces. This is why comparisons should
only be made between molecules of similar
molecular masses. - B—At equilibrium the reverse reaction is going
at a steady rate (equal to that of the forward
reaction and not equal to 0). A sudden decrease
in pressure will cause the rate of the reverse reac-
tion to increase to generate more gas to increase
the pressure. It will eventually slow and become
constant again. - A—Only the slow step in a mechanism leads to
the rate law. There is one H 2 O 2 and one I- in
the slow step; therefore, one of each of these will
be in the rate law. - B—This must be true according to the Law
of Conservation of Energy (First Law of
Thermodynamics). Answers A and C violate
the Law of Conservation of Energy. For the
final temperature to be the average, the metals
would need to have equal masses and equal
specific heats. The masses given in the problem
are clearly different, and it is extremely unlikely
that two different metals would have the same
specific heat.
20-Moore_PE01_p307-340.indd 327 31/05/18 1:58 pm