1 The Effects of Temperature and Catalyst on Reaction Rate 15.1Activation Energy and Arrhenius Equation 15.2Interpretation of Rates of Gaseous Reactions.

1

The Effects of The Effects of Temperature and Temperature and

Catalyst on Catalyst on Reaction RateReaction Rate

15.115.1 Activation Energy and Arrhenius EquationActivation Energy and Arrhenius Equation

15.215.2 Interpretation of Rates of Gaseous Interpretation of Rates of Gaseous

Reactions at Molecular LevelReactions at Molecular Level

15.315.3 Energy Profile Energy Profile

15.415.4 Effect of Catalysts on Rates of ReactionsEffect of Catalysts on Rates of Reactions

1515

2

Activation Activation Energy and Energy and Arrhenius Arrhenius EquationEquation

3

Activation EnergyActivation Energy

Exothermic reaction

Activation energy, Ea

= energy required to start the reaction

related to the rate of reaction

4

Endothermic reaction

Activation EnergyActivation Energy

related to the rate of reaction

5

Most reactions have positive Ea since energy is absorbed to break bonds in reactant particles.

6

Arrhenius Arrhenius EquationEquation

Since rate = k[A]a[B]b...

RTEa-

Aek

At fixed concentrations, rate depends on k which in turn depends on

A depends on the nature of the reaction and varies with T

temperature (T) andthe nature of the reaction (A and Ea)

7

Q.23

R(298K)Ea

298298 eAk 298K, At

R(308K)Ea

308308 eAk 308K, At

K) R(298Ea

eA

K) R(308Ea

eA

kk

298

308

298

308

Assume A is a constant

3081

2981

REa

e = 1.92 2

A 10 K in T doubles the rate

R = 8.31 J K1 mol1

8

Arrhenius Arrhenius EquationEquationRate of reaction

exponentially with temperature

RTEa

eRate

9


RTEa

AeRate

T ,

A TA

(Minor effect) Rate

10


RTEa

AeRate

T ,

(Major effect)

RTEa (less

negative)

e RTEa

Rate

(more positive)

11


RTEa

AeRate

T ,

A TA

(Minor effect) Rate

12


RTEa

AeRate

T ,

(Major effect)

RTEa (more

negative)

e RTEa

Rate

(less positive)

13

Determination of Activation Determination of Activation EnergyEnergy

RTEa

Aek

T1

RE

-AlogAelogklog ae

RTE

ee

a

T1

2.303RE

-AlogAelogklog a10

RTE

1010

a

14

1/T (K1)

logek

RE-

slope a

logeA

Determination of EDetermination of Eaa by Graphical by Graphical MethodMethod

T1

RE

-Alogklog aee

15

Q.24

logek 1/T (K1)

14.9 1.80103

9.4 1.55103

6.8 1.43103

3.2 1.28103

16

1/T (K1)

logek

Ea = -slope R 182 kJ mol1

logeA 24.8

< E(H – I)

before the H – I bond is completely broken (refer to p.32)

H – H and I – I bonds are formed

17

If logerate is plotted against 1/T,

since rate = k[A]x[B]y…

logerate = logek + loge[A]x[B]y…

= logek + constant

RTE

constant Alog ae

y-intercept

18

Determination of Activation Determination of Activation Energy Using Two Rate Energy Using Two Rate ConstantsConstants

1

a

RTE

11 eAk

2

a

RTE

22 eAk

12

a

T1

T1

RE

2

1

2

1 eAA

kk

12

a

T1

T1

RE

e

12

a

T1

T1

RE

2

1 lnekk

ln

12

a

T1

T1

RE

19

Interpretation Interpretation of Rates of of Rates of Gaseous Gaseous

Reactions at Reactions at Molecular Molecular

LevelLevel

20

The Kinetic Theory of GasesThe Kinetic Theory of Gases - Develop - Developed by Maxwell and Boltzmaned by Maxwell and Boltzman

1. Gas particles are in a state of constant and random motion in all directions,

undergoing frequent collisions with one another and with the walls of the container. 2. The pressure exerted on the container is due to the collisions between gas particles and the walls of the containers.

