Answer Key, Problem Set 2 - oakton.edu

Chemistry 122 Mines, Spring, 2020

PS2-1

Answer Key, Problem Set 2

1. 15.49; 2. NT1; 3. NT2; 4. NT3; 5. NT4; 6. 15.78; 7. NT5; 8. NT6; 9. 15.62; 10. 15.64; 11. NT7; 12. 15.73*; 13. NT8; 14. 15.82; 15. NT9

--------------------------------------- Integrated Rate Laws (and Half Life)

1. 15.49. The data below show the [AB] versus time for the reaction represented by AB(g) → A(g) + B(g):

[Data (and rest of problem) not shown in key]

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 the “rationale for 15.47 and 15.48” after the END OF SET line at the end of this key for detailed reasoning for this assertion).

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

Long way:

2nd order (by default). No real need to plot 1/[AB] vs t, although one could.

not 0th order b/c not linear

not 1st order b/c not linear


PS2-2

(b) 2nd order integrated rate law: [AB]

1

[AB]

1

0

ktt

[substituting in point (100 s, 0.302 M)]

s) (100 M 852.2 2..M51.0 s) (100.1..M13.3 M 0.950

s) (100. M 0.302

1 1-1-1- kkk 1

1-1sM 0.0226 11--1

sM 8..50.022 s 100

M 852.2 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 [AB]

1

[AB]

1

0

k tt

yields:

7..M11.6 M 0.950

1s) (25.sM 850.022

[AB]

1 1-11-

s 25

M 0.618 0.6183..7..M11.6

1 [AB]

1-s 25

2. NT1. 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

kttkt

t ee0

0[A]

[A] [A][A] to solve for t when the fraction remaining is 1/4 or 1/8, but it is much

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.

3. NT2. 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) Draw a picture that shows the number of A and B molecules present at t = 3 min.

Answers: (a) 0th order; (b) (see right )

CHECK: Answer is reasonable—is between 0.459 M and 0.950 M (t = 50 and t = 0 points)

(b) t = 3 min


PS2-3

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 molecules convert into B molecules between 0 and 1 min, and two A's also convert into B’s between 1 and 2 minutes).

Thus the change in moles of A (mol) and change in concentration of A ([A]) are also the same in each time interval (since the volume of the box is the same throughout). That means

the change in concentration per change in time ([A]/t) 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 a species changes), the process

must be zeroth order in that species. 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 above (as well as problems 15.50 and 15.52, assigned in Mastering for PS2_Mastering_Part 1)! 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, specifically, analogous to this one (i.e., is zero order).

4. NT3. 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

5. NT4 (a) A student is asked on an exam “Is this mechanism valid?” and answers: Yes, because the sum of the

elementary steps equals the overall equation. Why did the student earn fewer than half the points on that question (i.e., they failed that question)?

Answer: There are two criteria that must be met in order for a given mechanism to be considered

“valid” (see part (b) discussion). The “summing of the elementary steps equaling the overall equation” is just one of them, and it is the “trivial” one. The more substantial criterion is that the rate law that is predicted by the mechanism must be consistent with (i.e., the same as) the rate law that is determined by experiment. So this student only gave the “trivial” criterion, and therefore only earned a small fraction of the total points for this problem. Not only does one need to recognize that there is a second criterion, one must then show that s/he knows how to assess whether that criterion is met or not (i.e., s/he must demonstrate the ability to predict a rate law from a mechanism before s/he can compare it to the experimental rate law) in order to earn full credit on such a problem on an exam.


PS2-4

(b) 15.75. 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 1k AB2 + A (slow)

AB2 + C 2k AB + BC (fast)

Answer: Yes

Reasoning/Strategy:

Two criteria (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.

So, one must verify each criterion separately:

1) Criterion #1: To check if the sequence of steps sums to the overall reaction equation:

Cross out any formulas for substances that are identical on both sides of the arrow (in the whole sequence)

Copy down any remaining formulas on both sides of the arrow.

Compare the result to the overall reaction equation and make a conclusion:

o It matches exactly or o It does not match exactly

NOTE: If it does not match exactly, there is no need to proceed to verify the 2nd criterion.

2) Criterion #2: To check if the predicted rate law matches the observes one, one much first determine the rate law predicted by a given mechanism:

i) Recognize that: “The overall rate of any step-wise process can be no faster than its slowest step.”

which ultimately means the following:

Overall reaction rate rate of the slowest step OR

Roverall Rslow

So, identify the slow step of the mechanism and look only at that for right now (in fact, it might be good for you to write down just that step somewhere, separate from the rest of the mechanism (so you don’t get confused by the rest of the steps).

ii) Write the rate law for (just) the slow step.

