1 rate equations and order of reactions 14.1rate equations and order of reactions 14.2zeroth, first...

1

Rate Equations and Rate Equations and Order of ReactionsOrder of Reactions

14.114.1 Rate Equations and Order of ReactionsRate Equations and Order of Reactions

14.214.2 Zeroth, First and Second Order ReactionsZeroth, First and Second Order Reactions

14.314.3 Determination of Simple Rate Equations from Initial Determination of Simple Rate Equations from Initial Rate MethodRate Method

14.414.4 Determination of Simple Rate Equations from DiffereDetermination of Simple Rate Equations from Differential Rate Equationsntial Rate Equations

14.514.5 Determination of Simple Rate Equations from IntegraDetermination of Simple Rate Equations from Integrated Rate Equationsted Rate Equations

1414

2

Rate Rate Equations and Equations and

Order of Order of ReactionsReactions

3

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

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 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

x = 0 zero order w.r.t. Ax = 1 first order w.r.t. Ax = 2 second order w.r.t. A

y = 0 zero order w.r.t. By = 1 first order w.r.t. By = 2 second order w.r.t. B

7


Rate k[B]2

Describe the reaction with the following rate law.

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


• Temperature-dependent

• Can only be determined from experiments

9

Rate k[A]x[B]y

units of k : -mol dm3 s1/(mol dm3)x+y or,mol dm3 min1 /(mol dm3)x+y


y3x3

13

yx )dm (mol)dm (mols dm mol

[B][A]rate

k

10

Rate k[A]0[B]0

units of k= mol dm3 s1/(mol dm3)0+0

= mol dm3 s1

= units of rate


11

Rate k[A][B]0


= s1


12

Rate k[A][B]


= mol1 dm3 s1


The overall order of reaction can be deduced from the units of k

13

Rate k[A]x[B]y[C]z…

For the reaction

aA + bB + cC + … products

units of k : -mol dm3 s1/(mol dm3)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 : -

1. Initial rate method (pp.17-18)

2. Graphical method (pp.19-26)

15

Determination of Determination of Rate Equations Rate Equations by Initial Rate by Initial Rate

MethodsMethods

16

5Cl(aq) + ClO3(aq) + 6H+(aq) 3Cl2(aq) +

3H2O(l)

Expt [Cl(aq)] / mol dm3

[ClO3(aq)]

/ mol dm3

[H+(aq)] / mol dm3

Initial rate / mol dm3 s1

1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

rate k[Cl(aq)]x[ClO3(aq)]y[H+

(aq)]z

17


[ClO3(aq)

] / mol dm3

[H+(aq)] / mol dm3


1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

zyx

zyx

5

5

(0.20)(0.08)(0.15)(0.40)(0.08)(0.15)

101.0104.0

From experiments 1 and 2,

4 = 2z

z = 2

= 2z

18


[ClO3(aq)

] / mol dm3

[H+(aq)] / mol dm3


1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

zyx

zyx

5

5

(0.40)(0.08)(0.15)(0.40)(0.16)(0.15)

104.0108.0


2 = 2y

y = 1

= 2y

19


[ClO3(aq)

] / mol dm3

[H+(aq)] / mol dm3


1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

zyx

zyx

5

5

(0.20)(0.08)(0.15)(0.20)(0.08)(0.30)

101.0102.0


2 = 2x

x = 1

= 2x

20

rate k[Cl(aq)][ClO3(aq)][H+

(aq)]2


[ClO3(aq)

] / mol dm3

[H+(aq)] / mol dm3


1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

From experiment 1,

1.0105 k(0.15)(0.08)(0.20)2

k = 0.02 mol3 dm9 s1

21

rate k[Cl(aq)][ClO3(aq)][H+

(aq)]2


[ClO3(aq)

] / mol dm3

[H+(aq)] / mol dm3


1 0.15 0.08 0.20 1.0105

2 0.15 0.08 0.40 4.0105

3 0.15 0.16 0.40 8.0105

4 0.30 0.08 0.20 2.0105

From experiment 2,

4.0105 k(0.15)(0.08)(0.40)2

k = 0.02 mol3 dm9 s1

22

Q.15 2C + 3D + E P + 2Q

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(a) rate k[C]x[D]y[E]z

23

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102


zyx

zyx

3

-2

(0.10)(0.10)(0.10)(0.10)(0.10)(0.20)

103.0102.4


8 = 2x

x = 3

= 2x

24

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102


zyx

zyx

3

-3

(0.10)(0.10)(0.10)(0.10)(0.20)(0.10)