21


3. Gas particles are treated as point masses because their volumes are negligible when compared with the volume of the container.

4. There is no interaction among gas particles except collisions.

22


5. Collisions between gas particles are perfectly elastic, i.e. the total kinetic energy is conserved.

23

Transfer of K.E. among molecules

Distribution of molecular

speeds

Distribution of Molecular Speeds in a Distribution of Molecular Speeds in a GasGasConsider a sample of gas:

24

Distribution of Molecular Speeds in a Distribution of Molecular Speeds in a GasGas

Area under curve = total no. of gas molecules

25


The mean kinetic energy of a sample of gas particles is proportional to its absolute temperature (T).

Tcm21 2

26


Tcm21 2

...nnn...cncncn

321

233

222

211

velocity square meanc2

velocity square mean rootc2

n1 = no. of molecules with velocity c1 and

n1 + n2 + n3 + ... = n (total no. of molecules)

27

2c

The distribution of velocity is not symmetrical the average velocity of a gas sample is best represented by the root mean square velocity2c

28


Tcm21 2

For a sample of gas containing n molecules,

2cmn31

RTPV n

where m is the absolute mass of the gas molecule

29

Variation in the distribution of molecular Speeds with Variation in the distribution of molecular Speeds with TT

As T ,

• Molecular speeds

• Curve becomes flattened

• Wider distribution of molecular speeds at a higher temp

• Area under the curve remains unchanged.

30

M : molar mass of gas in Kg

M r.m.s velocity

2cmn31

PV RTn

mn3RT

c2 If n = 1M

3RT

31

M3RT

c2

The areas underneath the curves are the same

The lighter molecules are more spread out in molecular speeds.

32

Q.25

3-2

102.02738.3143

M3RT

c

H2

3-2

1044.02738.3143

M3RT

c

CO2

= 1845 ms1

= 393 ms1

The lightest gases (H2, He) can escape from the gravitational pull of small planets

Very rare in the Earth’s atmosphere

33

Simple Collision TheorySimple Collision Theory

For a reaction to occur, the reactant particles must collide with

(1) kinetic energy Ea

(2) proper orientation.

34

Proper Orientation

HCl(g) + NH3(g) NH4Cl(s)

35

Improper Orientation

36

Improper Orientation

37





No. of effective collisions =

peZ RTEa

Z = collision frequency

Effective collision

38





Effective collision


peZ RTEa

aRTE

E K.E. with collisions of fraction ea

39





Effective collision


peZ RTEa

p = fraction of collisions with proper orientation

40

Theoretically, (from collision theory and kinetic theory)


peZ RTEa

peZrate RTEa

Experimentally,

...[Y][X]Ae...[Y]k[X]rate baRTE

baa

41

peZrate RTEa


baa

ZpA

If molarities of X, Y,… are fixed

= no. of collisions with proper orientation

TA

42

Interpretation of the Effect of Interpretation of the Effect of Temperature Change on Rate of Temperature Change on Rate of ReactionReaction

T

speed of reactant particles

collision frequency (Z)

A

rate (minor effect)


baa

TZpA

43

Interpretation of the Effect of Interpretation of the Effect of Temperature Change on Rate of Temperature Change on Rate of ReactionReaction

T

K.E. of reactant particles

fraction of collisions with K.E. Ea


baa

e i.e. RT-Ea

rate exponentially (major effect)

44

speed/ K.E.

No. of molecules

concave

convex

45

Fraction of particles with K.E. > E = area total

area shadedE

The shaded area = no. of particles with K.E. > E

RTE

0

enn

46

Fraction of particles with K.E. > Ea =

area totalarea shaded

Ea

RTE

0

a

enn

If E = Ea

47

As T , the fraction of particles with K.E. > Ea

increases exponentially.

RTE

0

a

enn

Rate increases exponentially with T

48

Limitations of Collision Theory

In aqueous phase, the interactions between the reactant particles and the solvent molecules have to be considered.

