Chem Complete U4 Notes

7/25/2019 Chem Complete U4 Notes

1/75

A Level Chemistry

UNIT 4

GENERAL PRINCIPLES OFCHEMISTRY 1

NOTES


2/75

Intr!"#tin

This unit includes the following. A quantitative study of chemical kinetics.

Further study of organic reaction mechanisms.

Entropy and equilibrium and their role in determining the direction and extent ofchemical change.

In the rganic !hemistry section" carbonyl compounds" carboxylic acids and their

derivatives will be studied.

#nowledge from $nits % and & will be necessary for $nit '

Assessment

The $nit examination will be %hour '( minutes. It will carry )( marks. It willcontain three sections" A" * and !.

Se#tin Ais an ob+ective test

Se#tin $short,answer and extended answer questions.

Se#tin Cwill contain data questions" requiring students to extract relevant datafrom the data booklet.

-uality of written communication will also be assessed in either section * or !.


3/75

%INETICS & R'tes ( re'#tin

Intr!"#tinIn kinetic studies the rate of chemical reactions is investigated.ome of the factors which can affect the rate of a reaction are/

The surface area of the reactants

The concentration of the reactants

The temperature at which the reaction is carried out

The presence of a catalyst

0ight is important for photochemical reactions

E)*eriment'l Meth!s (r Fll+in, Re'#tinsThe rate of a reaction can be determined by following some aspect of the materials whichchanges as the reaction proceeds. 1hen the effect of concentration on the rate of reaction isbeing investigated" it is important to keep other factors" such as the temperature" constant duringthe process.

M'ss #h'n,es2ass changes that take place during a reaction can be followed by placing the reaction vessel ona balance and reading the mass at specific intervals.

-l"me #h'n,es1hen a gas is given off in a reaction" that reaction can be followed by measuring the volume ofgas collected at various times. To do this a reaction flask could be connected to a gas syringe.

Press"re #h'n,esne useful way of following the reactions of gases if a pressure change takes place is to connectthe reaction flask to a pressure meter and measuring the pressure at given intervals.

Titr'tinThis could be used for following the reactions of an acid or alkali. It can also be used for followingreactions involving iodine 3using sodium thiosulphate4. 5ipette samples of the reaction mixturewould be removed from the reaction vessel" placed in a flask and the reaction quenched. It canthen be titrated to find the concentration of the reaction mixture at a particular time.

Pl'rimetry uitable for reactions involving optically active substances.

Me's"rin, #n!"#tivityuitable for reactions producing or consuming ions.


4/75

Clrimetri# me's"rementIf a reactant or product is coloured the reaction can be followed by using a colorimeter. Thismeasures changes in colour intensity.

A narrow beam of light is passed through the solution being investigated.

The filter allows an appropriate colour of light to be selected. The start and finish solutions mightbe tried with different filters to see which give the largest change in reading.The meter is usually calibrated to measure the amount of light absorbed by the solution.

Fin!in, r'teThe r'teof a chemical reaction is measured by how rapidly the #n#entr'tinof a substance ischanging. The substance may be a reactant or a product" and the sign shows whether thesubstance is disappearing or being formed.The rate is the instantaneous rate of fall in concentration of a reactant" or the instantaneous rateof increase in concentration of a product.

Cn#entr'tin 'n! R'teIt might be thought that the rate of reaction is always directly proportional to the concentration ofthe reactants" but this is not the case. It is only the substances that are present in the sloweststep 3the rate determining step4 of a reaction that actually affect the rate of the reaction.

To provide information about how the rate depends upon the concentration of reactants orcatalysts the rate equation is used.

5hotocell*eam oflight 2eterolution

under testFilter

R'te 't *int ./ ,r'!ient (

t'n,entconc.of

reactant

conc.

ofproduct

QQ


5/75

The R'te E0"'tinThe extent to which concentration determines the rate of a reaction is expressed as the rateequation. These values have to be determined by experiment6 they cannot be deduced from achemical equation 3as equilibrium equations are4.

uppose that we were investigating the reaction6

A 7 *

!then we would carry out experiments to see the effect of changing concentration of A and *.

uppose we obtain the following results for A.

!oncentration of A 8moldm,9 Initial rate 8moldm,9s,%

(.% (.(%

(.& (.(&

(.9 (.(9

(.' (.('

o we can write6 :ate ;A< The rate is said to be first order with respect to A.

If we found the following results for *.

!oncentration of * 8moldm,9 Initial rate 8moldm,9s,%

(.% (.(%

(.& (.('

(.9 (.()

(.' (.%=

:ate ;*


6/75

Units ( r'te 'n! 2:ate is measured in concentration8time/ usually mol dm,9s,%.

$nits of k depend on the order of reaction/

For a 3erorder reaction" rate > 6 so also has units of mol dm,9s,%.

For a (irst order reaction" rate > ;concentration ;concentration


7/75

!alculation/ assume r'te / A5m

$5n

!omparing 5 and -/ concentration of A doubles" and rate also doubles.

Therefore reaction is (irst r!erin A 3m> %4.

Gote that we cannot find a direct comparison in which only * changes. 1e must allow for the

effect of changing A first.

!omparing 5 and :/ concentration of A increases by a factor of three" so this alone would

change the rate to 9 H &.& > =.=.In additionthe concentration of * doubles" and rate goes up by a further factor of ' 3i.e.from =.= to &=.=4.Therefore" reaction is se#n! r!er in $" 3since rate ;& k ;A


8/75

Cn#entr'tin&time ,r'*hs 'n! h'l( li(e2The h'l( li(eof a reactant which is not in excess is the time taken for its concentration todecrease to half its initial value.For a (irst r!erreaction" the half,life is always constant/

?ere it takes %( min for the % of reactant remainingto fall from %(( to (" and another %( min tofall from ( to & 3or from & to '%" etc4.

For a second order reaction successive half,lives double.

Gote that if the graph of concentration against time is str'i,ht" order is 3er. If it is curved 3asabove" or falling more steeply at first4 you need to check successive half lives to find if they areconstant 3order %4 or if they double 3order &4.

E)'m*les ( #n#entr'tin7r'te st"!iesne method of determining rate is to follow the reaction over a period of time.

*romine and methanoic acid react as follows/*r&3aq4 7 ?!&?3aq4 &?*r3aq4 7 !&3g4

The reaction can be followed by calorimeter" measuring the colour intensity of the bromine. Acalibration plot of calorimeter readings for known concentrations would be made so that it ispossible to convert calorimeter readings to bromine concentration.

In this experiment the effect of changing the bromine concentration" a high concentration ofmethanoic acid is used" so that its concentration changes very little as the concentration of thebromine changes significantly.

100

% of reactant

remaining

first order50

25

12.5

time/min302010


9/75

:eadings for such an experiment are shown below6

Time 8s ;*romine< 8moldm,9

( (.(%((

9( (.(()(

=( (.((%

)( (.((9

%&( (.((9%( (.((9

&'( (.((''

9=( (.((&

'( (.((&(

=(( (.((%9

5lot a graph of bromine concentration against time.

Jraw tangents at intervals to fill in the table below and find the rate of change of the

bromine concentration at these times.

Time8s

;*r& k ;propanone


12/75

In general" in a one,step reaction" the rate of reaction is proportional to the product of theconcentrations of each species colliding. The number of species 3atoms" molecules or ions4 whichcollide in a one,step reaction is called its mle#"l'rity6

e.g. A products6 rate ;A


13/75

!"

!nerg#

$eaction coordinate

ranition tate

The Tr'nsitin St'te TheryThe transition state theory says that when molecules collide and react" they move through a stateof instability or high potential energy. This state of high potential energy is called the transitionstate or the activated complex. The energy required to attain the transition state is the activationenergy.

An example of this is the hydrolysis of a primary halogenoalkane.:,M 7 ?, ;?,:,M


14/75

'

-

-

$

*

$

$

+

$$

$

'-

:ate and :eaction 2echanism in :eal ystems:esults from kinetics studies provide information about reaction mechanism.

The rate equations for the hydrolysis of primary and tertiary halogenoalkanes are different.

For hydrolysis of primary halogenoalkanes the rate equation is/ :ate > k;:M k;:M ln A 7 ln e

ln k > ln A EA8:T

-!"/$


18/75

This can be used to find a value of the Activation Energy.A graph of

y > a bx"appears as follows

The log version of the Arrhenius equation can be plotted in a similar way

y > a bx

ln k > ln A EA8: 3%8T4

1hen a graph of ln k is plotted against %8T" the slope of the line is EA8:

If k is calculated for different values of T then a plot of ln k against %8Tgives a line of gradient > , Ea8:.

If k is calculated for different values of T then a plot of l, against %8Tgives a line of ,r'!ient / 7 E';


19/75

>etermin'tin ( the A#tiv'tin Ener,y (r ' re'#tin

odium thiosulphate reacts with dilute hydrochloric acid according to the equation/

&9&,

3aq4 7 &?7

3aq4 &3g4 7 3s4 7 ?&3l4

The sulphur forms a suspension and the mixture becomes progressively more cloudy.

The acid and the thiosulphate are mixed at a particular temperature and the time taken for thesulphur formed to obscure a mark behind the tube is noted.

The reaction is then carried out at different temperatures. :esults are shown below.

Temperature8o!

