unit 45 an introduction to industrial chemistry -...

Industrial Chemistry

Unit 45 An introduction to industrial chemistry

Unit 46 Factors affecting the rate of a reaction

Unit 47 Industrial processes

Unit 48 Green chemistry

Topic14

KeyC o ncepts

Factors affecting the rate of a reaction

• Activation energy• Maxwell-Boltzmann distribution curve• Arrhenius equation• Characteristics of catalysts

Green chemistry• Sustainable development• Principles of green chemistry• Atom economy• Practice of green chemistryprinciples

in industry

Industrial Chemistry

An introduction to industrial chemistry

• Importance of chemicalprocesses inindustry

• Rate equation• Order of reaction

Industrial processes• Conversion of ammonia to fertilizers• Chloro-alkali industry• Conversion ofmethane tomethanol

Topic 14� Industrial Chemistry �Unit 45 An introduction to industrial chemistry

45.1 – 45.9

Summary

1 Acatalyst is a substancewhichalters the rateof reactionwithout itselfundergoinganypermanent chemical changes.

2 For a general reactionof the type:

aA+bB products

the rateof equation takes the form

Rate =k[A]x[B]y

wherek is the rate constant;

x is theorderof reactionwith respect to the reactantA;

y is theorderof reactionwith respect to the reactantB.

Theoverall orderof reaction isx +y.

3 The following diagrams show how the rate varies with reactant concentration forzero, first and secondorder reactions.

zero order reaction rate = k

Concentration of reactant(mol dm–3)

Rat

e of

rea

ctio

n second order reaction rate = k[reactant]2


Rat

e of

rea

ctio

n

first order reaction rate = k[reactant]


Rat

e of

rea

ctio

n

4 The following diagrams show how the reactant concentration varies with time forzero, first and secondorder reactions.

zero order reaction rate = k

Time

Con

cen

trat

ion

of

reac

tan

t

first order reaction rate = k[reactant]

Time

Con

cen

trat

ion

of

reac

tan

t

second order reaction rate = k[reactant]2

Time

Con

cen

trat

ion

of

reac

tan

t

45.1 What is chemical industry?

45.2 Products from the chemical industry

45.3 The operationof a chemicalprocess in industry

45.4 Petrochemical industry

45.5 Production of vitaminC

45.6 The useof catalysts

45.7 The effect of change in concentration on the rate of areaction

45.8 Order of reaction

45.9 Units of rate constant k for reactionsofdifferentorder

45.10 Experimentaldeterminationoftherateequationforareaction– themethodof initial rate

45.11 Usingtheiodineclockmethodtodeterminetherateequationfor a reaction

Unit 45 An introduction to industrial chemistry


Example

The following reaction takes place at 750°C in a close container with a fixedvolume.

2H2(g) + 2NO(g) 2H2O(g) +N2(g)

a) SuggestanexperimentalmethodthatcanbeusedtofollowtheconcentrationofN2(g)in the reactionmixture. Briefly explain theprincipleof your suggestedmethod. (3marks)

b)The tablebelow lists three sets of experimental dataof the reaction at 750°C.

ExperimentInitial concentration (mol dm–3) Initial rate of formation

of N2(g) (mol dm–3 s–1)[H2(g)] [NO(g)]

1 0.0100 0.0250 0.500

2 0.00500 0.0250 0.250

3 0.0300 0.0125 0.375

Deduce the rate equation for the reaction, and calculate its rate constant at750°C. (5marks)

Answer

a) Monitor thepressureof the reactionmixture in a closedvessel. (1)

4molesofgasreacttogive3molesofgas.Thus,thepressureofthereactionmixturedecreases as the reactionproceeds. (1)

(P0 –Pt) is proportional to thenumberofmolesofN2(g) formed. (1)

b)The rate equation for the reaction canbe expressed as

rate =k[H2(g)]x[NO(g)]y

whenk is the rate constant.

