CHEMISTRY 104 – Help Sheet #8
Chapter 11 (Part I); Sections 11.1-11.3 - Kinetics
Do the topics appropriate for your lecture
Prepared by Dr. Tony Jacob
http://www.chem.wisc.edu/areas/clc (Resource page)
Nuggets: Average Rates; Initial Rates; Relative Rates; Rate Laws; Rate Constant and Units; Evaluating Rate
Laws – Initial Rates Method, Graphs, Description, Raw Data; Integrated Rate Laws
FACTORS THAT AFFECT RATE: To increase rate:
1. Molecular structure
2. Concentration ↑ → #collisions ↑ → reaction rate ↑
3. Temperature ↑ → KE ↑ → Ecollisions ↑ → more molecules have E > Ea (i.e., have sufficient energy;
Ea = activation energy) → reaction rate ↑ (note: k, rate constant, also changes as T changes)
4. Catalyst → Ea (activation energy) is lowered; more molecules have E > Ea; (i.e., have sufficient energy)
→ reaction rate ↑ (Note: both forward and reverse reaction rates ↑ )
5. Particle size (heterogeneous rxns) → particle size (solids) ↓ → #collisions ↑ since there is more
surface area for greater contact → reaction rate ↑
Δ[ ]
RATES – defined as a change in concentration over a change in time; Rate =
Δt
By convention, rates are defined as positive. Since Δ[ ]/Δt has a
negative value when reactants are used (reactant concentrations ↓ ⇒
negative value), a negative sign is added to yield a positive value.
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Δ[A]
Rate = −
where A is a reactant, A → B (rate units = M/time)
Δt
When products are used, no negative sign is added as the rate will yield
a positive value (product concentration ↑ ⇒ positive value);
Δ[B]
(rate units = M/time)
€ Rate =
Δt
•
•
•
Average rate is over a finite time: Determined by using the 2 closest
times that bracket the time of interest and their corresponding
concentrations;
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$ [A] t 2 − [A] t1 '
Δ[A]
Average Rate = −
= −&
);
Δt
% t 2 − t1
(
Initial rate is the rate from time = 0 to the next time data point;
$ [A] t 2 − [A] t1 '
Δ[A]
Initial Rate = −
= −&
) where t 1 = 0 ;
Δt
t
−
t
%
(
2
1
€
Instantaneous rate is the rate at an instant in time (see last page of Help
Sheet to estimate instantaneous rate)
Example: For €
the reaction: A → B, the data below was collected. a. What is the initial rate of the reaction of A? b. What is the
average rate of consumption of A between 20 and 40 sec? c. What is the average rate of consumption of A at 50 sec?
Answer:
[A] (M) time (s)
a. rate = -Δ[A]/Δt = -(0.905 – 1.00)/(10.0 – 0) = –(–0.0095) = 0.00950M/s
1.00
0.00
a
b. rate = -Δ[A]/Δt = -(0.670 – 0.819)/(40.0 – 20.0) = –(–0.00745) = 0.00745M/s
0.905 10.0
c. rate = -Δ[A]/Δt = -(0.549 – 0.670)/(60.0 – 40.0) = –(–0.00605) = 0.00605M/s
0.819 20.0
0.741 30.0
0.670 40.0
c
0.607 50.0
0.549 60.0
b
RELATIVE RATES – comparing rates using stoichiometry
Relating the consumption of one reactant to the consumption of another reactant or to the formation of a
product. The inverse of the stoichiometric coefficient are placed in front of the rate of that chemical; the rates
are the expressions within the parenthesis. For the reaction: 2A + B → 3C the relative rates are written:
1
1
1
( rate A) =
( rate B) =
( rate C) ; substituting
coeffA
coeffB
coeffC
1 $ Δ[A] ' 1 $ Δ[B] ' 1 $ Δ[C] '
&−
) = &−
)= &
)
2 % Δt ( 1 % Δt ( 3 % Δt (
the coefficients and rate expressions →
(Note: Negative signs for reactants; no negative sign for product.)