Temperature3T4 8#

Time 8 s %8T8 #,%

%8time8 s,%3rate4

ln %8time

=( 999 %& (.((9((9 (.(9999 ,&.'')%

& 9& %.= (.((9( (.(9=9 ,&.)&9%=

''. 9%. & (.((9%( (.(9%' ,9.99&&

'( 9%9 9= (.((9%) (.(& ,9.9&9'. 9(. ( (.((9&& (.(&(((( ,9.)%&(&

& 9(% ' (.((99&& (.(%9%' ,'.9('(

&% &)' %%= (.((9'(% (.((=&% ,'.9)

5lot a graph of ln k against %8T.

Find the gradient

$se the relationship" Kradient > ,EA8: to calculate the activation energy for the reaction

given that : > .9 L#,%mol,%.

ar, on ide of ea,er

$eaction mixtreater at


20/75

C't'lysts 'n! r'teIn $nit &" it was seen that catalysts work by providing an alternative path for the reaction at alower activation energy.

The effect of this can be shown diagrammatically in a 2axwell,*oltmann distribution curve.

The effect of lowering the activation energy even by proportionally small amounts is illustrated inthe table in the NArrhenius EquationO section. 2any catalysts alter the activation energyconsiderably.

!"7ncatal#ed

reaction!"atal#ed

reaction


21/75

Entr*y

E)thermi# 'n! En!thermi# Re'#tins2ost chemical reactions give out heat energy as they take place" so the products have lessenergy 3and so are more stable4 than the reactants. These aree)thermi# reactions.

1hen ethanoic acid is added to sodium hydrogencarbonate" a reaction takes place and thetemperature decreases" so heat energy is being taken in. This is an example of an endothermicreaction in which the products have more energy 3are energetically less stable4 than thereactants.

An endothermic reaction can be compared to a ball spontaneously rolling uphill or a pencil lyingdown springing upright.

1hat enables chemical reactions to do what is energetically unfavourablePEnergy considerations by themselves are clearly not sufficient to explain why chemical reactionstake place.

The other factor that is important for chemical reactions is entr*y. Essentially materials go totheir most likely condition6 the most probable situation , that is where there is the maximum

freedom.

The greater the freedom molecules have" the greater the entropy. As molecules become morerandomly distributed their entropy increases. 1hen molecules gain energy they gain freedom ofmovement and so their entropy increases.The symbol used for entropy is .

olids have restricted movement for the molecules in them" so they have lower entropies thanother states. 5erfect crystals at ( #elvin have Cero entropy.In liquids" the molecules have greater freedom of movement than solids" so liquids have higherentropies than solids.This means that dissolved substances will have higher entropies than the solid.

The molecules in gases have much greater freedom of movement and so have comparativelyhigh entropy values.

!omplex molecules generally have higher entropies than simple molecules as there are moreways they can arrange themselves.


22/75

The table below gives entropy values of selected substances at &)#.

S"?st'n#e St'te Entr*y ;@%71ml71

Aluminium olid &.9

ilicon dioxide olid '%.

1ater 0iquid (.(Argon Kas %'.

!arbon dioxide Kas &%9.=It can be seen from this table that the more ordered the state" the lower the value of entropy. Thesolids have the lowest values. 0iquids have higher values and gases have still higher values.

The table below gives entropy values of certain alkanes.

Al'ne St'te Entr*y ;@%71ml71

2ethane Kas %=.&Ethane Kas &&).

5ropane Kas &=).)*utane Kas 9%(.%

5entane 0iquid &=%.%

The table of alkanes shows that as the complexity of the gaseous molecules increases so do theentropy values. 5entane has a lower value than butane because it is liquid rather than gas.

1hen a chemical reaction takes place new substances are formed. The products of a reactionwill have different entropy values to the reactants" so the entropy of Nthe systemO changes. Thechange in entropy can be found using the following equation/

SSYSTEM / SPRO>UCTS 7 SREACTANTS

For spontaneous change to take place the entropy must increase.!hanges that incur an increase in entropy include/

Formation of gas molecules Jissolving of a solid

Jecomposition of a substance 3as the resulting components have greater freedom of

movement4 Kases becoming more randomly distributed 3diffusion4

1hen magnesium burns the magnesium atoms combine with oxygen molecules to form amagnesium oxide ionic lattice.

&2g3s4 7 &3g4 &2g3s4

In this change a gas is converted to a solid" so the entropy in this change seems to decrease andbreak our rule about spontaneous change.Indeed" the entropy change is ,&&%.L#,%mol,%.?owever as the reaction takes place energy is given out and so the surrounding molecules gainenergy and so gain entropy.


23/75

-8'

1hen considering entropy in changes" it is necessary to consider what happens to thesurroundings as well as the system 3chemical reaction4 itself.

STOTAL / SSYSTEM 9 SSURROUN>INGS

Any entropy increase in the surroundings comes from the energy which the system gives out.

The increase in entropy depends upon the proportional increase in energy of particles in thesurroundings. If for example the surroundings gain &(kL of energy" the proportional increase isgreater if the surroundings originally have low energy. Temperature is a measure of the energy ofmolecules" so the change in entropy in the surroundings is6

SSURROUN>INGS>

It can be seen from this that as the temperature becomes higher" the value of Q$::$GJIGKdecreases.

A reaction is feasible if is QTTA0positive. This will depend upon the change in entropy of thesystem and the change in entropy of the surroundings.

STOTAL / 7 H;T 9 SSYSTEM

The table below shows the various possibilities.

Q@TE2 Q$::$GJIGK !omparative values QTTA0 Feasibility ofreaction

9 9 9 @es7 9 Q@TE2R Q$::$GJIGK 7 Go

7 9 Q@TE2S Q$::$GJIGK 9 @es9 7 Q@TE2R Q$::$GJIGK 9 @es9 7 Q@TE2S Q$::$GJIGK 7 Go

7 7 7 Go

For the example of burning magnesium Q? > ,%&('kLmol,%at &)#.

o Q$::$GJIGK> ,3,%&('48&) > 7'.('(9kL#,%mol,%

$sing QTTA0 > Q@TE2 7 Q$::$GJIGK

!onverting 7'.('(9kL#,%mol,%to 7'"('(.9 L#,%mol,%

QTTA0> 3,&&%.%4 7 37'"('(.94 > 9"%. L#,%mol,%

It can now be seen that the overall entropy change is positive" showing that the reaction isfeasible.

otice tat te nit

of : are in ;


24/75

+-+ +

+

+

+++

-

----

-

-+-

+ +

+

+

+++

-

----

--

Other e)'m*les

!ombustion of carbon

!3s4 7 &3g4 !&3g4

Q$::$GJIGK> ,3,9)9.48&) > 7%.9&(' kL#,%

mol,%

3or 7%"9&(.' L#,%

mol,%

4


QTTA0> 379.(&4 7 37%"9&(.'4 > 7%"9&9.') L#,%mol,%

The positive value shows that this reaction is feasible.

Jecomposition of calcium carbonate!a!93s4 !a3s4 7 !&3g4

Q$::$GJIGK> ,37%48&) > ,(.) kL#,%mol,%3or ,) L#,%mol,%4


QTTA0> 37%=4 7 3,)4 > ,'9& L#,%mol,%

The negative value shows that this reaction is not feasible at room temperature.

olutions1hen an ionic substance is placed in water" the water molecules" being highly polar" are attractedto the ions. The oxygen in the water molecule carries a partial negative charge and is attracted tocations. The hydrogen in the water molecule carries a partial positive charge and is attracted toanions.

The process of water molecules linking to ions is called hydration of ions. The water moleculesare vibrating" so as they bond to the ions they shake the ions free from the lattice.

The process of dissolving is shown below.

ome ionic compounds do not dissolve in water because the electrostatic attraction between theions is too great for the water molecules to overcome.

There are two key energy processes taking place as an ionic substance dissolves. The latticehas to be broken apart. This is an endothermic process. As new bonds form between the water

8::=:! >3.02 ;1?@ ,;mol-1


25/75

>(g) > *-(g) > aA >(aA) > *-(aA)

molecules and the ions an exothermic process takes place. The energy change that takes placewhen a solution dissolves is a balance of these two energy changes. This is illustrated in thediagram below.

Jefinitions for Enthalpy !hange in olution

0attice Enthalpy"The heat energy given out 3exothermic enthalpy change4" when one mole of a crystal lattice isformed from separate" gaseous ions at an infinite distance apart under standard conditions. For one mole

Enthalpy of ?ydration"

The heat energy given out when one mole of gaseous ions dissolve in an excess of water to forman infinitely dilute solution under standard conditions. For one mole

Enthalpy of olution"The solution enthalpy is a measure of the amount of heat energy change when one mole ofsubstance is dissolved in excess water.

>(g) > *-(g)

8'Bat

*()

aeo ion

Battice

ental#

!ntal#8''#d'#dration

ental#

onic olid8':oln

>(aA) > *-(aA)!ntal# of :oltion '#drated ion

>(g) > *-(g) *()

>(g) > aA >

(aA)

*-(g) > aA *-(aA)


26/75

0attice EnthalpyThe lattice enthalpy is a measure of the strength of an ionic lattice. ome values of latticeenthalpy are given below.

Cm*"n! L'tti#e ener,y; @ml71

GaF ,)&9

Ga!l ,=

Ga*r ,'&GaI ,=))

#!l ,(%

:b!l ,=

!s!l ,='

2g!l& ,&')9

!a!l& ,&&9

2g ,9)99

!a ,9%9

It can be seen that as the siCe of the anion increases" the value of the lattice energy drops. This

is because the sum of the radii increases 3distance between the centres of charge increases4"and so electrostatic attraction decreases as and the lattice enthalpy decreases a consequence.Increasing siCe of cation also causes a decrease in the lattice enthalpy.

