Chemistry 122
Mines, Spring, 2016
Answer Key, Problem Set 2—Complete and with full explanations
1. 13.45 [in Mast]; 2. 13.46 [in Mast]; 3. MP; 4. MP; 5. NT1; 6. 13.48 [in Mast]; 7. 13.50 [in Mast]; 8. 13.53 [in Mast]; 9. NT2;
10. NT3; 11. NT4; 12. MP; 13. MP; 14. MP; 15. NT5; 16. NT6; 17. NT7; 18. MP; 19. NT8; 20. NT9; 21. NT10;
22. MP; 23. MP; 24. MP; 25. NT11; 26. MP; 27. NT12; 28. NT13; 29. NT14; 30. NT15; 31. MP
--------------------------------------Integrated Rate Laws (and Half Life)
1. 13.45 [in Mast]; Indicate the order of reaction consistent with each observation.
(a) A plot of the concentration of the reactant versus time yields a straight line.
(b) The reaction has a half-life that is independent of initial concentration.
(c) A plot of the inverse of the concentration versus time yields a straight line.
2. 13.46 [in Mast];
Indicate the order of reaction consistent with each observation.
(a) The half-life of the reaction gets shorter as the initial concentration is increased
(b) A plot of the natural log of the concentration of the reactant versus time yields a
straight line.
(c) The half-life of the reaction gets longer as the initial concentration is increased.
Approach / Rationale for 13.45 and 13.46:
For 0th order, the [A] vs t plot is linear ([A]t  -kt + [A]0), since the rate (=|slope|) is constant (not
dependent on [A]). This means 13.45(a) matches 0th. 13.46(c) also describes 0th because if the
rate is constant, then if you start with a greater concentration of A and the rate of loss is constant, it
will take more time for half of it to react away. An Analogy: If you are driving at a constant rate of
50 mph, it will take you more time to drive 100 miles than to drive 30 miles, right? So if you were
going to drive 200 miles, it would take you more time to get to “halfway” (100 miles) than if you
were going to drive 60 miles (where getting to “halfway” would be driving 30 miles). The greater
your overall distance, the more time it takes to get to “half that distance”.)
You could also plug in the point (t1/2, ½ [A]0) into [A]t  -kt + [A]0 and derive that for 0th order
1
[A] 0
, which shows that t1/2 is proportional to [A]0 (0th order).
t1 / 2  2
k
Or you could just make a plot like I did in class, and convince yourself that if [A] vs. t is linear, each
time you drop the [A] by half and go to the right until you hit the line (keep making “triangles”), the
triangle gets smaller and smaller (very geometrical here). That is, the half life decreases each time.
For 1st order, the [A] vs t plot is an exponential decay function: [A] t  [A] 0 e  kt . If you take the ln of
both sides, you get (after a bit of algebra): ln [A]t  -kt + ln [A]0. Thus, 13.46(b) is consistent with
1st order. You can derive for yourself (see notes, text) or perhaps memorize the fact that for 1st
ln 2 0.693
. This shows that t1/2 does not depend on [A] for 1st order

order, the half life equals
k
k
(consistent with 13.45(b) being consistent with 1st order).
For 2nd order, you should be able to rationalize that the [A] vs t plot takes “longer” to get to zero
relative to a 1st order case because R is more sensitive to changes in [A] in 2nd order than 1st. Thus
as [A] decreases, the rate gets “more smaller” in 2nd vs. 1st order. Thus 13.46(a) matches 2nd order
(as [A] increases, rate increases by a greater factor than in 1st order, so the half life should be
shorter). To get 13.45(c) correct, you (unfortunately) have to memorize that the integrated rate law
1
1
1
 kt 
(or at least that it’s
vs t that is a straight line). If you know
for 2nd order is:
[A] t
[A] 0
[A] t
1
that, you could substitute in the point (t1/2, ½ [A]0) and derive that t1 / 2 
for 2nd order (to
k [A] 0
answer 13.46(a) without the analysis given above).
PS2-1
Answer Key, Problem Set 2
3. MP. Mastering Problem—no answer in key. However, see analysis above, as the ideas are very similar!
Hint: Focus in on what “a rate that is observed to slow down as the reaction proceeds” means
(in terms of which orders could be consistent with that).
4. MP. Mastering Problem—no answer in key. However, this problem applies the ideas from the prior
problems (conceptual) to actual numerical data, so you should be able to make a connection and
use the ideas above to help in this problem. For the “calculation of k” question, remember that
once you have the order, you should be able to write the integrated rate law equation for the
reaction. Then you can substitute in any (t, [A]t) point into that equation and solve for k.
5. NT1. 13.47.
The data below show the [AB] versus time for the reaction represented by AB(g) → A(g) + B(g):
Time (s)
0
50
100
150
200
250
300
350
400
450
500
[AB] (M)
0.950
0.459
0.302
0.225
0.180
0.149
0.128
0.112
0.0994
0.0894
0.0812
Determine the order of the reaction and the value of the rate constant. Predict the concentration of AB at 25 s.
Answers: (a) order is 2; (b) k = 0.0226 M-1s-1;
(c) [AB]25 s  0.618 M
Strategy:
(a) Short way: Find the initial half life (approximately) and the 2nd (or any other later) half life.
 If the t1/2 gets shorter with time (decreasing [reactant]), it’s 0th order.
 If it stays the same, it is 1st order.
 If it gets longer with time (decreasing [reactant], it’s 2nd order (see prior problem for
rationale here).
Long way: Make a plot of [AB] vs t (or take differences [’s] “by hand” between time
intervals).
 If the plot is a line (or if the ’s are the same), it’s 0th order.
 If not, it’s either 1st or 2nd order
 Plot ln [AB] vs time. If it’s linear, it’s 1st order. If not, it’s 2nd order (since in this class,
there are only 3 options for this kind of Q: 0, 1, or 2
 (If you really wanted to verify 2nd order, plot 1/[AB] vs time. If it’s linear, it’s 2nd order.
(b) Once order is determined, substitute [AB]0 and any (t, [AB]) point into the proper integrated
rate law for that order. Solve for k.
NOTE: Mastering points out that you could also find k by determining the slope of the
appropriate linear plot. But that assumes you’ve actually made plots (and seems like
more work to me, unless you have a program calculate the best fit line for you), so I
did not show that approach in this key.
(c) Once k is known, substitute t = 25 s into the integrated rate law (with “known” k value now
included!) with [AB]0 and solve for [AB]t=25!
Execution of Strategy:
(a) Short way. Half of 0.95 is about 0.48, so the first half life is a little less than 50 s. Half of
0.459 is about 0.23. So the 2nd half life is about 100 seconds (150 – 50 = 100).
 2nd order
PS2-2
Answer Key, Problem Set 2
Long way:
not 1st order
b/c not linear
not 0th order
b/c not linear
 2nd order (by default). No real need to plot 1/[AB] vs t, although one could.
(b) 2nd order integrated rate law:
1
1
 kt 
 [substituting in point (100 s, 0.302 M)]
[AB] t
[AB] 0
1
1
 k (100. s) 
 3.311..M -1  k (100. s)  1.052..M -1  2.258 M -1  k (100 s)
0.302 M
0.950 M

