1. Rate Equations and Order of Reactions14
Rate Equations and Order of Reactions
14.1 Rate Equations and Order of Reactions
14.2 Zeroth, First and Second Order Reactions
14.3 Determination of Simple Rate Equations from Initial Rate Method
14.4 Determination of Simple Rate Equations from Differential Rate Equations
14.5 Determination of Simple Rate Equations from Integrated Rate Equations

2. Rate Equations and Order of Reactions

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

4. 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 and y can be  integers or fractional
x  y is the overall order of reaction.

5. Rate  k[A]x[B]y For the reaction aA + bB  cC + dD
For multi-step reactions,
x, y have no direct relation to the stoichiometric coefficients and can ONLY be determined experimentally.
For single-step reactions (elementary reactions),
x = a and y = b (refer to p.35)

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

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

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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+…

14. 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 (pp.17-18)
Graphical method (pp.19-26)

15. Determination of Rate Equations by Initial Rate Methods

16. 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

17. 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

18. 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

19. 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

20. 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

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 2,
4.0105  k(0.15)(0.08)(0.40)2
k = 0.02 mol3 dm9 s1

22. (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

23. (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

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 3,
= 2y
1 = 2y
 y = 0

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 4,
= 3z
9 = 3z
 z = 2

26. (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

27. (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

28. 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

29. 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

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 4,
= 2y
2 = 2y
 y = 1

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 3 and 4,
= 2z
2 = 2z
 z = 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
Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]
= k[CH3COCH3(aq)][H+(aq)]

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
(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

34. Determination of Rate Equations by Graphical Methods

35. Two types of rate equations : -
Differential rate equation
Integrated rate equation

36. (Differential rate equation)
A  products
(Differential rate equation)
shows the variation of rate with [A]
Two types of plots to determine k and n

37. [A]
rate k
n = 0
rate = k

38. Examples of zero-order reactions : -
2NH3(g) N2(g) + 3H2(g)
Fe or W as catalyst
2HI(g) H2(g) + I2(g)
Au as catalyst
Decomposition of NH3/HI can take place only on the surface of the catalyst.
Once the surface is covered completely (saturated) with NH3/HI molecules at a given concentration of NH3/HI, further increase in [NH3]/[HI] has no effect on the rate of reaction.

39. [A]
rate k
n = 0
rate = k

40. [A]
rate
n = 1
linear
slope = k

41. [A]
rate
n = 2
parabola
k cannot be determined directly from the graph

42. [A]
rate
n = 2
n = 1
n = 0

43. slope
log10[A]
log10rate
y-intercept
n = 2
slope = 2
n = 1
slope = 1
log10k
n = 0
slope = 0

44. Derivation not required
(Differential rate equation)
If n = 0
Derivation not required
[A]t = [A]0 – kt
(Integrated rate equation)

45. [A]t = [A]0 – kt constant rate shows variation of [A] with time
(Integrated rate equation)
shows variation of [A] with time
time
[A]t
[A]0
constant rate

46. (Differential rate equation)
If n = 1, ln
loge[A]t – loge[A]0 = kt
loge[A]t = loge[A]0  kt
Or [A]t  [A]0 ekt
(Integrated rate equation)

47. Two types of plots to determine k and n
Or [A]t  [A]0 ekt
loge[A]t = loge[A]0  kt
Two types of plots to determine k and n
time
loge [A]t
linear  n = 1
loge [A]0
slope = k

48. Two types of plots to determine k and n
Or [A]t  [A]0 ekt
loge[A]t = loge[A]0  kt
Two types of plots to determine k and n
time
[A]t
[A]t varies exponentially with time
constant half life  n = 1

50. loge[A]t = loge[A]0  kt

51. = 6.9103 s1

52. Q.17
sucrose  fructose + glucose
Rate = k[sucrose] k = h1 at 298 K
(a)

53. Q.17
sucrose  fructose + glucose
Rate = k[sucrose] k = h1 at 298 K
(b)
[A]t  [A]0 ekt
87.5% decomposed  [A]t = 0.125[A]0
0.125 = ekt
= e0.208t
ln0.125 = 0.208t
t = 9.99 h

54. (Differential rate equation)
If n = 2, Or
(Integrated rate equation)

55. Or
time
Linear  n = 2
Slope = k

56. [A]t  more rapidly with time in the early stage
Variable half life

57. Plotting based on integrated rate equations
time
[A]t
n = 1
n = 2
n = 0
More common because [A]t and time can be obtained directly from expereiments.