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Example: What is the rate of production of C in the reaction, 3A → B + 2C, if the rate of consumption of A is 3.5 x 10-2M/hr?
⎛ Δ[C] ⎞
⎛ Δ[C] ⎞ 2 ⎛
1 ⎛ Δ[ A] ⎞ 1 ⎛ Δ[C] ⎞
1⎛
−2 ⎞ 1 ⎛ Δ[C] ⎞
−2 ⎞
−2
Answer: ⎜ −
⎟ = rate of production of C = 2.3x10 M/hr
⎟= ⎜
⎟ → ⎜⎝ 3.5x10 ⎟⎠ = ⎜
⎟→ ⎜
⎟ = ⎜⎝ 3.5x10 ⎟⎠ → ⎜
⎝ Δt ⎠
3 ⎝ Δt ⎠ 2 ⎝ Δt ⎠
3
2 ⎝ Δt ⎠
⎝ Δt ⎠ 3
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RATE LAW: rate as a function of concentration; the rate law is written:
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unknown exponent
Rate = k[reactant€1]unknown exponent[reactant
where k is the rate constant;
2]
n is the order of reactant1; m is the order for reactant2; and n + m = reaction order
Common exponent values (i.e., orders) are 0 (zero order), 1 (first order), and 2 (second order)
Using 2A + B → 3C, the rate law is: Rate = k[A]n[B]m ; n, m, and k are determined experimentally
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order units of k
0
M/s
1
1/s
UNITS of k: units vary depending on the order of the reaction
Example: Determine the units of k for a second order reaction.
Δ[ A]
2
= k[ A] 2
Δt
M
M
Substitute units (time units can be s, min, hrs, etc.); for Rate use
and for [A] use M:
= kM 2
s
s
⎛ 1 ⎞⎛ M ⎞
⎛ 1 ⎞
-1 -1
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Solve for k: ⎜
⎟⎜ ⎟ = k ; k = ⎜
⎟ also written: k = M s
2
Ms
s
⎝
⎠
⎝
⎠
⎝M ⎠
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1/Ms
Answer: Start by writing a general rate law for a second order reaction: Rate = −
DIFFERENT METHODS TO DETERMINE RATE LAW (i.e., find the order(s) and k)
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I. Initial
rates method
1. Set up rate ratio from 2 experimental sets of data:
rate1
k[A]1a [B]1b[C]1c
=
rate 2
k[A]a2[B]b2[C]c2
choose experiments so only 1 chemical changes and all the rest cancel out
2. Cancel out k and other chemicals with the same concentrations; solve for exponent of the one
remaining chemical whose concentration is
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3. Repeat for other chemicals/exponents until all exponents (i.e., orders) are evaluated
4. Choose any set of experimental data, plug in values with now evaluated exponents and solve for k
Exp [A], M [B], M initial rate (M/s)
Example: Write the rate law for 2A + B → 3C and determine the orders and k.
Answer: Start by writing the general rate law; Rate = k[A]n[B]m
1
1. Choose 2 exps in which only 1 concentration changes (1 and 3; or 1 and 2)
2. Solve for n by choosing Exp 1 and Exp 3:
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1
2
3
0.10
0.10
0.20
0.10
0.20
0.10
2.0 x 10-2
8.0 x 10-2
4.0 x 10-2
n
m
⎛ 0.20 ⎞ n ⎛ 0.10 ⎞ m
rate 3 k[A] 3[B] 3
4.0x10 -2 [0.20] n[0.10] m
n
n m
=
; cancel k’s and insert #’s:
; 2=⎜
=
⎟ ⎜
⎟ = 2 1 ; 2 = 2 ; n =1
-2
n
m
0.10
0.10
rate1 k[A] n[B] m
⎝
⎠
⎝
⎠
2.0x10
[0.10] [0.10]
1
1
n
m
⎛ 0.10 ⎞ n ⎛ 0.20 ⎞ m
rate 2 k[A] 2[B] 2 8.0x10 -2 [0.10] n[0.20] m
m
n m
=
3. Repeat for m with Exp 1 and Exp 2:
;
; 4 =⎜
=
⎟ ⎜
⎟ =1 2 ; 4 = 2 ; m= 2
rate1 k[A] n[B] m 2.0x10 -2 [0.10] n[0.10] m
⎝ 0.10 ⎠ €
⎝ 0.10 ⎠ €
€
1
1
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4. Rate law: Rate = k[A]1[B]2 and plug in values from any exp (I chose Exp 1): 2.0 x 10-2 = k(0.10)1(0.10)2; solve for k; k = 20M-2s-1
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2
DIFFERENT METHODS TO DETERMINE RATE LAW (i.e., find the order(s) and k) - continued