Fromexperiments 1 and2,

0.500=k(0.0100)x(0.0250)y

0.250=k(0.00500)x(0.0250)y

0.5000.250

= ( 0.01000.00500 )x (0.5)

x= 1 (0.5)

∴ the reaction is first orderwith respect toH2(g).

Fromexperiments 1 and3,

0.500=k(0.0100)(0.0250)y

0.375=k(0.0300)(0.0125)y

0.5000.375

= ( 0.01000.0300 )( 0.0250

0.0125 )y (0.5)

y= 2 (0.5)

∴ the reaction is secondorderwith respect toNO(g).

ExamtipsExamtipsExamtipsExamtips ♦ Questions often give the initial reactant concentrations and initial rateof a reaction and ask students to derive the rate equation for thereaction.

♦ To deduce the order of a certain reaction with respect to one ofthe reagents, only the concentration of that reagent should vary.Concentrations of other reagents can be made much higher so thattheir concentrations canbe considered as constant.

♦ Consider an experiment used to study the kinetics of the followingacid-catalyzed reaction:

I2(aq)+CH3COCH3(aq)H+(aq)

CH3COCH2I(aq)+H+(aq)+ I–(aq)

Excess CH3COCH3(aq) and H2SO4(aq) were used in the experiment.Samplesofthereactionmixturewerewithdrawnatregulartimeintervalsand titratedagainst standardNa2S2O3(aq).Thegraphbelowshows theplot of the titre against time.

14

0

15

16

17

18

19

20

0 5 10 15 20 25 30 35

Time (min)

Titr

e (c

m3 )

The rate equation for the reaction canbe expressed as

rate =k[I2(aq)]x[CH3COCH3(aq)]y[H+(aq)]2

AsexcessCH3COCH3(aq)andH2SO4(aq)wereused,theirconcentrationscanbe considered as constant.

The rate equation canbe simplified as

rate =k’[I2(aq)]x

The titre decreases with time linearly, i.e. the rate of reaction isindependent of I2(aq). Thus, the reaction is zero order with respect toI2(aq).

(–Slope) of the line represents the rate of the reaction.


3 Todeterminetheinitialrateofthereactionbetweenperoxodisulphateionandiodideion:

S2O82–(aq) + 2I–(aq) 2SO4

2–(aq) + I2(aq)

add a small, fixed amount of thiosulphate ion and starch solution to the reactionmixture at the start. Measure the time for the reaction mixture to turn dark blue(the iodine clockmethod).

Initial rate∝ 1time for adarkblue colour to appear

ExamtipsExamtipsExamtipsExamtips ♦ TodeducetheorderofthereactionbetweenNa2S2O3(aq)andH2SO4(aq)with respect toNa2S2O3(aq),different volumesofNa2S2O3(aq)areusedtoprepare reactionmixturesof equal volume. Thus, the concentrationof Na2S2O3(aq) in each reaction mixture is directly proportional to thevolume ofNa2S2O3(aq) used.

Reaction mixture

Volume used (cm3)

1.0 mol dm–3

H2SO4(aq)H2O(l)

0.040 mol dm–3

Na2S2O3(aq)

1 10.0 25.0 5.0

2 10.0 20.0 10.0

3 10.0 15.0 15.0

4 10.0 10.0 20.0

Notice that different volumes of water are used in different reactionmixture tomake the total volume of all reaction mixtures equal.

