Topic 5.4: Chemical kinetics II Topic 5.4: Chemical kinetics II

recall that rates of reaction may be expressed by empirical rate equations of the form: recall that rates of reaction may be expressed by empirical rate equations of the form: rate = k[A]

rate = k[A]mm_{[B]}_{[B]}nn_{,}_{,}_{where}_{where}_{m}_{m}_{and}_{and} _{n}_{n}_{are 0, 1 or 2}_{are 0, 1 or 2}

define the terms

define the termsrate constant rate constant andandorder of reactionorder of reactionand understand that these are experimentally determinedand understand that these are experimentally determined the concept of molecularity is

the concept of molecularity isnotnotrequiredrequired

deduce rate equations from given experimental initial rate deduce rate equations from given experimental initial rate datadata recall that reactions with a lar

recall that reactions with a large activation energy will have a ge activation energy will have a small rate constantsmall rate constant students will be expected to be

students will be expected to be familiar with the Arrhenius equation butfamiliar with the Arrhenius equation but notnotto recall itto recall it

understand that many reactions take place in several steps, one of which will be the rate-determining step understand that many reactions take place in several steps, one of which will be the rate-determining step

understand that it is sometimes possible to deduce information regarding the mechanism of a chemical reaction from kinetic data understand that it is sometimes possible to deduce information regarding the mechanism of a chemical reaction from kinetic data understand that many reactions proceed through a

understand that many reactions proceed through a transition statetransition state select and describe a suitable experimental technique for fol

select and describe a suitable experimental technique for following a given reactionlowing a given reaction present and interpret the results of kinetic measurements in graphical form

present and interpret the results of kinetic measurements in graphical form define the term

define the termhalf-lifehalf-lifeand recall that this is constant for any given first-order reaction.and recall that this is constant for any given first-order reaction. questions requiring a knowledge of the

questions requiring a knowledge of the products of the radioactive decay willproducts of the radioactive decay will notnotbe askedbe asked Rate equations

Rate equations A + B

A + BProducts Products If If the the rate rate of of reaction reaction depends depends upon upon the the conc. conc. of of A A and and B: B: Reaction Reaction Rate Rate = = k[A]k[A]mm[B][B]nn (m and n(m and n are 0, 1 or 2)

are 0, 1 or 2)

Rate constant and order of reaction Rate constant and order of reaction If the reaction Rate = k[A]

If the reaction Rate = k[A]mm[B][B]nn then reaction is of order m with respect to A and of order n with then reaction is of order m with respect to A and of order n with respect to B. The overall order of respect to B. The overall order of reaction is (m+n). The proportionality constant k is called the rate constant for the reaction. The rate constant and order o

reaction is (m+n). The proportionality constant k is called the rate constant for the reaction. The rate constant and order o f reactionf reaction are experimentally determined.

are experimentally determined.

Rate equations from experimental data Rate equations from experimental data Rate equations are of the form rate = k[A]

Rate equations are of the form rate = k[A]mmwhere k is a proportionality constant. A graph of rate of reaction againstwhere k is a proportionality constant. A graph of rate of reaction against [concentration]

[concentration]mmis plotted and the gradient of the graph wis plotted and the gradient of the graph will give you the constant of the reaction kill give you the constant of the reaction k. . A data table may yield a rateA data table may yield a rate equation.

equation. E.g.

E.g. A reacts with A reacts with B to form C. B to form C. From the table beFrom the table below find the rate elow find the rate equation and calculaquation and calculate the rate conste the rate constant.tant. experiment

experiment [A]/mol[A]/mol dm dm-3-3 [B]/mol [B]/mol dm dm-3-3

initial rate/ mol initial rate/ mol dm dm-3-3ss-1-1 1 1 1.00 1.00 1.00 1.00 4.004.00 2 2 2.00 2.00 1.00 1.00 8.008.00 3 3 1.00 1.00 2.00 2.00 16.016.0