The fraction of collisions with proper orientation (the steric factor, p) cannot be predicted. It can only be determined experimentally.

Collision theory is based on the calculations from kinetic theory of ideal gases. Thus, it is ONLY applicable to reactions in gas phase.

49

Consider the 2nd order single-step gas phase rxR(g) + R(g) products

Given : k = 1.00102 mol1 dm3 s1 at 473 K,

[R(g)]initial = 1102 mol dm3, L = 6.021023 mol1

Gas constant = 8.31 J K1 mol1, Ea = 100 kJ mol1

Initial collision frequency(Z) = 7.771032 s1

(a) Estimate

(i) The no. of effective collisions per m3 per second.

(ii) The no. of collisions with K.E. Ea per m3 per second.

(b) Hence, deduce the steric factor, p of the reaction.

Q.26

50

= (1.0010-2 mol1 dm3 s1)(1.0010-2 mol dm3)2

= 1.0010-6 mol dm3 s-1

= 1.0010-6 mol 6.021023 mol-1 dm-3 s-1

= 6.021017 molecules dm-3 s-1

= 6.021020 molecules m-3 s-1

6.021020 molecules of R are decomposed per cubic meter per second

(a)(i)2initialk[R]rate Initial

51

Consider the 2nd order single-step gas phase rx

R(g) + R(g) products

Rate = 6.021020 molecules m-3 s-1

One effective collision leads to decomposition of Two molecules of R.

Thus, no. of effective collisions per cubic meter per second

= 3.011020

(a)(i)

52


peZ RTEa

= 3.011020 m-3 s-1

No. of collisions with K.E. Ea

RTEa

eZ

K) )(473mol K J (8.31

mol J 1000100

1332 1-1-

-1

esm107.77

= 7.771032 m-3 s-1 (8.9310-

12)

= 6.941021 m-3 s-1

(a)(ii)

53


peZ RTEa

= 3.011020 m-3 s-1

No. of collisions with K.E. Ea = 6.941021 m-3 s-1

aEK.E. with collisions of no.collisions effective of no.

p

0.0434sm106.94sm103.01

p 1321

1320

= 4.34 %

(b)

54

Q.27


baa

If Ea 0

1 eRTEa

babaRTE

[Y]A[X] ...[Y][X]Aeratea

Rate is independent of T (A-level)

55

K.E.

No. of molecules

Ea

56

K.E.

No. of molecules

EaEaEa


peZ RTEa

Z p

57

Energy Energy ProfileProfile

Transition Transition State TheoryState Theory

58

Transition State Theory

- focuses on what happens after the collisions have started.

59

Energy profile

- shows the variation of the potential energy of the reaction mixture as the reaction proceeds.

reaction coordinate

P.E.

60

Consider the one-step reaction,

A–B + X A + B–X

P.E. of the reaction mixture are calculated for any A–B and B–X distances, and the results are plotted on a contour diagram

61


A–B + X A + B–X

At R,

A-B distance is short

B-X distance is long

before reaction

62


A–B + X A + B–X

The valley at R represents the potential energy for the initial state of the system, i.e. A–B and X

63


A–B + X A + B–X

At P,

A-B distance is long

B-X distance is short

after reaction

64


A–B + X A + B–X

The valley at P represents the potential energy for the final state of the system, i.e. A and B–X

65


A–B + X A + B–X

The energy contours rise in all directions from the valleys at R and P,

but the ‘easiest’ path is shown by the bold line RTP

66

The transition state is like a col (山坳 ) in a mountain region

67

RP

T

68

In the transition state,

• Bond between A and B is partially broken

• Bond between B and X is partially formed

A-B + X ABX A + B-X

Thus, Ea is lower than E(A-B)

69

Transition state (Activated complex) is the least stable arrangement of the system in the most probable reaction pathway.