Example

In an experiment to study the kinetics of the reaction:

2I–(aq) + S2O82–(aq) I2(aq) + 2SO4

2–(aq)

the time (t) for the formationofaverysmallbut fixedamountof I2 fromdifferentmixtures of KI(aq) and K2S2O8(aq) was measured. The results of three experiments arelistedbelow:

ExperimentInitial concentration (mol dm–3)

t (s)[S2O8

2–(aq)] [I–(aq)]

1 0.100 0.100 136

2 0.200 0.100 68

3 0.200 0.200 34

a) Describe and explainhow time (t) canbe found experimentally. (3marks)

b)DeducefromtheaboveinformationtherateequationforthereactionofI–(aq)withS2O8

2–(aq). (3marks)

The rate equation for the reaction is rate =k[H2(g)][NO(g)]2 (1)

Use the results of experiment1,

0.500moldm–3s–1 =k(0.0100moldm–3)(0.0250moldm–3)2

k = 8.00 x 104 (1) dm6mol–2s–1 (1)

➤Questions may give the units of the rate constant of a reaction and askstudents to deduce theoverall order of the reaction.

➤Students shouldbe familiar with theunits of rate constant.

Overall order of reaction

Expression for rate constant k

Units of rate constant k

0 k = rate moldm–3s–1

1 k =rate

concentrations–1

2 k =rate

concentration2 dm3mol–1s–1

Studentsmaybeaskedtocalculatethevalueofkfromgivendata.Rememberto state also theunits.

RemarksRemarks*

45.10 – 45.11

Summary

1 The initial rateof reaction is the rateof reaction at the start.

2 To determine the initial rate of the reaction between sodium thiosulphate solutionanddilute sulphuric acid:

Na2S2O3(aq) +H2SO4(aq) Na2SO4(aq) + SO2(g) +H2O(l) + S(s)

measure the time to reach theopaque stage.

Initial rate∝ 1time to reach theopaque stage

Na2S2O3(aq) + H2SO4(aq)

white paper

cross

beaker

Topic 1410 Industrial Chemistry 11Unit 46 Factors affecting the rate of a reaction

Answer

a) Add a small, fixed amount of sodium thiosulphate solution and starch solution tothe reaction mixture at the start. Record the time for a dark blue colour to appear(iodine clockmethod). (1)

The thiosulphate ion consumes any iodine formed from the reaction. (1)

Freeiodineinthereactionmixturegivesadarkbluecomplexwiththestarchwhenall the thiosulphate ionhasbeenusedup. (1)

b)The rate equation for the reaction canbe expressed as

rate = k[S2O82–(aq)]x[I–(aq)]y

wherek is the rate constant.

From experiments 1 and 2, the time t becomes half when the concentration ofS2O8

2–(aq) ion is doubled. (0.5)

Thus, the reaction is first orderwith respect to S2O82–(aq) ion. (0.5)

Fromexperiments2and3,thetimetbecomeshalfwhentheconcentrationofI–(aq)ion is doubled. (0.5)

Thus, the reaction is first orderwith respect to I–(aq) ion. (0.5)

The rate equation for the reaction is rate =k[S2O82–(aq)][I–(aq)] (1)

➤Students shouldbe familiar with the ‘iodine clock experiment’.

➤Besidesmeasuringtimet,theprogressofthereactioncanalsobefollowedby the two methods below:

– using a colorimeter;

– titration of the iodine formed with standard sodium thiosulphatesolution.

RemarksRemarks*

46.1 The effect of change in temperature on the rate of areaction

46.2 A further look at the collision theory

46.3 The energydistribution curve

46.4 Rate equation and reactionmechanism

46.5 The Arrhenius equation

46.6 Catalysts

46.7 How does a catalystwork

46.8 Heterogeneous catalysis

46.9 Homogeneous catalysis

46.10 Enzyme catalysis

Factors affecting the rate of a reactionUnit 46

Topic 1412 Industrial Chemistry 13Unit 46 Factors affecting the rate of a reaction

46.1 – 46.5

Summary

1 Activationenergy is theminimumkineticenergy requiredbycollidingparticles fora reaction tooccur.

2 The following diagram shows the energy profile for an exothermic reaction A–B +C A+B–C.

∆H = enthalpy change of reaction

Reaction coordinate

Pote

nti

al e

ner

gy

A + B–C

A–B + C

EA = activation energy for the forward reaction

activated complexA----B----C

3 Thedistributionof kinetic energy amongmolecules in a gas is called theMaxwell-Boltzmanndistributionofenergies.Thecurvebelowshowsthenumberofmoleculesat each energyvalue at temperatureT.