The higher the charge on either or both of the ions" the greater the lattice enthalpy. As the chargeon the ion increases the electrostatic attraction also increases" so the lattice energy becomesgreater.

?ydration EnthalpyThe hydration enthalpy is a measure of the attraction between an ion and water molecules.ome values of hydration enthalpy are given below.

In Hy!r'tin enth'l*y;@ml71

In Hy!r'tin enth'l*y;@ml71

0i7 ,')) F, ,'

Ga7 ,9)( !l, ,9%

#7 ,9( *r , ,9%

2g&7 ,%)% I, ,9(

!a&7 ,%=&

Al97 ,'=%9

It can be seen that as the siCe of the cation increases" the value of the hydration enthalpy drops.This is because the distance between the centre of ionic charge and the water moleculeincreases" and so electrostatic attraction decreases as and the hydration enthalpy decreases aconsequence. Increasing siCe of anion also causes a decrease in the hydration enthalpy frosimilar reasons.

The higher the charge on the ion" the greater the hydration enthalpy. As the charge on the ionincreases the electrostatic attraction also increases" so the hydration enthalpy becomes greater.


27/75

Enthalpy of olutionThe enthalpy of solution is related to the lattice and hydration enthalpies as follows/

Q?soln > , Q?latt 7 Q?hyd

!onsidering sodium chloride

Q?soln > , Q?latt 7 Q?hyd

Q?soln3Ga!l4 > , 3,=4 7 3,9)( 7 ,9%4 > 7kLmol,%

It can be seen that the enthalpy of solution for sodium chloride is endothermic. To understandwhy the process of dissolving occurs spontaneously it is necessary to look at the entropy changesinvolved in the process.

QTTA0 > Q@TE2 7 Q$::$GJIGK

Although this is an endothermic reaction making Q$::$GJIGKnegative" the Q@TE2has a highpositive value because the process of dissolving means that a highly ordered ionic latticebecomes a much less ordered solution of ions.

The dissolving of ammonium chloride is even more endothermic with a Q?solnof 7%= kLmol

,%

.

a>(g) > l-(g)

8''#d8'Bat

a>(aA) > l-(aA)

!ntal#

8':olnal()

'4>

(g) > l-(g)

8''#d8'Bat

'4>

(aA) > l-(aA)!ntal#

8':oln >16 ,;mol-1

'4l()


28/75

The values of system entropy and enthalpy of solution are shown below.

o Q$::$GJIGK> ,37%=48&) > ,(.(9 kL#,%mol,%3or ,9. L#,%mol,%4

!alculating QTTA0.

QTTA0 > Q@TE2 7 Q$::$GJIGK

QTTA0 > 37&.4 7 3,9.4 > 7&&'. L#,%mol,%

The positive value of QTTA0allows us to predict that ammonium chloride will be soluble.

8::=:! >[email protected] ;


29/75

E0"ili?ri"m

>yn'mi# E0"ili?ri"m

Equilibrium involves reactions which do not go to completion. These reactions are reversible.

If we consider a reaction between A and * to form ! and J which is reversible. 1hen A and *are mixed" the molecules will form ! and J. ?owever" as soon as molecules of ! and J areformed and collide they can also react to become A and *. uch a reaction is written6

A 7 * ! 7 J

The reaction reaches a point at which the proportion of each chemical becomes constant. This isdescribed as equilibrium.

The forward and backward reactions continue" but in balance" the same number of moleculesreacting in each direction" so a dynamic equilibrium exists.

."'ntit'tive E0"ili?ri'In this section we deal with the quantitative aspects of equilibria. -uantities of materials involvedin equilibrium reactions can be expressed as a concentration in moles per dm9" or for gases as apartial pressure.

In a vessel containing a mixture of gases" the *'rti'l *ress"reof one gas is the pressure it would

exert if it occupied the vessel alone.

The total pressure in a mixture of gases is the sum of all the partial pressures.It follows that/ *'rti'l *ress"re ( A / mle (r'#tin ( A B tt'l *ress"re

For example" if one gas makes up % by volume of the mixture 3or % of the molecules4 itspartial pressure will be (.% H 5" where 5 is the total pressure.

The mle (r'#tinof a gas in a mixture is the fraction of the total number of moles of gaspresent/

mle (r'#tin ( A / moleofno.total

"ofmoleofno.

The sum of all the mole fractions must always be %.(.

If nitrogen and hydrogen are mixed in a %/9 molar ratio" and together they have a partial pressureof (( 25a" then/

partial pressure of nitrogen > U H (( > &(( 25apartial pressure of hydrogen > V H (( > =(( 25a.

Equilibrium is when a reaction has a constant concentration of reactants and products.

Jynamic equilibrium is when a reaction has a constant concentration of reactants and

products exist as the forward and backward reactions takes place in both directions at


30/75

E)'m*les& moles of nitrogen" 9 moles of oxygen and %mole of carbon dioxide were placed in a vessel.The total pressure was atmospheres.!alculate the partial pressure of each of the gases in the mixture.

&.g of carbon monoxide" .g of carbon dioxide and &.9g of nitrogen dioxide were placed in avessel. The total pressure was atm.!alculate the partial pressure of each of the gases in the mixture.

E0"ili?ri"m e)*ressin

0e !hatelier will give us an idea of the direction of change" but" where concentrations areinvolved" we can calculate precisely what happens. In general" for a reversible reaction between 'moles of substance A" and ?moles of substance * etc/

'A 7 ?* #! 7 ! J

when the system has come to equilibrium" the ratio given by #c will be constant/

#c >

C][a

["]

dD][

c[)]

where ;!< means concentration at equilibrium" normally in mol dm9

i.e. equilibrium constant > 3product of products4 divided by 3product of reactants4.

The main mistakes students make over this are to forget that the right hand side of the euationgoes on to!" and to forget that concentrations are multipliedtogether" not added.

#cis called the #n#entr'tin e0"ili?ri"m #nst'nt.

For gases" it is often more convenient to express this constant in terms of partial pressures/

#p >

C

a

" EE

d

DE

c

)E

where )E

is the partial pressure of ! at equilibrium

#pis called the *ress"re e0"ili?ri"m #nst'nt2

Gote that the both types of equilibrium constant will have units" unless 3c 7 d4 > 3a 7 b4. Theytherefore normally have different numerical values" for the same position of equilibrium.

Total moles > & 7 9 7 % > =5G&> & 8 = x > %.=atm

5&> 9 8 = x > &.atm5!&> % 8 = x > (.9atm

2oles ! > &. 8 & > (.% mol"2oles !&> . 8 '' > (.& mol"2oles G&> &.9 8 '= > (.( mol"Total moles > (.% 7 (.& 7 (.( > (.9mol

5!> (.% 8 (.9 x > &.&)atm5!&> (.& 8 (.9 x > '.atm5G&> (.( 8 (.9 x > %.%'atm


31/75

The equilibrium constant is only affected by temperature / it does not change if concentrationsor pressures are varied" nor in the presence of catalysts.

The!osition of euilibrium" however" isnormally affected by changes in concentration.

A large "aluefor #cor #pusually means there is a high concentration of products 3on top in theequations4 at equilibrium " and so the theoretical yield is high.

!onversely" a small "aluefor #cor #pusually means there is a very low theoretical yield possible"when the reaction has come to equilibrium.

Units ( the E0"ili?ri"m #nst'nt2

The units of the equilibrium constant depend upon the equilibrium expression and must becalculated by substituting in the units of concentration or pressure and cancelling down.

E)'m*les1rite #c expressions for the following and give the units

&Fe973aq4 7 &I,3aq4 &Fe

&73aq4 7 I&3aq4

G?93aq4 7 ?&3l4 G?'3aq4 7 ?,3aq4

!u&73aq4 7 'G?93aq4 !u3G?94'&7

3aq4

!?9!&?3aq4 7 !&??3aq4 !?9!&!&?3aq4 7 ?&3l4

1rite #p values for the following and give units 3based on pressure given in atm4G&3g4 7 9?&3g4 &G?93g4

&&3g4 7 &3g4 &93g4

[Fe2>]2[2]]2[-]2

['4]['-]


32/75

1hen a reaction contains materials which are all gases or aqueous solutions" the equilibrium issaid to be homogeneous. If a reaction contains solid or liquid components" the equilibrium is saidto be heterogeneous" and the expressions do not include these components.

!a!93s4 !a3s4 7 !&3g4

9Fe3s47 '?&3g4 Fe9'3s4 7 '?&3g4

5b!l&3s4 5b&7

3aq4 7 &!l,3aq4

E0"ili?ri"m C'l#"l'tins2

1e will normally find values for #cand #pfrom experimental data.

#$am!le 1 &alculate 'c for the esterification of ethanoic acid b( ethanol gi"en that for a 1dm3 ofthis homogeneous liuid euilibrium the amounts !resent are as shown below. CH=CO '.% 3no units for esterification reaction4

#$am!le 2;In the reaction gi"en below, ).1 mol of * is mi$ed with ).3 mol of +, dissol"ed in ).5dm3 of water, and allowed to come to euilibrium, when the amount of is found to be ).)- mol.ind the euilibrium constant, 'c.