1
2.258 M -1
 0.02258.. M -1s 1 
100 s
s
1
M
6
2
2
0
.
0
 k 
NOTE: If you used a different point than t = 100 s, the k value probably is a little bit different.
As long as you’re very close to my result, you should consider it fine if the work
seems analogous to mine. This applies (obviously) to the [AB] value below, also.
(c) Now substituting in t = 25 s with value of k (and [AB]0) back into
1
1
 kt 
[AB] t
[AB] 0
yields:
1
1
 0.02258 M-1s 1(25. s) 
 1.617..M-1
[AB] 25 s
0.950 M
 [AB] 25 s 
1
 0.6183..  0.618 M
1.617..M-1
CHECK: Answer is reasonable—is
between 0.459 M and 0.950 M (t = 50 and
t = 0 points)
6. 13.48 [in Mast] Mastering Problem—no answer in key. Use strategy in prior problem, but obviously if
the order is different, the integrated rate law will be different (but approach the same).
7. 13.50 [in Mast] Mastering Problem—no answer in key. Use strategy in prior problem, but obviously if
the order is different, the integrated rate law will be different (but approach the same).
8. 13.53 [in Mast]
The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.42 x 10-4 s-1 at a
certain temperature.
(a) What is the half-life for this reaction?
(b) How long will it take for the concentration of SO2Cl2 to decrease to 25% of its initial concentration?
(c) If the initial concentration of SO2Cl2 is 1.00 M, how long will it take for the concentration to decrease to 0.78 M?
(d) If the initial concentration of SO2Cl2 is 0.150 M, what is the concentration of SO2Cl2 after 2.00 x 102 s? After 5.00
x 102 s?
**Since this is in Mastering, the values may be different and I will not post a detailed solution here.**
However, the strategy / analysis should be something like this: Recognize that half life is related to k
via the integrated rate law. The precise relationship varies, depending on the order. You needed to
know these relationships for earlier problems in this set. In your analysis of the other parts, you
should recognize that either reactant concentrations and times are given or asked for, or “percent of a
reactant’s initial concentration” is given, and these are all either variables in an integrated rate law, or
can be made to relate to them (e.g., the “fraction of A remaining at time t” is equal to [A]t/[A]0, as
PS2-3
Answer Key, Problem Set 2
discussed in lecture, and % is just fraction x 100). So all of these problems are versions of
“integrated rate law calculation” problems.
**Please also note that “decreased to 25%” means something different than “decreased by 25%”.
9. NT2.
A reaction is first order in A. It takes 46 seconds for the concentration of A to decrease by one-half of its initial value.
How long will it take for the concentration of A to decrease to one-fourth of its initial value? to one-eighth of its initial
value?
Answers: 92 s (two half lives), 138 s (three half lives)
Reasoning. For a reaction that is first order with respect to a reactant, the half life is independent of
reactant concentration (and hence time) (see prior problem). Thus, after one half life, there is half of
the original concentration remaining. Considering this as your next “starting point”, after another half
life, the concentration will become half of “one half of the original amount” or one fourth of the original
concentration. After three half lives, the concentration becomes half of “one fourth of the original” or
one eighth. So it would take 92 seconds in this case (two half lives back to back) for the
concentration to decrease to one-fourth of its initial value and 138 seconds (three half lives) to reach
one-eighth.
Obviously you could use the t1/2 relation to solve for k and then use the integrated rate law