58. Plotting based on differential rate equations
Less common because rate cannot be obtained directly from expereiments.

59. Plotting based on differential rate equations
log10[A]
log10rate
n = 0
n = 1
log10k
n = 2
slope = 1
slope = 2
slope = 0
Less common because rate cannot be obtained directly from expereiments.

60. Summary : - For reactions of the type A  Products
mol1 dm3 s1 k
against t 2
s1 k
ln[A]t against t 1
mol dm3 s1
[A]t against t
[A]t  [A]0 – kt
Units of k
Slope
Straight line plot
Integrated rate
equation
Order

61. Examples of First Order Reactions
2H2O2(aq)  2H2O(l) + O2(g)
Rate = k[H2O2(aq)]

62. Examples of First Order Reactions
Rate equation
2N2O5(g)  4NO2(g) + O2(g)
Rate = k[N2O5(g)]
SO2Cl2(l)  SO2(g) + Cl2(g)
Rate = k[SO2Cl2(l)]
(CH3)3CCl(l) + OH-(aq)  (CH3)3COH(l) + Cl-(aq)
Rate = k[(CH3)3CCl(l)]
(SN1)
All radioactive decays
e.g. Rate = k[Ra]
SN1 : 1st order Nucleophilic Substitution Reaction

63. Examples of Second Order Reactions
For a reaction involving one reactant only:
2NOCl(g)  2NO(g) + Cl2(g)
Rate = k[NOCl(g)]2
2NO2(g)  2NO(g) + O2(g)
Rate = k[NO2(g)]2

64. Examples of Second Order Reactions
2. For a reaction involving one reactant only:
Reaction
Rate equation
H2(g) + I2(g)  2HI(g)
Rate = k[H2(g)][I2(g)]
CH3Br(l) + OH(aq)  CH3OH(l) + Br(aq)
Rate = k[CH3Br(l)][OH(aq)] (SN2)
CH3COOC2H5(l) + OH(aq)  CH3COO(aq) + C2H5OH(l)
Rate = k[CH3COOC2H5(l)][OH(aq)]
SN2 : 2nd order Nucleophilic Substitution Reaction

65. For a reaction involving two reactants: A + B  products
Rate = k[A][B]
To determine the rate equation, the concentration of one of the reactants must be kept constant (in large excess) such that the order of reaction w.r.t. the other reactant can be determined.

66. For a reaction involving two reactants:
A + B  products
Rate = k[A][B]
excess
When [B] is kept constant,
rate = k’[A] (where k’ = k[B]excess)

67. Rate = k[A][B]excess = k’[A]
k can be determined from k’ if [B]excess is known
Linear  first order

68. For a reaction involving two reactants:
A + B  products
Rate = k[B][A]
excess
When [A] is kept constant,
rate = k”[B] (where k” = k[A]excess)

69. Rate = k[A]excess[B] = k’’[B]
k can be determined from k’’ if [A]excess is known
Linear  first order

70. = (3.6103 s1)(150s) = 0.58 Q.18(a) 2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
= (3.6103 s1)(150s)
= 0.58

71. Q.18(b)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
= 0.01
ln0.01 = (3.6103 s1)t
t = 1279 s

72. Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the partial pressure of NO2 is 1.0 atm.
(Gas constant, R = atm dm3 K1 mol1)
= mol dm3

73. Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the partial pressure of NO2 is 1.0 atm.
(Gas constant, R = atm dm3 K1 mol1)
[NO2(g)] = mol dm3
Rate of reaction = k[NO2(g)]
= (3.6103 s1)(0.021 mol dm3)
= 7.6105 mol dm3 s1

74. = 2(7.6105 mol dm3 s1) = 1.5104 mol dm3 s1 Q.18(c)
2NO2(g)  2NO(g) + O2(g)
first-order reaction, k  3.6  103 s1 at 573 K
Calculate the rate of consumption of NO2 when the partial pressure of NO2 is 1.0 atm.
(Gas constant, R = atm dm3 K1 mol1)
= 2(7.6105 mol dm3 s1)
= 1.5104 mol dm3 s1

75. Mass of Ra remaining after 5 years = 0.297 g
Q.19
Half life = 6.67 years
= (0.104 y1)(5 y)
= 0.595
Mass of Ra remaining after 5 years = g

76. Q.20
Half life = 4.51  109 years
Let 1.000x be the mass of U left behind at time t
Mass of Pb produced at time t = 0.231x
Mass of U consumed at time t = 0.231x
Mass of U at t0 = 1.231x
= 0.208
kt = (1.541010 y1)t = 0.208
t = 1.35109 years

77. Q.21(a)

78. Q.21(b)
No. of moles of He formed
No. of moles of U decayed =
= (3.20103 mol)
= 4.00104 mol
No. of moles of Pb produced = 4.00104 mol
No. of moles of U at t0 = (4.0010 104) mol
= 8.40104 mol

79. Q.21(b)
= 0.647
kt = 0.647
(1.541010 y1)t = 0.647
t = 4.20109 y

80. P Q.22(a) 2N2O5(g)  4NO2(g) + O2(g) 1.1 2.7 4.4 5.9 7.2 8.9 10.9 13.3
/ kPa
1250
800
550
400
300
200
100
t / minute 5 2 O N P

81. Constant half life  350 minutes  1st order
Q.22(a)
2N2O5(g)  4NO2(g) + O2(g)
1.1
2.7
4.4
5.9
7.2
8.9
10.9
13.3
/ kPa
1250
800
550
400
300
200
100
t / minute 5 2 O N P
Constant half life  350 minutes  1st order

82. Q.22(a)

83. linear  first order k =  slope = 1.99103 min1 Q.22(a)/(b) 0.10
0.99
1.48
1.77
1.97
2.19
2.39
2.59
1250
800
550
400
300
200
100
t / minute
time
linear  first order
k =  slope = 1.99103 min1

84. 14.1 Rate Equations and Order of Reactions (SB p.27)
Check Point 14-1
The reaction between tyrosine (an amino acid) and iodine obeys the rate law: rate = k [Tyr] [I2]. Write the orders of the reaction with respect to tyrosine and iodine respectively, and hence the overall order.
Answer
(a) The order of the reaction with respect to tyrosine is 1, and the order of the reaction with respect to iodine is also 1. Therefore, the overall order of the reaction is 2.

85. Check Point 14-1 Answer Back
14.1 Rate Equations and Order of Reactions (SB p.27)
Check Point 14-1
Back
(b) Determine the unit of the rate constant (k) of the following rate equation: Rate = k [A] [B]3 [C]2 (Assume that all concentrations are measured in mol dm–3 and time is measured in minutes.)
Answer
(b)  k =
 Unit of k =
= mol-5 dm15 min-1

86. 14.2 Zeroth, First and Second Order Reactions (SB p.29)
Check Point 14-2
The initial rate of a second order reaction is 8.0 × 10–3 mol dm–3 s–1. The initial concentrations of the two reactants, A and B, are 0.20 mol dm–3. Calculate the rate constant of the reaction and state its unit.
Answer
8.0  10-3 = k  (0.20)2
 k = 0.2 mol-1 dm3 s-1
Back

87. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
Example 14-3
For a reaction between two substances A and B, experiments with different initial concentrations of A and B were carried out. The results were shown as follows:
Expt
Initial conc. of A (mol dm-3)
Initial conc. of B (mol dm-3)
Initial rate (mol dm-3 s-1) 1
0.01
0.02
0.0005 2 3
0.04

88. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
Example 14-3
(a) Calculate the order of reaction with respect to A and that with respect to B.
Answer

89. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
Example 14-3
(a) Let x be the order of reaction with respect to A, and y be the order of reaction with respect to B. Then, the rate equation for the reaction can be expressed as:
Rate = k [A]x [B]y
Therefore,
= k (0.01)x (0.02)y (1)
= k (0.02)x (0.02)y (2)
= k (0.01)x (0.04)y (3)
Dividing (1) by (2),
 x = 1

90. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
Example 14-3
(a) Dividing (1) by (3),
 y = 2

91. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.30)
Example 14-3
(b) Calculate the rate constant using the result of experiment 1.
(c) Write the rate equation for the reaction.
Answer
Using the result of experiment (1),
Rate = k [A] [B]2
= k  0.01  0.022
k = 125 mol-2 dm6 s-1
(c) Rate = 125 [A] [B]2
Back

92. Check Point 14-3 4 0.4 0.060 In the kinetic study of the reaction,
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
Check Point 14-3
In the kinetic study of the reaction,
CO(g) + NO2(g)  CO2(g) + NO(g)
four experiments were carried out to determine the initial reaction rates using different initial concentrations of reactants. The results were as follows:
Expt
Initial conc. of CO(g) (mol dm-3)
Initial conc. of NO2(g) (mol dm-3)
Initial rate (mol dm-3 s-1) 1
0.1
0.015 2
0.2
0.030 3 4
0.4
0.060

93. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
Check Point 14-3
(a) Calculate the rate constant of the reaction, and hence write the rate equation for the reaction.
Answer

94. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
Check Point 14-3
(a) Let m be the order of reaction with respect to CO, and n be the order of reaction with respect to NO2. Then, the rate equation for the reaction can be expressed as:
Rate = k [CO]m [NO2]n
Therefore,
0.015 = k (0.1)m (0.1)n (1)
0.030 = k (0.2)m (0.1)n (2)
0.030 = k (0.1)m (0.2)n (3)
Dividing (1) by (2),
 m = 1

95. 14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
Check Point 14-3
(a) Dividing (1) by (3),
 n = 1
 Rate = k [CO] [NO2]
Using the result of experiment (1),
0.015 = k (0.1)2
k = 1.5 mol-1 dm3 s-1
 Rate = 1.5 [CO] [NO2]

96. Check Point 14-3 Answer Back
14.3 Determination of Simple Rate Equations from Initial Rate Method
(SB p.31)
Check Point 14-3
(b) Determine the initial rate of the reaction when the initial concentrations of both CO( g) and NO2( g) are 0.3 mol dm–3.
Answer
Initial rate = 1.5  0.3  0.3
= mol dm-3 s-1
Back

97. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(a) Write a chemical equation for the decomposition of hydrogen peroxide solution.
Answer
(a) 2H2O2(aq)  2H2O(l) + O2(g)

98. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(b) Explain how you could find the rate of decomposition of hydrogen peroxide solution in the presence of a solid catalyst using suitable apparatus.
Answer

99. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
In the presence of a suitable catalyst such as manganese(IV) oxide, hydrogen peroxide decomposes readily to give oxygen gas which is hardly soluble in water. A gas syringe can be used to collect the gas evolved. To minimize any gas leakage, all apparatus should be sealed properly. A stopwatch is used to measure the time. The volume of gas evolved per unit time (i.e. the rate of evolution of the gas) can then be determined.