A 7 & * ! 7 9 J At start/ (.% (.9 ( (

At equilibrium" we know/ (.(=therefore/ (.%,(.(& (.9,(.(' (.(&

3since % A 9 J" (.(& A (.(= J6 (.(' * (.(= J " and (.(&! is formed4.

#c >2

3

C][["]

D][[]

>2

3

(0.26/0.5)(0.0@/0.5)

(0.06/0.5)(0.02/0.5)

> 0.2?040.16

[email protected]

> %.=(H%(9 mol dm9

Gote that the number of moles is divided by the total volume 3(.4 to obtain the concentration.


33/75

#$am!le 2 If at 55o& the !artial !ressure of nitrogen dio$ide in an euilibrium mi$ture is ).-/atmand the !artial !ressure of dinitrogen tetrao$ide in the mi$ture is ).33atm what is the "alue of '!for the reaction at this tem!erature0

N


34/75

Temperature fall/, position of equilibrium moves to right" #pand #cbecome bigger

En!thermi# re'#tins/

Temperature rise/, position of equilibrium moves to right" #pand #cbecome bigger.

Temperature fall/, position of equilibrium moves to left" #pand #cbecome smaller.


35/75

E0"ili?ri"m 'n! Entr*yEquilibrium is linked to entropy. For a reaction to be feasible the total entropy change 3the sum ofchange in system entropy and change in surroundings entropy4 must be positive. The higher thepositive total entropy change" the further an equilibrium is driven in that direction and the greaterthe equilibrium constant. The link is given by the following equation/

S / R ln %

?ere Q is the change in total entropy" ie QTTA0

The equation for determining total entropy is STOTAL / SSURROUN>INGS 9 SSYSTEM

ince Q$::$GJIGK > , Q?8T

Then STOTAL / 7 H;T 9 SSYSTEM

This equation shows us that as the temperature increases the value of , Q?8T or Q$::$GJIGKdecreases" so the effect is less pronounced on QTTA0.

In the entropy section" one example looked at was the decomposition of calcium carbonate.

Jecomposition of calcium carbonate!a!93s4 !a3s4 7 !&3g4

Q$::$GJIGK> ,37%48&) > ,(.) kL#,%mol,%3or ,) L#,%mol,%4


QTTA0> 37%=4 7 3,)4 > ,'9& L#,%mol,%

If this calculation is repeated for a temperature of %(((o!" the effect of the reduced Q$::$GJIGKcan be seen.

Q$::$GJIGK> ,37%48%&9 > ,(.%'( kL#,%mol,%3or ,%'( L#,%mol,%4


QTTA0> 37%=4 7 3,%'(4 > 7& L#,%mol,%

The value of is now positive" showing that the reaction is now feasible.In general raising the temperature of an endothermic reaction makes it more likely to take placeand drives an equilibrium to the right.

8::=:! >165 ;


36/75

A**li#'tin ( R'tes 'n! E0"ili?ri'

F'#trs e((e#tin, e0"ili?ri"m #nst'ntsThe effect of various factors 3such as temperature and use of a catalyst4 on the position ofequilibrium can be found in unit &.

Tem*er't"reIn $nit & it was seen that by 0e !hatelierBs principle" increasing the temperature of a reactioncauses a shift towards the endothermic reaction. Jecreasing the temperature of a reactioncauses a shift towards the exothermic reaction.The explanation for this was seen in the Entropy topic by considering entropy factors inequilibrium reactions.If the forward reaction is exothermic" an increase in temperature favours the reverse reaction"reducing the quantity of product and increasing the quantity of reactant" so the equilibriumconstant is reduced.If the forward reaction is endothermic" an increase in temperature favours the forward reaction"reducing the quantity of reactant and increasing the quantity of product" so the equilibriumconstant is increased.

An increase in temperature always produces an increase in rate. The equilibrium can shiftbecause the proportional increase in rate of the forward and backward reactions will depend uponwhether they are endothermic or exothermic.

Press"reIn $nit & it was seen that by 0e !hatelierBs principle" increasing the pressure of a gaseousreaction causes a shift towards the side with fewer molecules. Jecreasing the pressure of agaseous reaction causes a shift towards the side with the larger number of molecules.The effect of changing pressure on the equilibrium constant can be seen using a hypotheticalexample/ A J 7 E

If A is &( dissociated at pressure of % atm" the value of #p can be found.

A J E

2oles , start % ( (

2oles , equilibrium (. (.& (.&

5artial pressure (. 8 %.& x%> (.=atm

(.& 8 %.& x%> (.% atm

(.& 8 %.& x%> (.% atm

#p > 5*x 5J8 5A> (.% x (.% 8 (.= > (.('% atm

If the pressure is raised the equilibrium would shift to the left. 1hat will this do to the equilibriumconstantP

At a pressure of & atm the equilibrium shifts to the left so that A is only %'.9 dissociated.A J E

2oles , start % ( (

2oles , equilibrium (. (.%'9 (.%'9

5artial pressure (. 8 %.%'9 x &> %. atm

(.%'98 %.%'9 x &> (.& atm

(.%'98 %.%'9 x &> (.& atm

#p > 5*x 5J8 5A> (.& x (.& 8 %. > (.('% atm


37/75

It can be seen that although the equilibrium has shifted to the left" the equilibrium constant isunchanged.1hen pressure is increased the molecules are pushed closer together" so effectively theconcentration increases. This generally produces an increase in the rate of a reaction.

C't'lystThe presence of a catalyst increases the rate of a reaction of both forward and backwardreactions in proportion. This means that the use of a catalyst does not shift an equilibrium in anyparticular direction and does not change an equilibrium constant.

The table below gives a summary of the effect of changing various conditions on equilibrium.

!hange incondition

:eaction Effect onEquilibrium

positionEquilibrium

constant

Temperatureincrease

Exothermic hifts to left Jecreases

Temperaturedecrease

Exothermic hifts to right Increases

Temperatureincrease

Endothermic hifts to right Increases

Temperaturedecrease

Endothermic hifts to left Jecreases

5ressureincrease

2olecules on 0? R 2olecules on :? hifts to right Go change

5ressuredecrease

2olecules on 0? R 2olecules on :? hifts to left Go change

5ressure

increase

2olecules on 0? S 2olecules on :? hifts to left Go change

5ressuredecrease

2olecules on 0? S 2olecules on :? hifts to right Go change

!atalystadded

Go change Go change


38/75

In!"stri'l Pr#essesInformation on entropy change" enthalpy change and equilibrium constants can be used to selectconditions for industrial processes.

The reaction between nitrogen and hydrogen to produce ammonia is a reversible reaction.G& 7 9?& &G?9

If nitrogen and hydrogen are mixed and heated virtually no ammonia is produced.ince the value of QTTA0is positive it is clear that the reaction is feasible" so the choice ofconditions is key to the production of ammonia in this reaction.

It is an exothermic reaction" so a low temperature favours the formation of ammonia.

There are two molecules of gas on the right hand side and four on the left" so a high

pressure favours the formation of ammonia.

The problem in the production of ammonia is that at the low temperature required for a goodequilibrium position" the rate of reaction is so slow as to be non,existent.

A high pressure increases the concentration of the gases" and so increases the rate. Theproblem with creating a pressure is that it is very expensive in terms of building the plant and interms of maintenance.

?aber devised conditions whereby ammonia could be produced economically.

TEMPERATURE ?igh temperature favours good rate0ow temperature favours equilibrium

PRESSURE ?igh pressure favours good rate and good equilibrium0ow pressure is cheaper

TYPICAL CON>ITIONS 5ressure &( AtmTemperature '(o!!atalyst Iron

$nder these conditions ammonia can be produced economically. The use of a catalyst meansthat a good rate can be produced at a moderate temperature which allows about %( , % ammonia to be produced.

Although %( is not a particularly good yield" the unused gases are recycled" so there is nowaste" and a rapid reaction means that good quantities of ammonia are produced.

In this process as in other industrial processes" the system never actually attains equilibrium.This is because it is more economical to remove the reaction mixture from the reaction vessel

when a certain amount of product has formed" separate out the product and recycle to reactants.

!hoice of catalyst is very important for the chemical industry. The iron catalyst for the ?aberprocess becomes more effective when small quantities of other materials called promoters aremixed with the iron.


39/75

The table below shows the effect of promoters on the iron catalyst in the ?aber process at atemperature of '((o! and a pressure of &(( atm.

!atalyst 5romoter ammonia in exit gases

Iron Gone 9

Iron #& )

Iron #& 7 Al&9 %9 %'

A more effective catalyst means that lower temperatures can be used to achieve the same rate"with the result that the equilibrium yield can be increased. Alternatively a lower pressure could beused so making the process less expensive and safer.

The chemical industry takes various steps to make reactions more efficient" so saving resourcesand preventing wastage" thereby making them more sustainable.

For example in the ?aber process" with an exothermic reaction" the gases emerging form thereactor are cooled. Incoming gases can be used to do this" which saves energy as these gasesdo not need to be heated using fossil fuels.

The ?aber process has been used as an example of how conditions can be selected to controlthe reaction and how the process can be made more efficient. These are applicable to the wholechemical industry.

In general the chemical industry chooses conditions for a process which makes that process safeand economically effective.

Another way in which the creation of a product can be made more efficient is to choose analternative reaction with improved atom economy.