In experiment 1 and 2 doubling [A] multiplies rate by 2 so rate proportional to In experiment 1 and 2 doubling [A] multiplies rate by 2 so rate proportional to [A][A] In experiment 1 and 3 doubling [B] multiplies rate by 2

In experiment 1 and 3 doubling [B] multiplies rate by 222so rate proportional to [B]so rate proportional to [B]22 so rate = k[A][B]

so rate = k[A][B]22.. k = [A][B]

k = [A][B]22 /rate /rate = 1.00mol dm= 1.00mol dm-3-3* (1.00mol dm* (1.00mol dm-3-3))22/4.00mol dm/4.00mol dm-3-3ss-1-1 k = 0.250 mol dm

k = 0.250 mol dm-3-3ss

Activation energy and rate constant Activation energy and rate constant Some bonds in a molecule must break

Some bonds in a molecule must break before it can react and form new before it can react and form new bonds. bonds. Energy is needed to break these bonds Energy is needed to break these bonds is called theis called the activation energy.

activation energy. Reactant molecules mReactant molecules must be given ust be given enough energy to pass enough energy to pass the activation energy the activation energy barrier to react. barrier to react. The activationThe activation energy and the rate constant are

energy and the rate constant are linked by the Arrhenius equation.linked by the Arrhenius equation. k=Ae

k=Ae-Eact/RT-Eact/RT

where k=rate cons

where k=rate constant, tant, e = the base e = the base of natural logarithms, of natural logarithms, A is a conA is a constant for any given stant for any given reaction, Eact = the activation ereaction, Eact = the activation energy, nergy, RR = the gas c

= the gas constant, onstant, T = the tempeT = the temperature in K.rature in K. The Arrhenius equation shows that the r

The Arrhenius equation shows that the r ate constant (k) decreases if the activation ate constant (k) decreases if the activation energy (Eact) increases.energy (Eact) increases. A reaction will have a small rate constant if it has a large activation energy.

A reaction will have a small rate constant if it has a large activation energy. The activation energy for a

The activation energy for a reaction can be calculated as follows.reaction can be calculated as follows. ln k = ln Ae ln k = ln Ae-Eact/RT-Eact/RT ln k = ln A + ln e ln k = ln A + ln e-Eact/RT-Eact/RT ln k = ln A - Eact/RT ln k = ln A - Eact/RT log k=log A - Eact/2.3RT log k=log A - Eact/2.3RT log k=log A - Eact/2.3R * 1/T log k=log A - Eact/2.3R * 1/T

If k is calculated for different values of T the

If k is calculated for different values of T the n a plot of log k against 1/T gives a line of n a plot of log k against 1/T gives a line of gradient = - Eact/2.3R.gradient = - Eact/2.3R. The rate determining step in a reaction

The rate determining step in a reaction

Reactions often occur in several steps, one of which, the rate determining slow step, is likely to control the overall rate o

Reactions often occur in several steps, one of which, the rate determining slow step, is likely to control the overall rate o f reaction.f reaction. e.g. for an S

e.g. for an SNN1 reaction two steps are 1 reaction two steps are involvedinvolved

RX ---> R

RX ---> R+++ X+ X-- step step 1 1 slowslow R

The

The rate rate depends depends on on the the slow slow step step 1. 1. rate rate = = k[RX] k[RX] first first orderorder E.G. RX=(CH

E.G. RX=(CH33))33CBrCBr

For an S

For an SNN2 reaction there is a rate determining slow step involving two species2 reaction there is a rate determining slow step involving two species

RX + OH

RX + OH---> HO--R--X---> HO--R--X rate = k[RX][OH

rate = k[RX][OH--] ] second second orderorder E.G. RX = CH

E.G. RX = CH33BrBr

Mechanisms and kinetic data Mechanisms and kinetic data

The mechanism for a reaction can be proposed with help from kinetic data but some speculation is needed. The mechanism for a reaction can be proposed with help from kinetic data but some speculation is needed. 1. The rate equation gives us information about what reacts in the rate determining step.

