RES12 Chem Book Answers

1 Pearson Education Ltd 2008This document may have been altered from the original

Examination answers

1.1 Rings, polymers and analysis 1 (a) Structure:

Empirical formula: CH (b) Concentrated nitric acid and concentrated sulfuric

acid. (c) (i) NO2+

(ii) HNO3 + H2SO4 NO2+ + HSO4 + H2O (iii)

NO2O2N

H

NO2H

2 (a) (i) I: Br2, II: NaOH (ii) Dye, colouring or indicator. (iii) Add phenylamine to sodium nitrite and

hydrochloric acid below 10 C. Add this product to an alkaline solution of phenol.

(b)

C C

C

CH

H H

H

H

C

C

H

C C

C

CH

H H

H

H

C

C

H

C C

CCC CCC

Theporbitalsoverlapaboveandbelowtheplaneof the ring structure.

Theelectronsaredelocalisedandspreadacrossall six carbon atoms.

Inbenzeneallthebondsarethesamelengthbetween the lengths of the single and double bonds.

Phenolismorereactivebecausetheringisactivated.Thelonepairfromtheoxygenontheringis delocalised into the ring so electrophiles are more attractedtotheringthaninbenzene.

3 (a) (i) FeBr3 or AlBr3 (ii) Halogen carrier.

(iii)

Br2 HBr

Br

(iv) Bromobenzene (b)

Step 1 Step 2

Br

H

Br H

Intermediate

Br

(c) (i)

H

HH

H H

HH

H H

H H

H H

H H

H

p orbitalsoverlap

Cyclohexene

Benzene

p orbitalsoverlap togive a bondabove andbelow thetwo carbonatoms

See diagram00.00.00.00

H

HH

H H

HH

H H

H

(ii) Inbenzenetheelectronsarespreadoverthewhole ring structure, as they are delocalised. When cyclohexene is reacted with bromine the bromine is polarised by the double bond. However,whenbenzeneisreactedwithbromine, the bromine cannot be polarised by theringastheelectronsarespreadoverthemolecule. Electrophiles are less attracted to the benzene.Greaterenergyisneededtobreakthecloudinbenzenethanincyclohexene.

4 (a) (i) D = Propanone (ii) E = Propanal (b) (i) Reagent(s):2,4-dinitrophenylhydrazine.

Observation:orangeprecipitate. (ii) Reagents:Tollensreagent.ObservationforD:

nochange.ObservationforE:Silvermirrorformed.


5 (a) CH3CHO + 2[H] CH3CH2OH (b)

CH3 C

H

O

H

H

Organic product

O

CCH3

H

H

NaBH4Step 1

EthanalCH3 C

HH

O

H O

H

Step 2

Intermediate

(iii) An electron-pair donor (iv) Thenucleophileisattractedtothepositive

charge.Lonepairofelectronsformsadativebond. An electron pair from the double bond goestotheoxygenatom.The bond in the carbonyl bond breaks.

(c) Hydrogendoesnothavealonepairofelectrons. 6 (a)

O

Z-but-2-enalcis but-2-enal

E-but-2-enal(trans)

O

(b) (i) AddTollensreagent.Heatreactioninawaterbath.But-2-enalgivesasilverprecipitateorsilvermirror.

(ii) Aldehydes can be oxidised but ketones cannot. (c) (i) CH3CH=CHCH2OH (ii) Redox reaction/reduction or addition. (d) C4H6O + 5O2 4CO2 + 3H2O 7 (a) (i) Reagent:Tollensreagent.Observation:silver

mirror formed. Organic product: butanoic acid. (ii) Reagent:bromine.Observation:decolourises.

Typeofreaction:electrophilicaddition. (b) Recrystallise the product in order to purify the

orange precipitate. Measure the melting points of the crystals. Compare known melting points with data tables.