40/75

A#i! $'se E0"ili?ri'

Intr!"#tinAcids occur in natural systems. !itric acid is produced by a number of plants in their fruit. Earlyinvestigations of acids found them to have a sour taste. In the nineteenth century a wedish

chemistry by the name of Arrhenius suggested that acid were substances that dissolved in waterforming hydrogen ions" ?7. For example in hydrochloric acid" the dissolved hydrogen chlorideundergoes the following reaction/

?!l ?7 7 !l,

Arrhenius also proposed that strong acid were fully dissociated into ions" but that weak acidswere only partially dissociated.

As knowledge of atomic structure grew" it was understood that a hydrogen ions was simply aproton" and that it was unlikely that protons would exist independently in solution. !onsequently itwas proposed that the hydrogen ions +oin with water molecules in solution to form thehydroxonium ion" ?9

7. o the reaction taking place when hydrogen chloride dissolves should bewritten/

?!l 7 ?& ?97 7 !l,

This idea was taken further in the *ronstead 0owry theory of acids.

The $rnste'!7L+ry Thery1hen hydrogen bromide is dissolved in water" it forms an acid as the following reaction takesplace/

?*r 7 ?& ?97 7 *r,

In this reaction the hydrogen bromide transfers a proton to a water molecule. The *ronstead,0owry theory uses this idea to form a more general theory of acids.

According to *ronstead,0owry6An '#i! is ' *rtn !nr

and A ?'se is ' *rtn '##e*tr2

o in the reaction between hydrogen bromide and water the ?*r donates a proton to the water"so the ?*r is the acid and the water is acting as a base.

Examples6 ?G9 7 ?& ?97 7 G9

,

?!G 7 ?& ?97 7 !G,

!?9!&? 7 ?& ?97 7 !?9!&

,

G?9 7 ?& G?'7 7 ?,

!9&, 7 ?& ?!9

, 7 ?,

These reactions can be regarded as equilibrium. In the reverse reaction the proton moves backin the other direction" so the right hand side of the equation also has a proton donor and a protonacceptor 3or acid and base4.The proton donor and proton acceptor on the right hand side of the equation are called the#n",'te '#i!and #n",'te ?'se.


41/75

o?*r 7 ?& ?9

7 7 *r,

A!IJ *AE !GL !GL A!IJ *AE

Examples A!IJ *AE !on+ugate !on+ugateA!IJ *AE

!?9!&? 7 ?& ?97 7 !?9!&

,

?& 7 G?9 G?'7 7 ?,

?& 7 !9&, ?!9

, 7 ?,

The *H #n#e*tince it is the hydroxonium ion" ?9

7" that causes a material to be acidic" the higher theconcentration of this ion" the greater the acidity.

The concentration of this ion is measured on the p? scale.This is a log scale defined as follows6

*H / 7l,H95

For a strong acid" it is assumed that all the molecules form ?7ions.

Find the p? of3a4 (.% mol dm,9?!l 4log ).1 1

3b4 (.(% mol dm,9?G9 4log ).)1 2

3c4 (.& mol dm

,9

?!l 4log ).2 )./

3d4 (.((% mol dm,9 ?I 4log ).))1 3

Nte For strong acids a %( fold ditultion 3(.% mol dm,9 to (.(% mol dm,94 results in a p? changeof % unit.

ome acids" such as sulphuric are !i?'si#6 this means they can release two hydrogen ions fromeach molecule. o a % mol dm,9 solution of H


42/75

% )%x%(,% 1

%x%(,& 2%x%(,9 3

%x%(,' %x%(, 5

%x%(

,=

-%x%(, /

%x%(,

%x%(,) 8

%x%(,%( 1)%x%(,%% 11

%x%(,%& 12%x%(,%9 13

%x%(,%' 1

The !iss#i'tin #nst'nt Dini# *r!"#t %+ (r +'ter21ater molecules dissociate ?& 7 ?& ?97 7 ?,and the concentration of ?9

7ions at &o! is % x %(,mol dm,9" that is p? " and this is taken asneutral.This equilibrium exists in any solution in water.If the material is an acid and increases the ?9

7 concentration" the concentration of ?,decreases correspondingly" so that when the ?9

7 concentration is multiplied by the ?,concentration the same value is always obtained.

In water the concentration of ?97and ?,are both % x %(,mol dm,9.

o when the two are multiplied % x %(, x % x %(, > % x %(,%'mol& dm,=

This value is always the same for any solution in water.It is called the +'ter !iss#i'tin #nst'nt %+2

#wallows us to find the concentration of ?97in alkalis and consequently to calculate their p?.

%+ / H=O95 ) OH75 / 1 ) 1714ml< !m7

And ;?97< >

o for sodium hydroxide solution of concentration (.(%moldm,9

;?97< > > % x %(,%&

p? > ,log3% x %(,%&4 > %&

Examples1. For potassium hydroxide solution of concentration (.&mol dm,9.


43/75

[6397] > > 5 $ 1)41

!6 4log 5 $ 1)41 13.3

&. For sodium hydroxide solution of concentration (.(mol dm,9.

[6397] > > 2 $ 1)413

!6 4log 2 $ 1)413 12./

Strn, 'n! +e' '#i!s 'n! ?'sesThe acidity of a solution is measured using the p? scale.If the same concentration of hydrochloric acid and ethanoic acid were taken" they would nthavethe same p? value. This is because the hydrochloric acid !iss#i'tes3splits up4 #m*letelyinto ?9

7and !l,ions" whereas only a small fraction of ethanoic acid molecules dissociate.

1hen an acid is fully or near fully dissociated" it is said to be a strong acid" but one which is onlyslightly dissociated is said to be a weak acid. This should not be confused with concentration of

the acid.The same idea applies to bases. A strong base is one in which the particles !iss#i'te#m*letelyto form hydroxide ions.

Sl"tin Cn#entr'tin ( sl"tin; ml!m7=

Cn#entr'tin ( H=O9 ;

ml!m7=*H

?ydrochloric acid (.% (.% %

Ethanoic acid (.% (.((%9 &.)

?ydrofluoric acid (.% (.(&' %.=?ydrocyanic acid (.% (.((((& '.

Enth'l*y ( Ne"tr'lis'tin1hen an acid is added to an alkali" there is a temperature change. This is due to the energy

produced from the reaction6 ?7 7 ?, ?& the enthalpy for this reaction is ,.9kL mol,%.

o whenever a strong acid and a strong base are added together this is the enthalpy change forthe reaction. 1ith a weak acid or base the enthalpy change for the reaction is less exothermicthan this" as some energy is used in dissociating the acid.

St'n!'r! Ml'r Enth'l*y ( Ne"tr'lis'tin DHn is the enth'l*y #h'n,e *er mle ( +'ter

(rme! in the ne"tr'lis'tin ( 'n '#i! ?y 'n 'l'li D


44/75

The dissociation of an acid is an equilibrium process" and the dissociation constant is derivedfrom the equilibrium constant.

The equilibrium expression for the dissociation of acid" ?A" is/?A 7 ?& ?9

7 7 A,

The equilibrium constant is'c

In a dilute solution the concentration of the water is not going to change significantly during thedissociation process" and so for these reactions it can be taken as constant.o6

This new constant is the acid dissociation constant" #a

#a>'"(aA)][

]aA)(["aA)]('[ -+

The lower of the acid dissociation constant" #a" the weaker the acid.

This equation can then be used to find the p? of a weak acid.

C'l#"l'tin, the *H ( ' +e' '#i!The acid dissociation expression can be rearranged/

#a ;?A< > ;?97< ;A, middle of vertical section > about 4 and quickly flattens

out again" heading towards p?>%9.

'3'/a'Gea, acid/

trong ae14

enoltalein - ok12

enoltalein- ok '10

@

6

met#l orange - nomet#l orange- ok 4

2

10&olme of ae added/cm3


48/75

f the two most common indicators" methyl r'n,echanges colour from about 9.,.(

3vertical line to left4 and *henl*hth'lein from about .,%(/ either is suitable as itchanges completely over the vertical section around %( cm9.

n the right" for !?9!?8Ga?" the curve starts higher at &.) since the acid is weak"

then over the buffer regionit rises slowly.

The end,point is around p?>."then the latter part of the curve is similar to the first one. nly *henl*hth'leinis suitable to detect the endpoint/ methyl orange would change

very slowly over the buffer region.


49/75

2

46

@

10

12

14

10

&olme of ae added/cm3

'

'l/'3trong acid/

Gea, ae

n the left" strong acid against weak base can be titrated using methyl r'n,e.

1hile" on the right the weak acid and base give no sharp endpoint and there is no suitable

indicator.

ummary/

using a weak base" like ammonia" phenolphthalein is unsuitable because it doesnBt start tochange until after the endpoint.

using a weak acid" like ethanoic acid" methyl orange is unsuitable because it changes

steadily" and change is complete before the endpoint.

>eterminin, *%' 'n! %' (rm titr'tin #"rves

1hen a weak acid is titrated with a strong base" the following curve is obtained.

p#a > p? at half neutraliCation

#a > antilog , p#a

'3'/'3Gea, acid/

Gea, ae14

enol-

talein- no

12

enoltalein- no'

10

@

6

met#l orange nomet#l orange- ok 4

2

10

&olme of ae added/cm3

In the reaction/?A 7 ?, A, 7 ?&

At half way to the end point" ;?A< > ;A, 12K ) 17 ml !m7=

Chi#e ( in!i#'tr (rm *%Inv'l"es

odium ethanoate is the salt formed from ethanoic acid a weak acid" and sodium hydroxide astrong base. 1hen it is placed in water it splits up completely into ions. ome water moleculesalso are dissociated" so the solution contains Ga7" !?9!&

," ?7and ?,.