[A] t kt 
 [A]t  [A] 0 e kt 
e  to solve for t when the fraction remaining is 1/4 or 1/8, but it is much
[A] 0


faster (and helpful, generally speaking!) to realize that you can just take two, three, or however many
half lives you wish to get to a certain power of ½ remaining.
10. NT3.
The pictures to the right represent the
progress of the reaction A  B in which A
molecules (represented, overly simplistically, as
open spheres) are converted to B molecules
(represented, overly simplistically, as black
spheres).
(a) What is the order of the reaction?
(Remember, always give reasoning!)
(b) t = 3 min
(b) Draw a picture that shows the number of A and B molecules present at t = 3 min.
Answers:
(a) 0th order; (b) (see right )
Reasoning:
(WARNING: THIS IS NOT AS SIMPLE AS YOU MIGHT THINK. YOU REALLY NEED TO LOOK CLOSELY AT THE
PICTURES IN PROBLEMS LIKE THIS!) The time interval between the "snapshots" is the same, and
the change in the number of particles is also the same (two A's convert to B between 0 and 1
min, and two A's also convert to B between 1 and 2 minutes). Since the volume is the same,
that means the change in concentration per change in time is the same, and that means that
the rate is the same between times 0 and 1 and 1 and 2. In order for the rate to remain the
same as time goes by (and as the concentration of the species changes), the process
must be zeroth order. This can be seen from the rate law: Rate = k[A]0  Rate = k 
Rate doesn't depend on concentration. Note that another approach here could be to make a
plot of the number of A’s as a function of time (analogous to a [A] vs t plot!). Note that you’d
get a straight line in this problem, which would also tell you it is zeroth order!
NOTE: This kind of problem is essentially a nanoscopic picture version of the kinds of problems
in 5, 6, and 7 above! Here you “count reactant particles” whereas there you had to look
at reactant concentrations (or concentration differences). Try to go back and see which
one of those problems is analogous to this one (i.e., is zero order).
PS2-4
Answer Key, Problem Set 2
11. NT4. Consider a reaction with only one reactant, A. How will doubling the concentration of A affect
the (i) (initial) rate and the (ii) (initial) half life of the reaction in the following cases?
(a) reaction is 0th order in A; (b) reaction is 1st order in A
Answers: (a) In 0th order, doubling [A] does not affect rate, but it doubles initial half life
(b) In 1st order, doubling [A] doubles rate, but half life stays the same.
Reasoning:
These ideas were essentially already covered earlier in the problem set (integrated rate
laws). I apologize for the repetition here. Hopefully it was a good “review” of sorts of “early”
PS2 material.
Mechanisms
12. MP. Mastering Problem—no answer in key. Read the introductory material in the problem for help
(and also look at class notes and text!)
13. MP. Mastering Problem—no answer in key. Read the introductory material in the problem for help
(and also look at class notes and text!)
14. MP. Mastering Problem—no answer in key. Read the introductory material in the problem for help
(and also look at class notes and text!)
15. NT5 13.73.
Consider the overall reaction, represented by the following equation, and which is experimentally observed to be
second order in AB and zero order in C:
AB + C → A + BC
Is the following mechanism valid for this reaction?
AB + AB
k

AB2 + A (slow)
AB2 + C
k

AB + BC (fast)
1
2
Answer: Yes
Reasoning:
Two requirements must both be met by a proposed mechanism in order to consider it “valid”
(i.e., “possible”) for an overall reaction:
1) The sequence of elementary steps must sum to the equation for the overall reaction.
2) The rate law that the mechanism predicts must match the observed rate law.
Proof of (1) for this mechanism:
Sum of the equation for the two steps yields:
AB + AB + AB2 + C

AB2 + A + AB + BC
Cancelling out identical species on both sides yields:
AB + C → A + BC
(which matches the overall reaction equation)
Demonstration of (2) for this mechanism:
Strategy and Approach
To determine the rate law that a mechanism predicts, remember this key concept:
“The overall rate of any step-wise process can be no faster than its slowest step.”
This ultimately means the following:
Overall reaction rate  rate of the slowest step OR
Roverall  Rslow
Now use the fact that the rate law for an elementary reaction is “known” (i.e., does not
PS2-5
Answer Key, Problem Set 2
need to be determined experimentally), because of theoretical considerations. Namely,
since each reactant in an elementary step is involved in the collision that leads to
reaction (because an elementary step occurs in one collisional event only!), the rate is
proportional to the collisional frequency of those reactants.
Hopefully, a bit of thought will convince you that
the frequency of collisions will be proportional to the concentration of each “body”
that is involved in the collision.
For example, if A collides with B in an elementary step, then the collisional frequency will
be proportional to [A] and [B]:
If you double [A] (keeping [B] the same), the collisional frequency will double.
If you double [B] (keeping [A] the same), the collisional frequency will also double.
If you double both [A] and [B], the collisional frequency will quadruple.
If A collides with another A, then each “body” counts “separately”—it’s like doubling [A]
and [B] in the prior paragraph—the collisional frequency will quadruple.
Okay, I’ve taken up enough space in this key on this topic! See Table 13.3 and the
discussion just prior to it for more insight on this. The bottom line is that every
elementary step has a known rate law based on the coefficients in the elementary step’s
equation, and so one can write:
Roverall  Rslow = kslow[reactant #1 in slow step]its coefficient[reactant #2 in slow step]its coefficient
where the boxed equation is just the rate law for the slow (rate-determining) step.
Thus, to determine the rate law a mechanism predicts:
1) Find the slow step
2) Write the rate law for the slow step
3) Set this equal to the overall rate.
Important Note: The above (alone) “works” for reactions with a slow first step. If the slow
step is not the 1st step, more work needs to be done, but I am not going to make you
responsible for doing the calculational work for such a mechanism.
Execution of Strategy
Roverall  Rslow k1[AB][AB]  k1[AB]2 (predicted by this mechanism)
Since the overall reaction’s rate  k1[AB]2, the reaction is predicted by this mechanism to
be 2nd order in AB and 0th order in C.
Since this matches the experimentally observed behavior (and the two steps sum to the overall
equation as shown earlier), this mechanism is “valid” (i.e., possible).
16. NT6 13.76.
Consider this two-step mechanism for a reaction:
NO2(g) + Cl2(g)
k

