1. For Reaction Rates and Mechanisms
Practice Problems
For Reaction Rates and Mechanisms

2. Init. Rate of Formation of C (M min-1)
Example 1
 Rate data were obtained for following reaction:
A + 2B ---> C + 2D
What is the rate law expression for this reaction?
Exp.
Initial A (mol/L)
Initial B (mol/L)
Init. Rate of Formation of C (M min-1) 1
0.10
3.0 x 10-4 2
0.30
9.0 x 10-4 3 4
0.20
0.40
6.0 x 10-4

3. 1) compare exp. 1 and exp. 3. A remains constant and B is tripled
1) compare exp. 1 and exp. 3. A remains constant and B is tripled. The rate from 1 to 3 remains constant. Conclusion: B is not in the rate law expression.
2) compare exp 1 to exp 2. The concentration of A triples (and we don't care what happens to B). The rate triples. Conclusion: first order in A.
3) we can also show first order in A by comparing exp 1 to exp 4. The concentration of A doubles and the rate doubles. Remember, B is not part of the rate law, so we don't pay any attention to it at all.
rate = k[A]

4. Example 2
For the reaction A + B --> products, the following initial rates were found. What is the rate law for this reaction?
Trial 1: [A] = 0.50 M; [B] = 1.50 M; Initial rate = 4.2 x 10-3 M/min
Trial 2: [A] = 1.50 M; [B] = 1.50 M; Initial rate = 1.3 x 10-2 M/min
Trial 3: [A] = 3.00 M; [B] = 3.00 M; Initial rate = 5.2 x 10-2 M/min

5. 1) Order with respect to A:
Look at trial 1 and trial 2. B is held constant while A triples. The result is that the rate triples. Conclusion: A is first order.2) Order with respect to B:
Look at trials 2 and 3. The key to this is that we already know that the order for A is first order. Both concentrations were doubled from 2 to 3 and the rate goes up by a factor of 4. Since A is first order, we know that a doubling of the rate is due to the concentration of A being doubled.
So, we look at the concentration change for B (a doubling) and the consequent rate change (another doubling - remember the overall increase was a factor of 4 - think of 4 as being a doubled doubling).
Conclusion: the order for B is first order.
The rate law is rate = k [A] [B]

6. Init. Rate of Formation of products (mM min-1)
Example 3
The following data were obtained for this chemical reaction: A + B ---> products
(a) Determine the rate law for this reaction.  (b) Find the rate constant.
Exp.
Initial A (mmol/L)
Initial B (mmol/L)
Init. Rate of Formation of products (mM min-1) 1
4.0
6.0
1.60 2
2.0
0.80 3
3.0
0.40

7. 3) The rate law is this: rate = k [A] [B]2
1) Look at experiments 2 and 1. From 2 to 1, we see that A is doubled (while B is held constant). A doubling of the rate with a doubling of the concentration shows that the reaction is first order with respect to A.
2) Now compare experiments 1 and 3. The concentration of A is held constant while the concentration of B is cut in half. When B is cut in half, the overall rate is cut by a factor of 4 (which is the square of 2). This shows the reaction is second order in B.
3) The rate law is this: rate = k [A] [B]2

8. 4) Note that the comparison in (2) can be reversed
4) Note that the comparison in (2) can be reversed. Consider that the concentration of B is doubled as you go from exp. 3 to exp. 1. When the concentration is doubled, the rate goes up by a factor of 4 (which is 22).
5) We can use any set of values to determine the rate constant:
rate = k [A] [B]21.60 mM min-1 = k (4.0 mM) (6.0 mM)2
k = mM-2 min-2
the units on k can be rendered in this manner:
k = L2 mmol-2 min-1

