Answer Key, Problem Set 1, Written

Chemistry 122 Mines, Spring, 2020

PS1-1

Answer Key, Problem Set 1, Written

1. 15.32; 2. 15.34 (add part (e): Estimate / calculate the initial rate of the reaction); 3. NT1; 4. NT2; 5. 15.37; 6. 15.39; 7. 15.41; 8. NT3; 9. NT4; 10. 15.46; 11. NT5

------------------------------ Rates of Formation / Loss (and Relation to Coefficients), Average & Instantaneous Rates, and Rates of Reaction

1. 15.32. Consider the reaction (represented by): NO2(g) → NO(g) + 2

1 O2(g) (rest of problem not copied here)

Answers: (a) 0.0047 M∙s-1 and 0.0038 M∙s-1 ; (b) 0.0019 M∙s-1

Reasoning:

(a) tt

−=

−=

][NO][NO 22

1

1Reaction of Rate Avg. here because the coefficient of NO2 is 1. Thus,

From 10 s to 20 s, 1-22 sM 0.0047s 10.

M 0.047

s 10s 20

M 0.951M 0.904][NO][NO ==

−

−−=

−

−−=

if

if

ttRate.Avg

From 50 s to 60 s, 1-22 sM 0.0038s 10.

M 0.038

s 50s 60

M 0.778M 0.740][NO][NO ==

−

−−=

−

−−=

if

if

ttRate.Avg

(b) R(form, of O2) = ½ R(loss of NO2) because of the coefficients in the equation (1 : ½ for NO2 : O2). Thus, the

(average) rate of formation of O2 between 50 s and 60 s = ½ x 0.0038 M∙s-1 (from (a)) = 0.0019 M∙s-1

Answer Key, Problem Set 1

PS1-2

2. 15.34. Consider the reaction (represented by):

2 H2O2 (aq) → 2 H2O (l) + O2 (g)

The graph below….(rest of problem not copied here)

Answers: (my results): (a) 0.0095 M∙s-1; (b) 0.0063 M∙s-1; (c) 0.0033 M∙s-1; (d) 0.59 mol O2; (e) 0.014 M∙s-1

(a) Strategy/Explanation:

Avg. rate of reaction equals the rate of loss of a reactant (or rate of formation of a product) divided by its coefficient in the balanced chemical equation (see Eq. 15.5 in Tro, or handout). Since we have data for H2O2, the appropriate expression for the rate of reaction is:

Rate of rxn = t

−

]O[H 22

2

1 because the coefficient of H2O2 is a 2. So,

1) Find the values of [H2O2] at t = 10 and 20 s.

2) Calculate t

−

]O[H 22

2

1

Execution of Strategy: Note: I scanned the plot and enlarged it in this key for better precision.

At 10. s, [H2O2] = 0.75 M; at 20. s, [H2O2] = 0.56 M (see plot above; points added by me). Thus,

1-sM 0.0095 =

−=

−=

−=

s 10.

M 0.19-

s 10. - 20.

M 0.75 - 0.56]O[Hrxn) (of 22

2

1

2

1

2

1

tRate

(b) Strategy/Explanation:

“Instantaneous rate” means the rate at a single point in time, rather than over a time interval. The instantaneous rate of loss of H2O2 can be estimated by finding the slope of the tangent to the curve for [H2O2] at a point in time and then taking its opposite (to make the value positive). Since this question asks for instantaneous rate of reaction, one must divide the rate of loss of H2O2 by the coefficient of H2O2 in the balanced equation (see (a)). So in this problem one must:

1) Draw a tangent to the curve at the point representing t = 30 s; and


PS1-3

2) Find the slope of the tangent line by picking any two points on the line and using the definition of

slope x

y

=

t

]O[H means here which 22 .

3) Multiply the result in (2) by -½ to get rate of reaction

−

t

]O[H determine end, the in i.e., 22

2

1

Execution of Strategy:

1) I created a tangent line on the plot below at t = 30 s by “adjusting” a line to make it go through the

point on the curve at t = 30 s, and also have a slope equal to the slope of the curve at 30 s (see short-dashed line on plot):

2) and 3): Looking at the dotted line and picking as my two points, the “intercepts” (just for convenience—

you could use any two points), I get the points (0 s, 0.78 M) and (62 s, 0 M). Thus:

1-sM 0.0063 ==

−=

−=

−= 1-22 sM 920.006

s 62.

M 0.78-

s 0. - 62.