How? Well, for any elementary step, the rate law has the form:

Rslow = kslow[reactant #1 in slow step]its coefficient[reactant #2 in slow step]its coefficient

(See Table 15.3 for specific examples of the generalized equation above).

NOTE: The equation above “works” for an elementary step precisely because an elementary step occurs in one “collisional event”. Thus, every collision “matters”. This isn’t the case for an overall reaction that occurs in multiple steps. For such (overall) reactions, the rate law can only be determined by experiment (i.e., you cannot assume that the order is equal to a reactant’s coefficient).

iii) Combine the two relationships above to write:

Roverall kslow[reactant #1 in slow step]its coefficient[reactant #2 in slow step]its coefficient


PS2-5

Thus, in summary, to determine the rate law predicted by a given mechanism:

i) Find the slow step

ii) Write the rate law for the slow step

iii) Set this equal to the overall rate.

Important Note: The above procedure, as stated, only technically “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.

iv) Compare the (orders of each reactant in the) rate law just determined (for the slow step) to the observed (experimental) rate law and make a conclusion:

o It matches exactly (i.e., the orders of each reactant are identical in both rate laws) or

o It does not match exactly (i.e., at least one order of one reactant is not identical in both rate laws).

Make the Final Conclusion:

If either of the above criteria is not met, the mechanism is not “valid”—it cannot be the mechanism for the overall reaction equation.

If both of the above criteria are met, the mechanism is “valid”—it is at least possible that it is the mechanism for the overall reaction equation (although it also might not be). It is still “in the running” (it has not be eliminated as an option).

Execution of Strategy (for this mechanism):

Checking Criterion #1 for this mechanism:

Overall equation is: AB + C → A + BC

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

This sum exactly matches the overall equation (so proceed to verify Criterion #2)

Checking Criterion #2 for this mechanism:

i) The slow step is:

AB + AB 1k AB2 + A (slow)

ii) Since AB appears twice (and could be written as 2 AB rather than AB + AB), the order of AB is two. Since C does not appear in this step, its order is zero. Thus the rate law of the slow step is just:

Rslow k1[AB][AB] k1[AB]2

iii) Thus (since Roverall Rslow) :

Roverall k1[AB]2 (prediction of this mechanism)

iv) The problem states that the (overall) reaction AB + C → A + BC is “experimentally observed to be second order in AB and zero order in C”, the observed rate law is thus:

R k[AB]2[C]0 = k[AB]2 (observed/experimental)

Thus the predicted rate law exactly matches the experimentally observed rate law (both show the order of AB to be 2 and the order of C to be 0).

(matches)


PS2-6

Making the Final Conclusion:

Since the sequence of elementary steps sums to the overall equation, AND the rate law predicted by the mechanism matches experiment,

This mechanism is “valid”.

6. 15.78. Consider this two-step mechanism for a reaction:

NO2(g) + Cl2(g) 1k ClNO2(g) + Cl(g) (slow)

NO2(g) + Cl(g) 2k ClNO2(g) (fast)

Rest of question not retyped into this key.

Answers: (a) 2 NO2(g) + Cl2(g) 2 ClNO2(g)

(b) Cl is the only intermediate

(c) R = k1[NO2][Cl2] (i.e., 1st order in both NO2 and Cl2)

Reasoning (see prior problem for more detailed explanation of “strategy” for (a) and (c)):

(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

ClNO2’s remain on the product side. Hence: 2 NO2(g) + Cl2(g) 1k 2 ClNO2(g)

(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]. Thus, Roverall k1[NO2][Cl2] (predicted by this mechanism)

7. NT5. (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”.

(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)

8. NT6. (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


PS2-7

Answers:

(b) Ea,r (i) = +15 kJ/mol; Ea,r (ii) = +60 kJ/mol (See dotted arrows on plot)

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). 9. 15.62. The rate constant…(rest of problem not retyped into this key)

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) 273(32

molK

J 8.314

1-131- s 10 x 1.2s 0.055

a

a

E

RT

E

eAek

J/mol 5.732510 x 854.ln J/mol 5.7325

10 x 854.ln 10 x 854. 15-a

a15-J/mol 5.732515-

a

EE

e

E

-(-33.017…)(2535.7 J/mol)*

83721 83700 J/mol 83.7 kJ/mol

*NOTE: Significant figure rules are different for logs than for either multiplication/division or addition/subtraction. Since there are two sig figs in 4.58 x 10-15, the natural log of this number will have two significant figures after the decimal place [not “two sig figs”]. While you do not need to know this rule right now, it’s useful to note that the result makes sense here (4 SF in -33.017 rather than 2 SF) if you try ln(4.5 x 10-15) [=-33.034..] and ln(4.7 x 10-

15) [=-32.991..]. Clearly, the variation is not in the units place of -33.017—it is in the hundredths place (approximately ±0.02). Ask me if you have questions about this.

