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

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

Activation Activation Energy and Energy and Arrhenius Arrhenius EquationEquation

Activation EnergyActivation Energy

Exothermic reaction

Activation energy, Ea

= energy required to start the reaction

related to the rate of reaction

Endothermic reaction

Activation EnergyActivation Energy

related to the rate of reaction

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

Arrhenius Arrhenius EquationEquation

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

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)

R(298K)Ea

298298 eAk 298K, At

R(308K)Ea

308308 eAk 308K, At

K) R(298Ea

K) R(308Ea

Assume A is a constant

e = 1.92 2

A 10 K in T doubles the rate

R = 8.31 J K1 mol1

Arrhenius Arrhenius EquationEquationRate of reaction

exponentially with temperature

AeRate

(Minor effect) Rate

AeRate

(Major effect)

RTEa (less

negative)

e RTEa

(more positive)

AeRate

(Minor effect) Rate

AeRate

(Major effect)

RTEa (more

negative)

e RTEa

(less positive)

Determination of Activation Determination of Activation EnergyEnergy

-AlogAelogklog ae

2.303RE

-AlogAelogklog a10

1/T (K1)

slope a

Determination of EDetermination of Eaa by Graphical by Graphical MethodMethod

-Alogklog aee

logek 1/T (K1)

14.9 1.80103

9.4 1.55103

6.8 1.43103

3.2 1.28103

1/T (K1)

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

If logerate is plotted against 1/T,

since rate = k[A]x[B]y…

logerate = logek + loge[A]x[B]y…

= logek + constant

constant Alog ae

y-intercept

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

11 eAk

22 eAk

1 lnekk

Interpretation Interpretation of Rates of of Rates of Gaseous Gaseous

Reactions at Reactions at Molecular Molecular

LevelLevel

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.

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.

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

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:

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

Area under curve = total no. of gas molecules

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

Tcm21 2

...nnn...cncncn

velocity square meanc2

velocity square mean rootc2

n1 = no. of molecules with velocity c1 and

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

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

Tcm21 2

For a sample of gas containing n molecules,

2cmn31

RTPV n

where m is the absolute mass of the gas molecule

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.

M : molar mass of gas in Kg

M r.m.s velocity

2cmn31

PV RTn

c2 If n = 1M

The areas underneath the curves are the same

The lighter molecules are more spread out in molecular speeds.

102.02738.3143

1044.02738.3143

= 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

Simple Collision TheorySimple Collision Theory

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

(1) kinetic energy Ea

(2) proper orientation.

Proper Orientation

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

Improper Orientation

No. of effective collisions =

peZ RTEa

Z = collision frequency

Effective collision

peZ RTEa

E K.E. with collisions of fraction ea

Effective collision

peZ RTEa

p = fraction of collisions with proper orientation

Theoretically, (from collision theory and kinetic theory)

peZ RTEa

peZrate RTEa

Experimentally,

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

peZrate RTEa

If molarities of X, Y,… are fixed

= no. of collisions with proper orientation

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

speed of reactant particles

collision frequency (Z)

rate (minor effect)

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

K.E. of reactant particles

fraction of collisions with K.E. Ea

e i.e. RT-Ea

rate exponentially (major effect)

speed/ K.E.

No. of molecules

concave

convex

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

area shadedE

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

Fraction of particles with K.E. > Ea =

area totalarea shaded

If E = Ea

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

increases exponentially.

Rate increases exponentially with T

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.

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.

= (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

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)

peZ RTEa

= 3.011020 m-3 s-1

No. of collisions with K.E. Ea

K) )(473mol K J (8.31

mol J 1000100

1332 1-1-

esm107.77

= 7.771032 m-3 s-1 (8.9310-

= 6.941021 m-3 s-1

(a)(ii)

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.

0.0434sm106.94sm103.01

p 1321

= 4.34 %

If Ea 0

1 eRTEa

babaRTE

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

Rate is independent of T (A-level)

No. of molecules

EaEaEa

peZ RTEa

Energy Energy ProfileProfile

Transition Transition State TheoryState Theory

Transition State Theory

- focuses on what happens after the collisions have started.

Energy profile

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

reaction coordinate

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

A–B + X A + B–X

A-B distance is short

B-X distance is long

before reaction

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

A–B + X A + B–X

A-B distance is long

B-X distance is short

after reaction

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

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

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

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)

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

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.