1. The rate equation gives us information about what reacts in the rate determining step. 2. Sensible products must be suggested for

2. Sensible products must be suggested for the rate determining step.the rate determining step.

3. If more molecules of reactant remain and more product molecules are still to be formed more s

3. If more molecules of reactant remain and more product molecules are still to be formed more s teps must be proposed.teps must be proposed. e.g. What is the mechanism for the following reaction

e.g. What is the mechanism for the following reaction CH

CH33CHCH22I I + + NHNH33---> CH---> CH33CHCH22NHNH22+ HI+ HI

if

if rate rate = k= k[CH[CH33CHCH22I]I]

CH

CH33CHCH22I is a halogenoalkane so likI is a halogenoalkane so likely to take part in a nucleophilic suely to take part in a nucleophilic substitution. bstitution. NHNH33is a nucleophile because of its lone pair of is a nucleophile because of its lone pair of

electrons on the nitrogen atom.

electrons on the nitrogen atom. The rate equation shows thThe rate equation shows that it is first order so the slow step in the mechat it is first order so the slow step in the mechanism must involve onlyanism must involve only CH CH33CHCH22-I.-I. CH CH33CHCH22---I ---I ---> ---> CHCH33CHCH22 + + + + II

--A nucleophilic attack by ammonia is now possible in a fast step; A nucleophilic attack by ammonia is now possible in a fast step; NH NH33+ CH+ CH33CHCH22 + + ---> CH ---> CH33CHCH22NHNH33 + +

a final fast step might be loss of hydrogen ion; a final fast step might be loss of hydrogen ion; CH

CH33CHCH22HH22N----HN----H++---> CH---> CH33CHCH22NHNH22 + + HH++

For the

For the reaction reaction 2ICl(g) + 2ICl(g) + HH22(g) ----> 2HCl(g) + I(g) ----> 2HCl(g) + I22(g)(g)

Experiments show that rate = k[ICl(g)][H Experiments show that rate = k[ICl(g)][H22(g)](g)]

so the rate determining step involves 1 molecule of ICl and one of H so the rate determining step involves 1 molecule of ICl and one of H22

ICl(g) + H

ICl(g) + H22(g) ----> products(g) ----> products

possible products are HCl because it is a product of the overall reaction and HI because the elements hydrogen and iodine are left possible products are HCl because it is a product of the overall reaction and HI because the elements hydrogen and iodine are left over.

over. ICl(g) + H

ICl(g) + H22(g) ----> (g) ----> HCl(g) HCl(g) + H+ HI(g) I(g) slow slow stepstep

check full equation to see what is unaccounted for one molecule of ICl still has to react so check full equation to see what is unaccounted for one molecule of ICl still has to react so ICl(g) + HI(g) ----> HCl(g) + I

ICl(g) + HI(g) ----> HCl(g) + I22(g) (g) fast fast stepstep

The kinetic data has led to a 2 step mechanism for the reaction. The kinetic data has led to a 2 step mechanism for the reaction. The transition state

The transition state

When particles collide the break

When particles collide the breaking of bonds and the formation of new boning of bonds and the formation of new bonds may take place at the same ds may take place at the same time. time. Half way throughHalf way through the process an

the process an intermediate called the intermediate called the transition state is transition state is formed. formed. The transition state The transition state has more energy has more energy than the the reactathan the the reactants. nts. TheThe transition state has more energy than t

transition state has more energy than t he products. e.g. for the he products. e.g. for the reaction between a hydrogen molecule and a reaction between a hydrogen molecule and a chlorine radical.chlorine radical. H-H + Cl

H-H + Cl.. <=> <=> H- H- - - H H - - - Cl Cl ----> ----> HH..+ H-Cl+ H-Cl reactants

reactants energy energy = E = E transition transition state state energy energy > E > E products products energy energy < E< E for exothermic reaction

for exothermic reaction

Experimental techniques for following reactions Experimental techniques for following reactions Reactions can be followed by

Reactions can be followed by

* sampling, quenching and titrating (suitable for

* sampling, quenching and titrating (suitable for acid base reactions)acid base reactions) * measuring gas volumes (suitable for