ClNO2(g) + Cl(g) (slow)
NO2(g) + Cl(g)
k

ClNO2(g)
1
2
(fast)
(a) What is the overall reaction (equation)?
(b) Identify the intermediates in the mechanism.
(c) What is the predicted rate law?
Answers:
(a) 2 NO2(g) + Cl2(g)  2 ClNO2(g)
(b) Cl is the only intermediate
(i.e., 1st order in both NO2 and Cl2)
(c) R = k1[NO2][Cl2]
PS2-6
Answer Key, Problem Set 2
Reasoning:
(a) Cl appears on the right side of step one, but on the left side of step two. Thus, is cancels out
of the overall equation. Two NO2’s and a Cl2 remain on the left side (reactants), and two
k
2 ClNO2(g)
ClNO2’s remain on the product side. Hence: 2 NO2(g) + Cl2(g) 
1
(b) An intermediate is “a species produced during some step in a mechanism and then used up
in a later step”. Cl was produced in Step 1 and used in Step 2 so it meets the definition.
Although NO2 is also used up in the 2nd step, it is “only” a reactant because it was not
produced by any prior step in the mechanism.
(c) By the strategy in the last problem, we can write:
Roverall  Rslow R1st Step (since the 1st Step is the slow step)
Since the coefficients of NO2 and Cl2 are both 1 in the 1st Step, we can write
Rslow k1[NO2][Cl2].
17. NT7.
Thus, Roverall k1[NO2][Cl2] (predicted by this mechanism)
(a) How is the rate of an overall reaction related to the rate of (any of the) steps in its mechanism?
(b) In order for a reaction to be 0th order in a reactant A, what must be true about the mechanism for the reaction?
(c) Describe how the “condition” you described in (b) explains why the overall reaction rate does not depend on [A].
Answers:
(a) The overall rate of a reaction equals the rate of the slowest step in its mechanism. The slow
step is the “bottleneck”. (Sorry for the duplication here. This idea was needed for Q9 and Q10.)
(b) In order for the reaction to be 0th order in A, reactant A must appear as a reactant only in a step
after the slow step in the mechanism.
(c) If A appears as a reactant only after the slow step in the mechanism, then changing its
concentration (within a reasonable range) will not affect the rate at all because that step is
already faster than the slow step. It’s like getting “faster driers” when the rate limiting step is the
washing of the dishes—having faster driers won’t make the dishes go into the cupboard any
faster since the drier must wait to get the clean dish from the washer.
Temperature Dependence & Related Ideas (PE Diagrams, the Arrhenius Equation [Law] and Collision Theory)
18. MP. Mastering Problem—no answer in key. If you are struggling with this problem after using your
regular resources, try the next problem and then look at my key for that problem to see if that helps.
19. NT8.
(a) Draw a fairly accurate potential energy (rxn progress) diagram for each of the following cases &
(b) State the activation energy for the reverse reaction (in each case)
(i) H (or E)  +10 kJ/mol, Ea = 25 kJ/mol
(ii) H (or E)  -10 kJ/mol, Ea = 50 kJ/mol
NOTE: Assume these are both 1-step
(i.e., elementary) reactions
Answers:
(a)
Ea = +50
Ea = +25
Products
(Potential)
Energy
(kJ/mol)
Ea,r = +60
Ea,r = +15
E = +10
Reactants
Reactants
(i)
(ii)
Reaction Progress
(b) Ea,r (i) = +15 kJ/mol;
E = -10
Products
Reaction Progress
Ea,r (ii) = +60 kJ/mol
PS2-7
(See dotted arrows on plot)
Answer Key, Problem Set 2
Explanation:
(a) The E (or H) refers to the difference between the products’ (collective) potential energy and
the reactants’ (collective) potential energy. The Ea refers to the difference between the
transition state’s PE and the reactants’ (collective) PE. The two quantities are essentially
independent of one another.
(b) The activation energy for the reverse reaction is the difference between the transition state’s
PE and the products’ (collective) PE. Its value is most easily “seen” by looking at the distance
on the PE curve plots, although it also can be shown that: Ea,r  Ea,f - E (i.e., E  Ea,f – Ea,r).
20. NT9.
The rate constant of a reaction at 32°C is 0.055 s-1. If the frequency factor is 1.2 x 1013 s-1, what is the activation
barrier (activation energy)?
Answer: 83700 J/mol or 83.7 kJ/mol
Reasoning:
NOTE: Recall that T(in K) = T(in C) + 273.15, and that 1 kJ = 1000 J
Assume Arrhenius behavior: k  Ae
 Ea
 Ea
RT


 0.055 s-1  1.2 x 1013 s-1 e


 4.58 x 10 -15  e 25 35.7 J/mol  ln 4.58 x 10 -15 
 Ea
J 

 8.314
(32  273 K)
K mol 



 Ea
 E a  ln 4.58 x 10 -15 2535.7 J/mol 
2535.7 J/mol
83721  83700 J/mol  83.7 kJ/mol
21. NT10.
The rate constant (k) for a reaction was determined as a function of temperature. A plot of ln k versus 1/T (in
K) is linear and has a slope of -1.01 x 104 K. Calculate the activation energy for the reaction.
Answer: 8.40 x 104 J/mol or 84.0 kJ/mol
Reasoning:
Because of the form of the Arrhenius equation, on a plot of lnk vs. 1/T, the slope (m) equals
E
 a (so that Ea  -mR) and the y-intercept equals ln A:
R
k  Ae