9. Init. Rate of Formation of products (M s-1)
Example 4
 The following data were obtained for the chemical reaction: A + B ---> products
(a) Determine the rate law for this reaction.  (b) Find the rate constant.  (c) What is the initial rate of reaction when [A]o = 0.12 M and [B]o = 0.015
Exp.
Initial A (mol/L)
Initial B (mol/L)
Init. Rate of Formation of products (M s-1) 1
0.040
9.6 x 10-6 2
0.080
1.92 x 10-5 3
0.020

10. 1) Examine exps. 2 and 3. A remains constant while B is doubled in concentration from 3 to 2. The result of this change is that the rate of the reaction doubles. We conclude that the reaction is first order in B.
2) Now we look at exps 1 and 2. B remains constant while the concentration of A doubles. As a result of the doubled concentration, the rate also doubles. Conclusion: the reaction is first order in A.

11. 3) The rate law for this reaction is:
rate = k [A] [B]
4) We can use any set of data to calculate the rate constant:
9.6 x 10-6 M s-1 = k (0.040 M) (0.040 M)
k = M-1 s-1
Comment: remember that M-1 s-1 is often written as L mol-1 s- 1.
5) We use the rate constant along with the data in part (c) of the question:
rate = ( M-1 s-1) (0.12 M) (0.015 M)rate = 1.08 x 10-5 M s-1

12. Example 5
Consider the reaction that occurs when a ClO2 solution and a solution containing hydroxide ions (OH¯) are mixed, as shown in the following equation:
2ClO2(aq) + 2OH¯(aq) ---> ClO3¯(aq) + ClO2¯(aq) + H2O(l)
When solutions containing ClO2 and OH¯ in various concentrations were mixed, the following rate data were obtained:

13. Determination #1: Determination #2: Determination #3:
[ClO2]o = 1.25 x 10¯2 M;
[OH¯]o = 1.30 x 10¯3 M 
Initial rate for formation of ClO3¯ = 2.33 x 10¯4 M s-1
Determination #2:
[ClO2]o = 2.50 x 10¯2 M;
Initial rate for formation of ClO3¯ = 9.34 x 10¯4 M s-1
Determination #3:
[OH¯]o = 2.60 x 10¯3 M 
Initial rate for formation of ClO3¯ = 1.87 x 10¯3 Ms-1

14. (a) Write the rate equation for the chemical reaction. 
(b) Calculate the rate constant, k. 
(c) Calculate the reaction rate for the reaction when [ClO2]o = 8.25 x 10¯3 M and [OH¯]o = 5.35 x 10¯2 M.

15. 1) Compare #1 and #2. The concentration of ClO2 doubles (hydroxide remains constant) and the rate goes up by a factor of four (think of it as two squared). This means the reaction is second order with respect to ClO2.2) Compare #2 and #3. The concentration of hydroxide is doubled while the [ClO2] remains constant. The rate doubles, showing that the reaction is first order in hydroxide.
3) The rate law is: rate = k [ClO2]2 [OH¯]
4) Calculation for the rate constant:
1.87 x 10¯3 M s-1 = k (2.50 x 10¯2 M)2 (2.60 x 10¯3 M)k = 1.15 x 103 M¯2 s-1

16. Often the rate constant unit is rendered thusly: L2 mol-2 s-1.
Note that the overall order of the rate law is third order and that this is reflected in the unit associated with the rate constant.
5) Calculation for part (c):
rate = (1.15 x 103 M¯2 s-1 ) (8.25 x 10¯3 M)2 (5.35 x 10¯2 M)rate = 4.19 x 10¯3 M s-1

17. Initial rate of formation of C
Example 6
2A + B ---> C + D
The following data about the reaction above were obtained from three experiments:
Calculate the rate expression in terms of [A] for experiment 1.
Exp.
[A]
[B]
Initial rate of formation of C 1
0.6
0.15
6.3 x 10-3 2
0.2
2.8 x 10-3 3
7.0 x 10-4