103.0103.0


1 = 2y

y = 0

= 2y

25

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102


zyx

zyx

3

-2

(0.10)(0.10)(0.10)(0.30)(0.10)(0.10)

103.0102.7


9 = 3z

z = 2

= 3z

26

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(a) rate k[C]3[D]0[E]2

= k[C]3[E]2

27

Expt [C] / mol dm3

[D] / mol dm3

[E] / mol dm3


1 0.10 0.10 0.10 3.0103

2 0.20 0.10 0.10 2.4102

3 0.10 0.20 0.10 3.0103

4 0.10 0.10 0.30 2.7102

(b) rate k[C]3[E]2

From experiment 1,

3.0103 k(0.10)3(0.10)2

k = 300 mol4 dm12 s1

28

Q.16H+

CH3COCH3(aq) + I2(aq) CH3COCH2I(aq) + H+(aq) + I(aq)

Initial rate/ mol dm3 s1

Initial concentration/ mol dm3

[I2(aq)] [CH3COCH3(aq)] [H+(aq)]

3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) rate k[I2(aq)]x[CH3COCH3(aq)]y[H+

(aq)]z

29




3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103


(aq)]z

z3-y1-x4-

z-3y-1x-4

5

-5

)10(5.0)10(2.0)10(1.5)10(5.0)10(2.0)10(2.5

103.5103.5


1 = 1.67x

x = 0

= 1.67x

30




3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103


(aq)]z

z3-y1-x4-

z-3y-1x-4

5

-5

)10(5.0)10(2.0)10(2.5)10(5.0)10(4.0)10(2.5

103.5107.0


2 = 2y y = 1

= 2y

31




3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103


(aq)]z

z3-y1-x4-

z-2y-1x-4

5

-4

)10(5.0)10(4.0)10(2.5)10(1.0)10(4.0)10(2.5

107.0101.4


2 = 2z z = 1

= 2z

32




3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(a) Rate = k[I2(aq)]0[CH3COCH3(aq)][H+(aq)]

= k[CH3COCH3(aq)][H+(aq)]

33




3.5 105 2.5104 2.0101 5.0103

3.5 105 1.5104 2.0101 5.0103

1.4 104 2.5104 4.0101 1.0102

7.0 105 2.5104 4.0101 5.0103

(b) Rate = k[CH3COCH3(aq)][H+(aq)]

From experiment 1,

3.5105 k(2.0101)(5.0103)

k = 0.035 mol1 dm3 s1

34

Determination of Determination of Rate Equations Rate Equations

by Graphical by Graphical MethodsMethods

35

Two types of rate equations : -

(1) Differential rate equation

(2) Integrated rate equation

36

A products

nk[A]dt

d[A] Rate

(Differential rate equation)

shows the variation of rate with [A]

Two types of plots to determine k and n

37

[A]

rate

nk[A]dt

d[A] rate

n = 0

k

rate = k

38

Examples of zero-order reactions : -

2NH3(g) N2(g) + 3H2(g)

Fe or W 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.

2HI(g) H2(g) + I2(g) Au as catalyst

39

[A]

rate

nk[A]dt

d[A] rate

n = 0

k

rate = k

40

[A]

ratek[A]

dtd[A]

rate

n = 1

slope = k

linear

41

[A]

rate

2k[A]dt

d[A] rate

n = 2

k cannot be determined directly from the graph

parabola

42

nk[A]dt

d[A] rate

[A]

rate n = 2 n =

1

n = 0

43 log10[A]

log10rate

nk[A] rate n

1010 k[A]log ratelog

slope

y-intercept

n = 0

n = 1

log10k

n = 2

[A]nlogklog 1010

slope = 1

slope = 2

slope = 0

44

nk[A]dt

d[A] (Differential rate equation)

kdtd[A]

t

0 0

A

A

t

tdtkd[A]

kt[A][A] 0t

[A]t = [A]0 – kt (Integrated rate equation)

If n = 0

Derivation not required

45

[A]t = [A]0 – kt (Integrated rate equation)

shows variation of [A] with time

time

[A]t

rate kdt

d[A]slope

[A]0

constant rate

46

nk[A]dt


(Integrated rate equation)

If n = 1, k[A]dt

d[A]

kdt[A]d[A]

t

0 0

[A]

[A]

t

tdtkd[A]