Kinetic energy E

Nu

mbe

r of

mol

ecu

les

wit

h k

inet

ic e

ner

gy E T

4 A reactiongoes faster at ahigher temperaturebecause

• the reactantparticleshavemore kinetic energy and collidemoreoften;

• alargerportionofthereactantparticleshaveenergyequaltoorgreaterthantheactivation energy and thus can reactupon collision.

5 a) TheArrhenius equation is

k =Ae– EA

RT

where k is the rate constantof the reaction; A is a constantwhich is independentof temperature; e is thebaseof thenatural logarithm; EA is the activation energy for the reaction in Jmol–1; R is the ideal gas constant (i.e. 8.31JK–1mol–1); and T is the temperature inKelvin.

b)logk = logA – EA

2.3R x 1

T

So, a graph of log k against 1T

will give a straight line of slope – EA

2.3R. We

can calculate the valueof the activation energy,EA, from the slope.

logk

1T

EA

2.3Rslope = –

ExamtipsExamtipsExamtipsExamtips ♦ The x-axis of an energy profile is reaction coordinate, NOT ‘reactioncoordination’ or ‘time’.

✘ ✘

♦ DoNOT confuse rate constants with equilibrium constants.

e.g.

Consider the following system which comprises two single stepreactions:

E(g) + F(g)k1

k–1 2G(g)

(k1 andk–1 are rate constants.)

Rate equation for the forward reaction =k[E(g)][F(g)]

Rate equation for thebackward reaction =k–1[G(g)]2

Equilibrium constantKc =[G(g)]2

[E(g)][F(g)]

♦ The following table summarizes the effect of various changes on achemical equilibrium system.

aA(g)+bB(g)k1

k–1 cC(g)+dD(g)

Kc =[C(g)]c[D(g)]d

[A(g)]a[B(g)]b

Topic 141� Industrial Chemistry 1�Unit 46 Factors affecting the rate of a reaction

Action on equilibrium

system

Rates of forward

and backward reactions

Rate constants

Attainment of

equilibrium

Shift of position of equilibrium

Value of Kc

Increasing theconcentration of

A(g) or B(g)

rate offorwardreactionincreases

no change faster to the right no change

Decreasing theconcentration of

A(g) or B(g)

rate offorwardreaction

decreases

no change slower to the left no change

Increasein

pressure(decrease

involume)

a +b>

c +d

rates of bothreactions

increase, butthe forwardreaction rate

increasesmore

no change faster

to the right

no changea +b

=c +d

same increasein ratesof bothreactions

no change

a +b<

c +d


increase, butthebackwardreaction rate

increasesmore

to the left

Decreasein

pressure(increase

involume)

a +b>

c +d


decrease, butthe forwardreaction rate

decreasesmore

no change slower

to the left

no changea +b

=c +d

samedecrease in


no change

a +b<

c +d


decrease, butthebackwardreaction rate

decreasesmore

to the right

Action on equilibrium

system

Rates of forward

and backward reactions

Rate constants

Attainment of

equilibrium

Shift of position of equilibrium

Value of Kc

Increase intemperature

rates of bothforward and

backwardreactions

increase, butnot to the

same extent

bothk1andk–1

increase,but not tothe same

extent

faster

forexothermicreaction,

to left; forendothermicreaction, to

right

forexothermicreaction,Kcdecreases;

forendothermicreaction,Kc

increases

Decrease intemperature


backwardreactions

decrease, butnot to the

same extent

bothk1andk–1

decrease,but not tothe same

extent

slower

forexothermicreaction, toright; for

endothermicreaction, to

left

forexothermicreaction,Kcincreases;

forendothermicreaction,Kcdecreases

Positive catalyst


backwardreactions

increase tothe same

extent

bothk1andk–1

increase,but their

ratiok1

k–1

remainsconstant

faster no change no change

♦ FortheenergydistributioncurvesofmoleculesinagasattemperatureTandT+100,thepeakoftheT+100curve isfurthertotherightoftheT curve, but with a lowermaximum.