II. Description
Example: Determine the rate law for 2A + B → 3C given
I. When [A] doubles, the rate increases by 4
II. When [B] doubles, the rate is unchanged.
Answer: Start by writing the general rate law; Rate = k[A]n[B]m
1. From info I, [A] is changing and since no info is provided for [B] it is assumed that [B] is constant; set up rate ratio and [B] cancels →
n
a
b
a
rate1 k[A]1[B]1 rate1 k[A]1 rate1 ⎛ [A]1 ⎞
=
=
;
;
;
=⎜
⎟
rate 2 k[A] a [B] b rate 2 k[A] a rate 2 ⎝ [A] 2 ⎠
2
2
2
n
4 ⎛ 2⎞
= ⎜ ⎟ ; 4 = 2n ; n = 2
1 ⎝ 1⎠
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€ is changing and
2. From info II, [B]
since no info is provided for [A] it is assumed that [A] is constant; set up rate ratio and [A] cancels →
Let rate2 and [A]2 = 1, and change rate1 and [A]1 to match the information provided in I and solve:
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m
a
b
m
rate1 k[A]1[B]1 rate1 k[B]1
rate1 ⎛ [B]1 ⎞
=
=
;
;
;
=⎜
⎟
rate 2 k[A] a [B] b rate 2 k[B] m rate 2 ⎝ [B] 2 ⎠
2
2
2
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Let rate2 and [B]2 = 1, and change the rate1 and [B]1 to match the information provided in II and solve:
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3. Rate = k[A]2[B]0; Rate = k[A]2
m
1 ⎛ 2⎞
= ⎜ ⎟ ; 1 = 2m ; m = 0
1 ⎝ 1⎠
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III. Graphs - Integrated Rate Laws
Rate Law: function of concentration (called the rate law);
Integrated Rate Law - function of time; when a question involves time use the Integrated Rate Laws
Order
Rate Law
Integrated Rate Law
0
rate = k[A]0 = k
[A]t = -kt + [A]o
y-axis vs x-axis
(y = mx + b)
[A]t vs t
1
rate = k[A]1
ln[A]t = -kt + ln[A]o
ln[A]t vs t
m = -k
2
rate = k[A]2
1
1
m=k
1
[A]t = kt + [A]o
[A]t
Slope
yt1/2 (half life)
intercept
m = -k
[A]o
t 1 = [A]o 2k
2
vs t
ln[A]o
€1[A]
o
t 1 = 0.693
2
Units of k
M/
time
1/
time
k
t 1 = 1 k[A] 1/
0 [(M)(time)]
2
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integrated rate law graphs:
" [A] t %
1st order rate law: ln[A]t = -kt + ln[A]o can also be written: ln$
' = − kt this rearranged equation is
# [A] 0 &
useful when given % remaining questions; the % remaining (not % consumed) is substituted for ([A]t/[A]0)
IV. Raw data: Two rates + two concentrations: Use two average rates and the corresponding concentrations at
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the same point in time; use initial rates methods (many instructors do not cover this type of problem)
HALF-LIFE t1/2 – The time it takes for half the substance to react or decay; [A] t = [A] o
2
all radioactive decays are 1st order with t1/2 = 0.693/k
" 1 %n
# 2&
Multiple half-lives: [A]t = $ ' [A]0 where n = #half-lives
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MATH REVIEW – USING LOG OR LN:
ln(xa/ya) = ln(x/y)a
ln(ax) = xln(a)
ln(xy) = ln(x) + ln(y)
ln(x/y) = ln(x) – ln(y)
1. Assume the following reaction occurs as follows: 2NO2(g) → 2NO(g) + O2(g)
time (s)
[NO2]
0
50
100
150
200
250
300
350
400
0.0100
0.0079
0.0065
0.0055
0.0048
0.0043
0.0038
0.0034
0.0031
a. What is the initial rate of consumption for NO2?