[A]1

loge[A]t – loge[A]0 = kt

Or [A]t [A]0 ekt loge[A]t = loge[A]0 kt

ln

47



time

loge [A]t

loge [A]0

slope = k

linear n = 1

48



time

[A]t

[A]t varies exponentially with time

constant half life n = 1

49

seconds 100tlife, half21

50

loge[A]t = loge[A]0 kt

21tt when 0t [A]

21

[A]

21kt[A]log[A]

21

log 0e0e

21kt[A]log[A]log

21

log 0e0ee

21kt

21

loge

21kt 2loge

k2log

t e

21

51

seconds 100tlife, half21

s 100ln2

k = 6.9103 s1

52

Q.17sucrose fructose + glucose

Rate = k[sucrose] k = 0.208 h1 at 298 K

(a) h 3.33h 0.208

ln2k

ln2t 12

1

53

Q.17sucrose fructose + glucose

Rate = k[sucrose] k = 0.208 h1 at 298 K

(b) [A]t [A]0 ekt

87.5% decomposed [A]t = 0.125[A]0

0.125 = ekt

ln0.125 = 0.208tt = 9.99 h

= e0.208t

54

nk[A]dt


(Integrated rate equation)

If n = 2,

kt[A]

1[A]

1

0t

tk[A]1

[A][A]

0

0t Or

55

kt[A]

1[A]

1

0t

tk[A]1

[A][A]

0

0t Or

0[A]

1

Linear n = 2

Slope = k

time

t[A]1

56

kt[A]

1[A]

1

0t

tk[A]1

[A][A]

0

0t Or

time

[A]t

n = 2

Variable half life

n = 1

0k[A]1

t21

[A]t more rapidly with time in the early stage

57

time

[A]t

n = 1n = 2

n = 0

Plotting based on integrated rate equations

More common because [A]t and time can be obtained directly from expereiments.

58

[A]

rate n = 2 n =

1

n = 0

Plotting based on differential rate equations

Less common because rate cannot be obtained directly from expereiments.

59 log10[A]

log10rate

n = 0

n = 1

log10k

n = 2

slope = 1

slope = 2

slope = 0

Plotting based on differential rate equations

Less common because rate cannot be obtained directly from expereiments.

60

mol1 dm3 s1k against t2

s1kln[A]t against t1

mol dm3 s1k[A]t against t[A]t [A]0 – kt0

Units of kSlopeStraight line

plot

Integrated rate

equationOrder

t[A]1

kt[A]

1[A]

1

0t

kt[A][A]

ln0

t

Summary : - For reactions of the type

A Products

61

2H2O2(aq) 2H2O(l) + O2(g)

Rate = k[H2O2(aq)]

Examples of First Order ReactionsExamples of First Order Reactions

62

Examples of First Order ReactionsExamples of First Order Reactions

Reaction 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

1. 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

Examples of Second Order Examples of Second Order ReactionsReactions

64

Examples of Second Order Examples of Second Order ReactionsReactions

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

2. For a reaction involving one reactant only:

65

2. 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


A + B products

Rate = k[A][B]

When [B] is kept constant,

excess

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


A + B products

Rate = k[B][A]

• When [A] is kept constant,

rate = k”[B] (where k” = k[A]excess)

excess

69

Rate = k[A]excess[B] = k’’[B]

k can be determined from k’’ if [A]excess is known

Linear first order

70

Q.18(a)

2NO2(g) 2NO(g) + O2(g)

first-order reaction, k 3.6 103 s1 at 573 K

kt[A][A]

ln0

t = (3.6103 s1)(150s)

0

t

[A][A]

= 0.58

71

Q.18(b)

2NO2(g) 2NO(g) + O2(g)


kt[A][A]

ln0

t

0

t

[A][A]

= 0.010

0

[A]0.01[A]

ln0.01 = (3.6103 s1)t

t = 1279 s

72

Q.18(c)

2NO2(g) 2NO(g) + O2(g)


RTnVP22 NONO

RT

P(g)][NO 2NO

2

(g)]RT[NORTV

nP 2

NONO

2

2

K) )(573mol K dm atm (0.082atm 1.0

113

= 0.021 mol dm3

Calculate the rate of consumption of NO2 when the partial pressure of NO2 is 1.0 atm.(Gas constant, R = 0.082 atm dm3 K1 mol1)

73

Q.18(c)

2NO2(g) 2NO(g) + O2(g)



[NO2(g)] = 0.021 mol dm3

Rate of reaction = k[NO2(g)]