Kinetic energy E

Num

ber

of m

olec

ules

w

ith k

inet

ic e

nerg

y E

T + 100

T

Example

Explainwhetherthefollowingexplanationprovidedbyastudentregardingtheincreasein reaction rate is appropriate.

‘Therateofagaseousreactionincreaseswithtemperaturebecausetheaveragekineticenergyof the reactantmolecules increaseswith temperature.’ (3marks)


Answer

The rateof a reactiondependson the effective collision frequency. (1)

When the temperature increases, the reactant molecules have more kinetic energyand collidemoreoften. (1)

A largerportionof the reactantmoleculeshaveenergyequal toorgreater than theactivation energy and thus can reactupon collision. (1)

➤Students should be able to explain why a reaction goes faster at a highertemperature in terms of activation energy.

RemarksRemarks*

46.6 – 46.10

Summary

1 Apositive catalyst is one that speedsupa reactionwhile anegative catalyst is onethat slowsdowna reaction.

2 Acatalystworksbyprovidinganalternativepathwayforthereactiontotakeplace.Apositive catalyst provides an alternativepathwaywith a lower activation energy.

Pote

nti

al e

ner

gy

Reaction coordinate

EA

EA

products

reactants

reaction pathway with a positive catalyst

∆H

uncatalyzedreaction pathway

’

3 Homogeneous catalyst — where the catalyst and the reactants are in the samephase.

Heterogeneous catalyst — where the catalyst and the reactants are in differentphases.

4 Enzymes are biological catalysts. They are usually homogeneous with the reactantsandpresent in the same aqueous solution.

5 Ethanol can be produced by the fermentation of sugars and starches. Enzymes inyeast can convert glucose to ethanol and carbondioxide.

C6H12O6(aq)enzymes in yeast

2C2H5OH(aq) + 2CO2(g)

glucose ethanol carbondioxide

Example

In the upper atmosphere, ozone is converted to oxygen according to the equationshownbelow.This is a single step reaction.

O3 +O 2O2

a) Write the rate equation for this reaction. (1mark)

b)The relationship between reaction rate constant k and absolute temperatureT canbe representedby theArrhenius equation:

k =Ae– EA

RT

where R is the ideal gas constant, A is theArrhenius constant, and EA is the activation energy for the reaction in Jmol–1.

The rate constant of the reaction at 266K is found to be 8 times of that at210K. Calculate the activation energy for the reaction. (Ideal gas constantR = 8.31JK–1mol–1) (2marks)

c) Iffreechlorineradicals(Cl•)arepresentintheupperatmosphere,theywillcatalyzethe conversion of ozone to diatomic oxygen, leading to ozone depletion. The rateequation for theCl• catalyzedozonedeplection is

rate =k[Cl•][O3]

i) Suggest, with reasons, which of the following two possible mechanisms fits therate equation.

Mechanism1 Cl• +O3 ClO•+O2 (fast)

ClO•+O Cl• +O2 (slow)

Cl• +O3 ClO•+O2 (slow)

ClO•+O Cl• +O2 (fast)Mechanism2

(2marks)

ExamtipsExamtipsExamtipsExamtips ♦ Students should be able to spell the terms ‘Haber process’ and‘homogeneous / heterogeneous catalysis’.