M 0.78 - 0.]O[H

2

1

2

1

2

1

tRate s) 30. at rxn (of

Uncertainty

probably about 0.0003 M∙s-1 due to slope estimation


PS1-4

(c) Strategy/Explanation:

This is similar to (b) except that the rate of formation of O2 is asked for (not rate of “reaction”). However, since the coefficient of O2 is 1, the “rate of formation of O2” equals the rate of reaction in this problem! Thus, you can actually do exactly the same thing you did in (b) [i.e., find the slope of the plot for H2O2 and take ½ of that] except do it at the point corresponding to t = 50 s.


1) I created what I believe to be a reasonably good tangent line to the curve at t = 50 s (long-dashed line on plot below).

2) and 3): Looking at the long-dashed line, the “y-intercept” is (0 s, 0.55 M), and another point is (60. s, 0.15 M) [I didn’t pick the “x-intercept” here because it was slightly off my scale]. Thus:

1-sM 0.0033 ==

−=

−=

−=

= 1-222

2 sM 3...30.003s 60.

M 0.40-

s 0. - 60.

M 0.55 - 0.15]O[H][O) formationO (of

2

1

2

1

2

1

1

1

ttRate

(d) Strategy/Explanation:

First of all, the problem’s wording is a bit tricky, since it says the “initial volume” is 1.5 L, perhaps implying that the volume might change with time. Hopefully you realize(d) that the volume of the container is just 1.5 L and this does not change during the reaction. As such, if you step back and really analyze this problem, hopefully you will recognize that it is actually just a stoichiometry problem (it has nothing to do with “rate” really)! You are asked for “moles of O2 formed” in the first 50 s, and you are effectively “given” the “concentration of H2O2 lost” during this same time interval (because you can get this from the graph by determining [H2O2]50 s – [H2O2]0 s). So, how do you relate the “moles of O2 formed” to the “molar of H2O2 lost” (when you know the volume is 1.5 L)? There are two basic ways:

One way:

molarity of H2O2 ([H2O2]) → moles of H2O2 → moles of O2

In this approach, the first step involves multiplying M by V (in L) to get moles. The second step involves

multiplying by a mole ratio from the balanced equation:

22

2

OH mol 2

O mol 1.

2nd way:

molarity of H2O2 ([H2O2]) → molarity of O2 ([O2]) → moles of O2


PS1-5

In this approach, the first step involves multiplying by a mole ratio from the balanced equation:

22

2

OH mol 2

O mol 1 [You could do this in an “ICF” type of table, or not]. The second step would involve

multiplying the resulting M (i.e., [O2] formed) by V (in L) to get moles (of O2 formed).


[H2O2]50 s = 0.22 M; [H2O2]0 s = 1.00 M (see plot below [points added by me]).

1st Approach:

Thus, [H2O2] (in 1st 50 s) = 0.22 M – 1.00 M = -0.78 M

0.78 M H2O2 was lost (in 1st 50 s)

2O mol 0.59 580.5lost OH mol 2

formed O mol 1 x L 51. x

L

OH mol 80.7

22

222 == (formed in 1st 50 s)

2nd Approach (with ICF table): 2 H2O2 (aq) → O2 (g) + 2 H2O (l) [I swapped the order on the right for convenience only]

[H2O2] (M) [O2] (M) [H2O] (M)

Initial 1.00 (not known/needed) (not known/needed)

Change 0.22 – 1.00 =

-0.78 - ½ (-0.78) = 0.39 M

(not asked for)

Final

(t = 50 s) 0.22 (not asked for) (not known/needed)

[H2O2] (1st 50 s) = 0.22 M – 1.00 M = -0.78 M H2O2

2O mol 0.59 mol 580.5 L 1.5 x formed O

lost OH mol 2

formed O mol 1 x

L

OH mol 80.7 2

22

222 === L

mol 90.3 (formed in 1st 50 s)

(e) Strategy/Explanation:

“Initial rate” means “the instantaneous rate at t = 0”. Thus, do the same thing as in (b) except at t = 0 s.


PS1-6


I “drew” an approximate tangent line to the curve at t = 0 s (dot-dash line; see below).

Looking at the dot-dash line, the “y-intercept” is (0 s, 1.00 M), and the “x-intercept” is (36. s, 0. M) Thus:

1-sM 0.014 ==

−=

−=

−= 1-22 sM 8...30.01

s 36.

M 1.00-

s 0. - 36.

