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Chemistry 122Mines, Spring, 2012

PS1-1

Answer Key, Problem Set 1Full Version

1. 13.26; 2. 13.32; 3. 13.34 (add part (e): Estimate / calculate the initial rate of the reaction); 4. NT1;

5. NT2; 6. 13.37; 7. 13.39; 8. 13.41; 9. NT3; 10. 13.44; 11. NT4------------------------------1. 13.26. 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 theaverage rate of the reaction over this time interval?

(c) Predict the rate of change in the concentration of N2Oover this time interval. In other words, what ist

O][N2 ?

Answers: (a)ttt

][O][NO][NreactionofRate 222

2

1

2

1

(b) 0.0020 Ms-1

(over the 1st 15 s)

(c) -0.0040 Ms-1

Explanation/Reasoning:

(a) Since the rates ofloss and formation of the reactants and products are proportional to the coefficients inthe balanced equation (i.e., these rates are not all equal), a single rate of reaction will be obtained ifeach rate ofloss orformation is divided by the species coefficient in the balanced equation. That is why

rate of reaction is defined astx

X][1if 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

M0.030L0.500

Omol0.015][O 22 )(over sst 151 O2 [b/c M moles / L]

Thus, 1-2 sM0.0020s15.0

M0.030][O

t

(c) 1-1-2222 sM0.0040-)sM(0.0020][OO][N

][OO][N

2

tttt

22

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 ofN2Olost : O2 formed in the balanced equation. The negative sign is needed here because rate of change of N2Owas asked for rather than rate of loss (and rates of loss are always the opposite of rates of formation).

2. 13.32. Consider the reaction (represented by): NO2(g) NO(g) +2

1O2(g)

The following data were collected for the concentration of NO2 as a function of time:

Time (s) 0 10 20 30 40 50 60 70 80 90 100

[NO2] (M) 1.000 0.951 0.904 0.860 0.818 0.778 0.740 0.704 0.670 0.637 0.606

(a) What is the average rate of the reaction between 10 and 20 s? Between 50 and 60 s?

(b) What is the rate of formation of O2 between 50 and 60 s?

Answers: (a) 0.0047 Ms-1

and 0.0038 Ms-1

; (b) 0.0019 Ms-1

Reasoning:

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Answer Key, Problem Set 1

PS1-2

(a)t

Rate.Avg

][NO 2 here because the coefficient of NO2 is 1. Thus,

From 10 s to 20 s, 1-22 sM0.0047s10.

M0.047

s10s20

M0.951M0.904][NO][NO

if

if

ttRate.Avg

From 50 s to 60 s, 1-22 sM0.0038s10.

M0.038

s50s60

M0.778M0.740][NO][NO

if

if

ttRate.Avg

(b) R(form, of O2) = R(loss of NO2) because of the coefficients in the equation (1 : for NO2 : O2). Thus, the(average) rate of formation of O2 between 50 s and 60 s x 0.0038 Ms

-1(from (a))

0.0019 Ms-1

3. 13.34. Consider the reaction (represented by):

2 H2O2(aq) 2 H2O (l) + O2(g)

The graph below (to the right in this key) shows the

concentration of H2O2 as a function of time.

Use the graph to calculate each quantity:

(a) The average rate of the reaction between 10 and 20 s.

(b) The instantaneous rate of the reaction at 30 s.

(c) The instantaneous rate of formation of O2 at 50 s

(d) If the initial volume of the H2O2 is 1.5 L, what total amount of O2

(in moles) is formed in the first 50 s of reaction?

(e) [added] Estimate / calculate the initial rate of the reaction.

Answers: (my results): (a)0.0095 Ms-1

; (b) 0.0063 Ms-1

; (c)0.0034 Ms

-1; (d) 0.59 mol O2; (e) 0.014 Ms

-1

Strategy/Explanation:

(a) Avg. rate of reaction here t

]O[H 222

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

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

2) Calculatet

]O[H 222

1

(b) Instantaneous rate means the rate at a single point in time, rather than over a time interval. It can beestimated by finding the (absolute value of the) slope of the tangent to a curve at a point in time. So inthis problem one must:

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

2) Find the (absolute value of the) slope of the tangent line by picking any two points on the line and

using the definition of slopex

y

(which here representst

]O[H 22 ).

3) Multiply the result in (2) by - to get rate of reaction (see (a)).

(c) This is similar to (b) except that the rate offormation 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 thisproblem! Thus, you can actually do exactly the same thing you did in (b) except do it at the pointcorresponding to t = 50 s.

(d) Once you step back and really analyze this problem, hopefully you will recognize that it is actually just astoichiometry problem involving concentration and volume! You are asked for moles of O2 formed inthe first 50 s, and you are effectively given the concentration of H2O2 lost during this same timeinterval (because you can get this from the graph by determining [H2O2]50 s [H2O2]0 s). So, the strategyhere is to go from

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

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Answer Key, Problem Set 1

PS1-3

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

OHmol2

Omol1.