8 (a)

C

H

H

C

H

H

C OH

Butan-1-ol

H C

H

H H

H

C

H

H

C

CH3

H

C OH

2-methylpropan-1-ol

H

H

H

C

H

C

OH

C H

2-methylpropan-2-ol

H

H

CH HH H

H

C

H

H

C

H

O

C H

Butan-2-ol

H

H C

H

H H

H

(b) (i) Butan-1-ol can be oxidised to produce butanal andfurtheroxidisedtogivebutanoicacid.Thisoccurs as primary alcohols can be oxidised. Butan-2-ol can be oxidised to produce butanone. No further oxidation occurs as ketones cannot be oxidised. 2-Methylpropan-2-ol cannot be oxidised as it is a tertiary alcohol and is not oxidised. 2-Methylpropan-1-ol can be oxidised to produce 2-methylpropanal and further oxidised togive2-methylpropanoicacid.

D: 2-methylpropan-1-ol. E: 2-methylpropanoic acid. (ii) (CH3)2CHCOOH + C2H5OH

(CH3)2CHCOOC2H5 + H2O (c)

CC

H

H

C

H

H

CHO C

H

H H

H

O

ONa

9 (a) (i) Methyl butanoate. (ii) Heat/boil/warm/reflux with aqueous HCl or

aqueous NaOH. (b) (i)

HO

(ii)

or

C2H5

HOCH2CH2 H

H

HO

C C


(iii)

O

CH2CH3CH2

C C

H

CH2

H

C

O

H3C

10 (a) (i) Concentrated H2SO4. (ii) Thisistopreventthelossofreactantsand

productsbyevaporation. (b) (i)

C

O

C HH

C

H

HH

C

H

HH

C O C HH C

H

H

(ii) Butan-2-ol. (c) Flavouringsorperfumes.11 (a) A compound containing nitrogen where the nitrogen

is attached to one carbon atom only. (b) C2H5NH2 + H+ C2H5N

+H3

(c) (i) StageI:Sn,conc.HCl.StageII:NaNO2, HCl. StageIII:Phenol,NaOH.

(ii)

N N

NaOH

Stage I

Stage III

CH3 OH NaCl H2O

6[H]

NO2

CH3

2 H2O

NH2

CH3

OHN2Cl

CH3

(iii) Dyes.

12 (a) Stage 1: React phenylamine with sodium nitrite and hydrochloricacidat10Corbelow.Thisformsadiazoniumsaltwhichisunstableandmustbekeptbelow10C.Stage2:Reactthediazoniumsaltwithan alkaline solution of phenol below 10 C.

(Diazonium salt)

2 HCI NaNO2 NaCl 2 H2O

N2ClNH2

NaCl H2O

NaOH

HO

N2Cl OH

N N

(b) (i)

E110SO3

Na

N

N SO3Na

OH

(ii) 16 carbon atoms and 10 hydrogen atoms. (c) NaOH(aq) (d)

H2N

Phenol

Amine

SO3Na

SO3Na

OH


13 (a) Diamino compound contains two amine groups 1,4 shows the position of the amine groups on the benzenering.

(b) (i) Reduction/redox reaction. (ii) TinandconcentratedHCl.Heatunderreflux. (iii)

NO2 12[H]

H2N

O2N

NH2 4H2O

(c) (i) TheaminogroupacceptsH+ using its lone pair or electrons on the nitrogen and forming a co-ordinate bond.

(ii)

ClH3N NH3Cl

14 (a)

CH2

C C

H

CH3(CH2)6CH2

H

(CH2)6COOH

(b) (i) Oleic acid: C18H34O2. Ethanol: C2H6O. Ethyl oleate: C20H38O2 (ii) C18H34O2 + C2H6O C20H38O2 + H2O (c) Hydrolysis using hot aqueous HCl.

1.2 Polymers and synthesis 1 (a) (i) Hydrochloric acid. Aqueous acid heated under

reflux. (ii) Amino acids. (iii)

CC

H

H

H N

H

H

(or equivalent)

O

O H

(iv) A reaction with water which breaks down a compoundintosmallersub-units.Inthereactionabovethepeptidebondisbroken.

(b) (i) For stereoisomerism in compound A, there must be a chiral centre, which is a carbon atom with four different groups attached to it. Stereoisomerism occurs when there is a differentspatialarrangementofthegroups.Twomirror images that are non-superimposable are observed.

(ii) Themoleculecontainstwochiralcentresandeach chiral centre has two isomers.

