1. 6.5
Rate Law

2. Video
CrashCourse on Kinetics

3. Key Words
Rate Law Equation Rate Constant Order of Reaction Total Order of Reaction

4. Rate Equations and Order of Reactions

5. rate law or rate equation
For the reaction aA + bB  cC + dD
Rate  k[A]x[B]y
rate law or rate equation

6. Rate  k[A]x[B]y For the reaction aA + bB  cC + dD
where x and y are the orders of reaction with respect to A and B
x  y is the overall order of reaction.

7. Rate  k[A]x[B]y For the reaction aA + bB  cC + dD
x = 0  zero order for A
x = 1  first order for A
x = 2  second order for A
y = 0  zero order for B
y = 1  first order for B
y = 2  second order for B

8. The reaction is zero order for A and second order for B.
For the reaction aA + bB  cC + dD
Describe the reaction with the following rate law.
Rate  k[B]2
The reaction is zero order for A and second order for B.

9. Rate  k[A]x[B]y k is the rate constant
For the reaction aA + bB  cC + dD
Rate  k[A]x[B]y
k is the rate constant
Temperature-dependent
Can only be determined from experiments

10. mol dm3 s1/(mol dm3)x+y or, mol dm3 min1 /(mol dm3)x+y
For the reaction aA + bB  cC + dD
Rate  k[A]x[B]y
units of k : -
mol dm3 s1/(mol dm3)x+y or,
mol dm3 min1 /(mol dm3)x+y

11. Rate  k[A]0[B]0 units of k = mol dm3 s1/(mol dm3)0+0
For the reaction aA + bB  cC + dD
Rate  k[A]0[B]0
units of k
= mol dm3 s1/(mol dm3)0+0
= mol dm3 s1
= units of rate

12. Rate  k[A][B]0 units of k = mol dm3 s1/(mol dm3)1+0 = s1
For the reaction aA + bB  cC + dD
Rate  k[A][B]0
units of k
= mol dm3 s1/(mol dm3)1+0
= s1

13. Rate  k[A][B] units of k = mol dm3 s1/(mol dm3)1+1 = mol1 dm3 s1
For the reaction aA + bB  cC + dD
Rate  k[A][B]
units of k
= mol dm3 s1/(mol dm3)1+1
= mol1 dm3 s1
The overall order of reaction can be deduced from the units of k

14. mol dm3 s1/(mol dm3)x+y+z+…
For the reaction
aA + bB + cC + …  products
Rate  k[A]x[B]y[C]z…
units of k : -
mol dm3 s1/(mol dm3)x+y+z+…

15. Determination of rate equations
To determine a rate equation is to find k, x, y, z,…
Rate  k[A]x[B]y[C]z…
Two approaches : -
Initial rate method
Graphical method

16. Determination of Rate Equations by Initial Rate Methods

17. rate  k[Cl(aq)]x[ClO3(aq)]y[H+(aq)]z
5Cl(aq) + ClO3(aq) + 6H+(aq)  3Cl2(aq) + 3H2O(l)
Expt
[Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
rate  k[Cl(aq)]x[ClO3(aq)]y[H+(aq)]z

18. From experiments 1 and 2, = 2z 4 = 2z  z = 2 Expt [Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
From experiments 1 and 2,
= 2z
4 = 2z
 z = 2

19. From experiments 2 and 3, = 2y 2 = 2y  y = 1 Expt [Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
From experiments 2 and 3,
= 2y
2 = 2y
 y = 1

20. From experiments 1 and 4, = 2x 2 = 2x  x = 1 Expt [Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
From experiments 1 and 4,
= 2x
2 = 2x
 x = 1

21. rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
Expt
[Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
From experiment 1,
1.0105  k(0.15)(0.08)(0.20)2
k = 0.02 mol3 dm9 s1

22. rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
Expt
[Cl(aq)]
/ mol dm3
[ClO3(aq)]
[H+(aq)]
Initial rate
/ mol dm3 s1 1
0.15
0.08
0.20
1.0105 2
0.40
4.0105 3
0.16
8.0105 4
0.30
2.0105
rate  k[Cl(aq)][ClO3(aq)][H+(aq)]2
From experiment 2,
4.0105  k(0.15)(0.08)(0.40)2
k = 0.02 mol3 dm9 s1

23. (a) rate  k[C]x[D]y[E]z
Q.15 2C + 3D + E  P + 2Q
Expt
[C]
/ mol dm3
[D]
[E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(a) rate  k[C]x[D]y[E]z

24. (a) rate  k[C]x[D]y[E]z
Expt
[C]
/ mol dm3
[D]
[E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(a) rate  k[C]x[D]y[E]z
From experiments 1 and 2,
= 2x
8 = 2x
 x = 3

25. (a) rate  k[C]x[D]y[E]z
Expt
[C]
/ mol dm3
[D]
[E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(a) rate  k[C]x[D]y[E]z
From experiments 1 and 3,
= 2y
1 = 2y
 y = 0

26. (a) rate  k[C]x[D]y[E]z
Expt
[C]
/ mol dm3
[D]
[E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(a) rate  k[C]x[D]y[E]z
From experiments 1 and 4,
= 3z
9 = 3z
 z = 2

27. (a) rate  k[C]3[D]0[E]2 = k[C]3[E]2 Expt [C] / mol dm3 [D] [E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(a) rate  k[C]3[D]0[E]2
= k[C]3[E]2

28. (b) rate  k[C]3[E]2 From experiment 1, 3.0103  k(0.10)3(0.10)2
Expt
[C]
/ mol dm3
[D]
[E]
Initial rate
/ mol dm3 s1 1
0.10
3.0103 2
0.20
2.4102 3 4
0.30
2.7102
(b) rate  k[C]3[E]2
From experiment 1,
3.0103  k(0.10)3(0.10)2
k = 300 mol4 dm12 s1

29. Initial concentration
Q.16 H+
CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I(aq)
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
(a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z

30. Initial concentration
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
(a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 1 and 2,
= 1.67x
1 = 1.67x
 x = 0

31. Initial concentration
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
(a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 1 and 4,
= 2y
2 = 2y
 y = 1

32. Initial concentration
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
(a) rate  k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]z
From experiments 3 and 4,
= 2z
2 = 2z
 z = 1

33. Initial concentration
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]
= k[CH3COCH3(aq)][H+(aq)]

34. Initial concentration
Initial rate
/ mol dm3 s1
Initial concentration
/ mol dm3
[I2(aq)]
[CH3COCH3(aq)]
[H+(aq)]
3.5 105
2.5104
2.0101
5.0103
1.5104
1.4 104
4.0101
1.0102
7.0 105
(b) Rate = k[CH3COCH3(aq)][H+(aq)]
From experiment 1,
3.5105  k(2.0101)(5.0103)
k = mol1 dm3 s1

35. Task
Review Questions 1, 3 and 4 on page 382.