RP

T

70

Advantages of Transition State Theory1. Ea and A can be calculated A Zp

the steric factor p can be predicted

2. It explains why the reaction pathway is specific.

3. It is applicable to gaseous and aqueous reactions.

71

Energy Profile : One-step Energy Profile : One-step MechanismMechanism

A-B + X ABX A + B-X

72

Example of One-step MechanismExample of One-step Mechanism

Rate = k[CH3Cl][OH]

Bimolecular One-step 2nd Order Nucleophilic Substitution Reaction

SN2

73

CH3Cl + OH CH3OH + Cl

74

Energy Profile : Multi-step Energy Profile : Multi-step MechanismMechanismE1 > E2

Step 1 is the rate determining step

75

Energy Profile : Multi-step Energy Profile : Multi-step MechanismMechanism

B][AeAB][Ak Rate RTE

11

1

76

Multi-step MechanismMulti-step Mechanism

Step 1: A ─ B A + B (intermediate)

Step 2: A + B + X A + B─ X

Overall reaction: A ─ B + X A + B ─ X

• Chemical reactions take place in two or more steps

• Formation of an intermediate

77

• Hydrolysis of 2-chloro-2-methylpropane

Example of Multi-step MechanismExample of Multi-step Mechanism

(1)

(2)

carbocation

78

Rate = k[C(CH3)3Cl]Unimolecular Two-step 1st Order Nucleophilic Substitution Reaction

SN1

79

+ OH

Rate = k[C(CH3)3Cl]

E1 E2

E1 > E2

Step 1 is r.d.s.

80

Reaction mechanism is the detailed sequence of steps that occur in a reaction .

Reaction mechanisms are theoretical proposals used to explain the experimentally determined rate laws.

Reaction Mechanism and Rate Law

81

Each of the steps in a mechanism is called an elementary step.


The number of reactant particles that takes part in each elementary step is called the molecularity of that step.

82


Unimolecular – one particle collides with the wall of the vessel or the excess solvent


83

Pseudo-1st order reaction

CH3COOCH3 + H2O CH3COOH + CH3OH

Rate = k[CH3COOCH3][H2O]

If H2O is used as solvent (in large excess)

[H2O] constant throughout the reaction

Rate = k’[CH3COOCH3]

Unimolecular reaction

84


Unimolecular – one particle collides with the wall of the vessel or the excess solventBimolecular – two particles collide together

Termolecular – three particles collide together simultaneously (very rare)


85

The slowest step in a particular mechanism is called the rate-determining step

Requirements for writing reaction mechanisms : 1. The sum of elementary steps must give the overall balanced equation for the reaction.2. The mechanism must agree with the experimentally determined rate law.

86

Consider the reaction

A + B + C D

Rate = k[A][B]Only one intermediate

R + C D (fast)

A + B R (slow) r.d.s

Proposed mechanism : -

87

A + B X (slow) r.d.s.

X Y (fast)

Y + C D (fast)

Q.28

88

Reaction coordinate

P.E.

A + B + C

X + C Y + C

D

E1E2

E3

E1 > E2 E3

Step one is the r.d.s

Q.28

Rate = k[A][B]

89

A + B X (fast)

X Y (slow) r.d.s.

Y + C D (fast)

Q.28

k1

k2

Rate = k[X]

[A][B]k[A][B]kkk

Rate 32

1

k

At equilibrium, k1[A][B] = k2[X]

90

Effect of Effect of Catalysts on Catalysts on

Rates of Rates of ReactionsReactions

91

Working Principle of Catalysts and Working Principle of Catalysts and their Effects on Reaction Ratestheir Effects on Reaction Rates

Catalysts alter the rates of reaction,

1. but remain chemically unchanged at the end of the reaction

2. by providing new, alternative reaction pathways with different activation energies.