M 1.00 - 0.]O[H

2

1

2

1

2

1

tRxnofRateInitial

3. NT1. Consider a reaction represented by the eqn aA + bB → cC and the following average rate data over some time period t:

1-sM 0.0080A

=

−

t 1-sM 0.0120

B=

−

t 1-sM 0.0160

C=

t

Determine the coefficients (a, b, and c) in the standard balanced chemical equation (i.e., lowest whole #’s).

Answer: 2 A + 3 B → 4 C (i.e., a = 2; b = 3; c = 4)

Explanation: The ratio of coefficients in a balanced chemical equation must be equal to the ratio of changes in mole values (in absolute value) of reactants and/or products during a chemical reaction. This is essentially the meaning of the coefficients in a balanced equation. Assuming a fixed volume and a fixed time interval

(t), the rate of change of a species is directly proportional to change in concentration and thus change in moles:

Rate of change of X (M/s) [X] (M) moles of X (mol)

This means that the ratio of coefficients (= ratio of moles) also equals the ratio of rates of change. For this problem, then, since 0.0080 : 0.0120 : 0.0160 equals 8 : 12 : 16 (divide all values by 0.0100) which equals 2 : 3 : 4, the set of lowest whole number coefficients would be 2, 3, and 4.

Uncertainty probably

about 0.001 M∙s-1 due to slope estimation


PS1-7

(Differential) Rate Laws / Meaning of Orders (in addition to ideas from the previous section)

4. NT2. The reaction represented by the following equation was found to have the rate law shown.

2 N2O5(g) → 4 NO2(g) + O2(g); Rate Law: Rate = k[N2O5(g)]

(a) What is the order of the reaction with respect to N2O5?

Answer: First order (because the exponent of [N2O5(g)] in the rate law is 1).

(b) How does the rate of formation of O2 compare with the rate of loss of N2O5? (State the correct answer from among the statements below and then explain your answer).

(i) O2 forms half as fast as N2O5 is lost.

−=

tt

]O[N][O i.e., 522

2

1

(ii) O2 forms twice as fast as N2O5 is lost. (iii) O2 forms four times as fast as N2O5 is lost. (iv) It depends on how much N2O5 you start with.

Explanation: Since the amount of moles of O2 produced is always half the amount of moles N2O5 used up (based on the 2 : 1 ratio of N2O5 : O2 in the balanced equation), the rate of formation of O2 must be half the value of the rate of loss of N2O5.

(c) A specific trial of the above reaction is carried out, and the concentration vs. time plot of one of the species involved is plotted below. Identify which species is depicted in the plot below, and then draw curves representing the other two species involved in the reaction (copy the plot onto your paper, and be careful to be reasonably quantitatively accurate!) Give brief reasoning

(one possible) Answer

Explanation:

The balanced chemical equation is: 2 N2O5(g) → 4 NO2(g) + O2(g)

What does this mean? It means that for every two moles of N2O5(g) that undergo reaction, four moles (i.e., twice as many) of NO2 and one mole (i.e., half as many) of O2 will be formed. Thus, the limiting value for the amount of NO2 produced (i.e., the value when essentially all of the N2O5 has reacted) is twice the initial value of N2O5, whereas the limiting value for the amount of O2 produced is half the initial value of N2O5. Note that the products’ concentrations should INCREASE with time since they are being produced; the reactant’s concentration should DECREASE with time since it is being “used up”. If you

assume that you started with no products at t = 0, the plots for NO2 and O2 would start at the origin, and the graph would look something like what is shown in the answer box (see above, right).

Note that the slope at any point in time (on a plot of concentration vs. time) represents the rate of change of that species’ concentration at that time. So this is yet another way to see that at any point in time during this reaction, the rate of increase of NO2 is 4 times the value of the rate of increase of O2, and the rate of formation of O2 is half the value of the rate of loss of N2O5 (see part (b) above).