♦ To show that the reaction below is catalyzedby Fe2+(aq) ions,

S2O82–(aq)+2I–(aq) 2SO4

2–(aq)+ I2(aq)

it is important to include a control experiment.


ii)Sketch the energyprofiles for theuncatalyzed and catalyzed reactions.(3marks)Po

ten

tial

en

ergy

Reaction coordinate

Uncatalyzed reaction

O3 + O

2O2

Pote

nti

al e

ner

gy

Reaction coordinate

Catalyzed reaction

Cl• + O3 + O

Cl• + 2O2

Answer

a) Rate =k[O3][O] (1)

wherek is the rate constant.

b)k =Ae– EA

RT

log 8kk

= EA

2.3R ( 1

210 – 1

266 ) log8= EA

2.3R ( 1

210 – 1

266 ) (1)

EA = 17.2kJmol–1 (1)

∴ the activation energy for the reaction is 17.2kJmol–1.

c) i) Mechanism2 (1)

Species in the rate equationmatch those in the slow stepof the reaction. (1)

ii)

Pote

nti

al e

ner

gy

Reaction coordinate

Uncatalyzed reaction

O3 + O

2O2

Pote

nti

al e

ner

gy

Reaction coordinate

Catalyzed reaction

Cl• + O3 + O

Cl• + 2O2

(1mark for an energy ‘hump’ (1 mark for two energy ‘humps’ between between reactants andproduct) reactants andproduct; 0.5mark for EA (catalyzed) <EA (uncatalyzed); 0.5mark for theEA of the first step greater than thatof the second step)

➤For part (b), the question may give the values of k, A and T, and askstudents to calculateEA.

➤Students should be able to draw the energy profiles of uncatalyzedand catalyzed reactions. Make sure that EA (catalyzed) is less than EA(uncatalyzed).

➤For the energy profile of the catalyzed reaction in part (c), the activationenergyof the slowstep (first step) shouldbegreater than thatof the faststep (second step).

➤Questionsmayaskstudentstoclassifythecatalyst involvedinareactionasa homogeneous catalyst or a heterogeneous catalyst.

e.g.

In the above reaction, Cl• is a homogeneous catalyst because O3, O andCl• are all in thegaseous phase.

➤Students may need to deduce the rate equation for a reaction based onthemechanism of the reaction.

e.g.

Consider the acid-catalyzed reaction:

X(aq)+Y(aq)H+(aq)

Z(aq)

The following mechanism wasproposed:

X(aq)+H+(aq) XH+(aq) (fast)

XH+(aq)+Y(aq) Z(aq)+H+(aq) (slow)

Basedon this mechanism, deduce the rate equation.

For the first step,

K =[XH+(aq)]

[X(aq)][H+(aq)]

For the second step,

rate=k[XH+(aq)][Y(aq)]

=kK[X(aq)][H+(aq)][Y(aq)]

RemarksRemarks*

Topic 1420 Industrial Chemistry 21Unit 47 Industrial processes

47.1 Building a chemicalplant in a city

47.2 What doplantsneed to growproperly?

47.3 Nitrogenous fertilizers

47.4 Ammonia – the key tonitrogenous fertilizers

47.5 Obtaining an economicyield in theHaberprocess

47.6 From ammonia tonitrogenous fertilizers

47.7 Working on large scaleproductionofnitrogenous fertilizers

47.8 NPK compound fertilizers

47.9 The pros and consof fertilizers

47.10 The chloro-alkali industry

47.11 Manufacture of chlorine

47.12 Uses ofmethanol

47.13 From natural gas tomethanol

47.14 Methanol – a green feedstock?

47.15 Choosing a site for a chemicalplant

Industrial processesUnit 47 47.1 – 47.15

Summary

1 There are threemainwaysof fixingnitrogen from the air:

a) a chemical reaction in the air during lightning flashes;

b)the actionof bacteria;

c) themanufactureof ammoniaby theHaberprocess.