Ea
RT
 lnk  
Ea
E 1
 lnA  lnk   a  lnA 
RT
R T

y = m
x + b
Here, the slope of such a plot was said to be -1.01 x 104 K, and thus:
Ea = -mR = -(-1.01 x 104 K)(8.314 J/K·mol) = 83971 J/mol = 84000 J/mol = 84.0 kJ/mol
22. MP. Mastering Problem—no answer in key. However, you may wish to try (or just see my solution
to) a similar problem, 13.70, the detailed solution of which is shown at very end of this answer
key.
23. MP. Mastering Problem—no answer in key. But this one has a video with it, so watch it! Also check
class notes, text, etc. I stressed this idea quite firmly in class (and edited the problem, even).
24. MP. Mastering Problem—no answer in key. Just make sure you don’t answer too hastily, since you
only get one shot at this one [since it’s a 50-50 multiple choice]. 
25. NT11
Draw a representation of the transition state (also called an activated complex) of the slow step of the mechanism in
each of the following problems. Use a dotted or dashed line to indicate each partially made/partially broken bond, and
a solid line to indicate each (fully made/intact) bond.
(a) 13.73 (15. NT5)
(b) 13.76 (16. NT6)
(c) 13.75
**NOTE: Only use the slow step from these problems (see instructions above). Ignore the rest of the text, questions or
steps in the problems themselves!**
PS2-8
Answer Key, Problem Set 2
Answers: (a) A—B - - B - - A
(i.e., The A-B bond in one A-B molecule (the left one in my
representation) remains intact, while the A-B bond in the other
A-B molecule is partially broken. A B-B bond is partially made.
(b) Cl - - Cl - - N—O
(i.e., the Cl-Cl bond is partially broken, and a Cl-N bond is
partially made. The two N-O bonds remain intact.)
O
Cl
(c) Cl
H
C
(i.e., the H-C bond is partially broken, and a Cl-H (or H-Cl)
bond is partially made. The C-Cl bonds remain intact.
Cl
Cl
Reasoning:
In order to write a proper structural representation of a transition state, you must determine which
bond(s) in the reactant(s) are breaking and which bond(s) in the product(s) are being formed in
the elementary reaction. This is not always as easy as it seems! DRAW SKELETON
STRUCTURES WHEN POSSIBLE.
(a) Equation for slow step is:
AB + AB  AB2 + A
Reactants: A—B and A—B
Products:
Bonds broken: one A—B (that’s it!)
A—B—B
&
A
Bonds made: one B—B (that’s it!)
Thus, recognize that the B of one A—B molecule must hit the B atom of the other A—B
molecule in order for these two bond making / breaking processes to happen during one
collision.
One way to do this is to orient the molecules such that one A—B is written as is, but the other
one is “flipped” so that the B is on the left: B—A. Hopefully if you write them this way next to
each other (A—B B—A) you can see that this is how they would be oriented during a
collision. Now draw dashed lines to represent the bonds partially made / broken in the TS:
A—B - - B - - A.
(a) Equation for slow step is:
Reactants: Cl—Cl
1
NO2(g) + Cl2(g) 
k
Products:
Cl—N—O
N—O
ClNO2(g) + Cl(g)
&
Cl
O
O
Bonds broken: Cl—Cl (that’s it!)
Bonds made: Cl—N (that’s it!)
Thus, recognize that one of the Cl atoms in Cl—Cl must hit the N atom of the NO2 in order
for these two bond making / breaking processes to happen during one collision.
Orient the molecules such that one Cl is between the other Cl and the N atom on NO2:
Cl—Cl
N—O
O
Then draw dashed lines to represent the bonds partially made / broken in the TS:
Cl - - Cl - - N—O
26. MP. Mastering Problem—no answer in key. (Hint: See prior problem to help visualize transition
states. You need to analyze the structure of the transition state in order to assess the
orientation factor.)
PS2-9
O
Answer Key, Problem Set 2
27. NT12 13.71.
Consider these two gas-phased [and elementary!] reactions:
(a) AA(g) + BB(g)
→ 2 AB(g)
(b) AB(g) + CD(g)
→ AC(g) + BD(g)
If the reactions have identical activation barriers and are carried out
under the same conditions, which one would you expect to have the
faster rate?
Answer: The first one (Reaction (a)). It has a larger orientation factor and thus a larger k and faster rate
Reasoning:
“Same conditions”  the concentrations of each reactant are the same, and the T’s are the same.
Since the elementary reactions are both bimolecular, if the concentrations are all equal, the
difference in the rates must be a result of a difference in k values. If the activation barriers are
identical, then the only thing left to affect the rate constant is the orientation factor (part of the
frequency factor).
So, this question really boils down to “Which reaction has the greater orientation factor”. To
address this, it is helpful to consider the structure of the transition states, as this will demonstrate
how “restrictive” the orientation of the reactants must be at the point of collision. The TS that is
less “restrictive” in the allowed orientations is the one that will have the greater orientation factor,
and (all other things being equal, as in this problem) the greater k value.
The transition states (plus R’s and P’s) for each of the elementary reactions in this problem are:
A
B
A
B
A
B
A
C
A
C
A
C
A
B
A
B
A
B
B
D
B
D
B
D
Reactants
(travelling toward
one another)
Transition
State
Reactants
(travelling toward
one another)
Products
Transition
State
Products
Reaction (b)
Reaction (a)
Hopefully, you can see that because the molecules in Reaction (a) are symmetrical, its orientation
demands are less restrictive (bigger orientation factor) than those of Reaction (b). Specifically, if you
imagine the C-D molecule “flipping” its orientation 180 (see below), the collision with A-B would not
yield products,
A
D
A
C
A
C
B
C
B
D
B
D
Given products could not form
(A-D and B-C would result instead!)
whereas if you “flipped” B-B, the same transition state would ensue and products could be formed.
Catalysis
[Related to both Mechanisms, PE Diagrams / Collision Theory]
& other Multi-concept Problems
28. NT13. (i) State whether each of the (hypothetical) changes stated below [e.g., (a) – (e)] would increase,
decrease, or have no affect on the rate of a chemical reaction (assume Arrhenius behavior), and
(ii) Describe how each change achieves this effect on the rate.
(a) an increase in T
For these two parts, copy down a graph like the one in Fig. 13.14 and refer to it in your
(b) an increase in Ea
explanations. Make sure you know the physical significance of the term “exponential factor”
(c) a doubling of both T and Ea
(d) a doubling of T when Ea triples
Don’t do (ii) for these two since you should have discussed T and Ea in
parts (a) and (b) already. Just give your reasoning for your answer to (i).
PS2-10
Answer Key, Problem Set 2
(e) a catalyst is added (If you give a short, 1-sentence answer to this, I expect a “correction”
once you look at the key. This is not quite as simple as most students think.)
(f) an increase in the concentration of a reactant species (be careful here!)
Answers: (with ultra-condensed responses to part (ii))
(a) increases rate; increases avg. KE of collisions  Greater fraction of collisions have KEcollision > Ea
(b) decreases rate; barrier larger  Smaller fraction of collisions have KEcollision > Ea
(c) rate unaffected; effects essentially cancel out
(d) decreases rate; factor of increase in Ea (x3) is greater than factor of increase in T (x2), so effect is
same as in (b) (smaller fraction of collisions have KEcollision > Ea
(e) increases rate; provides a new pathway that does not go through the original “slow step”. So new
pathway likely has lower overall Ea (and thus larger k)
(f) increases rate (unless order of reactant is zero); increases number of collisions per second
Before I get to each answer specifically, I want to take this opportunity to remind you of MY VIEW IN A NUTSHELL (think of
three fundamental variables):
A rate law has the general form:
Rate = k[A]m[B]n....