Time

Concentration

Time

Concentration

NO2

O2 N2O5


PS1-8

NOTE #2: There is really no need to assume that you start without any products present (although that is what is often done, for convenience, when we begin the discussion of kinetics). In the next chapter we will see many problems in which some “products” are present at the start of some reaction. So IF you were to assume that you started with [NO2]0 = ½ [N2O5]0 and [O2]0= 1.5 x [N2O5]0, the plot would look as shown below (all curves have the same exact shape and size as above, but those for NO2 and O2 are just moved up relative to where they were in the prior plot):

5. 15.37. What are the units of k for…(rest of problem not copied in key)

(a) s-1

(b) M-1∙s-1 Answers

(c) M∙s-1

Reasoning / “Proof”

Write an “example” rate law for each “case” and then solve for k with units only (no numbers):

1st order: R = k[A] 11

−−

=

⎯⎯⎯⎯⎯ →⎯== sM

sMRk units of terms in

[A]

2nd order: R = k[A]2 111

2

1−−

−−

==

⎯⎯⎯⎯⎯ →⎯== sMM

s

M

sMRk units of terms in

2[A]

0th order: R = k[A]0 = k 1−⎯⎯⎯⎯⎯ →⎯== sMRk units of terms in

6. 15.39. A reaction in which A, B, and C react to form products is…(rest of problem not copied in key)

Answers: (a) R = k[A][B]2[C]0 (or just R = k[A][B]2) (b) 3 (c) factor of 2 (d) factor of 4 (e) factor of 1 (no change) (f) factor of 8

Explanations:

(a) An “order” is an exponent of a concentration term. If the reaction is “first order in A”, that means that the exponent of [A] in the rate law is a “1”. If it is “second order in B, that means that the exponent of [B] in the rate law is a “2”. If it is zero order in C, that means that the exponent of [C] is zero (but since anything to the zero power is one, that term can be left off entirely).

(b) Overall order = sum of individual orders = 1 + 2 + 0 = 3

Time

Concentration

NO2

O2

N2O5

Obviously, you could have started with any initial concentrations of NO2 or O2, so I can’t write all of the possible correct answers here! ☺

Another possible correct answer


PS1-9

(c) Whatever factor of change there is in the concentration, that value raised to the order is the factor of change in the rate. When the order is “1”, the factor of changes are equal, so if concentration doubles, rate doubles.

“Proof”: R = k[A]1. Let the first [A] be [A](1); then after the concentration is doubled, [A](2) = 2 x [A](1)

R(1) = k[A](1), and after doubling the concentration, R(2) = k (2 x [A](1))1 = 2 x k[A](1) = 2 x R(1).

The last mathematical equation means “Rate after doubling [A] = 2 times the original Rate”

(d) Whatever factor of change there is in the concentration, that value raised to the order is the factor of change in the rate. When the order is “2”, the factor of change of the rate equals the square of the factor of change of the concentration, so if concentration doubles, rate quadruples.

“Proof”: R = k[B]2. Let the first [B] be [B](1); then after the concentration is doubled, [B](2) = 2 x [B](1)

R(1) = k[B](1)2, and after doubling the concentration, R(2) = k (2 x [B](1))2 = 22 x k[B](1)2 = 22 x R(1).

The last mathematical equation means “Rate after doubling [B] = 4 times the original Rate”

(e) If a reaction is zeroth order in C, that means that the rate is independent of [C]. I.e., if you double [C], the rate remains the same.

(f) If more than one change in concentration occurs, the factor change in the rate is the product of the factors of changes that would have occurred if each change had been the only change that occurred.

“Proof”: R = k[A][B]2. Let the first [A] be [A](1); then after the concentration is doubled, [A](2) = 2 x

[A](1). Let the first [B] be [B](1); then after the concentration is doubled, [B](2) = 2 x [B](1)

R(1) = k[A](1) [B](1)2, and after doubling both concentrations,

R(2) = k(2 x [A](1)) (2 x [B](1))2 = 2 ∙22 x k[A](1)[B](1)2 = 2∙22 x R(1) = 8 x R(1)

The last mathematical equation means “Rate after doubling [A] and [B] = 8 times the original Rate” (Note: I left off [C] here for simplicity. We know that the rate is unaffected by a change in [C] because it is zero order in C.)

Determining (Differential) Rate Laws / Method of Initial Rates

7. 15.41. Consider the data below showing the initial rate of a reaction (A → products)…(rest of problem not copied in this key)

Answers: order is 2; R = k[A]2, where k = 5.3 M-1∙s-1 or 5.25 M-1∙s-1

Reasoning:

To find the order:

Verbal: Going from Trial 1 to Trial 2, the (initial) [A] doubles (0.100 M to 0.200 M), and the (initial) rate

quadruples (0.053 to 0.210 M/s; 0.210 / 0.053 = 3.96 4). That means the order of A must be 2 (2m =

4 m = 2). This can be checked / confirmed by noting that in going from Trial 1 to Trial 3, the [A]

triples, and the rate goes up by 0.473 / 0.053 = 8.92 9 times, which equals 32. Thus (again), the order

is 2.