* measuring gas volumes (suitable for reactions involving a change in gas volume)reactions involving a change in gas volume) * polarimetry (suitable for r

* polarimetry (suitable for reactions involving optically active substances)eactions involving optically active substances) * measuring conductivity (suitable for reactions producing o

* measuring conductivity (suitable for reactions producing o r consuming ions)r consuming ions) * colorimetry (suitable for

* colorimetry (suitable for reactions involving coloured substances)reactions involving coloured substances) * dilatometry (suitable for reactions involving r

* dilatometry (suitable for reactions involving r eactions in which liquids change volumes)eactions in which liquids change volumes) Presenting and interpreting kinetics graphs

Presenting and interpreting kinetics graphs

From a graph of concentration against time the initial slope of the

From a graph of concentration against time the initial slope of the graph is the initial rate of reaction.graph is the initial rate of reaction. A graph of rate against concentration will be a straight line through the origin if the rate is pr

A graph of rate against concentration will be a straight line through the origin if the rate is pr oportional to the concentration.oportional to the concentration. Half-life

Half-life

The time taken for the reaction to go to half completion is called the half

The time taken for the reaction to go to half completion is called the half -life of the reaction t-life of the reaction t1/21/2. The half life of a 1st ord. The half life of a 1st orderer

reaction is independent of the initia

reaction is independent of the initial concentration. l concentration. 1st order reactions have a 1st order reactions have a constant half life. tconstant half life. t1/21/2= 0.69/k = 0.69/k where k where k = the rate= the rate

constant for the reaction constant for the reaction 3.

concentration of reactants/products with time concentration of reactants/products with time

Sum of the powers to which theSum of the powers to which the

concentrations are raised in the rate equation concentrations are raised in the rate equation

OR number of species involved in (up to OR number of species involved in (up to and including) the rate determining step

and including) the rate determining step

OR sum of partial orders if illustrated OR sum of partial orders if illustrated with a general rate equation

with a general rate equation (b)

(b) gases gases A A and and B B react react according according to to the the equation equation A A + + 3B3B ABAB33

(i)

(i) State State the the order order of of reaction reaction with with respect respect to to each each of of the the reactants reactants ExptExpt 1 + 3: double [A], doubles rate so order

1 + 3: double [A], doubles rate so order 11

Expt Expt 1 + 2: double [B], four x rate so order

1 + 2: double [B], four x rate so order 22

OR

OR _{Double [A]}_{Double [A]}
keeping [B] constant doubles rate so order

keeping [B] constant doubles rate so order 11

Double [B] Double [B] keeping [A] constant four x rate so order

keeping [A] constant four x rate so order 22

(ii) Write the rate equation for the reaction between A and B. (ii) Write the rate equation for the reaction between A and B.

Rate = k [A] [B] Rate = k [A] [B]22 (iii) Use the experimental data from

(iii) Use the experimental data fromExperiment 1Experiment 1to calculate the rate constant, including to calculate the rate constant, including unitsunits

= =

(iv) Suggest a possible mechanism for the reaction between A and B, leading to the formation of AB

(iv) Suggest a possible mechanism for the reaction between A and B, leading to the formation of AB33. Identify the rate-. Identify the rate-determining step. determining step. OR OR OR OR

(c) The rate constant,

(c) The rate constant, k k , for the reaction in (b) was, for the reaction in (b) was measured at different temperatures.

measured at different temperatures. Plot a graph of log

Plot a graph of log1010k k against 1/Tagainst 1/T

(ii)

(ii) The The Arrhenius Arrhenius equation equation can can be be written written wherewhere

Calculate the gradient of the

Calculate the gradient of the graph and hence calculate the value of graph and hence calculate the value of activation energy,activation energy, E E aa..

=

=--5750 5750 (K) (K) EEaa==

(+)5750 x 2.30 x 8.31= (+)110 kJ mol (+)5750 x 2.30 x 8.31= (+)110 kJ mol-1-1

2.