That means that there are two factors that directly affect rate: concentrations and k. If k can be expressed by the
Arrhenius equation, then it, in turn, is dependent on two more variables: Ea and T (actually three, if you include A [see
below]) In my opinion, then, there are really 3 major variables affecting rate: reactant concentrations, Ea, and T. (A, the
Arrhenius constant is a fourth factor of importance, although it is often not discussed much for some reason). The
presence of a catalyst, as discussed below, affects rate by affecting one or more of these “existing” variables, and thus
should not be considered a separate variable in my opinion.
Okay, back to the actual questions:
Detailed Reasoning:
(a) An increase in T generally will increase the rate of reaction. How does it achieve this effect?
Conceptually, the increase in T results in an increased fraction of collisions having a KE greater
than or equal to the activation energy required for reaction to occur. This is so because (from
Kinetic Molecular Theory) average kinetic energy depends only on T. If you increase T,
you increase the average KE of the nanoscopic particles in the sample, and so a larger
fraction of the collisions will have the energy needed to reach the transition state and react
(see below). NOTE: Ea is unaffected by T—it is a “barrier” of sorts, and the barrier height is
fixed. But when T is increased more collisions now have the energy to get over that barrier.
#
c
o
l
l
i
s
i
o
n
s
T1
T2
Ea
Increased fraction of
collisions that have the
energy needed to reach
the transition state (after T
increase) [~400% larger area
than before the T increase,
whereas the average KE only
increased by ~20-30%]
KE
Region where KEcollision > Ea
Original fraction of
molecules that have
the energy needed
to reach TS (and
react)
Mathematically, assuming the rate constant follows the Arrhenius equation, you can say that the
effect is achieved via increasing the exponential factor, and thus the value of the rate
PS2-11
Answer Key, Problem Set 2
 Ea
constant, k. As T increases, k increases (if k  Ae RT ), and thus the rate of reaction will
increase (for a given set of concentrations; see general form of rate law!).
(Note to the curious: Although an increase in T does not increase the concentration of reactants, it DOES increase the
number of collisions per second. I imagine that this effect (minor compared to the exponential term) is achieved
mathematically by an increase in the value of A, the Arrhenius constant. In other words, I believe that A is also dependent
on T although this is often apparently ignored/assumed negligible.)
(b)
IMPORTANT NOTE: There is no possible way to actually increase the activation energy for a particular reaction!! That’s
why I stated in the problem that these changes were “hypothetical”. It would be better to imagine comparing two reactions
with identical A values, one with an Ea twice as large as the other. But “raising the Ea” is much more concise!
An increase in Ea would decrease the rate of reaction. It would achieve this effect by
decreasing the fraction of molecules (“collisions”) at a given temperature that would have
enough energy to get over the barrier (see next paragraph). Mathematically, the decrease
happens by decreasing the exponential factor, and thus k (similar reasoning as for T, above,
except in the other direction).
** On the plot above in part (a), pick the curve for just one temperature. Now imagine moving
the line denoting Ea to the right until it is twice as big (!) Look at how many fewer
collisions now have the energy to reach the transition state! That means the k will be
dramatically smaller, and the rate dramatically slower.
(c) If the T doubles and the activation energy (could somehow) double, then the effects would
essentially cancel out. The collisions would have a greater average KE, but the activation energy
barrier would be greater as well (by the same factor). Mathematically, Ea is in the numerator of
the exponent in the exponential factor and T is in the denominator, so doubling them both will not
change the value of the exponent).
(d) If the T doubles and the activation energy (could somehow) triple, then the effects would not
cancel out. The collisions would have a greater average KE, but the activation energy barrier
would be greater by a larger factor [here, 1.5x]). So even though the collisions would have a
greater average KE, since the Ea is bigger by a greater factor, there will actually be a smaller
fraction of collisions with KEcollision > Ea. Mathematically, Ea/T would be 3/2, so the exponent
would increase in absolute value, and since the exponent is negative, the value of the
exponential factor would decrease (decreasing k).
(e) The presence of a catalyst will increase the rate. The way that a catalyst affects the rate of a
reaction is by changing the pathway of the reaction (to one that doesn’t go through the
original slow step). This causes there to be a different rate law with a different k and with the
concentration of catalyst added:
Rate (with catalyst present) = kcatalyst[A]m’[B]n’[catalyst]p
In the new pathway, the activation energy is typically much lower and thus the rate constant
(kcatalyst) is much larger than the original one (uncatalyzed k). So as stated above, the major way
that a catalyst mathematically affects rate is via increasing the rate constant, k, by providing a new
path with lower Ea.
(f) Unless the order (exponent) of a reactant is zero (or negative, which we have not discussed so don’t worry
about it; but it IS possible under certain circumstances!), the rate will increase as the concentration of a
reactant increases. The way this is achieved is via an increase in the number of collisions per
second. (Note that if a reactant species does not appear in the mechanism until after the ratedetermining step, then its number of collisions with things will not affect rate, and that is how its
order can be zero.) Note that the larger the order (exponent) of a given reactant, the greater is
the effect that a change in its concentration will have on the rate.
29. NT14 13.80.
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme
increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower must the activation barrier
be when sucrose is in the active site of the enzyme? (Assume the frequency factors for the catalyzed and
uncatalyszed reactions are identical and a temperature of 25C)
Answer: 34.2 kJ/mol lower (i.e., the new Ea would be 74 kJ/mol)
Reasoning:
PS2-12
Answer Key, Problem Set 2
NOTE: This problem really is about catalysis. The fact that they mention “enzyme” and “active
site” is really a non-issue as far as the problem goes. An enzyme is a biological catalyst. As such,
when the problem says “when sucrose is in the active site of the enzyme”, that is the equivalent of
saying “in the catalyzed mechanism or pathway”. So basically, this problem is asking “If the rate is 1
million times faster when the catalyst is present, how much lower is the Ea in the new (catalyzed)
pathway?”
One more note: Although they say the “rate” is 1 million times faster, they really mean that the
“rate constant” is 1 million times larger. That is, you should assume the same initial
concentration of reactants here (water and sucrose), such that R is proportional to k.
Once the above is recognized, hopefully you can see that this problem is very similar to Q22 (a
Mastering Problem) and 13.70 (see end of set [extra practice item]) in that one must create a “ratio
k 
of rate constants”,  2  (and this ratio equals 1 million in this problem) However, here instead of
 k1 
there being two different T’s, there are two different Ea’s: one for the uncatalyzed reaction (Ea,uncat),
and one for the catalyzed reaction (new mechanism) (Ea,cat). As such, we write kuncat (instead of k1)
kcat