(iii) Use a naturally occurring amino acid. (iv) Ahigherdosewouldberequired.Theother

stereoisomermayhavesideeffects. 2 (a) (i) RCH(NH2)COOH (ii)

C

O

O

C

CH H

C

H

C C H

H

H

HH H

H

N

H H

H

(b) (i) Anionwhichhasbothapositiveandanegativecharge.Theoverallchargeiszero.

(ii) A proton is transferred from the acidic COOH group to the basic NH2 group.

C

O

O

C

R

H

N

H H

H

C

O

O

C

R

N

H

H H

H

(c) (i) Heat with HCl (aq). (ii) Hydrolysis. (d) (i) Ammonia in ethanol. (ii) Leucine synthesised in the laboratory contains a

mixture of two optical isomers. Leucine from meat (a natural source) contains only one of the isomers.

3 (a) For L:

H

H

H

C C

For M:

C

O HO

CO

H C

OH

H

C

H

OH

H

H O

(b)

O C

O

O C

O

O C

O


(c) Itisacondensationreaction.Whenthereactionoccurs a small molecule such as water is lost.

(d) Fibres/clothing. 4 (a) (i)

H HHHHH

H3C

CC

CCC C

CH3 H3C CH3 H3C CH3

(ii) CH2 (iii)

H

C C

CH3

H

H

(b) (i) Name of functional group: peptide.

C

O

N

H

(ii) Condensation. (iii) Displayed formula of H:

O

C CC

H

H

NC

H

H

N

H O

O HH

H

Skeletal formula of H:

N

H

OH

O

H2N

O

(iv) No. of moles of glycine = 1.40/75.0 = 0.0187 moles. Expected no. of moles of dipeptide = 0.0187/2 = 9.33 103 mol. Expected mass = 9.33 103 132 = 1.232 g. % = (1.10/1.232) 100. Percentage yield 89.3%

(v)

H CC

HH

H

NCl

H O

OH

5 (a) Inadditionpolymerisation,theC=Cdoublebondinan unsaturated molecule (monomer), such as ethene, breaks open and the monomer adds on to agrowingpolymerchain.Themonomersaddoneat a time to form a long polymer such as poly(ethene), HDPE.

Incondensationpolymerisation,apolymersuchasPETisformedwhenmonomersinthiscaseethane-1,2-diolandbenzene-1,4-dicarboxylicacid join together and lose a small molecule for each bondthatforms.Inthecaseofthispolyester,awater molecule forms for each ester linkage that forms between OH and HOOC.

C

O

HO

H

H

H

C C

H

H

Cn C

H

H

Addition: Monomer ethene

H n

H

H

H

C C

H

H

H

O C

n

H

C O

H

Condensation

C

H

H

C OH

H

H

C

O

C

O

C

O

OHHO

H2O

(b) (i)

O

C C

H

H

H

C O

CH3

(ii) Reagents: HCl (aq). Conditions: Heat and reflux (iii) CH3COOH (c)

O

C

HO C

H

H

C

H

H

OH

OH

O

CHO

6 (a)

HCIN

H

O

C

Cl

(CH2)4 C

O

Cl

H

N

H

(CH2)6 N

H

HHexanedioyl dichloride

Repeat of unit polymer

1,6-diaminohexane

Otherproduct

C

O

(CH2)6(CH2)4 NC

O

H


(b) (i)

O

COH

O

CHO

H

NH

H

NH

(ii) The intermolecular forces in Nomex are stronger than in nylon.

7 (a)

NO2

(b) CH3COOH and CH3OH (c)

C

O

OH

O

C

O

CC OH

C6H5

H

NC

H

CH3

N

HH

H

O

C C

C6H5

H

NC

H

CH3

N

HH

H

8 (a) (i) NaOH (aq) (ii) Substitution. (iii) C6H5CH2Cl + NaOH C6H5CH2OH + NaCl (b) (i)

O

C C

H

H

OC

H

H

H

(ii) C6H5CH2OH + CH3COOH CH3COOCH2C6H5 + H2O

(iii) Flavourings.

(iv)

C

O

ClC

H

H

C

H

H

HO

C

O

O

H

H

CCH3 Cl

9 (a) C10H10O5 (b) (i) Reagent(s): K2Cr2O7 (aq).