2.Two gases,Two gases,AAandandBB, react according to the equation, react according to the equation AA(g) + 2(g) + 2BB(g)(g)→→ABAB22(g)(g)

A series of kinetics experiments performed at

A series of kinetics experiments performed at constant temperature gave the following results:constant temperature gave the following results:

(a) (i) Calculate, showing your working, the order of reaction with respect to

(a) (i) Calculate, showing your working, the order of reaction with respect to AAand to B.and to B.

Expt Expt 1 1 + + 2: 2: as as [B] [B] doubles doubles rate rate x4 x4 so so second second order order (wrt (wrt B) B) OR OR As As [B] [B] doubles doubles withwith

[A] constant rate x4 so second order (wrt B) [A] constant rate x4 so second order (wrt B)

Expt Expt 1 1 + + 3: 3: as as [A] [A] doubles doubles rate rate x2 x2 so so first first order order (wrt (wrt A) A) OR OR As As [A] [A] doubles doubles withwith

[B] constant rate x2 so first order (wrt A) [B] constant rate x2 so first order (wrt A) (ii) Write the rate equation for the reaction. (ii) Write the rate equation for the reaction. rate = k [A] [B] rate = k [A] [B]22 (iii) Calculate the rate constant,

(iii) Calculate the rate constant, k k , for the reaction in, for the reaction inexperiment 3experiment 3, , stating stating its its units. units. k k == 0.000195/0.1x0.1

0.000195/0.1x0.122 = 0.195mol= 0.195mol-2-2dmdm66ss-1-1

(b) (i) Explain, in terms of collision theory, why the rate of r

(b) (i) Explain, in terms of collision theory, why the rate of r eaction increases with an increase in temperature.eaction increases with an increase in temperature.

Increasing T means molecules have/collide with greater Increasing T means molecules have/collide with greater energyenergy

so a greaterso a greater proportionproportion /more /moreof of the molecules collide with/have E > Ethe molecules collide with/have E > Eaa /the activation /the activation

energy energy

so a greater proportionso a greater proportion of the collisionsof the collisionsare successful OR moreare successful OR moreof the collisionsof the collisionsareare

successful/more successful collisions in a given time successful/more successful collisions in a given time

(ii) Suggest, with an explanation, the least number of steps which is

(ii) Suggest, with an explanation, the least number of steps which is likely likely to exist in the mechanism of the reaction betweento exist in the mechanism of the reaction between AA and

andBB..

(at least) two steps(at least) two steps

three particles is unlikely three particles is unlikely

OR valid mechanism e.g. OR valid mechanism e.g. A+B

A+B →→ AB AB fast fast AB AB + + BB →→ ABAB22slowslow

OR OR A+B

A+B →→ AB slow AB + BAB slow AB + B →→ABAB22fastfast

(c) The variation of the rate constant,

(c) The variation of the rate constant,k k , with change in temperature is given by the, with change in temperature is given by the

Arrhenius

Arrhenius equation: equation: wherewhereAAis a constant.is a constant. In a series of experiments performed at

In a series of experiments performed at various temperaturesvarious temperatures T T to determine the rateto determine the rate constant,

constant,k k , for the decomposition of a gas, for the decomposition of a gas XX, a graph of ln, a graph of lnk k against 1/ against 1/ T T gave agave a straight line of slope

-straight line of slope -Ea/R Ea/R

Use the graph to calculate the value of the activation energy, in kJ mol

Use the graph to calculate the value of the activation energy, in kJ mol – – 11, for the, for the

decomposition of

decomposition of XX. The value of the gas constant. The value of the gas constant R R = 8.31 J K= 8.31 J K – – 11molmol – – 11..

