Kg
mol
500
.
0
518
.
0
(1)
=
3.44988
o
C
this is the CHANGE
∆
T
bp
=
T
new
T
−
normal
T
new
=
∆
T
bp
+
T
normal
=
110.6
+
3.44988
o
C
≈
114
o
C
ans (d)
1
6.
answer (e) is correct
. A rate law cannot be determined by inspection of the overall balance equation for the bulk
chemical reaction.
Answers (a), (b), (c), and (d) all treat the overall reaction equation like a mechanism step, and try to take the rate law
relationships from the stoichiometry. This is not appropriate because the mechanism is not provided.
7. Use the integrated rate law for a 1
st
order reaction,
ln
kt
amt
starting
left
amt

=
All of the variables are provided except amt left. Substitute the known values and solve.
Ln
=
0314
.
0
left
amt
(6.4 x 10
−
3
−
min
1
−
)(62.0 min)
It is helpful to recall that
Ln
=
0314
.
0
left
amt
ln (amt left)
ln (0.0314)
−
ln (amt left)
(
3.46095)
=
0.3968
−
−
−
amt left
=
0.02112 M
ans. (d)
8.
Let “cat” mean “catalyst”
and “no cat” means “no catalyst”
eqn (i)
ln k
with cat
=
ln A
−
(
29
cat
with
a
E
RT
,
1
eqn (ii)
ln k
no cat
=
ln A
−
(
29
cat
no
a
E
RT
,
1
Subtract (ii)
from (i):
ln k
with cat
ln k
−
no cat
=
0
−
(
29
cat
with
a
E
RT
,
1
+
(
29
cat
no
a
E
RT
,
1
recall that
ln k
with cat
ln k
−
no cat
=
ln
cat
no
cat
with
k
k
which is the ratio requested.
ln
cat
no
cat
with
k
k
=
(
29
kJ
kJ
K
K
mol
kJ
x
55
64
)
673
)(
/
10
3145
.
8
(
1
3

⋅

ln (ratio)
=
1.698
ratio =
e
1.698
=
4.9947
≈
5.0
ans (a)
9.
The plot of 1/[X] vs. time is linear with positive slope. This is diagnostic for 2
nd
order kinetics.
If
rate is only
dependent on [X] then the term [Y] does not need to appear in the rate law but if it did, it would have an
exponent of 0.
Only
answer (b) is correct
.
10. (a) wrong. Increasing temperature on a reaction not at equilibrium has no effect on the concentration of
reactants. (Had equilibrium been established, and had Temp THEN been increased, for an exothermic reaction
more reactants would form over time. But that is not this question. This question is about kinetics.)
(b)
Wrong. Increasing Temp. means that more molecules will have the threshold E
a
for reaction, but the Ea itself is
a constant that depends on the molecular structure, which includesmlowest energy configuration, and probable
alternative configurations. Molecular structure does not change with temperature.
(c)
Correct
Collision frequency increases because the average particle velocity is greater at higher temps, they
travel faster and run into each other more often.
(d) Correct
. The fraction of collisions with sufficient energy increases as more particles have energy
> Ea
2
(e)
wrong. See answer (b)
(f)
wrong. It is true that the rate of internal rotations may increase, but the statistical probability that a given
molecule has a given orientation at any point in time remains the same.
11.
Perform algebraic manipulation on the equations (i) and (ii) until their sum yields equation (iii).
Keep eqn. (i)
as is, and its Keq as is.
Keep eqn. (ii)
as is; add the two equations and multiply their K’s, to find K
overall
.
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 Spring '08
 LEMASTER
 Chemistry, Mole, Reaction, Solubility, Chemical reaction, Keq.