b. What is the average rate of consumption of NO2 between 50sec and 150sec?
c. What is the concentration of NO2 at 150sec?
d. What is the average rate of consumption of NO2 at 150sec?
e. What is the concentration of NO2 at 350sec?
f. What is the average rate of consumption of NO2 at 350sec?
g. Explain why the rate of consumption of NO2 is dropping.
2. For the reaction N2(g) + 3H2(g) → 2NH3(g), with the rate of consumption of N2, rate = -Δ[N2]/Δt, is
6.0 x 10-2M/min. a. What is the rate of consumption for H2, rate = -Δ[H2]/Δt?
b. What is the rate of production for NH3, rate = Δ[NH3]/Δt?
3. If the rate of formation of O2, rate = Δ[O2]/Δt, is 3.25 x 10-5M/min, what is the rate of consumption of O3,
rate = -Δ[O3]/Δt? 2O3(g) → 3O2(g)
4. The following reaction occurs with a Pt catalyst: H2 + CH2=CH2 → CH3-CH3
How will the following change over time? (write I for increase, D for decrease, or NC for no change)
a. [H2] b. [CH3-CH3] c. amount Pt d. the rate constant, k e. the order of the reaction
5. A general reaction written as A + 2B → C is studied and yields the following data:
Exp.
[A]0
[B]0
Rate (M/s)
1
0.10 M
0.10 M
2
0.20 M
0.10 M
4.0 x 10-5
1.6 x 10-4
3
0.10 M
0.20 M
4.0 x 10-5
a. Write the general form of the rate law from the reaction above.
b. From the data above, determine the orders in the rate law for A and B?
c. What is the value of k? Include units.
d. What is the order of the reaction?
e. What happens to the rate when the concentration of B doubles.
6. The following reaction, CO + Cl2 → COCl + Cl was studied by the method of initial rates.
Exp.
[CO]0
[Cl2]0
Rate (M/s)
1
2
1.00 x 10-2 M
2.00 x 10-2 M
1.00 x 10-2 M
1.00 x 10-2 M
6.60 x 10-3
1.32 x 10-2
3
1.00 x 10-2 M
2.00 x 10-2 M
2.64 x 10-2
a. Write the general form of the rate law from the reaction above.
b. Determine the orders in the rate law for CO and Cl2?
c. What is the value of k (include units)?
d. What is the order of the reaction?
e. If the concentrations of both reactants were doubled, what would be the change in rate of reaction?
7. The reaction of NO with H2 was followed and the data collected shown below.
2NO(g) + H2(g) → N2O(g) + H2O(g)
Exp.
[NO]
[H2]
Rate (M/s)
1
2
1.50 x 10-4 M
4.50 x 10-4 M
2.00 x 10-4 M
2.00 x 10-4 M
1.26 x 10-9
1.13 x 10-8
3
3.00 x 10-4 M
6.00 x 10-4 M
1.51 x 10-8
a. Write the general form of the rate law from the reaction above.
b. Determine the orders in the rate law for NO and H2.
c. Determine the value of the rate constant; include units
d. What are the reaction orders with respect to each reactant?
e. What is the order of the reaction? f. If the concentration of NO was doubled and the concentration of H2 cut
in half, what would the change in rate of reaction be?