PS1-10

“Brute Force” Approach:

Set up the rate law: R = k[A]m (1)

Substitute values from Trials 1 and 2 into the rate law (1) above:

Trial 1: 0.053 M/s = k(0.100 M)m

Trial 2: 0.210 M/s = k(0.200 M)m

Now look at the ratio of rates and then substitute into the equation using the “right-hand side” expressions for the rates (from above):

(Note: Although not necessary, I prefer to divide the equation having the larger rate in it by the one with the smaller rate

since I find it easier to work with values greater than one rather than fractions.)

So, dividing Trial 2’s equation by Trial 1’s, you get:

m

m

m

0.100

0.200

0.053

0.210

M) (0.100

M) (0.200

M/s 0.053

M/s 0.210

(1)

(2)

===

k

k

R

R

( ) 2 m 2.004 693. m

==

To find k:

Use any trial and just substitute in the values for initial rate and [A]. Now that the order (m, here) is known, k will be the only unknown!

Trial 1: 0.053 M/s = k(0.100 M)2 111

2

1−−

−−

==

== sMM

s

M

sMk 5.3

5.3

0.0100

0.053

M) (0.100

M/s 0.0532

As a check:

Trial 2: : 0.210 M/s = k(0.200 M)2 111

2

1−−

−−

==

== sMM

s

M

sMk 5.25

5.25

0.0400

0.210

M) (0.200

M/s 0.2102

(NOTE: These two values are consistent with one another. The best thing to do in principle [e.g., if this were an actual set of experiments] would be to calculate all three “k’s” and then average them [this is what you will do for Experiments 20 and 21 in lab], but I will not do that here. On an exam, for a similar problem, you may choose any ONE trial and leave it at that.)

8. NT3. In each table, [A], [B], & [C] represent the initial concentrations of reactants A, B, & C (in M) for a differ-

ent reaction (type), and “Rate” represents the initial rate of reaction for each trial (in M/s). For each table:

(a) State which two trials must be compared first in order to determine an order for one of the reactants, A, B, or C.

(b) State the reactant (A, B, or C) whose order will be determined by this comparison (in (a)), and state how you know (i.e., explain).

(c) Determine the order of the reactant you identified in (b) above. Show work and/or give reasoning.

(d) Do NOT attempt to determine the other orders in these reactions. This problem was designed, rather, to help you practice how to begin any initial rates problem. State clearly how you determine which two trials to compare first in any “initial rates” kind of problem!

Trial [A] [B] [C] Rate

1 5 2 8 10 2 10 2 32 160 3 10 2 4 20 4 15 6 8 2430

Trial [A] [B] [C] Rate

1 4 15 6 36 2 20 60 12 11520 3 16 15 12 576 4 4 5 6 4

(note the cancellation of the k’s, and units in the numerator and denominator of both sides)

(i) (ii)


PS1-11

Answers (w/o reasoning):

(i) (a) Trials 2 and 3. (ii) (a) Trials 1 and 4.

(b) C (b) B

(c) order = 1 (i.e., 1st order) (c) order = 2 (i.e., 2nd order)

(d)(same for (i) and (ii): For the first “pairing”, always pick two trials that have the following property: Only one reactant’s concentration value is different in the two trials—all other reactant concentration values are unchanged in the two trials.

Strategy/Reasoning:

Since an “order” of a reactant is the exponent of the concentration of that reactant in the rate law, its value determines how changes in concentration of that reactant affect the rate. As such, when “initial rates” data are given/available for a reaction in which there is more than one reactant, one can only determine an initial order by comparing two trials in which only one reactant’s concentration differs—all other reactant’s concentration values much remain unchanged.

This is analogous to trying to determine whether dieting or exercising causes you to lose weight. If you diet and exercise, and you lose weight, how do you know how much, if any, of the weight loss is due to the change in diet, and how much, if any, is due to the change in exercise routine? You don’t. But if you go a month in which you keep your diet identical, but change only your exercise routine, then any weight change you experience can be attributed to the exercise (assuming you haven’t changed anything else in your life! :))

Once two such trials are found, you can determine the order as in the prior problem. Namely, you can calculate the factor of change in the concentration and the factor of change in the rate, and then compare them, considering the mathematical form of a rate law, to determine the order. This can be done verbally or with the “brute force” mathematical approach (see prior problem).