Conditions: heat with H+ and distil product immediately. Organic product:

CH2CH2 C C

OO

OHOC

O

H

(ii) Reagent(s): HCl (aq). Conditions: aqueous and heat under reflux. Organic products:

CH2CH2 C C

OO

OHHOHO OH

(iii) Reagent(s): NaOH (aq). Conditions: none. Organic product:

CH2CH2 C C

OO

ONaOHO

10 (a) HOOC(CH2)4COOH and H2N(CH2)6NH2. (b) condensation polymerisation small molecule, H2O,

is eliminated. (c) Structural similarity Both have peptide or amide links Both form H-bonds between molecules. Chemical similarity Both can be hydrolysed back to monomers by

reaction with hot aqueous acid or alkali: e.g. CO(CH2)4CONH(CH2)6NH

HOOC(CH2)4COOH + H2N(CH2)6NH2 Differences Protein can be water-soluble, nylon-6,6 not. Proteins are biodegradable, nylon-6,6 is not. Nylon-6,6 has a regular chain of alternating

monomers. Proteins are made from 20 different amino acids.


11 (a) (i)

CH3

H

H

C C

H

CH3

C C

H

H

H n

(ii) Addition:ThemonomerhasaC=Cdoublebond.Thedoublebondbreaksandnosubstance is lost.

Condensation: A small molecule such as water is lost.

(b)

CH2

CH3

Cl

O CH2

CH3

O

O O

CH CHC

(c) (i)

CH2

CH3

Cl

O CH2

CH3

O

O O

CH CHC

(ii) H reacts with NaOH, poly(propene) does not. H is an ester and is polar so will be hydrolysed

by NaOH. Poly(propene) is non-polar. 12 (a)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COONa

OHNa

O2N N

N OH

COOH

O2N NH2

O2N N2Cl

O2N N

N

OH 2 H2O

COOH

H2N N

N

OH 6[H]

COOH

Below 10 C

NaNO2/HCI

(b)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COONa

OHNa

O2N N

N OH

COOH

O2N NH2

O2N N2Cl

O2N N

N

OH 2 H2O

COOH

H2N N

N

OH 6[H]

COOH

Below 10 C

NaNO2/HCI

(c)

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COONa

OHNa

O2N N

N OH

COOH

O2N NH2

O2N N2Cl

O2N N

N

OH 2 H2O

COOH

H2N N

N

OH 6[H]

COOH

Below 10 C

NaNO2/HCI

(d) Tinandconcentratedhydrochloricacid,reflux.

O2N N

N OH

COOH

OH

COOH

NaOH

O2N N

N

COONa

OHNa

O2N N

N OH

COOH

O2N NH2

O2N N2Cl

O2N N

N

OH 2 H2O

COOH

H2N N

N

OH 6[H]

COOH

Below 10 C

NaNO2/HCI

1.3 Analysis1 (a) (i) Paper, column or thin-layer chromatography. (ii) TheRfvalue. (iii)

x

yRf

x y

Solvent front

Start point

(b) (i) Retention time. (ii)

Retention time

Dete

ctio

n

Start

(c) (i) Relativesolubilityinliquidstationaryphaseoradsorption to solid stationary phase.

(ii) Each component distributes itself between the mobile phase (carrier gas) and the stationary phase.

Different components are held more or less strongly on the stationary phase and this leads to separation.

Thisdependsuponmolecularsize,volatilityandextent of bonding to the stationary phase.


2 (a) Adsorption at 1700 cm1 shows the presence of C=O. Adsorption at 1200 cm1 shows the presence of CO. No broad adsorption at 25003500 cm1 so does not

contain OH. (b) The mass spectrum shows an M+ peak at 116 therefore 116 32 (2 oxygen atoms) = 84 therefore 6 carbon atoms maximum. Molecular formula of X = C6H12O2. Base peak = 57 CH3CH2C=O+. NMR. peak areas show 12H atoms and four different

proton environments. = 0.9 ppm suggests CH3 next to CH2 (triplet) = 1.2 ppm suggests 2 CH2 next to CH (as split

into a doublet). = 2.3 ppm CH2 next to CH3 (quartet) Other CH2 next to C=O = 4.1 ppm is CH next to 2 CH3.