10. 15.64. The rate constant (k)…(rest of problem not retyped into this key)

Answer: 8.40 x 104 J/mol or 84.0 kJ/mol

(a)

Reaction Progress

(Potential) Energy (kJ/mol)

Reactants

Reaction Progress

Reactants

(i) (ii)

Products E = +10

Ea = +25

Products

E = -10

Ea = +50

Ea,r = +15 Ea,r = +60


PS2-8

Reasoning:

Because of the form of the Arrhenius equation, on a plot of lnk vs. 1/T, the slope (m) equals

R

Ea (so that Ea -mR) and the y-intercept equals ln A:

ATR

EkA

RT

EkAek aaRT

Ea

ln 1

ln ln ln

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

11. NT7 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) 15.75 (b) 15.78 (c) 15.77

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) (i.e., the Cl-Cl bond is partially broken, and a Cl-N bond is partially made. The two N-O bonds remain intact.)

(c)

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: A—B—B & A

Bonds broken: one A—B (that’s it!) 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.

**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!**

O

Cl - - Cl - - N—O

C

Cl

Cl

Cl Cl H (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.


PS2-9

(b) Equation for slow step is: NO2(g) + Cl2(g) 1k ClNO2(g) + Cl(g)

Reactants: Cl—Cl Products: Cl—N—O & Cl

N—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:

Then draw dashed lines to represent the bonds partially made / broken in the TS:

12. 15.73. Consider these two gas-phased [and elementary!] reactions:

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:

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,

whereas if you “flipped” B-B, the same transition state would ensue and products could be formed.

Rest of question not typed into this key.

A

A

A

A

B

B

A BB

B A B

Reactants (travelling toward

one another)

Products Transition State

Reactants (travelling toward

one another)

Products Transition

State

B D

A CA

B

C

D

C

D

A

B

Reaction (a) Reaction (b)

Given products could not form

(A-D and B-C would result instead!)

B D

A CA

B

C

D

D

C

A

B

O

O

O

Cl - - Cl - - N—O

O

Cl—Cl N—O


PS2-10

Catalysis [Related to both Mechanisms, PE Diagrams / Collision Theory] & other Multi-concept Problems

13. NT8. (i) State whether each of the (hypothetical) changes stated below [e.g., (a) – (f)] would increase, decrease, or have no effect on the rate of a chemical reaction (assume Arrhenius behavior), and

(ii) Describe how the change in (a), (b), (e), and (f) achieves this effect on the rate.

NOTE: You must copy down/sketch a graph like the one in Fig. 15.14 on your paper to earn credit for parts (a) and (b) of the problem below. Also, make sure to refer to this graph in your explanations for parts (ii)(a) and (ii)(b). Make sure you know the physical significance of the term “exponential factor”.

(a) an increase in T

(b) an increase in Ea

(c) a doubling of both T and Ea

(d) a doubling of T when Ea triples

(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 quantities or factors):

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 quantities: Ea

and T (actually three, if you include A [see below]) In my opinion, then, there are really 3 major factors 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” quantities, and thus should not be considered a separate factor 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.

Don’t do part (ii) for these two since you should have discussed T and Ea in parts (ii)(a) and (ii)(b) already. Just make sure to provide your reasoning for your answer to Part (i)(a) and (i)(b) [which should have been done already].


PS2-11

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

constant, k. As T increases, k increases (if aE

RTk Ae

), 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 involve 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

# co l l i s i o n s

Ea T1

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%]

Original fraction of collisions that have the energy needed to reach TS (and react)

KE

Region where KEcollision > Ea

T2


PS2-12

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 rate-determining 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.

14. 15.82. The activation barrier for the hydrolysis…(rest of question not typed into this key)

Answer: 34.2 kJ/mol lower (i.e., the new Ea would be 74 kJ/mol)

Reasoning:

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 a Mastering problem and 15.72 (see end of set [extra practice item]) in that one must create a “ratio of rate

constants”, 2

1

k

k

(and this ratio equals 1 million in this problem) However, here instead of 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) and

kcat (instead of k2) and the rate ratio is then: cat

uncat

k

k

Interestingly enough, your text does not give you an analogous equation to Equation 15.27 (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 that Mastering problem!). Namely, just substitute into the Arrhenius equation for each k and simplify:

, cat, cat , uncat

, uncat

-

aa a

a

EE E

RTRT RTcat

E

uncat RT

k Aee

kAe

Taking the ln of both sides yields:

, cat , uncat -

, cat , uncatln ln -

a aE E

RT RT a acat

uncat

E Eke

k RT RT

, uncat , cat

, uncat , cat

1

a a

a a

E EE E

RT RT RT

, uncat , cat

1 ln cat

a a

uncat

kE E

k RT

, uncat , cat ln cata a

uncat

kRT E E

k

Note: Ea,uncat – Ea,cat is exactly what is asked for

in this problem!