Catalysis Catalytic action

92

Positive catalyst:

• Provides an alternative reaction pathway with a lower activation energy


93

• Lower Ea

Ea’

Greater fraction of molecules withK.E. greater than or equal to Ea

Reaction proceeds faster

94

Negative catalyst:

• Provides an alternative reaction pathway with a higher activation energy


95

• Higher Ea

Ea”

Smaller fraction of molecules withK.E. greater than or equal to Ea

Reaction proceeds slower

96


With catalysts, the contour diagrams and thus the energy profiles are totally different from those without

97

CatalystCatalyst

Homogeneous Catalyst

Homogeneous Catalyst

HeterogeneousCatalyst

HeterogeneousCatalyst

Reactants & catalyst are in the

same phase

Reactants & catalyst are NOT in the same

phase

98

Characteristics of Catalysts

1. For a given reversible reaction,

Reactants Products

k1

K-1

catalysts affect the rates of forward reaction and backward reaction to the same extent.

99

Q.29

Reactants Products

k1

k-1

Without catalyst

With catalyst

Reactants Products

'1k'1-k

Show that

1

'1

1

'1

kk

kk

100

)E(ERT1

RTE

1

RTE

'1

1

'1 11

1

'1

e

eA

eAkk '

)E(E

RT1

RTE

1-

RTE

'1-

1-

'1-

'1-1-

1-

1-

e

eA

eAkk

'

E1 E-1

E’1

E’-1

=

P.E.

Reaction coordinate

'11

'11 EEEE

101

11

'1

1

'1 mol kJ 50)E(E if

kk

Calculate

)E(ERT1

RTE

1

RTE

'1

1

'1 11

1

'1

e

eA

eAkk '

Given : T = 298 K, R = 8.31 J K1 mol1

K) )(298mol K J (8.31

mol J 1050)E(E

RT1

1

'1 11

13'11

eekk

= 5.9108

102


2. Catalysts are chemically unchanged at the end of reactions, but may undergo physical changes.

E.g. Lumps of MnO2 used in the decomposition of H2O2 become powdered at the end of the reaction.

103


3. Only small quantity is sufficient to catalyze a reaction because catalysts can be

regenerated.

However, if the catalysts are involved in the rate equation, higher concentrations may affect the rate more.

104


4. The effect of heterogeneous catalysts depends on the surface area available for the catalytic action.

Surface area of solid catalyst number of reaction sites catalytic activity

E.g. Finely divided Fe powder is used as the catalyst in Haber process.

105


5. Catalytic actions are specific especially in biological systems.

E.g. Enzymatic actions are highly specific.

106


6. The efficiency of a catalyst is often enhanced by adding promoters. Promoters

have no catalytic actions on their own.

E.g. Fe2O3, KOH, Al2O3 in Haber process

107


7. The efficiency of a catalyst can be lowered by adding poisons or inhibitors. Catalyst poisons are specific in action.

E.g. Arsenic impurities may poison Pt but not V2O5 in Contact process

108


8. Transition metals or compounds/ions containing transition metals show marked

catalytic activities.

E.g. Pt, Ni, Fe, V2O5, MnO2, Mn2+ Fe3+

The catalytic actions are due to the presence of low-lying partially filled d-orbitals.

109

Heterogeneous Catalysis – Heterogeneous Catalysis – AdsorptionAdsorption

Occur on the surface of the catalyst.

1. Reactants are adsorbed on the surface, forming new bonds with the catalyst while

weakening bonds in reactants2. Products, once formed, are desorbed

from the surface,

110

Examples of heterogeneous catalysis2H2O2(aq) 2H2O(l) + O2(g)

MnO2(s)

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

C8H18(g) C4H10(g) + C4H8(g)Al2O3/SiO2(s)

CH2=CH2(g) + H2(g) CH3–CH3(g)

Ni(s)

111

fast

CH2=CH2 H–H

fast

slowH2C CH2H H


Ni(s)

112

fast

CH2=CH2 H–H

fast

slowH2C CH2H H


Ni(s)

113

fast

CH2=CH2 H–H

fast

slowH2C CH2H H


Ni(s)

114

Q.3Q.300

CH2=CH2 + H2

CH3-CH3

321 aaa EEE

115

Homogeneous catalysts participate in certain stages of reactions and are regenerated at the end or later stages of reactions.