C

C

H

CH3

CH3

C

O

O

C

H

H

H

H

H

3 (a) Compound has a molecular mass of 74. Relative empirical formula mass = 74

(3 12 + 6 + 2 16) therefore empirical formula = molecular formula

= C3H6O2.

100

80

60

40

20

010 15 20 25 30 35 40 45 50 55 60 65 70 75

m/z

Rela

tive

inte

nsity

(b) (i) Compound X is not an aldehyde or a ketone. (ii) Compound X does not contain C=C and is not a

phenol. (c) Structure 1: ethyl methanoate. Structure 3: propanoic acid. (d) Peak at 1750 cm1 indicates presence of C=O. Peak at 1250 cm1 indicates presence of CO. No peak at 25003500 cm1 means it does not

contain OH group. It cannot be structure 3.

(e) Correct structure of compound Y:

O

C HC

H

H

OC

H

H

H

Peak at = 2.1 ppm is due to COCH3 Peak at = 3.7 ppm is due to OCH3 Relative peak areas 1:1 as both groups have the

same number of carbon atoms. Peaks show no splitting as there are no protons on

the neighbouring carbon atoms.4 (a) Low boiling point. (b) Add 2,4-DNP (2,4-dinitrophenylhydrazine) to the

aldehyde or ketone. Purify or recrystallise the orange precipitate. Measure the melting point of the crystals. Compare the results with known values from a data table.

(c) (i)

H

HC

C

C

C

H

H

H

HH

OH

O

Abso

rptio

n of

ene

rgy

11 10 9 8 7 6 5 4 3 2 1 0Chemical shift, /ppm

n.m.r. spectrum of 3-hydroxybutanone

1 1 3 3

(ii) Re-run the spectrum having added D2O to the sample. The peak due to OH will disappear.

(iii) Peak at = 1.4 ppm (doublet): (1:1) due to one proton on adjacent carbon atom. peak at = 4.3 ppm (quartet): (1:3:3:1) due to three protons on the adjacent

carbon atom. (iv) See graph answer to (c)(i). (v) The number of protons in the same chemical

environment.5 (a) (i)

C

O

C

O

HO

C

H

H

C*

OH

H OH

(ii)

CH2COOH

CCOOHH

HO

CH2COOH

CHHOOC

OH


(b) (i)

100

80

60

40

20

040 60 80 100 120 140

Rela

tive

abun

danc

e (%

)

m/e

(ii) Molecular ion peak would be at 90. (c) (i)

Abso

rptio

n of

ene

rgy

12 11 10 9 8 7 6 5 4 3 2 1 0Chemical shift/

1

3

(ii) 4asOHprotonswouldnowbevisible. (d) (i)

C

H

H

H

C

C

H

H

C

O

O

C

O H

H

H

H

H

(ii) More fruity because of ester/less sour as acid is used up.

6 (a)

C

H

C

H

C HO

H

CH HH H

H

O

CC C

H

H

H

H

H

(b) CH3CH2COOH + (CH3)2CHCH2OH CH3CH2COOCH2CH(CH3)2 + H2O

Alcohol: 2-methylpropan-1-ol Carboxylic acid : propanoic acid. Heat the alcohol and the carboxylic acid with some

concentrated sulfuric acid in a water bath. (c) Mass spectrometry. Use the molecular ion peak, the m/zvalueforthe

peak at highest mass. (d) (i) Thevalueorchemicalshiftindicatesthetypeof

protonorchemicalenvironment.Thenumberofpeaksgivesthenumberofdifferenttypesofproton or the different numbers of chemical environments.Therelativeratioofpeakareasshows the number of protons of each type. Splittingpatternsgiveinformationaboutthenumber of neighbouring hydrogen atoms

(protons)onthecarbonunderinvestigation:thesplitting pattern (singlet, doublet, triplet or quartet) is one greater than the number of adjacent protons.

D2O is used to identify the presence of OH groups.

(ii)

CH3CH2COOCH2CH(CH3)2

S Q P R T

7 (a) (i) Find the m/zvalueofthepeakfurthesttotherightwith highest m/zvalue.