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

Execution of Strategy:

I scanned the plot from the text and enlarged it in this key for better precision. I adjusted a line to make it go through the point on the curve at t = 30 s, and which appeared to

have the slope of the curve at that point (see dotted line on plot). I did the same for a line throughthe points at t = 50 s (dashed line) and at t = 0 s (dot-dash line).

(a) At 10. s, [H2O2] 0.75 M; at 20. s, [H2O2] 0.56 M. Thus,

1s. 0095

s10.

M0.19-

s10.-20.

M0.75-0.56]O[H 222

1

2

1

2

1

tRate

(b) Looking at the dotted line and picking as my two points, the intercepts (just for convenienceyou could useany two points), I get the points (0 s, 0.78 M) and (62 s, 0 M). Thus:

1s. 0063

1-22 sM920.006

s62.

M0.78-

s0.-62.

M0.78-0.]O[H

2

1

2

1

2

1

tRate

(c) Looking at the dashed line, the y-intercept is (0 s, 0.55 M), and another point is (70. s, 0.08 M) [I didntpick the x-intercept here because it was slightly off my scale]. Thus:

1s. 0034

1-22 sM57...30.003

s70.

M0.47-

s0.-70.

M0.55-0.08]O[H

2

1

2

1

2

1

t

Rate

(d) [H2O2]50 s 0.22 M; [H2O2]0 s 1.00 M. Thus, [H2O2] -0.78 M 0.78 M H2O2 was lost (in 1st

50 s)

2Omol0.59580.5lostOHmol2

formedOmol1xL51.x

L

OHmol80.7

22

222 (formed in 1st 50 s)

(e) 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:

1s. 014

1-22 sM8...30.01

s36.

M1.00-

s0.-36.

M1.00-0.]O[H

2

1

2

1

2

1

tRate

Uncertainty probably

about 0.0003 Ms-1due to slope estimation

Uncertainty prob

bly about 0.00Ms

-1due to sloestimation

Uncertainty probably

about 0.001 Ms-1due to slopeestimation

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Answer Key, Problem Set 1

PS1-4

4. NT1. Consider a reaction represented by the equation aA + bB cC and the following average rate data over some time period

1-sM0.0080A

t 1-sM0.0120

B

t 1-sM0.0160

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 changesin mole values (in absolute value) of reactants and/or products during a chemical reaction. This isessentially 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 concentration and thus moles:

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

This means that the ratio ofcoefficients also equals the ratio ofrates 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 setoflowest whole numbercoefficients would be 2, 3, and 4.

5. 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 thestatements below and then explain your answer).

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

tt

]O[N][Oi.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 halfthe amount of moles N2O5used 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 plottebelow. Identify which species is depicted in the plot below, and then draw curves representing the other two species involvein 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, fourmoles(i.e., twice as many) of NO2 and one mole (i.e., half as many) of O2 will be formed. Thus, the limitingvalue for the amount of NO2 produced (i.e., the value when essentially all of the N2O5 has reacted) istwice the initial value of N2O5, whereas the limiting value for the amount of O2 produced is halfthe initial

Time

Concentration

Time

Concentration

NO2

O2N2O5

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Answer Key, Problem Set 1

PS1-5

value of N2O5. Note that the products concentrations should INCREASE with time since they are beingproduced; the reactants 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, andthe 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 rateofchange of that species concentration at that time. So this is yet anotherway to see that at any point intime 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 halfthe value of the rate of loss of N2O5 (see part (b) above).

NOTE #2: There is really no need to assume that you start without any products present (although that is what is often done, forconvenience, when we begin the discussion of kinetics). In the next chapter we will see many problems in which some productsare present at the start of some reaction. So IF you were to assume that you started with [NO2]0 = [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 O2are just moved up relative to where they were in the prior plot):

6. 13.37. What are the units ofk for each type of reaction?

(a) first-order reaction s-1

(b) second-order reaction M-1s

-1Answers

(c) zero-order reaction Ms-1

Reasoning / Proof

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

1st

order: Rk[A] 11

sM

sMRk unitsoftermsin

[A]

2nd

order: Rk[A]2 111

2

1

sMM

s

M

sMRk unitsoftermsin

2[A]

0th

order: Rk[A]0k 1 sMRk unitsoftermsin

7. 13.39. A reaction in which A, B, and C react to form products is first order in A, second order in B, and zero order in C.

(a) Write a rate law for the reaction.

(b) What is the overall order of the reaction?

(c) By what factor does the reaction rate change if [A] is doubled (and the other reactant concentrations are held constant)?