= (3.6103 s1)(0.021 mol dm3)= 7.6105 mol dm3 s1

74

dt(g)]d[NO

21

reaction of rate 2

Q.18(c)

2NO2(g) 2NO(g) + O2(g)



reaction of rate2dt

(g)]d[NO2

= 2(7.6105 mol dm3 s1)

= 1.5104 mol dm3 s1

75

Q.19

ktt at Ra of masst at Ra of mass

ln[Ra][Ra]

ln00

t

= (0.104 y1)(5 y)

g 0.50 year5 after remaining Ra of mass

= 0.595

eAcRa 01

22889

22888 Half life = 6.67

years

1 y0.104 y6.67

ln2tln2

k21

Mass of Ra remaining after 5 years = 0.297 g

76

Q.20 eHePbU 01

42

20682

23892 68

Half life = 4.51 109 years

110-9 y101.54 y104.51

ln2tln2

k21

1.2311.000

ln1.231x1.000x

lnt at U of masst at U of mass

ln[U][U]

lnkt-00

t

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 = (1.541010 y1)t = 0.208 t = 1.35109 years

77

Q.21(a)

eHePbU 01

42

20682

23892 68

y104.50 y101.54

ln2k

ln2t 9

1-10-21

78

Q.21(b)

eHePbU 01

42

20682

23892 68

No. of moles of U decayed =

No. of moles of He formed8

1

= (3.20103 mol)81

= 4.00104 mol

No. of moles of Pb produced = 4.00104 mol

No. of moles of U at t0 = (4.00104 + 4.40104) mol

= 8.40104 mol

79

0

t

[U][U]

lnkt-

Q.21(b)

eHePbU 01

42

20682

23892 68

mol 108.40mol 104.40

ln 4-

-4

= 0.647

kt = 0.647

(1.541010 y1)t = 0.647

t = 4.20109 y

80

Q.22(a)

2N2O5(g) 4NO2(g) + O2(g)

1.12.74.45.97.28.910.9

13.3

/ kPa

1250

8005504003002001000t /

minute

52ONP

81

Q.22(a)

2N2O5(g) 4NO2(g) + O2(g)

1.12.74.45.97.28.910.9

13.3

/ kPa

1250

8005504003002001000t /

minute

52ONP

Constant half life 350 minutes 1st order

t against (g)]Oln[N plot line Straight t52

kt(g)]Oln[N(g)]Oln[N 052t52

82

(g)]RTO[NRTV

nP 52

ONON

52

52

RT

P(g)]O[N 52ON

52

kt(g)]Oln[N(g)]Oln[N 052t52

ktRT

Pln

RT

Pln

0

ON

t

ON 5252

ktlnRT)ln(PlnRT)ln(P 0ONtON 5252

kt)ln(P)ln(P 0ONtON 5252

plot line straight t against )ln(P tON 52

Q.22(a)

83 time

tON )ln(P52

k = slope = 1.99103 min1

linear first order

Q.22(a)/(b)

0.100.991.481.771.972.19

2.392.59

1250

8005504003002001000t /

minute

tON )ln(P52

84

14.1 Rate Equations and Order of Reactions (SB p.27)

(a)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

14.1 Rate Equations and Order of Reactions (SB p.27)

(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

23 ]C[]B][A[Rate

63-

-1-3

)dm (molmin dm mol

Back

86

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.

14.2 Zeroth, First and Second Order Reactions (SB p.29)

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)

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 0.02 0.02 0.001 0

3 0.01 0.04 0.002 0

88


(a)Calculate the order of reaction with respect to A and that with respect to B.

Answer

89


(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,

0.0005 = k (0.01)x (0.02)y .......................... (1)

0.0010 = k (0.02)x (0.02)y .......................... (2)

0.002 0 = k (0.01)x (0.04)y .......................... (3)

Dividing (1) by (2),

x = 1

x)02.001.0

(0010.0

5 0.000

90


(a) Dividing (1) by (3),

y = 2

y)04.002.0

(0010.0

5 0.000

91


(b) Using the result of experiment (1),

Rate = k [A] [B]2

0.000 5 = k 0.01 0.022

k = 125 mol-2 dm6 s-1

(c) Rate = 125 [A] [B]2

Back

(b) Calculate the rate constant using the result of experiment 1.

(c)Write the rate equation for the reaction.

Answer

92

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 0.1 0.015

2 0.2 0.1 0.030

3 0.1 0.2 0.030

4 0.4 0.1 0.060

93

(a) Calculate the rate constant of the reaction, and hence write the rate equation for the reaction.