NOTE: Once one of the orders for a reactant is determined, then it is possible to compare trials later in which this reactant’s concentration and another have both changed, since you’ll already know how one of their changes will affect the rate, but that is for another problem.

Application of strategy:

For Table (i):

(a) If one looks closely at evey possible pair of trials (1, 2; 1, 3; 1, 4; 2, 3; 2, 4; and 3, 4), you wll find that the only pair in which only one reactant’s concentration changes is the pair of Trials 2 and 3. Specifically, Trials 2 and 3 have two reactants (A and B) with the same values of concentration ([A] = 10 M in both Trials 2 and 3, and [B] = 2 M in both Trials 2 and 3), while [C] is different in the two (32 M in Trial 2 and 4 M in Trial 3). So the change in rate between Trials 2 and 3 must be a result of the change in the concentration of C, and so the order of C can be determined.

In contrast, Trials 1 and 2 both have [B] = 2 M, but the concentrations of A and C are both different

between them (i.e., [A] = 5 M in Trial 1 and 10 M in Trial 2, and [C] = 8 M in Trial 1 and 32 M in Trial 2). So the “cause” of the change in rate between Trials 1 and 2 cannot be unambiguously determined—it might be all due to the change in [A], it might all be due to the change in [C], or it might be due to a combination of the changes in both. So you can’t get the order of either A or C by comparing these two trials!

(b) As stated above, since only [C] changes between Trials 2 and 3, it is C’s order that can be determined by comparing these two trials.

(c) Verbal reasoning: The rate of Trial 2 is 160 M/s, and that of Trial 3 is 20 M/s. Thus, the rate of Trial 2 is 8 times that of Trial 3 (160/20 = 8). The concentration of C is 32 M in Trial 2 and 4 M in Trial 3. Thus, the concentration of C in Trial 2 is 8 times that of Trial 3. Since the rate becomes 8 times bigger when [C] becomes 8 times bigger, that means rate is proportional to [C], which means the order of C is 1.


PS1-12

“Brute force” approach:

Set up the rate law: R = k[A]m[B]n[C]p (1)


Trial 2: 160 M/s = k(10 M)m(2 M)n(32 M)p

Trial 3: 20 M/s = k(10 M)m(2 M)n(4 M)p





𝑅(2)

𝑅(3)=

160 M/s

20 M/s=

𝑘(10 M)𝑚(2 M)𝑛(32 M)𝑝

𝑘(10 M)𝑚(2 M)𝑛(4 M)𝑝 ⇒

160

20= (

10

10)

𝑚

(2

2)

𝑛

(32

4)

𝑝

⇒ 8 = (1)𝑚(1)𝑛(8)𝑝

⇒ 8 = (8)𝑝 ⇒ 𝑝 = 1

For Table (ii):

(a) If one looks closely at evey possible pair of trials, the only two that have two out of three concentrations the same (and thus one different) is the pair of trials 1 and 4. A and C’s concentrations don’t change from 1 to 4, but [B] is different in those two trials.

(b) As stated above, since only [B] changes between Trials 1 and 4, it is B’s order that can be determined by comparing these two trials.

(c) Verbal reasoning: The rate of Trial 1 is 36 M/s, and that of Trial 4 is 4 M/s. Thus, the rate of Trial 1 is 9 times that of Trial 4 (36/4 = 9). The concentration of B is 15 M in Trial 1 and 5 M in Trial 4. Thus, the concentration of B in Trial 1 is 9 times that of Trial 4. Since the rate becomes 9 times bigger when [B] becomes 3 times bigger, that means rate is proportional to [B] squared ([B]2), which means the order of B is 2.

“Brute force” approach:

Set up the rate law: R = k[A]m[B]n[C]p (1)


Trial 1: 36 M/s = k(4 M)m(15 M)n(6 M)p

Trial 4: 4 M/s = k(4 M)m(5 M)n(6 M)p





𝑅(1)

𝑅(4)=

36 M/s

4 M/s=

𝑘(4 M)𝑚(15 M)𝑛(6 M)𝑝

𝑘(4 M)𝑚(5 M)𝑛(6 M)𝑝 ⇒

36

4= (

4

4)

𝑚

(15

5)

𝑛

(6

6)

𝑝

⇒ 9 = (1)𝑚(3)𝑛(1)𝑝

⇒ 9 = (3)𝑛 ⇒ 𝑛 = 2




PS1-13

9. NT4. The following data were determined for the reaction represented by the equation:

2 ClO2(aq) + 2 OH-(aq) → ClO3

-(aq) + ClO2

-(aq) + H2O(l)

The temperature was the same in all trials.