8. For the following reaction, 2A + B + C → 5D
a. Given the following information, write the rate law for the above reaction with the specific orders for A, B,
and C:
I. When the concentration of A is tripled the rate is tripled.
II. When the concentration of B is double the rate is unchanged.
III. When the concentration of C is tripled the rate is increased by a factor of 9.
b. If the initial concentrations are A = 1.0 M, B = 2.0 M, and C = 5.0 M and the rate is 1.0 x 10-4 M/s, what is
the rate constant for this reaction (include the proper units)?
c. If the concentration of A, B, and C are doubled, by what factor is the rate changed?
9. The reaction A → 2B + C is found to be second order with respect to the reactant, A.
a. Given this information, write the rate law.
b. What are the units of the rate constant, k?
c. When the concentration of A is 0.01M the rate of consumption of A is 1.0 x 10-2 M/min. What is the value
of the rate constant?
d. When the concentration of A is 0.01M the rate of consumption of A is 1.0 x 10-2 M/min. What is the rate of
production of B at this point in time?
e. When the concentration of A is increased to 0.05M, what is the new rate of loss of A?
10. The reaction 3A → 2B is found to be first order with respect to the reactant, A.
a. Write the rate law.
b. What are the units of the rate constant?
c. When the concentration of A is 0.15M the rate of consumption of A is 3.7 x 10-3 M/min, what is the value of
the rate constant?
d. When the concentration of A is 0.15M the rate of consumption of A is 3.7 x 10-3 M/min. What is the rate of
production of B at this time?
e. When the concentration of A is increased to 0.95M, what is the new rate of loss of A?
11. The following reaction, 2NO + Cl2 → 2NOCl, was analyzed.
It was found that when the concentration of NO was tripled the rate tripled. In another experiment, when the
concentration of NO was cut in half and the concentration of Cl2 was doubled, the rate doubled.
a. What is the order of NO?
b. What is the order of Cl2?
c. What is the overall rate law?
d. When [NO] = 0.050M and [Cl2] = 0.12M, the rate was 1.5 x 10-2. What is the rate constant?
12. A 5.00-L container is 1.00M N2O5 which decomposes to NO2 and O2. The reaction is:
2N2O5(aq) → 4NO2(aq) + O2(g)
The reaction with the N2O5 is run and the concentration of this chemical is recorded and the results tabulated.
time (s)
[N2O5]
a. What is the concentration of N2O5 at 400.0 seconds?
0
1.00
b. What is the average rate of consumption of N2O5 at 400.0 seconds?
200.0
0.880
c. What is the average rate of production of O2 at 400.0 seconds?
400.0
0.780
d. What is the concentration of N2O5 at 1000.0 seconds?
600.0
0.690
800.0
0.610
e. What is the average rate of consumption of N2O5 at 1000.0 seconds?
1000.0
0.540
f. What is the average rate of production of NO2 at 1000.0 seconds?
1200.0
0.480
(Not all instructors use raw data to solve rate law questions; skip parts “g-i”
1400.0
0.430
1600.0
0.380
if not covered.)
1800.0
0.340
g. Write the rate law for the loss of N2O5 by determining the order for N2O5.
2000.0
0.300
h. What is the order of the reaction?
i. What is the value of k at 1000.0 seconds?
13. A reaction A + B → C has a first order rate law rate = -Δ[A]/Δt = k[A] with k = 1.50 x 10-4 s-1.
a. If [A]0 = 1.55 M, what is the concentration of [A] after 10.0 minutes?
b. How long (in seconds) will it take to use up 99.9% of A?
14. The following reaction, 2A → B, has a rate law that is first-order with respect to reactants, and has a rate
constant of 1.50 x 10-2s-1. How long will it take for 35.0% of A to be consumed?
15. Given the following reaction, A → 2C has a rate law of –Δ[A]/Δt = k[A]1
After 21.75 hours, the concentration of A has decreased from 0.157M to 0.0966M. What is the rate constant?