Ethanoate ions and hydrogen ions tend to +oin to form undissociated ethanoic acid" leaving anexcess of hydroxide ions.

This means that a solution of sodium ethanoate isin fact alkaline.This is the case for any salt made from a weak acid and a strong base.In the same way a salt made from a strong acid and a weak base is acidic in solution.This means that at the point of neutralisation the p? is not .

1hen carrying out titrations using weak acid and strong base an appropriate indicator must beselected.

Titr'tin In!i#'tr

trong acid 7 strong alkali *romothymol blue1eak acid 7 strong alkali 5henolphthalein

trong acid 7 weak alkali 2ethyl orange

In a titration" when the moles of acid and moles of alkali are exactly the same" it is said to be atthe e0"iv'len#e *int2 The point at which the indicator changes colour is the en! *int.It is clearly important that for a particular titration the end point should be the same as theequivalence point.

An indicator is in fact a weak acid" ?In. 0ike other weak acids" it dissociates" so it forms ?7andIn,ions. The ?In will be one colour and the In,will be a different colour.

?In ?7 7 In,

!olour % !olour &

In an acid solution" the high concentration of ?7will cause the equilibrium to move to the left" andso it will be colour %.

'32'a>'32

-a>

'-'-

'>


51/75

1hen in alkali solution" the ?7will be removed and the equilibrium will shift to the right andbecome colour &.

1hen the amount of ?in and are exactly balanced the colour will be in between the two colours.ne drop of acid or alkali" and the equivalence point should be able to change the colour of theindicator.


52/75

['>][n-]

['n ]

The end point for a particular indicator can be found.As a weak acid" the #a" or #Inexpression can be written

#In >

At the end point" ;In,< > ;?In mid way between & colours of indicator 3a property of the indicator4.

E0"iv'len#e *int> when the stoichiometric amounts of acid and alkali have been added.

End point and equivalence point must coincide for an effective titration.

Cl"rs/


53/75

Phenl*hth'lein/ colourless from % to ." turns pink from . to %(" remains pink above %(.

Methyl r'n,e6red from % to 9." then changes from red through orange to yellow between

9. and .(6 yellow from .( to %'.


54/75

$"((er Sl"tins

A buffer solution contains a weak acid or weak base and one of its salts.It resists dramatic changes in p? if small quantities of acid or alkali is added to it.

Examples6

3i4 a weak acid with its sodium salt 3or similar4 e.g. ethanoic acid and sodium ethanoate"giving high concentrations of !?9!? molecules and !?9!ions.

3ii4 a weak base with one of its salts e.g. ammonia and ammonium chloride" giving highconcentrations of G?9molecules and G?'

7ions.

!onsidering the dissociation of ethanoic acid" in the buffer we have deliberately made theconcentrations of !?9!? molecules and !?9!

ions large" and this determines theconcentration of ?7ions" and so the p?.

CH=COOH H9 9 CH=COO

&

1hen a little additional strong acid is added" most of the ?7ions react with some of theethanoate to form ethanoic acid/

H9 9 CH=COO& CH=COOH

Therefore the concentration of ?7in the solution only rises slightly" and there is a very

small drop in p?. Although the ;CH=COO&< decreases" it only does so by a small amount

compared to the siCe of the reservoir of CH=COO& in the buffer" so the p? remains

relatively constant.

!onversely" if some alkali is added" most of the ?ions react with !?9!? molecules/

OH& 9 CH=COOH H


55/75

C'l#"l'tin, the *H ( ' ?"((er sl"tin

:i %.' H %(mol dm9

from which ;?7< > &. H %(mol dm9" and p? > 42

In the alternative method" the Hen!ersn7H'ssel?'#h equation is derived as follows from the

dissociation constant expression for the weak acid used in the buffer 3representing the acid as ?A4/

#a>'"][

]["]'[ -+

Taking logs gives log #a > log;?7< 7 log '"][

]["

and so log #a > log;?7

< log '"][

]["

or p#a > p? log '"][

]["

from which p? > p#a 7 log'"][

]["

ince the acid is weak it is present almost entirely in the form of molecules" and virtually all the A

comes from the salt present6 so" to a good approximation" ;A

< is the same as the concentrationof the salt and ;?A< is the same as the concentration of the acid" giving

*H / *%' 9 l, acid]ofionconcentrat[

]altoftion[concentra

from which it is clear that if the concentration of salt is equal to the concentration of acid" the p?of the buffer is equal to p#afor the acid.


56/75

Examples using pKa

:ii


57/75

'

'2

Chir'lity C'r?nyls 'n! C'r?)yli# A#i!s

Hetion on ti nit ma# inclde material from 7 2 ee #lla

Ismerism

Str"#t"r'l ismerism2tructural isomerism was dealt with in $GIT &.

All isomers are compounds with the same molecular formula. e.g. !'?%(or !&?=.

tructural isomers have atoms arranged in different orders. They have similar bptYs.

e.g. !?9!?&!?&!?93butane4 and !?9!?3!?94!?9 3&,methylpropane4

!?9!?&? 3ethanol4 and !?9!?93methoxymethane4

Stereismerism2

tereoisomers have the same molecular formula and the same structural formula.

Thesame atoms are arranged in the same order but with different orientations in space.

Gemetri#" cis,trans" or E,Z isomerism , also dealt with in $GIT & , is one form ofstereoisomerism.

Another form is *ti#'lisomerism.

Chir'lity!hirality leads to optical isomerism. ptical isomerism occurs when two compounds have thesame molecular formula" but are not superimposable on each other.If a compound contains a carbon atom bonded to four different groups or atoms" it can exist intwo forms which are mirror images of each other.

Example !?9!?3?4!&?

&63&6:96&926


58/75

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

''

'

'

'

'

'

'

'

'

The two isomers affect polarised light by rt'tin, the *l'ne ( *l'ris'tin ( *l'ne *l'rise!mn#hrm'ti# li,ht in **site !ire#tinsthis is where the term [optical[ comes from.

ptical isomers differ only in the extent to which they rotate the plane of polarised monochromaticlight. ptical isomers exist in two forms called en'ntimers. The dextrorotatory 374 form rotateslight to the right 3clockwise4 but the laevorotatory 3,4 rotates light to the left.

A sample of an optically active substance may contain both optically active isomers.

An equimolar mixture does not rotate light at all as equal and opposite rotations cancel. Thisoptically inactive mixture is called the r'#emi#mi)t"re or racemate.

The carbon atom with four different groups around it 3the #hir'l #entre4 is said to be 'ssymetri#2The two mirror image molecules are said to be #hir'l.

C'r?nyl Cm*"n!s

Intr!"#tin!arbonyl compounds contain the !> group.1hen this group occurs at the end of a carbon chain" the compound is an aldehyde 3:!?4" thename ending in al. 1hen group occurs within the carbon chain" the compound is a ketone3:!:%4" the name ending in one.

The carbonyl group in polar because of the electronegative oxygen atom.

The geometry around the carbonyl group is planar" with bond angles of about %&(o.

Physi#'l Pr*erties!arbonyl compounds are much more volatile than the corresponding alcohol because" unlikealcohols" they do not have any hydrogen bonding. They are less volatile than an alkane of similarformula mass because of the polarity of the molecules.

!ompound Formula Formula mass *oiling point 8o!

5ropane !9? '' ,'&

Eentan-3-oneEentanal

3-et#ltan-2-one2-et#ltanal


59/75

'3

'

> '2

'3

'3

> '2

2'5

'

> '2

Ethanal !?9!? '' &(Ethanol !&?? '=

Although carbonyl compounds do not have a hydrogen which is directly connected to the oxygen"and therefore they have no hydrogen bonding" when placed in water the oxygen in the carbonyl isable to form a hydrogen bond with the hydrogen in the water molecules. This makes the carbonylcompounds" especially those with short carbon chains" very soluble in 1ater.

Chemi#'l Pr*ertiesAldehydes and ketones are both attacked by nucleophiles" and can both be reduced to alcohols.?owever" only aldehydes can be readily oxidised" and this is the basis for tests to distinguishbetween them.

Aldehydes and ketones are obtained by oxidation of primary and secondary alcohols"respectively.

The test (r ' #'r?nyl ,r"*2

All carbonyl compounds react with &"',dinitrophenylhydraCine 3*radyYs reagent4.1hen a solution of &"',dinitrophenylhydraCine is added to a carbonyl" a reaction takes place atroom temperature producing orange crystals.This is used as the test for the presence of the carbonyl group.

Ethanal and &"',dinitrophenylhydraCine

5ropanone and &"',dinitrophenylhydraCine

5ropanal and &"',dinitrophenylhydraCine


60/75

O)i!'tinAldehydes are reducing compounds and can react with some oxidising agents.ince ketone cannot be oxidised" they do not take part in oxidation reactions.This is used as a test for distinguishing between aldehydes and ketones.

Aldehydes will react when heated with ammoniacal silver nitrate solution 3Tllens re',ent4.This is a reaction in which the aldehyde is oxidised" and the silver ions are reduced to silver.1hen carried out in a clean test tube it forms a silver mirror.