PS2-13

Substituting in 610cat

uncat

k

k and T 25C 298 K (and recalling R 8.314

J

mol K):

6 8.314 298 K ln 10 2477.5.. 13.81.. 34229J J

mol K mol

34.2 kJ/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:

, cat, cat

, uncat

(298)

, cat6 6

108000

(298)

10800010 ln(10 )

(298) (298)

aa

a

EE

RRTacat

E

uncat RRT

Ek Ae e

k R RAe e

, cat

1 13.81.. +108000

(298)aE

R

, cat 13.81.. 8.314 298 +108000a

JK E

mol K

, cat , cat 34229 +108000 108000 34229 74000 J/mola a

JE E

mol

(This means that the Ea,cat is 34000 J/mol (or 34 kJ/mol) lower)

15. NT9. 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.

(a) What is the balanced equation for the overall process (chemical reaction) that occurs?

Answer: 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): B—C and D (ii) (every) product (of the overall process): B and C—D (iii) (every) intermediate (if applicable): A—C (made in Step 1; used in later step) (iv) (every) catalyst (if applicable): A (used in Step 1; reproduced in later step) (v) the transition state of Step 1: A - - C - - B (or B - - C - - A) (vi) the transition state of Step 2 A - - C - - D (or D - - C - - A)

A + BC AC + B

AC + D A + CD -----------------------------

BC + D B + CD

Answers:


PS2-14

(c) 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):

(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

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

============================== END OF SET =====================================

EXTRA PROBLEMS AND/OR EXPLANATION TO SOME PROBLEMS ASSIGNED IN MASTERING ARE FOUND BELOW FOR EXTRA PRACTICE/UNDERSTANDING.

15.47 AND 15.48, Approach / Rationale:

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 15.47(a) matches 0th. 15.48(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”.)

Reaction Progress

(Potential) Energy

Reactants (except for A, the

catalyst)

Products (except for A, the catalyst)

A + B—C (+ D)

A—C + B (+ D)

E

Ea,Step 2

A - - C - - B (+ D)

A - - C - - D (+ B)

A + C—D (+ B) Intermediate

Ea, Step 1

= Ea, overall

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)


PS2-15

You could also plug in the point (t1/2, ½ [A]0) into [A]t -kt + [A]0 and derive that for 0th order

kt /

0

212

1[A]

, which shows that t1/2 is proportional to [A]0 (0th order).

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: kt

t e 0[A][A] . If you take the ln of

both sides, you get (after a bit of algebra): ln [A]t -kt + ln [A]0. Thus, 15.48(b) is consistent with 1st order. You can derive for yourself (see notes, text) or perhaps memorize the fact that for 1st

order, the half life equals k

.

k

ln 69302 . This shows that t1/2 does not depend on [A] for 1st order

(consistent with 15.47(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 15.47(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 15.48(c) correct, you (unfortunately) have to memorize that the integrated rate law

for 2nd order is: 0

1

[A]

[A]

1 k t

t

(or at least that it’s t[A]

1 vs t that is a straight line). If you know

that, you could substitute in the point (t1/2, ½ [A]0) and derive that 0

21

1

[A]kt / for 2nd order (to

answer 15.48(a) without the analysis given above).

15.55, Strategy (without numbers)

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 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%”.

15.72. If a temperature increase…(rest of question not retyped into this key)

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 2

1

3k

k

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:


PS2-16

2

2 1

1

2

1

a

a a

a

EE E

RTRT RT

E

RT

k Aee

kAe

(because 2 1

a

2 1 b and , and

a aE E

RT RT a bek Ae k Ae e

e

Taking the ln of both sides yields:

2

1 2 1 1 2 1 2

1 1ln a a a a aE E E E Ek

k RT RT RT RT R T T

(which is Eq. 15.27!)

4) After converting the two T’s into Kelvin (by adding 273.15), and recalling that 2

1

3k

k (#1

above), substitute into the equation and then solve for Ea:

2

1

1 1ln ln3

J 20.0 + 273.15 35.0 + 273.158.314

K mol

aEk

k

-1 -1J1.0986.. 8.314 0.003411 0.003245 K 0.000166..K

K mola aE E

-1

J1.0986.. 8.314

K mol 55006 J/mol

0.000166..Ka aE E

55 kJ/mol

Documents

Answer Key, Problem Set 2 - oakton.edu