2 Ammonia is thekey tonitrogenous fertilizers.The following flowdiagramoutlinestheHaberprocess for its synthesis.

N2 + H2

N2 + H2

heat exchanger

catalytic converter

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

400 – 450 °C,200 atmospheres

unreactedN2 + H2

to storage tanks

liquid NH3

NH3 + N2 + H2

NH3 + N2 + H2

N2 + H2

purifier and drier

(removal of CO, CO2,

etc.)

hydrogen

nitrogen

compressor

recyclingpump

ammoniacondenser

3 Themanufactureofnitric acid fromammonia involves three stages:

Stage A 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ΔH = –909kJmol–1

900°C, about10 atmospheres, and aplatinum-rhodiumcatalyst

Stage B 2NO(g) +O2(g) 2NO2(g) ΔH = –115kJmol–1

2NO2(g) N2O4(g) ΔH = –58kJmol–1

Low temperatures (about40°C) and7 – 12 atmospheres

Stage C 3N2O4(g) + 2H2O(l) 4HNO3(aq) + 2NO(g) ΔH = –103kJmol–1

coldwater

4 Thechloro-alkaliindustryisbasedontheelectrolysisofconcentratedsodiumchloridesolution.Theelectrolysisprocessgivesthreeimportantindustrialchemicals—chlorinegas,hydrogengas and sodiumhydroxide solution.

5 The threemain typesof cell used for the electrolysis of brine are:

a) mercury electrolytic cell;

b)diaphragmcell;

c) membrane cell.

Topic 1422 Industrial Chemistry 23Unit 47 Industrial processes

ExamtipsExamtipsExamtipsExamtips ♦ Questions often ask about theHaber process.

e.g.

How the feedstocks are obtained.

– Nitrogen canbeobtained by the fractional distillationof liquid air.

– Hydrogen canbeobtained fromnatural gas (methane).

CH4(g) +H2O(g) 3H2(g) +CO(g)

♦ Students should be able to spell the terms ‘bleach’ and ‘sterilizingagent’.

♦ Students need to remember the reaction conditions and chemicalequations for the

– synthesis of ammonia in theHaber process;

– manufacture of nitric acid fromammonia;

– syngas generation frommethane;

– synthesis ofmethanol from syngas.

6 Methanol is produced from syngas (a mixture of primarily carbon monoxide andhydrogen)obtained frommethane (natural gas).

Syngas generation— steam reformingofmethane

CH4(g) +H2O(g)nickel oxide catalyst

730°C,30 atmospheres

CO(g) + 3H2(g)

Methanol synthesis

CO(g) + 2H2(g)

copper and zincoxidecatalyst

300°C,50 – 100 atmospheres

CH3OH(g)

Example

The following diagram shows an electrolytic cell used in the manufacture of sodiumhydroxide.

gas X gas Y

anode cathode

NaCl(aq) of lower concentrationflowing out

membrane

NaCl(aq) of higher concentrationflowing in

NaOH(aq) of higher concentrationflowing out

NaOH(aq) of lower concentrationflowing in

➤Questions often ask about themembrane cell.RemarksRemarks*

a) Write an ionichalf-equation for the anodic reaction. (1mark)

b)Write an ionichalf-equation for the cathodic reaction. (1mark)

c) i) What typeof ions canpass through themembrane? (1mark)

ii)Explainwhy sodiumhydroxide canbeobtained from the electrolysis of brine. (2marks)

Answer

a) 2Cl–(aq) Cl2(g) + 2e– (1)

b)2H+(aq) + 2e– H2(g) (1)

c) i) Sodium ions (1)

ii)As hydrogen ions are discharged at the cathode, water dissociates to replace thehydrogen ions. (1)

Thus,hydroxide ions remain in the solution. (1)

Sodiumhydroxide solution canbeobtained.

Topic 142� Industrial Chemistry 2�Unit 48 Green chemistry

48.1 Sustainable development andgreen chemistry

48.2 Feedstocks from renewable resources

48.3 Atom economy

48.4 Energy efficiency

48.5 Manufacture of acetic acid in industry

48.6 Green chemistry and the life cycleof aproduct

Green chemistryUnit 48 48.1 – 48.6

Summary

1 Sustainable development can be defined as ‘development meeting the needs ofthe present without compromising the ability of future generations to meet theirneeds’.