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

A-B + X ABX A + B-X

Example of One-step MechanismExample of One-step Mechanism

Rate = k[CH3Cl][OH]

Bimolecular One-step 2nd Order Nucleophilic Substitution Reaction

CH3Cl + OH CH3OH + Cl

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

Step 1 is the rate determining step

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

B][AeAB][Ak Rate RTE

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

• Hydrolysis of 2-chloro-2-methylpropane

Example of Multi-step MechanismExample of Multi-step Mechanism

carbocation

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

Rate = k[C(CH3)3Cl]

E1 > E2

Step 1 is r.d.s.

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

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.

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

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

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

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)

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.

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 : -

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

X Y (fast)

Y + C D (fast)

Reaction coordinate

A + B + C

X + C Y + C

E1 > E2 E3

Step one is the r.d.s

Rate = k[A][B]

A + B X (fast)

X Y (slow) r.d.s.

Y + C D (fast)

Rate = k[X]

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

Rate 32

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

Effect of Effect of Catalysts on Catalysts on

Rates of Rates of ReactionsReactions

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

Positive catalyst:

• Provides an alternative reaction pathway with a lower activation energy

• Lower Ea

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

Reaction proceeds faster

Negative catalyst:

• Provides an alternative reaction pathway with a higher activation energy

• Higher Ea

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

Reaction proceeds slower

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

CatalystCatalyst

Homogeneous Catalyst

HeterogeneousCatalyst

Reactants & catalyst are in the

same phase

Reactants & catalyst are NOT in the same

Characteristics of Catalysts

1. For a given reversible reaction,

Reactants Products

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

Reactants Products

Without catalyst

With catalyst

Reactants Products

'1k'1-k

Show that

)E(ERT1

eAkk '

E1 E-1

E’-1

Reaction coordinate

'11 EEEE

'1 mol kJ 50)E(E if

Calculate

)E(ERT1

eAkk '

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

K) )(298mol K J (8.31

mol J 1050)E(E

= 5.9108

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.

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.

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.

5. Catalytic actions are specific especially in biological systems.

E.g. Enzymatic actions are highly specific.

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

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

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.

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,

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)

CH2=CH2 H–H

slowH2C CH2H H

CH2=CH2 H–H

slowH2C CH2H H

CH2=CH2 H–H

slowH2C CH2H H

Q.3Q.300

CH2=CH2 + H2

CH3-CH3

321 aaa EEE

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

• Acid-catalyzed esterification of ethanoic acid and methanol

CH3COOH(l) + CH3OH(l)

CH3COOCH3(l) + H2O(l)H+

r.d.s.

Uncatalyzed esterification

Rate = k[CH3COOH][CH3OH]

nucleophilic attack

r.d.s.

CH3+ H2O

Uncatalyzed esterification

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

Most probable resonance structure

Carbonyl C becomes more electron-deficient

More easily attacked by nucleophile

r.d.s.

Rate = k[RCOOH][H+]

r.d.s.

step 2

r.d.s.

step 2step 3

step 4step 5 step 6

r.d.s.

step 2step 3

step 4 step 6

r.d.s.

step 2step 3

step 4step 5

r.d.s.

step 2step 3

step 4step 5 step 6

H+ is regenerated

r.d.s.

step 2step 3

step 4step 5 step 6

For simplicity, steps 3 to 6 are combined

+ I2 + HIH+

Rate = k[CH3COCH3][H+]

H+ C OH

+ H+ slow

+ I2 C C OH

+ HI fastC C O

H+ C OH

+ H+ slow

At equilibrium,

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

Rate = k3[ ]

]][HCOCH[CHkkk

= k[CH3COCH3][H+]

Homogeneous Catalysis Using Transition Metal Ions

Principle : - Transition metals exhibit variable oxidation states

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

2SO42(aq)

The reaction is slow because

colliding particles carry like charges

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

2SO42(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.

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

2SO42(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

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

2SO42(aq)

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

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

2SO42(aq)

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

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

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

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

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

Catalytic Converters in Car Exhaust SystemsCatalytic Converters in Car Exhaust Systems

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)

Enzymes in the Production of Alcoholic Drinks

C6H12O6(aq)

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

Fermentation

The END

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:

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

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

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

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.

13K1035.08.4

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

)100273

1100273

(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

12 60 112 203 292

Rate constant (dm6 mol-2 s-

2.34 13.2 52.5 316 1000

Answer

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)

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

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

3a 1048.3E

(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

Ea = 33 206 J mol-1

= 33.2 kJ mol-1

400.0096.0

(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.

(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.

(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.

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

15.3 Energy Profile (SB p.60)

Answer

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

(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.

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

(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

Answer

(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.

(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.

(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)

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

rate of reaction slide

positive slide

temperature slide

arrhenius equation slide

ea slide

reaction rate

minor effect rate slide

molecular level slide

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