(d) By what factor does the reaction rate change if [B] is doubled (and the other reactant concentrations are held constant)?

(e) By what factor does the reaction rate change if [C] is doubled (and the other reactant concentrations are held constant)?

(f) By what factor does the reaction rate change if the concentrations of all three reactants are doubled?

Time

Concentration

NO2

O2

N2O5

Obviously, you could have startedwith any initial concentrations ofNO2 or O2, so I cant write all of thepossible correct answers here!

Another possible correct answer

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Answer Key, Problem Set 1

PS1-6

Answers: (a) R = k[A][B]2[C]

0(or justR = 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 thatthe 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 sinceanything to the zero power is one, that term can be left off entirely).

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

(c) Whatever factor of change there is in the concentration, that value raised to the order is the factor ofchange 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 ofchange in the rate. When the order is 2, the factor of change of the rate equals the square of thefactor 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 independentof [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 thefactors of changes that would have occurred if each change had occurred on its own.

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)2 222 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.)

8. 13.41. Consider the data below showing the initial rate of a reaction (A products) at several different concentrations of A.(a) What is the order of the reaction? (b) Write a rate law for the reaction including the value of the rate constant, k.

Experiment (trial) [A]0 (M) Initial Rate (M/s)

1 0.100 0.053

2 0.200 0.210

3 0.300 0.473

Answers: order is 2; R = k[A]2, where k = 5.3 M

-1s

-1or 5.25 M

-1s

-1

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Answer Key, Problem Set 1

PS1-7

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 (2m4 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 orderis 2.

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 sideexpressions 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 2s equation by Trial 1s, you get:

m

m

m

0.100

0.200

0.053

0.210

M)(0.100

M)(0.200

M/s0.053

M/s0.210

(1)

(2)

k

k

R

R

2m2.004693. m

To find k:

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

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

11

1

2

1

sMM

s

M

sM

k 5.3

5.3

0.0100

0.053

M)(0.100

M/s0.0532

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/s0.2102

(NOTE: These two values are consistent with one another. The best thing to do in principle [e.g., if this were an actual setof experiments] would be to calculate all three ks and then average them [this is what you will do for Experiments 20 and21 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.)

9. NT3. 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)?

(note the cancellation of the ks, and units in thenumerator and denominator of both sides)

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Answer Key, Problem Set 1

PS1-8

Answers: (a) Rk[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). Thatmeans the order of OH

-must be 1.

Brute Force Approach:

mk

k

R

R2.004~

0.100

0.100

0.0500

0.100393.9

M)(0.100M)(0.0500

M)(0.100M)(0.100

M/s10x2.88

M/s10x1.15

(1)

(2)nm

nm

nm

2-

-1

2 m

nk

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/s10x5.75

M/s10x1.15

(3)

(2)nm

nm

nm

2-

-1

1 n

Substituting into (1) with the two (now known) orders yields: (a) Rk[ClO2]2[OH

-]

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

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

2.88 x 10-2

Ms-1 = k(0.0500 M)2(0.100 M)1

M)(0.100M)(0.0500

sM10x2.88

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

-2s

-1) (0.175 M)

2(0.0844 M)

= 0.2977.. = 0.298 M/s

10. 13.44. The data below were collectted for a reaction represented by this equation:

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

Trial [CH3Cl]0 (M) [Cl2]0 (M) Initial Rate (M/s)

1 0.050 0.050 0.014

2 0.100 0.050 0.029

3 0.100 0.100 0.041

4 0.200 0.200 0.115

(a) Write an expression for the reaction rate law and calculate the value of the rate constant, k.

(b) What is the overall order of the reaction?

Answers:

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

;

k = 1.29 M-1/2

s-1

(b) 1.5

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PS1-9

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. Thatmeans 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 onenotices that 1.41 is very close to 2

1

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

Brute Force Approach:

mk

k

R

R2.002~

0.050

0.050

0.0500

0.100702.

M)(0.050M)(0.0500

M)(0.050M)(0.100

M/s0.014

M/s0.029

(1)

(2)nm

nm

nm

1 m

nk

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/s0.029

M/s0.041

(2)

(3)nm

nm

nm

2

1

n2.00

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)xln(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 ofR and concentrations

(along with the values of the orders). Here, Ill use Trial 4 (just to get 3 SFs):

0.115 Ms-1 = k(0.200 M)(0.200 M)1/21/2

-1

M)M)(0.200(0.200

sM0.115

k

11/2-

1/2

-1

1.29))((

57..81.2

sMMM

sM

(b) The overall order is 1 + = 1.5

11. NT4. 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).

(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:

7/27/2019 Ans PS01 122 S12_full

10/10

Answer Key, Problem Set 1

PS1-10

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!

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 heres yet a third way:change first to zero !