100. Initial rate (10-4 mol dm-3 s-1)
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(c) The table below shows the initial rates of decomposition of hydrogen peroxide solution of different concentrations. Plot a graph of the initial rate against [H2O2(aq)].
[H2O2(aq)] (mol dm-3)
0.100
0.175
0.250
0.300
Initial rate (10-4 mol dm-3 s-1)
0.59
1.04
1.50
1.80
Answer

101. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(c)

102. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(d) From the graph in (c), determine the order and rate constant of the reaction.
Answer

103. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(d) There are two methods to determine the order and rate constant of the reaction.
Method 1:
When the concentration of hydrogen peroxide solution increases from 0.1 mol dm–3 to 0.2 mol dm–3, the reaction rate increases from 0.59 × 10–4 mol dm–3 s–1 to about 1.20 × 10–4 mol dm–3 s–1.
∴ Rate  [H2O2(aq)]
Therefore, the reaction is of first order.
The rate constant (k) is equal to the slope of the graph.
k =
= 6.0  10-4 s-1

104. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
(d) Method 2:
The rate equation can be expressed as:
Rate = k [H2O2(aq)]x
where k is the rate constant and x is the order of reaction.
Taking logarithms on both sides of the rate equation,
log (rate) = log k + x log [H2O2(aq)] (1)
-3.74
-3.82
-3.98
-4.23
log (rate)
-0.523
-0.602
-0.757
-1.000
log [H2O2(aq)]

105. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
A graph of log (rate) against log [H2O2(aq)] gives a straight line of slope x and y-intercept log k.

106. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.34)
Example 14-4
Slope of the graph =
=1.0
 The reaction is of first order.
Substitute the slope and one set of value into equation (1):
-4.23 = log k + (1.0) (-1.000)
log k = -3.23
k = 5.89  10-4 s-1
Back

107. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
Decide which curve in the following graph corresponds to
(i) a zeroth order reaction;
(ii) a first order reaction.
Answer
(i) (3)
(ii) (2)

108. Concentration of N2O5 (mol dm-3) Initial rate (mol dm-3 s-1)
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(b) The following results were obtained for the decomposition of nitrogen(V) oxide.
2N2O5(g)  4NO2(g) + O2(g)
Concentration of N2O5 (mol dm-3)
Initial rate (mol dm-3 s-1)
1.6  10-3
0.12
2.4  10-3
0.18
3.2  10-3
0.24

109. Check Point 14-4 Answer (i) Write the rate equation for the reaction.
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(i) Write the rate equation for the reaction.
Answer
The rate equation for the reaction can be expressed as:
Rate = k [N2O5(g)]m
where k is the rate constant and m is the order of reaction.

110. Check Point 14-4 Answer (ii) Determine the order of the reaction.
14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(ii) Determine the order of the reaction.
Answer

111. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(ii) Method 1:
A graph of the initial rates against [N2O5(g)] is shown as follows:

112. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
As shown in the graph, when the concentration of N2O5 increases from 1.0  10–3 mol dm–3 to 2.0  10–3 mol dm–3, the rate of the reaction increases from mol dm–3 s–1 to 0.15 mol dm–3 s–1.
 Rate  [N2O5(g)]
The reaction is of first order.
Then, the rate constant k is equal to the slope of the graph.
k =
= 75 s-1
The rate equation for the reaction is:
Rate = 75 [N2O5(g)]

113. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
Method 2:
Taking logarithms on both sides of the rate equation, we obtain:
log (rate) = log k + m log [N2O5(g)] (1)
A graph of log (rate) against log [N2O5(g)] gives a straight line of slope m and y-intercept log k.
-0.62
-0.74
-0.92
log (rate)
-2.50
-2.62
-2.80
log [N2O5(g)]

114. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
Slope of the graph =
=1.0
The reaction is of first order.
Substitute the slope and one set of value into equation (1):
-0.92 = log k + (1.0) (-2.80)
log k = 1.88
k = s-1
The rate equation for the reaction is:
Rate = [N2O5(g)]

115. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(iii) Determine the initial rate of reaction when the concentration of nitrogen(V) oxide is:
(1) 2.0 × 10–3 mol dm–3.
(2) 2.4 × 10–2 mol dm–3.
Answer

