Theory and examples of solving seven types of chemical kinetics problems that may be found on standardized tests.

1. Solving Kinetics Problems
Writing Rate Expressions from Balanced
Equations
Finding Reaction Order from [A]-time Data
Finding Reaction Order from Initial Rate
Finding k from [A] & Rate Data
Calculating ½ Life from k
Drawing Reaction Profile from Data
Creating Reaction Mechanisms from Rate
Law & Finding the Slow Step

2. Writing Rate Expressions from
Balanced Equations



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The rate expression is not the rate law
It tells us what we are looking for in a rate
law experiment
Example: H2 (g) + I2 (g) → 2 HI (g)
Rate expression could be rate of
disappearance of hydrogen It tells you in
−
Rate = ΔH2 = 1 ΔHI
Δt
2 Δt
the lab what
to measure

3. Your turn

The equation is
−
−

Cu (s) + 2 Ag+ (aq) → Cu2+ (aq) + 2 Ag(s)
Write at least two expressions by which
we could measure the rate
Answers:
−
Rate = ΔCu = 1 ΔAg+ = ΔCu2+ = 1ΔAg
Δt
2 Δt
Δt
2 Δt

4. Finding Reaction Order from
[A]-time Data


Needed: a chart of concentration of a reactant we
want to study vs. time
Product: a fast graph of Ln[A] vs time or 1/[A] vs
time

If the Ln[A] v time graph is linear, it's first order

If the Ln[A] v time graph is a curve, it's 2 nd order

If [A] v time goes down in a linear fashion, it's
zero order (rare)

5. Your turn



Reactant A changes concentration with
time. Here is the data:
What is the order of the reaction in A?
Hint: find the natural log of each [A] and
graph on your graphing calculator

6. Solution

Plotting LN[A] v. time we get
The straight line plot
suggests a first order
equation Rate = k[A]

7. Your Turn
Butadiene changes concentration with time.
Here is the data:
Time (s)
0
[Butadiene] 0.01000
Mol/L
1000
1800
2800
3600
4400
0.00625
0.00476
0.00370
0,00313
0.00270
What is the order of the reaction in butadiene?
Again, find the natural log of each [A] and graph
on your graphing calculator.

8. Solution
Plotting LN[butadiene] vs time we get
This is not a straight line as shown by the straight line between
the first and last points, so the reaction must be 2nd order.

9. Finding Reaction Order from Initial
Rate
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
This is a favorite of test writers!
You are given concentrations of all
reactants and the rate of reaction for each
set of conditions
You must identify the controls and variables
and determine how the rate is affected
You are looking for doubling of rate when
concentration doubles (1st), or quadrupling
of rate when concentration doubles (2 nd)

10. Your Turn
SO2 + O2 → SO3
Given the following data, determine the
order of reaction in SO2 and O2
Look for doubling of concentrations with
other concentration held constant!

11. Solution
Given the following data, determine the
order of reaction in SO2 and O2
In 2 and 1, oxygen concentration doubles while SO2 is
held constant

12. Solution
Given the following data, determine the
order of reaction in SO2 and O2
In 2 and 1, rate of formation of the trioxide goes from
0.60 to 1.20, also a doubling

13. Solution
Given the following data, determine the
order of reaction in SO2 and O2
That means the rate is directly proportional to the
concentration of the oxygen gas, so the reaction is
first order in O2

14. Solution
Given the following data, determine the
order of reaction in SO2 and O2
When we look at Experiments 1 and 3, we see the
oxygen concentration is held constant, and the SO 2
concentration is doubling.

15. Solution
Given the following data, determine the
order of reaction in SO2 and O2
But at the same time, the rate of trioxide formation
goes from 1.2 to 4.8, which is a quadrupling. That is 2 2
times the initial rate, so the rate is going up faster
than the concentration.

16. Solution
Given the following data, determine the
order of reaction in SO2 and O2
This means that Rate = k[SO2]2
The reaction is 2nd order in SO2

17. Predicting Concentrations, Rates
Now that we have reaction order, let's see
if we can fill in the table.
The reaction is 1st order in O2 and 2nd order in SO2

18. Solutions
To find the oxygen in Exp 4, we see that
the rate is 17% lower than in Exp 1
The reaction is 1st order in O2 and 2nd order in SO2

19. Solutions
And the [SO2] is 33% lower than in Exp 1
That is predictable since Rx is 2nd order
The reaction is 1st order in O2 and 2nd order in SO2

20. Solutions
The change in sulfur dioxide accounts for
all the rate change, so oxygen is 0.20 M
0.20 M
The reaction is 1st order in O2 and 2nd order in SO2

21. Solutions
Now predict the rate of forming trioxide in
experiment 5
0.20 M
The reaction is 1st order in O2 and 2nd order in SO2

22. Solutions
Both concentrations change. Let's do
oxygen first
0.20 M
The reaction is 1st order in O2 and 2nd order in SO2

23. Solutions
The oxygen is 50% higher than in Exp 2,
so rate should be 50% higher, or 9.0 x
10-3
0.20 M
The reaction is 1st order in O2 and 2nd order in SO2

24. Solutions
But SO2 is 17% higher than in Exp 2, so
rate from that is 33% higher yet, 1.2 x 10 2
0.20 M
1.2 x 10-2 M/s
The reaction is 1st order in O2 and 2nd order in SO2

25. Finding k from [A] & Rate Data
Let's go with a first order reaction we have
already looked at:
Rate = -0.240[A] -2.49, from the equation of line
And k = negative of slope, or 0.240 here