16. One of the reactions in a breeder reactor involves the first order conversion of 239U to 239Np
→ 239Np + eThe half-life of 239U is 23.5 minutes. How long does it take for 35% of the 239U to be consumed?
239U
17. The reaction, A → B, is first order with a half-life of 2.75days. What fraction of A remains after
13.75days?
a. 1/4
b. 1/5
c. 1/16
d. 1/8
e. 1/32
18. What factors that can change the rate of a reaction?
19. When heated, tetrafluoroethylene dimerizes to form octafluorocyclobutane: 2C2F4 → C4F8
Data from a kinetics experiment was collected and plotted below.
a. What is the rate law for this reaction? b. What is the value of the rate constant?
c. What was the initial concentration of C2F4? d. What is the concentration of C2F4 at 600s?
ANSWERS
1. a. 4.2 x 10-5 M/s
{rate = -Δ[NO2]/Δt = -(0.0100-0.0079)/(0-50) = 4.2 x 10-5M/s}
b. 2.4 x 10-5 M/s {rate = -Δ[NO2]/Δt = -(0.0079-0.0055)/(50-150) = 2.4 x 10-5M/s}
c. 0.0055M {read it off the table}
d. 1.7 x 10-5 M/s {rate = -Δ[NO2]/Δt = -(0.0065-0.0048)/(100-200) = 1.7 x 10-5M/s}
e. 0.0034M {read it off the table}
f. 0.70 x 10-5 M/s {rate = -Δ[NO2]/Δt = -(0.0038-0.0031)/(300-400) = 0.70 x 10-5M/s}
g. As the concentration of a chemical drops, its rate of reaction will also drop.
2. a. rate = -Δ[H2]/Δt = 1.8 x 10-1M/min {(1/1)(-Δ[N2]/Δt) = (1/3)(-Δ[H2]/Δt); 3(6.0 x 10-2) = 1.8 x 10-1M/min}
b. rate = Δ[NH3]/Δt = 1.2 x 10-1 M/min
{(1/1)(-Δ[N2]/Δt) = (1/2)(Δ[NH3]/Δt); 2(6.0 x 10-2) = 1.2 x 10-1M/min}
{1/2(-Δ[O3]/Δt) = 1/3(Δ[O2]/Δt); (-Δ[O3]/Δt) = 2/3(3.25 x 10-5) = 2.17 x 10-5M/s}
3. rate = -Δ[O3]/Δt = 2.17 x 10-5 M/min
4. a. D (reactants decrease) b. I (products increase) c. NC (catalysts are not consumed)
d. NC (k is not a function of time) e. NC (order is not a function of time)
5. a. rate = k[A]n[B]m
b. rate law = k[A]2 {exps 2 over 1: (1.6 x 10-4/4.0 x 10-5) = (k/k)(0.20/0.10)n(0.10/0.10)m; 4 = 2n; n = 2;
exps 3 over 1: (4.0 x 10-4/4.0 x 10-5) = (k/k)(0.10/0.10)2(0.20/0.10)m; 1 = 2m; m = 0}
k = 4 x 10-3M-1s-1 {use any exp; exp 1: 4.0 x 10-5 = k(0.10)2(0.10)0; k = 4.0 x 10-3M-1s-1}
c.
d. 2 {n + m = 2 + 0 = 2}
e. no change; not part of the rate law
6. a. rate = k[CO]n[Cl2]m
b. rate = k[CO][Cl2]2 {exps 2 over 1: (1.32 x 10-2/6.60 x 10-3) = (k/k)(0.02/0.01)n(0.01/0.01)m; 2 = 2n; n = 1;
exps 3 over 1: (2.64 x 10-2/6.6 x 10-3) = (k/k)(0.01/0.01)1(0.20/0.10)m; 4 = 2m; m = 2}
k = 6.60 x 103 M-2s-1 {use any exp; exp 1: 6.6 x 10-3 = k(0.01)1(0.01)2; k = 6.60 x 103M-2s-1}
c.