116. 14.4 Determination of Simple Rate Equations from Differential Rate Equations
(SB p.36)
Check Point 14-4
(iii) The rate equation, rate = 75 [N2O5(g)], is used for the following calculation.
(1) Rate = 75 [N2O5(g)]
= 75 s–1  2.0  10–3 mol dm–3
= 0.15 mol dm–3 s–1
(2) Rate = 75 [N2O5(g)]
= 75 s–1  2.4  10–2 mol dm–3
= 1.8 mol dm–3 s–1
Back

117. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.39)
Example 14-5A
The half-life of a radioactive isotope A is years. How long does it take for the radioactivity of a sample of A to drop to 20% of its original level?
Answer

118. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.39)
Example 14-5A
Back
As radioactive decay is a first order reaction,
= 3.47  10-4 year-1 
 t = 4638 years
 It takes 4638 years for the radioactivity of a sample of A to dropt to 20 % of its original level.

119. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
Check Point 14-5A
(a) At 298 K, the rate constant for the first order decomposition of nitrogen(V) oxide is 0.47 × 10–4 s–1. Determine the half-life of nitrogen(V) oxide at 298 K.
N2O5  2NO O2
Answer

120. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
Check Point 14-5A
Let the half-life of nitrogen(V) oxide be .
 The half-life of nitrogen(V) oxide is s.

121. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
Check Point 14-5A
(b) The decomposition of CH3N = NCH3 to form N2 and C2H6 follows first order kinetics and has a half-life of minute at 573 K. Determine the amount of CH3N = NCH3 left if 1.5 g of CH3N = NCH3 was decomposed for minute at 573 K.
Answer

122. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.40)
Check Point 14-5A
(b)
Let m be the amount of CH3N=NCH3 left after minute.
m = g
Back

123. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
Example 14-5B
In the decomposition of gaseous hydrogen iodide, the following experimental data were obtained.
Determine the order of decomposition of gaseous hydrogen iodide graphically. You may try to plot graphs of [HI(g)] against time, ln[HI(g)] against
time and against time.
0.100
0.125
0.167
0.250
0.500
[HI(g)] (mol dm-3)
480
360
240
120
Time (min)
Answer

124. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
Example 14-5B
10.000
-2.303
0.100
480
8.000
-2.079
0.125
360
5.988
-1.790
0.167
240
4.000
-1.386
0.250
120
2.000
-0.693
0.500
1/[HI(g)]
(mol-1 dm3)
ln [HI(g)]
[HI(g)] (mol dm-3)
Time (min)

125. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
Example 14-5B
The order of decomposition can be determined by plotting:
[HI(g)] against time,

126. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.42)
Example 14-5B
(b) ln [HI(g)] against time,

127. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Example 14-5B
(c) against time,

128. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Example 14-5B
In graph (a), the plot of [HI(g)] against time is not a straight line, thus the decomposition reaction is not of zeroth order.
Similarly, in graph (b), the plot of ln [HI(g)] against time is not a straight line, thus the decomposition reaction is not of first order.
However, in graph (c), the plot of against time gives
a straight line, thus the decomposition of gaseous
hydrogen iodide is of second order.
Back

129. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Check Point 14-5B
The change in concentration of substance X as it decomposed at 698 K was recorded in the following table:
Determine the order of the reaction graphically.
0.056
0.063
0.072
0.083
0.100
[X] (mol dm-3)
200
150
100 50
Time (s)
Answer

130. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Check Point 14-5B
17.86
-2.88
0.056
200
15.87
-2.76
0.063
150
13.89
-2.63
0.072
100
12.05
-2.49
0.083 50
10.00
-2.30
0.100
1 / [X] (mol-1 dm3)
ln [X]
[X] (mol dm-3)
Time (s)

131. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Check Point 14-5B
As the graph of [X] against time is not a straight line, the reaction is not of zeroth order.

132. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Check Point 14-5B
Similarly, the plot of ln [X} against time is not a straight line, thus the reaction is not of first order.

133. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations
(SB p.43)
Back
Check Point 14-5B
The plot of
Against time gives a straight line, therefore the reaction is of second order.