Homogeneous Catalysis – Intermediate Homogeneous Catalysis – Intermediate FormationFormation

Stage 1: A + catalyst A ─ catalyst

Stage 2: A ─ catalyst + B A ─ B + catalyst

Overall reaction: A + B A ─ B

intermediate

116


• Acid-catalyzed esterification of ethanoic acid and methanol

CH3COOH(l) + CH3OH(l)

CH3COOCH3(l) + H2O(l)H+

117


118

CH3C

OH

O

O

CH3

H

C

O

H3C

OH

O

H

CH3

r.d.s.

Uncatalyzed esterification

Q.31

Rate = k[CH3COOH][CH3OH]

+

nucleophilic attack

119

C

O

H3C

OH

O CH3

-H+

CH3C

OH

O

O

CH3

H

C

O

H3C

OH

O

H

CH3

r.d.s.

H+

CH3C

O

O

CH3+ H2O

Uncatalyzed esterification

Q.31

+

120

Acid-catalyzed esterification

Protonation at carbonyl O rather than hydroxyl O since the former is more electron sufficient due to polarization of pi electron cloud

121


122


123


124


125

Most probable resonance structure

Carbonyl C becomes more electron-deficient

More easily attacked by nucleophile


126


127

r.d.s.

Rate = k[RCOOH][H+]


128

r.d.s.

step 2


129

r.d.s.

step 2step 3

step 4step 5 step 6


130

r.d.s.

step 2step 3

step 4 step 6


131

r.d.s.

step 2step 3

step 4step 5


132

r.d.s.

step 2step 3

step 4step 5 step 6

H+ is regenerated


133

r.d.s.

step 2step 3

step 4step 5 step 6

For simplicity, steps 3 to 6 are combined


134

C O

H3C

H3C

C O

H3C

IH 2C

+ I2 + HIH+

Rate = k[CH3COCH3][H+]

135

C O

H3C

H3C

H+ C OH

H3C

H3C

fast

C OH

H3C

H2C

H

H2C C

CH3

OH

+ H+ slow

H2C C

CH3

OH

+ I2 C C OH

I

CH3

I

H

H

fast

C O

H3C

IH 2C

+ HI fastC C O

I

CH3

I

H

H H

136

C O

H3C

H3C

H+ C OH

H3C

H3C

fast

C OH

H3C

H2C

H

H2C C

CH3

OH

+ H+ slow

At equilibrium,

k1[CH3COCH3][H+] = k2[ ]C OH

H3C

H3C

Rate = k3[ ]

C OH

H3C

H3C

]][HCOCH[CHkkk

332

13

= k[CH3COCH3][H+]

k1

k2

k3

137

Homogeneous Catalysis Using Transition Metal Ions

Principle : - Transition metals exhibit variable oxidation states

138

2I(aq) + S2O82(aq) I2(aq) +

2SO42(aq)

The reaction is slow because

colliding particles carry like charges

139

2I(aq) + S2O82(aq) I2(aq) +

2SO42(aq)

Fe3+

(aq)

2I(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) +

2SO42(aq)

Mechanism of catalyzed reaction : -

Both steps are fast because

colliding particles carry opposite charges.

140

2I(aq) + S2O82(aq) I2(aq) +

2SO42(aq)

Fe3+

(aq)

2I(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+(aq)

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) +

2SO42(aq)


The mechanism is made possible by the variable oxidation states of Fe

141

2I(aq) + S2O82(aq) I2(aq) +

2SO42(aq)

Fe2+

(aq)

2I(aq) + 2Fe3+(aq) I2(aq) + 2Fe2+

(aq)

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) +

2SO42(aq)


Q.31

142

2Ce4+(aq) + Tl+(aq) 2Ce3+(aq) + Tl3+(aq)Mn2+

(aq)


Ce4+(aq) + Mn2+(aq) Ce3+(aq) + Mn3+

(aq)Ce4+(aq) + Mn3+(aq) Ce3+(aq) + Mn4+

(aq)Mn4+(aq) + Tl+(aq) Mn2+(aq) + Tl3+(aq)