d. 3 {n + m = 1 + 2 = 3}
e. an increase by a factor of 8
7. a. rate = k[NO]n[H2]m
{rate = (2)1(2)2 = 8}
b. rate = k[NO]2[H2]
{exps 2 over 1: (1.13 x 10-8/1.26 x 10-9) = (k/k)(4.5 x 10-4/1.5 x 10-4)n(2 x 10-4/2 x 10-4)m; 8.97 = 3n;
9 = 3n; n = 2; exps 3 over 1: (1.51 x 10-8/1.26 x 10-9) = (k/k)(3.0 x 10-4/1.5 x 10-4)2(6 x 10-4/2 x 10-4)m; 11.98 = (2)23m;
2.99 = 3m; 3 = 3m; m = 1}
k = 280. M-2s-1 {use any exp; exp 1: 1.26 x 10-9 = k(1.5 x 10-4)2(2 x 10-4)1; k = 280. M-2s-1}
c.
d. first order with respect to H2; second order with respect to NO
e. 3 {n + m = 2 + 1 = 3}
f. an increase by a factor of 2 {rate = (2)2(1/2)1 = 2}
8. a. rate = k[A]1[B]0[C]2 = k[A]1[C]2 {rate1/rate2 = ([A]1/[A]2)x; (3/1) = (3/1)x; 3 = 3x; x = 1;
[B]: (1/1) = (2/1)y; 1 = 2y; y = 0; [C]: (9/1) = (3/1)z; 9 = 3z; z = 2}
k = 4.0 x 10-6 1/M2s {1 x 10-4 = k(1)1(2)0(5)2}
b.
c. The rate increases by a factor of 8. {rate = (2)1(2)0(2)2 = 8}
9. a. rate = k[A]2
b. 1/(M x time) {rate = k[A]2; M/time = k(M2); solve for k; k = 1/(time x M)}
c. k = 1.0 x 102 M-1min-1 {1 x 10-2 = k(0.01)2; solve for k; k = 1.0 x 102}
d. 2.0 x 10-2M/min {(1/1)(-Δ[A]/Δt) = (1/2)(Δ[B]/Δt); 2(1.00 x 10-2) = 2.0 x 10-2M/min}
e. 0.25M/min {rate = k[A])2; rate = (100)(0.05)2 = 0.25M/min}
10. a. rate = k[A]
b. 1/min {rate = k[A]; M/min = k(M); solve for k; k = 1/(min)}
c. k = 0.025 min-1 {3.7 x 10-3 = k(0.15); solve for k; k = 0.0247}
d. 0.0025 M/min {(1/3)(-Δ[A]/Δt) = (1/2)(Δ[B]/Δt); (2/3)(3.7 x 10-3) = 0.00247 M/min}
e. 0.024 M/min {rate = k[A]); rate = (0.025)(0.95) = 0.0238M/min}
11. a. 1 {rate1/rate2 = ([NO]1/[NO]2)n; (3/1) = (3/1)n; 3 = 3n; n = 1}
{rate1/rate2 = ([NO]1/[NO]2)n([Cl2]1/[Cl2]2)m; (2/1) = ((1/2)/1)1(2/1)m; 2 = (1/2)(2)m; 4 = 2m; m = 2}
c. rate = k[NO]1[Cl2]2
d. 21 M-2s-1 {1.5 x 10-2 = k(0.05)1(0.12)2; solve for k; k = 20.83}
b. 2
12. a. 0.780M {read it off the table}
b. 0.000475M/s {rate = -Δ[N2O5]/Δt = -(0.88-0.69)/(200-600) = 4.75 x 10-4M/s}
c. 0.000238M/s {(1/2)(-Δ[N2O5]/Δt) = (1/1)(Δ[O2]/Δt); (1/2)(4.75 x 10-4) = 2.375 x 10-4M/s}
d. 0.540M {read it off the table}
e. 0.000325M/s {rate = -Δ[N2O5]/Δt = -(0.61-0.48)/(800-1200) = 3.25 x 10-4M/s}
f. 0.000650M/s {(1/2)(-Δ[N2O5]/Δt) = (1/4)(Δ[NO2]/Δt); (4/2)(3.25 x 10-4) = 6.50 x 10-4M/s}
g. rate = k[N2O5]1 {rate1/rate2 = ([N2O5]1/[N2O5]2)n; (4.75 x 10-4/3.25 x 10-4) = (0.780/0.540)n; 1.462 = 1.444n; n = 1}
h. 1st order rxn