18. rate1 / rate3 = k1[A1]x[B1]y / k3[A3]x[B3]y
k's will cancel and [B] will cancel
rate1 / rate3 = [A1]x / [A3]x
/ = (0.6)x / (0.2)x
9 = 0.6x / 0.2x
9 = 3x
rate = k[A]2

19. Example 7 Determine the proper form of the rate law for:
CH3CHO(g) ---> CH4(g) + CO(g)
Determine the proper form of the rate law for:
CH3CHO(g) ---> CH4(g) + CO(g)
Exp.
[CH3CHO]
[CO]
Rate (M s-1) 1
0.30
0.20
0.60 2
0.10
0.067 3

20. 1) Note experiments 2 and 3. The [CH3CHO] remains constant while the [CO] changes. However, no change of the reaction rate is observed. From this, we conclude that CO is not part of the rate law.2) Examine experiment 3 compared to experiment 1. from 3 to 1, the [CH3CHO] triples and the rate goes up by a factor of 9 (which is 32). Conclusion: the reaction is second order in CH3CHO.
3) The rate expression is this: rate = k [CH3CHO]2

21. Example 8
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. 
Q + X --> products
Trial
[Q]
[X]
Rate 1
0.12 M
0.10 M
1.5 x 10-3 M/min 2
0.24 M
3.0 x 10-3 M/min 3
0.20 M
1.2 x 10-2 M/min

22. 1) Examine trials 1 and 2. [X] is held constant while [Q] is doubled (from 1 to 2).
As a result, the rate doubles. We conclude that the reaction is first order in Q.
2) Examine trials 3 and 1. The concentration of Q does not change from 3 to 1 but the concentration for X is doubled.
When this happens, we observe an eight-fold increase in the rate of the reaction. We conclude that the reaction is third order in X. This arises from the fact that 23 = 8.
In more detail:

23. rate3 / rate1 = k3[Q3]x[X3]y / k1[Q1]x[X1]y
k's will cancel and [Q] will cancel
rate3 / rate1 = [X3]x / [X1]x
0.012 / = (0.20)x / (0.10)x
8 = 0.20x / 0.10x
8 = 2x
x = 3

24. Comment: note that, with a third order, when the concentration increases by a factor of 2, the rate goes up by a factor of 8 (which is 23).
3) value of the rate constant:
rate = k [Q] [X] M min-1 = k (0.12 M) (0.20 M)3
k = 12.5 L3 mol-3 min-1
Note that this reaction is overall fourth order.

25. Initial reaction rate, M/s
Example 9
With the following data, use the method of initial rates to find the reaction orders with respect to NO and O2.:
Trial
[NO]o
[O2]o
Initial reaction rate, M/s 1
0.020
0.010
0.028 2
0.057 3
0.040
0.227 

26. 1) Doubling O2 doubles reaction rate (trials 2 and 1).
2) Doubling NO increases reaction rate 8x (trials 2 and 3).
3) Conclusion: first order O2, third order NO.

27. Example 10 For the following reaction:
A + B ---> 2Cit is found that doubling the amount of A causes the reaction rate to double while doubling the amount of B causes the reaction rate to quadruple. What is the best rate law equation for this reaction?
(a) rate = k [A]2 [B]  (b) rate = k [A] [B]  (c) rate = k [A] [B]2  (d) rate = k [A]1/2 [B]

28. The rate doubling when the concentration is doubled is a hallmark of first-order. The rate going up by a factor of 4 (with is two squared) when the concentration is doubled is a hallmark of second-order. Answer choice (c) is the correct answer.

29. BONUS
A rate law is 1/2 order with respect to a reactant. What is the effect on the rate when the concentration of this reactant is doubled?

30. Let us examine the effect on the rate (symbolized as R) as we double the concentration from step to step:
R = k[1]0.5 = 1  R = k[2]0.5 = 1.4  R = k[4]0.5 = 2  R = k[8]0.5 = 2.8  R = k[16]0.5 = 4

31. http://www. napavalley
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