The mechanism is made possible by the variable oxidation states of Fe

143

Applications of CatalystsApplications of Catalysts

Industrial CatalystsIndustrial Catalysts

1. Iron is used in the Haber process

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

2.Platinum or vanadium(V) oxide is used in the Contact process

2SO2(g) + O2(g) 2SO3(g)Pt or V2O5

144

3. Nickel, platinium or palladium is used in the hydrogenation of unsaturated oils to make margarine


145


4. Nickel and nickel(II) oxide are used in the production of town gas in Hong Kong.

C5H12(g) + 5H2O(g) 5CO(g) + 11H2(g)

2CO(g) + 2H2(g) CO2(g) + CH4(g)

Ni or NiO

146

Catalytic Converters in Car Exhaust SystemsCatalytic Converters in Car Exhaust Systems


147


2CO(g) + 2NO(g) 2CO2(g) + N2(g)

CxHy(g) + ( x + y/4) O2(g)

xCO2(g) + y/2 H2O(g)

2CO(g) + O2(g) 2CO2(g)

Pt

Pt

Rh

148

Enzymes in the Production of Alcoholic Drinks

Enzymes in the Production of Alcoholic Drinks


C6H12O6(aq)

2C2H5OH(aq) + 2CO2(g)enzyme

Fermentation

149

The END

150

15.1 Activation Energy and Arrhenius Equation (SB p.51)

For the following reaction:

C6H5N2 +Cl–(aq) + H2O(l)

C6H5OH(aq) + N2(g) + H+(aq) + Cl–(aq)

the rate constants of the reaction at different temperatures were measured and recorded in the following table:

151


Temperature (K) Rate constant (10-5 s-1)

278.0 0.15

298.1 4.10

308.2 20.00

323.0 140.00

Determine the activation energy graphically.

(Given: R = 8.314 J K–1 mol–1)Answer

152


3.096 10-3-6.57

3.245 10-3-8.52

3.355 10-3-10.10

3.597 10-3-13.41

1/T (k-1)ln k

153


A graph of ln k against gives a straight line with slope .T1

RaE

154


Back

y = -11.8 – (-7)

= -4.8

x = (3.48 – 3.13) 10-3

= 0.35 10-3 K-1

Slope =

= -13.7 103 K = -13.7 103 K

Ea = 13.7 103 K 8.314 J K-1 mol-1

= 113.9 103 J mol-1

= 113.9 kJ mol-1

The activation energy of the reaction is 113.9 kJ mol-1.

RaE

13K1035.08.4

155

The rate constant for a reaction at 110°C is found to be twice the value of that at 100°C. Calculate the activation of the reaction.(Given : R = 8.314 J K-1 mol-1)


Answer

Since k110 oC = 2 k100 oC,

Ea = 82 327 J mol-1

=82.3 kJ mol-1

The activation energy of the reaction is 82.3 kJ mol-1.

)100273

1100273

1(

314.8

Eln a

C110

C100

o

o

k

k

)100273

1100273

1(

314.8

E

21

ln a

Back

156

(a) The reaction

2A(g) + B(g) C(g)

was studied at a number of temperatures, and the following results were obtained:

Determine the activation energy of the reaction graphically.

(Given: R = 8.314 J K–1 mol–1)


Temperature (o

C)

12 60 112 203 292

Rate constant (dm6 mol-2 s-

1)

2.34 13.2 52.5 316 1000

Answer

157


6.911.77 10-3565

5.762.10 10-3476

3.962.60 10-3385

2.583.00 10-3333

0.853.51 10-3285

ln k1 / T (K-1)T (K)

158


A graph of ln k against gives a straight line with slope .T1

RaE

159


Slope =

= -3.48 103

Ea = 3.48 103 8.314

= 28.93 kJ mol-1

The activation energy of the reactions is 28.93 kJ mol-1.

103.51 101.7785.091.6

3-3-

3a 1048.3E

R

160


(b) Determine the activation energy of the following reaction using the data provided only.