and kcat (instead of k2) and the rate ratio is then:
kuncat
Interestingly enough, your text does not give you an analogous equation to Equation 13.27 (Eqn.
used for the T change problem). So here you are “forced” to just use your brain (the way I
wanted you to do in Q22!). Namely, just substitute into the Arrhenius equation for each k and
simplify:

kcat
Ae

kuncat

Ae
Ea , cat
RT
e
Ea , uncat
 Ea , cat  Ea , uncat
  
RT
RT


 
 
 
Taking the ln of both sides yields:
RT


 k 
 ln  cat   ln e 
 kuncat 
Ea , cat
RT
 Ea , uncat
  RT

 
 
 
 E

  - a, uncat  
RT
 RT 
E
E
1
 a, uncat  a, cat 
 Ea, uncat  Ea, cat  
RT
RT
RT

Ea, cat
 k 
1
 ln  cat  
 Ea, uncat  Ea, cat 
k
RT
 uncat 
 k 
Note: Ea,uncat – Ea,cat is exactly what is asked for
 RT ln  cat    Ea, uncat  Ea, cat  
in this problem!
k
 uncat 
k
J
):
Substituting in cat  106 and T  25C  298 K (and recalling R  8.314
mol  K
kuncat
l
o
m
/
J
k
2
.
4
3
J 
J

  8.314


 298 K  ln 106  2477.5.. 13.81..  34229

mol  K 
mol


NOTE: Many of you probably just plugged in the numbers earlier in the problem and solved for
Ea,cat. So it might be useful to put up what the work for that solution to the problem would look
like:


kcat
Ae
 106 
kuncat

Ae
Ea , cat
RT
Ea , uncat
RT


e

e

Ea , cat
R (298)
108000
R (298)
 13.81.. 
 ln(106 )  
 108000 
 

R(298)  R(298) 
Ea, cat
1
 Ea, cat +108000 
R(298)
PS2-13
Answer Key, Problem Set 2
J 