RCHOD'0 9 A,DNH=i#hrm'te sl"tin ?eat aldehyde with a mixture ofpotassium dichromate solution andsulphuric acid

Turns from an orange solutionto a green solution

Al!ehy!es +ith Fehlin,s sl"tin D)i!'tin

Aldehydes 3but not ketones4 reduce !u&7 to !u7giving a red brown precipitate of copper 3I4 oxidein this test.

Jrops of the carbonyl compound are added to equal volumes of FehlingYs solutions A and *.

The mixture is warmed in a water bath. 3oxidation4

RCHOD'0 9


62/75

'3

'

> -

..-

'3

9

'

-

'>

.. '3

9

'

'

If the p? is higher than " the concentration of ? 7ions is too low for the second stage of themechanism to take place.

Re'#tin ( i!ine in 'l'liIodine in alkali reacts with a specific group" !?9!: to produce a yellow precipitate of !I9?.:!&is also formed in this reaction.

This is used as a test for the presence of the !?9!: group.

3Ethanal" ethanol or any methyl etne can react in this reaction4.

The iodine present in the testing reagent can oxidise alcohols" and so !?9!??: will initiallyform !?9!: this will then react to give the yellow crystals. The old name for these crystals oftriiodomethane is iodoform" and this reaction is often referred to as the iodoform reaction.


63/75

The test is carried out by adding aqueous sodium hydroxide to iodine solution until the mixture+ust turns colourless. The organic material is then added" and the mixture is warmed.

!?9!: and !?9!??: give a positive iodoform reaction 3where : can be a carbon chain orhydrogen4.1hich of the following will give a positive iodoform reactionP

!?9!?&!? >o !?9!?&!!?9 ?es

!?9!??!?9 ?es ?!? >o


64/75

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

'

$

'

$

'

C'r?)yli# '#i!s

Intr!"#tin!arboxylic acids are compounds containing the carboxyl group" !&?" which consists of the !>group and the ? group on the same carbon.The name carboxyl comes from a combination of the names of these functional groups6

!arbonyl 7 hydroxyl > !arboxyl

Physi#'l Pr*ertiesThe !arboxyl carbon contains two oxygen atoms both of which are electronegative leaving thecarbon with a partial positive charge. This allows carboxylic acids to form stronger hydrogen thanalcohols" and they therefore have higher boiling pints than alcohols of similar formula mass.

Cm*"n! Frm"l' RFM $ilin, *int ;C

5ropanol !?9!?&!?&? =( )Ethanoic acid !?9!&? =( %%

The structure of the carboxyl group allows carboxylic acid to form dimers

Ethanoic acid has a melting temperature of %o!" so if the temperature falls below this it freeCes"and the similarity of froCen ethanoic acid to ice has given the pure acid the common name of,l'#i'l eth'ni# '#i!.

The ability of carboxylic acids to form hydrogen bonds means that the lower members of thehomologous series 3those with up to ' carbon atoms4 are miscible in all proportions with water.The longer the carbon chain" the less soluble in water the carboxylic acid becomes.

*utanoic acid5ropanoic acid

9,2ethylbutanoic acid&,2ethylbutanoic acid


65/75

Pre*'r'tin ( #'r?)yli# '#i!s!arboxylic acids can be prepared by

xidation of primary alcohols

xidation of aldehydes

?ydrolysis of nitriles

Pre*'r'tin (rm *rim'ry 'l#hls1hen a primary alcohol is heated under reflux with potassium dichromate and sulphuric acid acarboxylic acid is produced.

:!?&? 7 &;< :!&? 7 ?&

Pre*'r'tin (rm 'l!ehy!es1hen an aldehyde is heated under reflux with potassium dichromate and sulphuric acid acarboxylic acid is produced.

:!? 7 ;< :!&?

Pre*'r'tin (rm nitriles!arboxylic acids can be formed by hydrolysis of nitrile 3:!G4 compounds.The hydrolysis can be carried out by heating the nitrile with acid or alkali.Hy!rlysis "sin, !il"te hy!r#hlri# '#i!2

:!G 7 &?& 7 ?!l :!&? 7 G?'!l

Hy!rlysis "sin, '0"e"s s!i"m hy!r)i!e *r!"#es the s'lt ( the #'r?)yli# '#i!2:!G 7 ?& 7 Ga? :!&Ga 7 G?9

The acid can be obtained from the salt by adding a strong acid. :!&Ga 7 ?!l :!&? 7 Ga!l

Chemi#'l Pr*erties

Re!"#tin!arboxylic acids are formed by the oxidation of primary alcohols" and can be converted back tothese compounds using lithium tetrahydridoaluminate3III4 3lithium aluminium hydride] as areducing agent. The acid is treated with lithium aluminium hydride in ether" followed by theaddition of water.

e.g. :eduction of propanoic acid. &63&62&926 7 [6] &63&62&6296 7 629

:eduction of methanoic acid. 6&926 7 [6] &6396 7 629

Re'#tin +ith 'l#hls!arboxylic acids react with alcohols in the presence of concentrated sulphuric acid to form waterand an ester. The carboxylic acid is mixed with alcohol and concentrated sulphuric acid is added.The mixture is then warmed. :!? 7 :\? :!:\ 7 ?&

e.g. 5ropanoic acid 7 ethanol &63&62&926 7 &63&6296 &63&62&99&62&63 7 629 Ethyl *r*'n'te


66/75

$

$J

$

l

Esters are named from the 'lylgroup of the alcohol and the&'tefrom the carboxylic acid.The esters formed contain the ester ("n#tin'l ,r"*or ester link. This has the structure shownbelow. :!? 7 :\? :!:\ 7 ?&

Esters have characteristic odours" which makes them useful for flavouring. 5ear drops andpineapple flavourings are derived from the appropriate ester. Esters are also useful as solvents.

Re'#tin +ith *hs*hr"s *ent'#hlri!eThe ,? group in the acid will react with halogenating reagents" such as phosphoruspentachloride in the same way as the ? group in alcohols. These reactions occur at roomtemperature.The organic product of these reactions are '#yl #hlri!es3or acid chlorides4. This functionalgroup has the structure shown below.

Ethanoic acid reacts with phosphorus pentachloride to produce ethanoyl chloride.!?9!&? 7 5!l !?9!!l 7 ?!l 7 5!l9

Ethanoyl chloride

5ropanoic acid reacts with phosphorus pentachloride to produce propanoyl chloride.

!?9!?&!&? 7 5!l !?9!?&!!l 7 ?!l 7 @9&l35ropanoyl chloride

Ne"tr'lis'tin re'#tins!arboxylic acids react with alkalis" carbonates and hydrogencarbonates to form salts.

Re'#tin +ith s!i"m hy!r)i!e :!&? 7 Ga? :!&,Ga7 7 ?&

Re'#tin +ith s!i"m #'r?n'te2 &:!&? 7 Ga&!9 &:!&,Ga7 7 ?& 7 !&

Re'#tin +ith s!i"m hy!r,en#'r?n'te :!&? 7 Ga?!9 :!&

,Ga7 7 ?& 7 !&

uch reactions" using titration with a known concentration of alkali and appropriate indicator" canbe used to determine the quantity of acid present. For example this technique can be used to findthe quantity of citric acid in fruit.

Ester group

Acyl chloride group


67/75

$

$J

>eriv'tives ( C'r?)yli# '#i!s

rganic compounds made from carboxylic acids are called derivatives of carboxylic acids.This includes '#yl #hlri!esand esters.

EstersThe ester functional group is

Esters can be hydrolysed by boiling with acid or alkali.1hen hydrolysed by acid" the alcohol and the carboxylic acid are reformed.This is a reversible reaction" so does not go to completion.

:%!:& 7 ?& :%!&? 7 :

&?

1hen hydrolysed by an alkali" the alcohol and the salt of the carboxylic acid are formed.ince the carboxylic acid is not formed" this reaction can go to completion.uch a reaction is called s'*ni(i#'tin.

:%!:& 7 Ga? :%!&,Ga7 7 :&?

e.g. Acid hydrolysis of !?9!?&!&!?&!?9!?9!?&!&!?&!?9 7 ?& !?9!?&!&? 7 !?9!?&?

?ydrolysis of !?9!?&!&!?&!?9by sodium hydroxide solution!?9!?&!&!?&!?9 7 Ga? !?9!?&!&

,Ga7 7 !?9!?&?

Acid hydrolysis of 3!?949!!!?9 3!?949!!!?9 7 ?& !?9!&? 7 3!?949!?

?ydrolysis of 3!?949!!!?9by sodium hydroxide solution 3!?949!!!?9 7 Ga? !?9!&

,Ga7 7 3!?949!?

The reaction of natural esters with sodium hydroxide solution is used to make soap.An animal fat is an ester formed from propan,%"&"9,triol and carboxylic acids with long chains.

!?&!!%?9 !?&? ?!!!%?9 7 9#? ?!? 7 9 !%?9!

,#7

!?&!!%?9 !?&?

*nimal fat gl(cerol stearate salt A9*@

The products on saponification are propan,%"&"9,triol and the salt of the carboxylic acid.


68/75

-+

The structure of the salt of the carboxylic acid is shown below.

The charges on the molecule give the useful properties of soap. The charged section is attractedto the polar water molecules is hy!r*hili#. The hydrocarbon section is repelled by the water

molecules is hy!r*h?i#. The soap molecule can be pictured as like a tadpole withNhydrophilic headO and Nhydrophobic tailO.