Trial [ClO2]0 (M) [OH-]0 (M) Initial Rate (M/s)

1 0.0500 0.100 2.88 x 10-2

2 0.100 0.100 1.15 x 10-1

3 0.100 0.0500 5.75 x 10-2

(a) Determine the rate law (using “k” to represent the rate constant) (b) Determine the (value and units of the) rate constant (at the temperature at which these data were collected). (c) What would be the initial rate for a trial with [ClO2]0 = 0.175 M and [OH-]0 = 0.0844 M (at the same T)?

Answers: (a) R = k[ClO2]2[OH-]; (b) k = 115 M-2 ∙s-1; (c) 0.298 M/s

Reasoning:

The rate law will be of the form:

Rate = k[ClO2]m[OH-]n (1)

To find the orders (m and n):

Verbal: Going from Trial 1 to Trial 2, the (initial) [ClO2] doubles (0.05 M to 0.10 M) while the (initial) [OH-

] is kept the same, and the (initial) rate goes up by (1.15 x 10-1 M/s)/(2.88 x 10-2 M/s) = 3.993 4 times.

That means the order of ClO2 must be 2. In going from Trial 3 to Trial 2, the [OH-] doubles while the

[ClO2] is kept the same, and the rate also doubles (1.15 x 10-1 M/s divided by 5.75 x 10-2 = 2.00). That

means the order of OH- must be 1.

Brute Force Approach:

( )m

k

k

R

R2.004~

0.100

0.100

0.0500

0.100393.9

M) (0.100M) (0.0500

M) (0.100M) (0.100

M/s 10 x 2.88

M/s 10 x 1.15

(1)

(2)nm

nm

nm

2-

-1

=

===

2 = m

( )n

k

k

R

R2.002.00

0.0500

0.100

0.100

0.1002.00

M) (0.0500M) (0.100

M) (0.100M) (0.100

M/s 10 x 5.75

M/s 10 x 1.15

(3)

(2)nm

nm

nm

2-

-1

=

===

1 = n

Substituting into (1) with the two (now known) orders yields: (a) R = k[ClO2]2[OH-]

(b) To get the value of the rate constant, k, pick a trial and substitute in the values of R and

concentrations (along with the values of the orders). Here, I’ll use Trial 1:

2.88 x 10-2 Ms-1 = k(0.0500 M)2(0.100 M)1

M) (0.100M) (0.0500

sM 10 x 2.88

2

-1-2 = k

12-

2

-1

115)()(

.2511 −=

= sMMM

sM

(c) To get the initial rate for an experiment with [ClO2]0 = 0.175 M and [OH-]0 = 0.0844 M, just substitute into the rate law (with k now known):

R = (115.2 M-2 ∙s-1)[ClO2]2[OH-] = (115.2 M-2 ∙s-1) (0.175 M)2(0.0844 M)

= 0.2977.. = 0.298 M/s


PS1-14

10. 15.46. The data below were collected…(rest of problem not copied in this key)

CH3Cl(g) + 3 Cl2(g) → CCl4(g) + 3 HCl(g)

The rate law will be of the form:

Rate = k[CH3Cl]m[Cl2]n (1)

(a) To find the orders (m and n):

Verbal: Going from Trial 1 to Trial 2, the (initial) [CH3Cl] doubles (0.050 M to 0.100 M) while the (initial)

[Cl2] is kept the same, and the (initial) rate goes up by (0.029 M/s)/(0.014 M/s) = 2.07 2 times. That means the order of CH3Cl must be 1. In going from Trial 2 to Trial 3, the [Cl2] doubles while the

[CH3Cl] is kept the same, and the rate goes up by (0.041 M/s divided by 0.029 =) 1.41 times. If one

notices that 1.41 is very close to 2

1

22 = it is clear that the order of Cl2 must be ½ .

Brute Force Approach:

( )m

k

k

R

R2.002~

0.050

0.050

0.0500

0.100702.