A + B C

(Given: R = 8.31 J K–1 mol–1)

0.400400

0.096350

Rate constant (mol dm-3 s-1)Temperature (K)

Answer

161


(b)

Ea = 33 206 J mol-1

= 33.2 kJ mol-1

)400

1350

1(

31.8

E

400.0096.0

ln a

Back

162

(a) Explain why not all collisions between reactant molecules lead to the formation of products.

15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level (SB p.58)

Answer(a) For a reaction to occur, colliding molecules must have kinetic

energy equal to or greater than the activation energy to

break the bonds in the reactants, so that new bonds can

form in the products. Moreover, the collision must be in the

right geometrical orientation, and the atoms to be transferred

or shared do not come into direct contact with each other, so

that the atoms can rearrange to form products. Products

cannot be formed if the kinetic energy of the reactant

molecules cannot overcome the activation energy, or the

collision orientation is not appropriate.

163

(b) Describe the effect of temperature on the distribution of molecular speeds in a gaseous system.


Answer(b) An increase in temperature will lead to an increase in the

most probable speed of the molecules. The peak of the

curve of Maxwell-Boltzmann distribution of molecular speeds

shifts to the right and the curve becomes flattened. This

indicates that the distribution of molecular speed becomes

wider and the number of molecules having the most

probable speed decreases.

164

(c)Explain why the rates of chemical reactions increase with temperature.


Answer

(c) As temperature rises, the proportion of fast-moving

molecules increases. The kinetic energy of the molecules

also increases. A greater fraction of molecules can

overcome the activation energy required for a reaction to

occur. Therefore, the number of effective collisions increases

and hence the rates of chemical reactions increase.

Back

165

Draw an energy profile of a typical single-stage endothermic reaction.

15.3 Energy Profile (SB p.60)

Answer

Back

166

The energy profile of a multi-stage reaction is shown below:


167


(a)Which stage is the rate determining step? Explain your answer.

(b) Is the reaction exothermic or endothermic? Explain your answer.

Answer(a) Stage 2 is the rate determining step.

It is because stage 2 has the greatest amount of activation

energy.

(b) The reaction is exothermic.

It is because the potential energy of the products is lower

than that of the reactants.

Back

168


Referring to the energy profiles below, answer the questions that follow.

A B

169


Referring to the energy profiles below, answer the questions that follow.

C D

170


(a) Which reaction(s) is/are exothermic?

(b) Which reaction is the fastest?

(c) Which reaction has the greatest amount of activation energy?

(a) A, B and C

(b) B

(c) D

Answer

Back

171

(a) Explain what a negative homogeneous catalyst is.

15.4 Effect of Catalysts on Rates of Reactions (SB p.69)

Answer

(a) A negative homogeneous catalyst is a catalyst that

slows down a reaction. It exists in the same phase

as the reactants and products in the reaction, and

involves in the formation of an intermediate in the

reaction.

172

(b) Explain what a positive heterogeneous catalyst is.


Answer(b) A positive heterogeneous catalyst is a catalyst that

speeds up a reaction but it is not in the same phase

as the reactant and products. It provides an active

surface for the reactant particles to adsorb in a

reaction.

173

(c) Give three applications of catalysts.


Answer(c) Iron used in the Haber process;

Platinum or vanadium(V) oxide used in the Contact process;

Nickel, platinum or palladium used in the hydrogenation of

unsaturated oils to make margarine;

Nickel and nickel(II) oxide used in the production of town gas;

Platinum (or palladium) and rhodium used in catalytic

converters;

Enzymes used in fermentation of glucose to produce ethanol;

Enzymes used in the manufacture of biological washing

powders.

(any 3)

Back

1 The Effects of Temperature and Catalyst on Reaction Rate 15.1Activation Energy and Arrhenius Equation 15.2Interpretation of Rates of Gaseous Reactions.

Documents

rate of reaction slide

positive slide

temperature slide

arrhenius equation slide

ea slide

reaction rate

minor effect rate slide

molecular level slide