PlyestersThe reaction of an alcohol and a carboxylic acid to form an ester can be used to form polymers inwhich the monomers are +oined by an ester link. uch a polymer is called a *lyester. This is anexample of a #n!ens'tin *lymerin which monomers +oin by e+ecting a small molecule 3waterin the case of a polyester4.

n?!?&!?&? 7 n?&!,!=?',!&? ,3,!?&!?&,,!,!=?',!,,4,n 7 n?&ethane,%"&,diol benCene,%"', Terylene dicarboxylic acid

Tr'nsesteri(i#'tinThe burning of diesel oil from petroleum is not an environmentally s"st'in'?le method of

providing energy. An alternative is to use natural oils from plants that are renewable.uch oils will only partially combust in a normal diesel engine" and will therefore cause clogging ofthe engine. Engines can be modified to burn this type of fuel.

An alternative is to convert the triglyceride in the fat or oil to a methyl ester. The methyl ester ismore volatile and can be used in a normal diesel engine. The methyl ester can be formed in aprocess called transesterification.2ost biodiesel is produced by ?'se7#'t'lyse! tr'nsesteri(i#'tin

'2)

')

'2)

(

(

(

)

)

)

(

(

(> 3)'3('

)

(

('3)3

Cio Dieel

'2)

')

'2)

('

('

('

a

The hydrophilic head is attracted into the water and thehydrophobic tail is repelled by water so the moleculesarrange themselves as shown here. This breaks downthe surface tension of the water and allows it to wet asurface.ater


69/75

$

l

A#yl #hlri!esThe acyl chloride functional group is

Acid chlorides are highly reactive compounds.They are readily hydrolysed at room temperature" and will fume in moist air due to this reaction.

Important reactions of acyl chlorides are6

?ydrolysis 3reaction with water4

:eaction with alcohols

:eaction with concentrated ammonia

:eaction with amines

Hy!rlysisIn hydrolysis the acid chloride is converted to the carboxylic acid and ?!l is produced.:eaction with water :!!l 7 ?& :!&? 7 ?!l

Re'#tin +ith 'l#hlsThey react with alcohols at room temperature to produce the ester. :!!l 7 :^? :!:^ 7 ?!l

Re'#tin +ith #n#entr'te! 'mmni'They react with concentrated ammonia at room temperature to produce '#i! 'mi!es.

:!!l 7 G?9 :!G?& 7 ?!l

Re'#tin +ith 'minesThey react with amines at room temperature to produce secondary substituted amides.

An amine is a a carbon chain attached to the G?&functional group.:!!l 7 :^G?& :!,G?,:

^


70/75

a&elengt decreaing

FreAenc# increaing

!nerg# increaing

Kiile ligtnfra red 7ltra &ioleticro-Ga&e$adio Ga&e

S*e#trs#*y 'n! Chrm't,r'*hy

Intr!"#tinDisible light is one very small part of the electromagnetic spectrum. The different properties of thevarious types of radiation depend upon their wavelength. The diagram below shows a crude

illustration of the various types of radiation of relevance to chemists

In the $nit & section on pectroscopy the use of in(r' re! s*e#trs#*yand m'sss*e#trs#*yin analysis were looked at. The use of infra red spectroscopy to determine theextent of a reaction involving a change in functional group was also examined. #nowledge andunderstanding of these aspects are included in this topic of $nit '.

N"#le'r M',neti# Resn'n#e S*e#trs#*y

?ydrogen atoms can be detected using this sort of spectrometry.

Any spinning charge generates a magnetic field" so the protons in a nucleus have a magneticfield. If there are two protons in a nucleus they will have opposite spins so the magnetic fieldscancel. Gay nucleus with an even number of protons will have no overall magnetic field" but anucleus with an odd number of protons will have a resultant magnetic field.1hen a nucleus with a resultant magnetic field is placed in a strong magnetic field it will alignitself with that field.

If the proton supplied with sufficient energy it can flip to a position opposing the external field.The energy required to do this lies in the radio frequency region of the electromagnetic spectrum.

:inning roton

Eroton field

!xternal field


71/75

The actual energy required depends upon the exact environment on the proton. In nuclearmagnetic resonance" NMR" spectroscopy a substance is placed in a strong magnetic field andsub+ected to a range of radio frequencies. The point at which a particular radio frequency isabsorbed will depend upon the environments of the protons present in that substance.

The detector will show which radio frequencies are absorbed. The system is calibrated" usuallyusing the compound tetr'methylsil'ne SiDCH=4 as the ( point and then other frequenciescompared to this as shift values.elected shift values are given in the table below.

?ydrogen environment shift

!H9,: (.)

:,!H&,: %.9

:9!H &.(,!>,!H9,: &.9

:,!H&,? 9.=:H '.

,!H ).

In a high energy position the magnetic field ofthe proton is opposed to the externalmagnetic field.

The energy required to flip the proton lies inthe radio frequency region.Energy

In a low energy position the magnetic field ofthe proton is aligned with the externalmagnetic field.

$adio freAenc# detector

:amle!xternal

magnetic

field

$adio freAenc# generator


72/75

01233456?@91011

12

3

NMR s*e#tr'2The number of protons that can be found in each environment are given by area under the lineand defined by the integration trace.

A simplified nmr for propanal is shown below.

?igh resolution G2: spectrum give further information about the proton environment.

The effects are shown in the table below6

This means that when looked at in high resolution the following pattern is seen

The peak at ).) shows ,!H

The peak at &.9 shows ,!>,!H9,:

The peak at (.) shows !H9,:

a''JThe field experiencedby this proton will beaffected by the twoprotons on the ad+acentcarbon.

These protons can affect the field at ?\according to whether they are alignedwith the field or against it.

'

A triplet at ratio %/&/% indicates thatthere are two protons on thead+acent carbon

a b Alignment Gumber with thisalignment

& against field %

% with field %against field &

% with field %against field

& with field %


73/75

If there are three protons on the ad+acent carbon

a b c Alignment Gumber with thisalignment

9 against the field %

& against the field % with the field 9

& against the field % with the field & against the field % with the field

% against the field & with the field 9

% against the field & with the field

% against the field & with the field

9 with the field %

Uses ( NMR!learly G2: is important for chemical analysis. It also has useful applications in medicine.

The principle of nuclear magnetic resonance is used in 2agnetic :esonance Imaging" 2:I" bodyscanners which detect the protons in water molecules in the body. This process" unlike the use ofM,rays is thought to be completely harmless to patients.

Guclear magnetic resonance can be used to determine the purity of pharmaceutical products.

A quadruplet at ratio %/9/9/%

indicates that there are three protonson the ad+acent carbon


74/75

:intered gla dic

!lent

Cand of earated comonent

adorent

Chrm't,r'*hy

imple paper chromatography can be used to separate a mixture of dyes.The principle on which this works are also useful for more sophisticated techniques.

Essentially any form of chromatography used a fixed material 3stationary phase4 and moving

substance 3mobile phase4. The separation occurs due to the equilibrium between the componentsin the mixture and the stationary and mobile phases.

The process continues in this way" the rate of movement determined by the equilibriummovement which depends upon the strength of the interaction of the material with the stationaryphase and its solubility in the solvent in the mobile phase.

Cl"mn #hrm't,r'*hy

If the component is coloured" it is clear when it emerged from the column. If can be collected andthe solvent evaporated to obtain the pure substance. If the substance is colourless" there areother ways of detecting their presence" for example certain materials glow in ultra,violet light.

A

:tationar# ae

oile ae The low concentration of thematerial in the liquid 3at A4 causesthe material to dissolve

The liquid carries the dissolved materialforward. The low concentration of the

material at * causes the material to beadsorbed into the stationary phase.

oile ae

:tationar# aeB

ne version of this technique is column chromatography.In this process a column is packed with an adsorbent solid"such as alumina. The mixture is then placed in the top ofthe column so that it is adsorbed onto the surface of thesolid. The solvent 3or el"ent4 is then poured into the topand allowed to trickle through the column.5artition of the solutes between the moving solvent andthe stationery phase takes place. The rate at which thesolute moves down the column depends upon its partitioncoefficient.


75/75

Hi,h7Per(rm'n#e Li0"i! Chrm't,r'*hyThe effectiveness of column chromatography can be improved by using a very fine powder as thestationary phase. $nder these circumstances however gravitation is insufficient to drive thesolvent through the system" so a pressure is used to drive the materials through the column. Thisprocess is called high,performance liquid chromatography" ?50!.The ?50! technique is used to separate mixtures and the components can then be analysed.

G's7Li0"i! Chrm'tr,r'*hyIn gas,liquid chromatography the mobile phase is an inert gas and the stationary phase a liquidcoating on a powdered inert solid. The powder fills a coiled tube which about &mm in diameterand up to %(m long. The coiled tube is situated in an oven which controls its temperature.

The vaporiCed sample is in+ected into the carrier gas which moves through the tube at a constantrate. Dolatile components of a mixture are carried through the tube fast" while those that aremore soluble in the mobile phase take longer to pass through the tube. The time a componentspends going through the tube is called the retentin time.The area under each peak from the recorder is a measure of the amount of that component.

Detector:#ringe containing

amle

$ecorder

olmn ac,ed Git liAidcoating on a oGdered inert olidarrier ga

ermotaticall#

controlled o&en

Chem Complete U4 Notes

Documents