Value of slope = -1.2Value of slope = -1.2××101044

Multiply by -8.31Multiply by -8.31

Divide by 1000 to give 104 (kJDivide by 1000 to give 104 (kJ

mol mol-1-1))

### 4.

### 4.

2-bromo-2-methylbutane reacts with aqueous sodium hydroxide in a 2-bromo-2-methylbutane reacts with aqueous sodium hydroxide in a substitution reactionsubstitution reaction(a) The rate of reaction can be followed by measuring the concentration of 2-bro

(a) The rate of reaction can be followed by measuring the concentration of 2-bro mo-2-methylbutane at various times.mo-2-methylbutane at various times. In one such experiment, a known amount of 2-bromo-2-methylbutane was added to a

In one such experiment, a known amount of 2-bromo-2-methylbutane was added to a largelargeexcess of aqueous sodium hydroxide.excess of aqueous sodium hydroxide. The following results were obtained

The following results were obtained

(i) Plot a graph of the concentration of 2-bromo-2-methylbutane on the

(ii) Show TWO successive half-life measurements on your graph and write their values below.

(ii) Show TWO successive half-life measurements on your graph and write their values below. 11ststhalf life 15min (±half life 15min (±

1min)

1min) 22ndnd half life 15min (± 1min)half life 15min (± 1min)

(iii) What is the order of reaction with respect to 2-bro

(iii) What is the order of reaction with respect to 2-bro mo-2-methylbutane? Give a reason for your answer.mo-2-methylbutane? Give a reason for your answer. 1

1ststorder torder t

½

½is constantis constant

(b) When the reaction is repeated using equal concentrations of 2-bromo-2-methylbutane and aqueous sodium hydroxide, the (b) When the reaction is repeated using equal concentrations of 2-bromo-2-methylbutane and aqueous sodium hydroxide, the same results are obtained.

same results are obtained.

(i) What is the order of reaction with respect to hydroxide ions? (i) What is the order of reaction with respect to hydroxide ions?

Zero Zero (ii) Write the rate equation for the reaction. (ii) Write the rate equation for the reaction.

Rate = k [2-bromo-2-methylbutane] Rate = k [2-bromo-2-methylbutane]

(iii) Write a mechanism for the reaction which is consistent with your rate equation (iii) Write a mechanism for the reaction which is consistent with your rate equation

(c) The reaction between 2-bromobutane, C

(c) The reaction between 2-bromobutane, C_{2}_{2}HH_{5}_{5}CHBrCHCHBrCH_{3}_{3}and NaOH(aq) proceeds by the same mechanism as in (b)(iii).and NaOH(aq) proceeds by the same mechanism as in (b)(iii).
Use the mechanism to explain why the reaction of a single optical isomer of 2-bromobutane produces a mixture that is no

Use the mechanism to explain why the reaction of a single optical isomer of 2-bromobutane produces a mixture that is no longerlonger optically active.

optically active.

The The intermediate intermediate carbocation carbocation isis

planar planar

Equal attack from either sideEqual attack from either side

Therefore racemic mixture produced Therefore racemic mixture produced 4.

4.(a) (a) Define Define the the terms terms (i) (i) Overall Overall order order of of reactionreaction

sum of the powers to which thesum of the powers to which the concentrationconcentration(terms) are raised in the rate equation(terms) are raised in the rate equation / number of species involved up to an

(ii) Rate constant

(ii) Rate constant constant (of proportionality) in the rate equation / numerically = rateconstant (of proportionality) in the rate equation / numerically = rate when all concs 1 mol dm

when all concs 1 mol dm-3-3 /correct example /correct example

((bb) ) IIn n a a kkiinneettiic c ssttuuddy y oof f tthhe e rreeaaccttiioonn AAt t a a cceerrttaaiin n tteemmppeerraattuurre e tthhe e ffoolllloowwiinng g ddaatta a wweerre e oobbttaaiinneedd::

(i) State the order of reaction with respect to

(i) State the order of reaction with respect to CHCH33I and with respect to OHI and with respect to OH – – ions. With reasons.ions. With reasons.