26. Finding k from [A] & Rate Data
But suppose we have the data, but no
equation, and know it's first order
Look at Exp 1 and 2 and assure yourself that
the reaction is first order in A

27. Finding k from [A] & Rate Data
Look at Exp 1 and 3 and see that the
reaction is 2nd order in B

28. Finding k from [A] & Rate Data
Look at Exp 3 & 4 and see that the reaction
is 2nd order in C

29. Finding k from [A] & Rate Data
Thus the rate law is
Rate = k[A][B]2[C]2

30. Finding k from [A] & Rate Data
So solve for k and plug in the numbers
from any of the fully known data lines
K = Rate
[A][B]2[C]2
Rate = k[A][B]2[C]2

31. Finding k from [A] & Rate Data
So solve for k and plug in the numbers
from any of the fully known data lines
K = 2.85 x 1012
This is harder than anything on a test
Rate = k[A][B]2[C]2

32. Your Turn
Find the value of the rate constant:
2 NO (g) + Cl2 (g) → 2NOCl (g)
Data:
[NO]o mol/L
[Cl2]o mol/L
Initial Rate
Mol/L-min
0.10
0.10
0.18
0.10
0.20
0.36
0.20
0.20
1.45
First you must find the order of the reaction in both
reactants, and write the rate law, then plug in to solve
for k.

33. Solution
Find the value of the rate constant:
The rate law is Rate = k [NO]2[Cl2]
So k = (Rate)/[NO]2[Cl2] = 0.18 x 10-3 L2mol-2min-1
[NO]o mol/L
[Cl2]o mol/L
Initial Rate
Mol/L-min
0.10
0.10
0.18
0.10
0.20
0.36
0.20
0.20
1.45
First you must find the order of the reaction in both
reactants, and write the rate law, then plug in to solve
for k.

34. Calculating ½ Life from k
We'll stick with first order reactions here
Remember that t ½ = 0.693
k
Suppose the rate constant for a 1 st order
reaction is 0.18 x 10-3s-1 What is the halflife? If the initial concentration is 2.0 M,
what will it be after the reaction runs for 770
seconds?

35. Solution
Suppose the rate constant for a 1st order reaction is 0.18 x 103 -1
s What is the half-life? If the initial concentration is 2.0 M,
what will it be after the reaction runs for 770 seconds?
T ½ = 0.693 = 0.693 = 385 s
K
0.0018 s-1
Now 770/385 = 2.0 so that's 2 half-lives
In the first half-life, the concentration goes to 1.0 M
In the second half-life, the concentration goes to 0.50 M

36. Your Turn
A certain first-order reaction is 45.0%
complete in 65 s. What are the rate
constant and half-life for this process?
This is #37 on page 605

37. Your Turn
A certain first-order reaction is 45.0% complete in 65 s. What
are the rate constant and half-life for this process?
If [A]o = 100.0, then after 65 s, [A] = 55.0. In 1 st order Rxn,
LN([A]/[A]o = -kt, LN(55.0/100.0 = -k(65 s)
You do the arithmetic, but
K = 9.2 x 10-3 s-1 and t ½ = 0.693/k = 75 s
This is #37 on page 605

38. Drawing Reaction Profile from
Data
This should be a little familiar to you if you
recall the diagrams of endothermic and
exothermic reaction
We just add the activation energy and
diagram it like a hill the reactants have to
get over by colliding at the right energy and
orientation.

39. Drawing Reaction Profile from
Data
Draw a reaction energy profile for an endothermic
reaction with ΔH = +34 kJ/mol and a forward
activation energy of 66 kJ/mol. Calculate the
activation energy in the reverse direction.

40. Drawing Reaction Profile from
Data
Draw a reaction energy profile for an endothermic
reaction with ΔH = +34 kJ/mol and a forward
activation energy of 66 kJ/mol. Calculate the
activation energy in the reverse direction.

41. Your turn
Draw the reaction energy profile of an exothermic
reaction with a forward activation energy of 116
kJ/mol and a ΔH of -225 kJ/mol. Calculate the
activation energy in the reverse direction.

42. Your turn
Draw the reaction energy profile of an exothermic
reaction with a forward activation energy of 116
kJ/mol and a ΔH of -225 kJ/mol. Calculate the
activation energy in the reverse direction.
Ea for
reverse
direction is
341 kJ/mol
From UC Davis

43. Creating Reaction Mechanisms from Rate
Law & Finding the Slow Step
The rate law gives us a mathematical
picture of the initial and time-related
concentrations or pressures at a given
temperature.
However, to control reactions we need to
understand how they run. So we derive
reaction mechanisms from the rate law,
whenever possible.

44. Example
For the reaction H2 (g) + 2 ICl → I2 + 2 HCl the rate law
is found to be Rate = k [H2][ICl]
What is the most rational two-step mechanism that fits all
the information given?
First, a termolecular collision is almost never seen. So a
two-step mechanism is very reasonable.
The rate law implies that both hydrogen and ICl are
involved in the slow step.

45. Example
For the reaction H2 (g) + 2 ICl → I2 + 2 HCl the rate law
is found to be Rate = k [H2][ICl]
The rate law implies that both hydrogen and ICl are
involved in the slow step.
So we can reasonably say
• H2 + ICl → HI + HCl
• HI + ICl → I2 + HCl
Overall:
(slow)
(fast)
H2 (g) + 2 ICl → I2 + 2 HCl

46. Another mechanism
Here's one I found on YouTube:
Organic mechanism
Click on the link above