Answer

94


(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

m)2.01.0

(030.0

0.015

95


(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]

n)2.01.0

(030.0

0.015

96


(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(b) Initial rate = 1.5 0.3 0.3

= 0.135 mol dm-3 s-1

Back

97

(a) Write a chemical equation for the decomposition of hydrogen peroxide solution.

14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34)

Answer

(a) 2H2O2(aq) 2H2O(l) + O2(g)

98

(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


(b) 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

(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)].


Answer

[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

101


(c)

102

(d) From the graph in (c), determine the order and rate constant of the reaction.


Answer

103


(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

3-

-1-3

dm mol 0) - .3000(s dm mol 0) - 4-10 (1.8

104


(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)

-3.74-3.82-3.98-4.23log (rate)

-0.523-0.602-0.757-1.000log [H2O2(aq)]

105


(d) A graph of log (rate) against log [H2O2(aq)] gives a straight line

of slope x and y-intercept log k.

106


(d) 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

)8.0(5.0)02.4(71.3

Back

107


(a)Decide which curve in the following graph corresponds to

(i) a zeroth order reaction;

(ii) a first order reaction.

(a) (i)

(3)

(ii)

(2)

Answer

108


(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


(i) Write the rate equation for the reaction.

Answer

(i) 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


(ii) Determine the order of the reaction.Answer

111


(ii) Method 1:

A graph of the initial rates against [N2O5(g)] is shown as follows:

112


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 0.075 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)]

1-3-3-

-1-3

s dm mol 0) - 10 .61(s dm mol 0)(0.12

113


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.

-0.62-0.74-0.92log (rate)

-2.50-2.62-2.80log [N2O5(g)]

114


Slope of the graph =

=1.0The 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 = 75.86 s-1

The rate equation for the

reaction is:

Rate = 75.86 [N2O5(g)]

)62.2(50.2)74.0(62.0

115


(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


(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

The half-life of a radioactive isotope A is 1 997 years. How long does it take for the radioactivity of a sample of A to drop to 20% of its original level?

14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39)

Answer

118


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.

kt

693.0

2

1

1997693.0k

kt)[A]

[A]( ln 0

t410473.)% 20% 100

( ln

Back

119


(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 2NO2 + O221

Answer

120


(a) Let the half-life of nitrogen(V) oxide be .

The half-life of nitrogen(V) oxide is 14 745 s.

2

1t

2

1

14 693.0s1047.0

t

s 745 142

1 t

121

(b) The decomposition of CH3N = NCH3 to form N2 and C2H6 follows first order kinetics and has a half-life of 0.017 minute at 573 K. Determine the amount of CH3N = NCH3 left if 1.5 g of CH3N = NCH3 was decomposed for 0.068 minute at 573 K.


Answer

122


(b)

Let m be the amount of CH3N=NCH3 left after 0.068

minute.

m = 0.094 g

1

2

1

min76.40min017.0

693.0693.0 t

k

0.068 40.76)1.5

( ln m

Back

123


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.

0.1000.1250.1670.2500.500[HI(g)] (mol dm-3)

4803602401200Time (min)

[HI(g)]1

Answer

124


10.000-2.3030.100480

8.000-2.0790.125360

5.988-1.7900.167240

4.000-1.3860.250120

2.000-0.6930.5000

1/[HI(g)]

(mol-1 dm3)

ln [HI(g)][HI(g)] (mol dm-3)

Time (min)

125


The order of decomposition can be determined by plotting:

(a) [HI(g)] against time,

126


(b) ln [HI(g)] against time,

127


(c) against time,)]g(HI[

1

128


In graph (a), the plot of [HI(g)] against time is not a straight line, thu

s the decomposition reaction is not of zeroth order.

Similarly, in graph (b), the plot of ln [HI(g)] against time is n

ot 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.

)]g(HI[1

Back

129


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)

200150100500Time (s)

Answer

130


17.86-2.880.056200

15.87-2.760.063150

13.89-2.630.072100

12.05-2.490.08350

10.00-2.300.1000

1 / [X] (mol-1 dm3)

ln [X][X] (mol dm-3)

Time (s)

131


As the graph of [X] against time is not a straight line, the reaction is not of zeroth order.

132


Similarly, the plot of ln [X} against time is not a straight line, thus the reaction is not of first order.

133


The plot of

Against time gives a straight line, therefore the reaction is of second order.

]X[1

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