2 Green chemistry is ‘the utilization of a set of principles that reduce or eliminatethe use or generation of hazardous substances in the design, manufacture and useof chemicalproducts’.

3 principles of green chemistry

• Prevention

• Atomeconomy

• Lesshazardous synthesis

• Designing safer chemicals

• Safer auxiliary substances

• Energy efficiency

• Useof renewable resources

• Reducingderivatives

• Catalysis

• Design fordegradation

• Use real-time analysis forpollutionprevention

• Accidentprevention

4 Percentage atomeconomy

= relativemolecularmassor formulamassofdesiredproducttotal relativemolecularmassor formulamassof reactants

x 100%

ExamtipsExamtipsExamtipsExamtips ♦ TheHaber process is an example of green chemistry because

– the reaction has 100% atomeconomy;

– both the reactants (N2andH2)andproduct (NH3)arenon-toxicandpose no harm to the environment;

– the reaction takes place in gas phase (no solvent is used);

– the product is separated by cooling (no separating agent is used);

– therawmaterial (N2) isavailable in largeamount intheatmosphere(depletion is not a problem);

– the reaction does not require theuse of anyderivatives;

– a catalyst is used in theprocess.

Topic 142� Industrial Chemistry 2�Unit 48 Green chemistry

♦ Questionsoftengiveanunfamiliarreactionandaskstudentstoexplainwhy the reaction is an example of green chemistry.

e.g.

Benzoin can be obtained by heating benzaldehyde with potassiumcyanide in a solvent-free condition.

OH

C

O

C

H

KCNCHO2

The conversion can be considered as an example of green chemistrybecause

– the conversion does not use solvent.

– the conversion has a high percentage atomeconomy.

– a catalyst (KCN) is used.

Example

Compounds X and Y react in the presence of a small amount of NaOH(s) to givecompoundZ.

NaOH(s)withoutsolvent OCH3

OCH3

X Y Z

+ H2O+

OO

OCH3

OCH3

O

H

a) The relative molecular masses of compounds X and Y are 132.0 and 166.0respectively.

Calculate thepercentage atomeconomyof the reaction. (2marks)

(Relative atomicmasses:H=1.0,O=16.0)

b)In a typical experiment, 3.30g of compound X and 4.15g of compound Y yield5.46gof compoundZ.

Calculate thepercentage yieldof compoundZ. (2marks)

c) Give FOUR reasons why this reaction is considered as an example of greenchemistry. (4marks)

Answer

a) Percentage atomeconomy= relativemolecularmassofdesiredproducttotal relativemolecularmassof reactants

x 100%

= 132.0+166.0 – 18.0132.0+166.0

(1)

= 280.0298.0

= 94.0% (1)

b)Numberofmolesof compoundXused = 3.30g132.0gmol–1

= 0.0250mol (0.5)

Numberofmolesof compoundYused = 4.15g166.0gmol–1

= 0.0250mol (0.5)

Numberofmolesof compoundZobtained = 5.46g280.0gmol–1

= 0.0195mol (0.5)

Percentage yieldof compoundZ= 0.0195mol0.0250mol

x 100%

= 78.0% (0.5)

c) Any fourof the following:

• The reactionhas ahighpercentage atomeconomy. (1)

• The reactionhas ahighyield. (1)

• The reactiondoesnotuse solvent. (1)

• A catalyst (NaOH(s)) is used. (1)

• Theotherproduct (H2O(l)) isnon-toxic. (1)

➤Distinguish clearly between ‘percentage atom economy’ and ‘percentageyield’.

➤Questionsoftengiveunfamiliar reactionsandask students tocalculate the‘percentage atom economy’ of the reaction. So memorize the formula forpercentage atomeconomy.

RemarksRemarks*

unit 45 an introduction to industrial chemistry -...

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