M) (0.050M) (0.0500

M) (0.050M) (0.100

M/s 0.014

M/s 0.029

(1)

(2)nm

nm

nm

=

===

1 = m

( )n

k

k

R

R2.001.41

0.0500

0.100

0.100

0.1001.41

M) (0.0500M) (0.100

M) (0.100M) (0.100

M/s 0.029

M/s 0.041

(2)

(3)nm

nm

nm

=

===

2

1 n 2.00 n2

1

== 2 (see below*)

*As will be shown in lab, one can also solve for n as follows:

( ) 0.546..90.431...90.6

358...40.3

ln(2.00)

ln(1.41) ln(2.00) x ln(1.41) ln(2.00) ln(1.41) 2.001.41

n ====== nnn

Substituting into (1) with the two (now known) orders yields: (a) R = k[CH3Cl][Cl2]1/2

To get the value of the rate constant, k, pick a trial and substitute in the values of R and concentrations

(along with the values of the orders). Here, I’ll use Trial 4 (just to get 3 SF’s):

0.115 Ms-1 = k(0.200 M)(0.200 M)1/2 1/2

-1

M) M)(0.200 (0.200

sM 0.115

= k

11/2-

1/2

-1

1.29))((

57..81.2 −=

= sMMM

sM

(b) The overall order is 1 + ½ = 1.5

(Differential) Rate Laws / Conceptual Ideas

11. NT5. State whether the statement is true or false AND IF FALSE, CHANGE A FEW WORDS TO MAKE IT CORRECT.

Note: I will ask a few questions like this on each of my exams! YOU MUST CORRECT ANY FALSE STATEMENT CORRECTLY

TO RECEIVE CREDIT ON AN EXAM FOR THESE!

(a) It is possible to change the rate constant for a reaction by changing the temperature.

TRUE (more discussion about this during Week 2).

Answers:

(a) R = k[CH3Cl][Cl2]1/2;

k = 1.29 M-1/2∙s-1

(b) 1.5


PS1-15

(b) The reaction rate remains constant as a first-order reaction proceeds at a constant temperature.

FALSE. The reaction rate will continuously decrease as a reaction proceeds because the

concentrations of reactants will continuously decrease. It is the rate constant that remains constant

during any chemical reaction as long as the T is kept constant. Thus, two good ways to correct this

statement are:

OR

(c) The rate constant for a reaction is independent of reactant concentrations.

TRUE. Notice that most of these questions are forcing you to distinguish between “rate” and “rate

constant”. It is also important for you to distinguish both of these from “rate law”. The rate of reaction

varies with concentrations, but the rate constant does not vary with concentrations!

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

Additional problems which may or may not have been assigned in Mastering 15.26. Consider the reaction (represented by):

2 N2O(g) → 2 N2(g) + O2(g)

(a) Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

(b) In the first 15.0 s of the reaction, 0.015 mol of O2 is produced in a reaction vessel with a volume of 0.500 L. What is the average rate of the reaction over this time interval?

(c) Predict the rate of change in the concentration of N2O over this time interval. In other words, what is tΔ

Δ O][N2 ?

Answers: (a) ttt

=

=

−=

][O][NO][N reaction of Rate 222

2

1

2

1

(b) 0.0020 M∙s-1 (over the 1st 15 s)

(c) -0.0040 M∙s-1

Explanation/Reasoning:

(a) Since the rates of loss and formation of the reactants and products are proportional to the coefficients in the balanced equation (i.e., these “rates” are not all equal), a single “rate of reaction” will be obtained if each rate of loss or formation is divided by the species’ coefficient in the balanced equation. That is why

“rate of reaction” is defined as tx

−

X][1 if X is a reactant, and

tx

X][1 if X is a product (with coefficient x).

(b) Average rate of reaction (expressed in terms of O2) = t

][O2

M 0.030L 0.500

O mol 0.015 ][O 2

2 ) (over == sst 151 O2 [b/c M = moles / L]

The reaction rate remains constant as a first-order reaction proceeds at a constant temperature.

The reaction rate remains constant as a first-order reaction proceeds at a constant temperature.

decreases

constant

OR here’s yet a third way: change “first” to “zero” !

NOTE: The answers here are those for the values given in the textbook problem. In Mastering, the given values may be different, which would obviously affect the final answers. Be careful!


PS1-16

Thus, 1-2 sM 0.0020s 15.0

M 0.030][O==

t

(c) 1-1-2222 sM 0.0040- )sM (0.0020 ][OO][N

][OO][N

2 =−=

−=

=

−

tttt2

2

1

I.e., the rate of change of N2O is twice the rate of formation of O2, which can be seen from the 2:1 ratio of N2O lost : O2 formed in the balanced equation. The negative sign is needed here because rate of “change” of N2O was asked for rather than rate of “loss” (and rates of loss are always the opposite of rates of formation).

Documents

Answer Key, Problem Set 1, Written