Both orders 1Both orders 1 Double concentration of one while other is Double concentration of one while other is constant and the rate doublesconstant and the rate doubles ORORreferrefer

to two specific experiments to two specific experiments

(ii) Write the rate equation for the reaction. (ii) Write the rate equation for the reaction.

rate =
rate =kk_{[CH}_{[CH}
3
3I][OHI][OH
--]]

(iii) Calculate the value of the rate constant for this reaction, stating its units. (iii) Calculate the value of the rate constant for this reaction, stating its units. e.g.

e.g.kk_{= rate/[CH}_{= rate/[CH}
3
3I][OHI][OH
--]
] soso kk_{= 1 x 10}_{= 1 x 10} ––33_{mol}_{mol}––11_{dm}_{dm}33_{ss} -1-1
(c)

(c) The The reaction reaction CH3CH2Br CH3CH2Br + + OHOH – – CH3CH2OH + BrCH3CH2OH + Br – – has an SN2 mechanism that proceeds through a has an SN2 mechanism that proceeds through a transition state.transition state. (i) Draw the mechanism, showing the

(i) Draw the mechanism, showing the structure of the transition state.structure of the transition state.

(ii) Draw a reaction profile for this exothermic reaction. Show the energy level of the transition state o

(ii) Draw a reaction profile for this exothermic reaction. Show the energy level of the transition state o n the profilen the profile

1.

1.(a) Consider the following table, which shows data for the reaction between reactants A and B.(a) Consider the following table, which shows data for the reaction between reactants A and B.

(i)

(i) Define Define the the term term order order of of reaction. reaction. The The sum sum of of the the powers powers to to which which thethe concentrations are raised in the rate equation

concentrations are raised in the rate equation

(ii) Determine, giving reasons, the orders of reaction with respect to A and B.

(ii) Determine, giving reasons, the orders of reaction with respect to A and B. Hence write the rate equation for the reactionHence write the rate equation for the reaction 1st order because rate halves as [A] halves in expt. 1 → 2 1st order because rate halves as [A] halves in expt. 1 → 2 or [B] constant

or [B] constant

2nd order because rate quadruples /

2nd order because rate quadruples / increases by 2increases by 222as [B]as [B] doubles in expt. 2 → 3 or [A] constant

doubles in expt. 2 → 3 or [A] constant

(iii) Calculate a value for the rate constant and give its units. (iii) Calculate a value for the rate constant and give its units.

k

k = 0.008mol= 0.008mol-2-2dmdm66 ss-1-1 (iv) State how, if at all, the value of the rate

(iv) State how, if at all, the value of the rate constant would change if the temperature were increased.constant would change if the temperature were increased. (k) increases

(k) increases

(b) (i) Draw a Maxwell-Boltzmann curve for a sample of a gas (b) (i) Draw a Maxwell-Boltzmann curve for a sample of a gas

(iii) The rate of a reaction can also be increased by raising the temperature.

(iii) The rate of a reaction can also be increased by raising the temperature. Describe how the Maxwell-Boltzmann curve at aDescribe how the Maxwell-Boltzmann curve at a higher temperature differs from the curve you have drawn in (i).

higher temperature differs from the curve you have drawn in (i).

more ) to the right

more ) to the right Peak lowerPeak lower

(iv) Transition metals are important industria

(iv) Transition metals are important industrial catalysts. l catalysts. Identify an industrial process involving a transition mIdentify an industrial process involving a transition metal catalyst andetal catalyst and name the catalyst used.

name the catalyst used.

Explain why many transition metals and

Explain why many transition metals and their compounds are successful catalysts.their compounds are successful catalysts. Manufacture

Manufacture of of ammonia ammonia IronIron OR

OR Hydrogenation Hydrogenation of of oils oils Ni/Pt/PdNi/Pt/Pd OR

OR Manufacture Manufacture of of HH_{2}_{2}from CHfrom CH_{4}_{4} NiNi
Explanation: uses d orbitals to

Explanation: uses d orbitals to bond with reactants( at active sites)bond with reactants( at active sites) (v)

(v) How How do do the the rate rate constants constants for for the the catalysed catalysed and and uncatalysed uncatalysed reactions reactions compare? compare? Catalysed Catalysed k k bigger/Higher