i. 6.02 x 10-4 s-1 {at 400s: 4.75 x 10-4 = k(0.78)1; solve for k; k = 6.09 x 10-4s-1; at 1000s: 3.25 x 10-4 = k(0.54)1; solve for k;
k = 6.02 x 10-4s-1; since this is “real” data, average the k’s → k = 6.06 x 10-4s-1; note: averaging k’s is not typical for quiz or
exam questions!}
13. a. 1.42 M {ln[A]t = -kt + ln[A]0 = -(1.5 x 10-4s-1)(10min)(60s/1min) + ln(1.55) = 0.3483; [A] = e0.3483 = 1.416M}
b. 4.61 x 104 s {with %remaining better to use: ln([A]t/[A]0) = -kt where [A]t/[A]0 = %remaining; ln(0.001) = -(1.5 x 10-4)t;
t = 46052s}
14. 28.7s {with %remaining better to use: ln([A]t/[A]0) = -kt where [A]t/[A]0 = %remaining; ln(0.65) = -(1.5 x 10-2)t;
t = 28.72s}
15. k = 0.0223hr-1 {ln[A]t = -kt + ln[A]0; ln(0.0966) = -(k)((21.75) + ln(0.157); solve for k; k = 0.0223hr-1}
16. 14.6 minutes {t1/2 = 0.693/k with t1/2 = 23.5 min → k = 0.0295; ln([A]/[A]o) = -kt; [A] is the amount that is left over;
[A]/[A]o = 65% = 0.65; ln(0.65) = -(0.0295)t; solve for t}
17. e {13.75/2.75 = 5.0 half-lives; [A]t = (1/2)5 = 1/32 of the original amount remains}
18. Temperature, catalysts, concentration, surface area
19. a. rate = k[C2F4]2 {Since a linear plot occurs when 1/[] vs time is plotted → 2nd order}
b. k = 4.4 x 10-2 M-1s-1
{slope = k; slope = k = Δy/Δx = (32-10)/(500-0) = 4.4 x 10-2M-1s-1; note: if this question were on a
quiz/exam, typically a range of values for k would be accepted because of how the graph is read}
c. 0.10M {y-intercept = 1/[C2F4]0 = 1/10 = 0.1M}
d. 0.027M {can try to extend the line; easier to use: 1/[C2F4]t = kt + 1/[C2F4]0 = (4.4 x 10-2)(600) + 1/(.1) = 36.4;
[C2F4] = 1/36.4 = 0.0275M}
Estimating Instantaneous Rate
Question
Using the graph below approximate the instantaneous rate at 30 seconds for the reaction: A → B
0.007
0.006
0.005
[A]
0.004
0.003
0.002
0.001
10
20
30
40
50
60
70
time
Answer
Step 1. Draw a line tangent to the point at 30 seconds.
0.007
0.006
0.005
[A]
0.004
0.003
0.002
0.001
10
20
30
40
50
60
70
time
Step 2: Choose 2 points on the tangent line and estimate their coordinates.
0.007
0.006
0.005
[A]
0.004
(14, 0.003) 0.003
0.002
0.001
10
20
30
40
50
60
70
time
(42, 0.001)
Step 3: Using the coordinates to determine the slope of the tangent line which will be an approximate
instantaneous rate at 30 seconds; rate = -Δ[A]/Δt.
 (0.003− 0.001 
 0.002 
−5
Rate = − 
 = − 
 = 7 x 10 M / s
 14 − 42

 −28 