298K   Ea, cat +108000 
 13.81..  8.314
mol  K 

J
 34229
  Ea, cat +108000   Ea, cat  108000  34229  74000 J/mol
mol
(This means that the Ea,cat is 34000 J/mol (or 34 kJ/mol) lower)
30. NT15.
(a)
Consider the following mechanism for a reaction (where A, B, C, and D are atoms),
along with the corresponding potential energy diagram (original diagram not shown
in order to save space) for the two-step process.
What is the balanced equation for the overall process (chemical reaction) that occurs?
Answer: BC + D  B + CD
A + BC  AC + B
AC + D  A + CD
-----------------------------
BC + D  B + CD
(See above. AC cancels out and A cancels out)
(b) Draw a structural formula (i.e., connect atoms with a solid line to indicate a full bond, and a dashed line to indicate a
partial bond) of each of the following (if applicable), and give brief reasoning:
(i) (every) reactant (of the overall process):
(ii) (every) product (of the overall process):
(iii) (every) intermediate (if applicable):
(iv) (every) catalyst (if applicable):
(v) the transition state of Step 1:
(vi) the transition state of Step 2
(c)
Answers:
B—C and D
B and C—D
A—C (made in Step 1; used in later step)
A
(used in Step 1; reproduced in later step)
A - - C - - B (or B - - C - - A)
A - - C - - D (or D - - C - - A)
Put each species that you drew in part (b) in the appropriate spot on the (potential) energy diagram (copy a large
enough version of the diagram onto your page that you can draw all of these species without it being overly crowded
and messy!).
Answers (c) and (d):
A - - C - - B (+ D)
A - - C - - D (+ B)
Ea, Step 1
= Ea, overall
(Potential)
Energy
Ea,Step 2
A—C + B (+ D)
Intermediate
Reactants
A + B—C (+ D)
A + C—D (+ B)
E
Products
(except for A,
the catalyst)
(except for A, the
catalyst)
Reaction Progress
(d) Indicate each of the following on your (potential) energy diagram:
(i) The activation energy for Step 1
(ii) The activation energy for Step 2
(iii) The activation energy for the overall reaction. Note: In this case, Ea,overall  Ea,Step 1
(iv) E (or H) for the reaction
(e)
Which step in the mechanism for this reaction (1st or 2nd) is likely to be rate determining? Give your reasoning.
Answer: Step 1
PS2-14
Answer Key, Problem Set 2
Reasoning: Step 1 has the significantly greater Ea, so given even remotely similar frequency
factors, the k for Step 1 will be smaller than that for Step 2, and so under “normal” conditions
of concentration, Step 1 would be slower. The slow(er) step is the rate determining one.
(f)
State whether each of the following is endothermic or exothermic
(i) Step 1
(ii) Step 2
(iii) The overall rxn
Answers: Step 1 is endothermic (its products are higher in PE than its reactants)
Step 2 is exothermic (its products are lower in PE than its reactants)
The overall reaction is endothermic (P’s higher in PE than R’s)
31. MP. Mastering Problem—no answer in key. This problem is basically a mechanism question, but has
mechanisms for an overall reaction both without and then with a catalyst. Review ideas from
problems 12-16 on this set.
-------------END OF PROBLEM SET 2---------------------------EXTRA PROBLEM FROM OLD KEY SHOWN BELOW FOR EXTRA PRACTICE/HELP-----------13.70.
If a temperature increase from 20.0°C to 35.0°C triples the rate constant for a reaction, what is the value of the
activation barrier (activation energy) for the reaction?
Answer: 5.5 x 104 J/mol or 55 kJ/mol
Strategy (and Execution):
1) Recognize that a “tripling of the rate constant” means that k2 = 3k1 
k2
 3
k1
2) Recognize that the Arrhenius equation relates k and T and you have two k’s and two T’s.
3) I suppose one could memorize Eq. 13.27 from the text, I think that is a poor use of your brain
power and time. Memorize the Arrhenius equation and then just realize that k2 is the rate
constant at T2 and that k1 is for T1. Then recognize that you can set up the ratio of rate
constants (k2/k1) and derive the relationship you need as follows:

Ea


k 2 Ae RT2

 e 
Ea

k1
Ae RT1
Ea  Ea  
 

RT2  RT1  
(because k 2  Ae

Ea
RT2
and k1  Ae

Ea
RT1
, and
ea
 ea  b 
eb
Taking the ln of both sides yields:
ln
Ea  Ea   Ea
Ea  E a  1 1 
k2

 



  
k1
RT2  RT1   RT1
RT2  R  T1 T2 
(which is Eq. 13.27!)
4) After converting the two T’s into Kelvin (by adding 273.15), and recalling that
k2
 3 (#1
k1
above), substitute into the equation and then solve for Ea:
ln
k2
 ln3 
k1
Ea
8.314
J
K  mol
1
1


 20.0 + 273.15  35.0 + 273.15  


J 

-1
-1
 1.0986..  8.314
  Ea 0.003411  0.003245 K  Ea 0.000166..K 
K
mol








J 

1.0986..  8.314
K  mol 


 Ea  Ea  55006 J/mol  55 kJ/mol
0.000166..K -1


PS2-15