(ii) C2H3O2 has an empirical mass of 59. therefore 118/59 = 2 2 empirical formula = C4H6O4. (b) (i) OH peak disappears. (ii) Peak at =3.3ppmidentifiesCHgroup.It

shows as a quartet splitting pattern showing presence of three protons on the adjacent carbon atom.Thereisonlyonehydrogeninthisenvironment.

Peak at = 1.2 ppm identifies CH3group.Itshows doublet splitting showing the presence of oneprotonontheadjacentcarbonatom.Therearethreeprotonsinthisenvironment.

Relativepeakareasshowthenumberofhydrogens(protons)intheenvironment.

C

H

H

CH H

C

O

C

O

HO OH

(iii) Therewouldbeonepeakwithnosplittingasallfourprotonsareinthesameenvironment.

8 (a) Infrared: X and YbothhaveC=Oat16801750cm1. X and YbothhaveCOat10001300cm1. X and YbothhaveOHinrange25003300cm1. Only Y has an absorption at 31003500 cm1 for NH. Bothhavedifferentfingerprintregions. Mass spectrum: X and YhavedifferentM peaks, X = 88 and Y = 89. X and Yhavesomesimilarfragments,e.g.CH3CH+

at m/z = 28, COOH+ at 45, CHCOOH+ at 58, CH3CHCOOH+ at 73.

X and Yhavesomedifferentfragments,e.g.X has CH3CHCH3+ at m/z = 43 or (CH3)2CHCO+ at 71; Y has H2NNCH+ at 29, H2NCHCH3+ at 44, CH3(NH2)CHCO+ at 72 or NH2+ at 16.


(b) Proton environment

Chemical shift, /ppm

Splitting pattern

Relative peak area

OCH3 3.34.3 singlet 3

CH2CO 2.02.9 quartet 2

CH3C 0.71.6 triplet 3

2.1 Rates, equilibrium and pH1 (a) H2O2+2I + 2H+ I2 + 2H2O (b) (i) Experiment2hastwice[I] of Experiment 1 and

the rate has quadrupled, soorder=2withrespecttoI. Experiment 3 has twice [H+] of Experiment 2 and

the rate is unchanged, so order = 0 with respect to H+. (ii) rate = k [H2O2][I]2

(iii) k = rate/[H2O2][I]2

Experiment 1: k = 1.15 106/(0.01)(0.01)2 = 1.15 102 dm6 mol2 s1

(c)

Rate

of r

eact

ion

[H2O2(aq)]/mol dm30

Straight line increasingThrough 0/0

(d) (i) 2H2O2 2H2O + O2 (ii) 1 dm3 H2O2 20 dm3 O2 amount of O2 = 20/24 mol concentration of H2O2 = 2 20/24

= 1.67 mol dm3

2 Fromthegraph,thereisaconstanthalf-life.Therefore,order = 1 with respect to [CH3COCH3].

From the table, the rate doubles when [H+] doubles. Thereforeorder=1withrespectto[H+].

Fromthetable,theratestaysthesamewhen[I2] doubles. Thereforeorder=0withrespectto[I2].

Therateequationis:rate = k[H+][CH3COCH3]

k = rate _______________ [H+][CH3COCH3]

= 2.1 109

_______________ 0.02 1.5 103

= 7.0 105 dm3 mol1 s1

Therate-determiningstepinvolvesthespeciesintherateequation: H+ and CH3COCH3.

A possible two-step mechanism is: Step 1: CH3COCH3 + H+ CH3COHCH3+ Step 2: CH3COHCH3++I2 CH3COCH2I+HI+H+ H+ is used up in the first step and is regenerated in the

second step so H+ is a catalyst.

3 (a) Kc = [CH3COOC2H5] [H2O] ___________________ [CH3COOH] [C2H5OH]

(b) (i) CH3COOH, 1.0 mol; C2H5OH, 7.5 mol CH3COOC2H5, 5.0 mol; H2O, 5.0 mol

(ii) Kc = 5 5 _______

1 7.5 = 3.3 (no units)

(c) Theexperimentcouldbeleftlongertoreachequilibrium.

Thecompositionsofthemixturescouldbeanalysedattimeintervalsuntilthecompositionsdonotchangeany more.

(d) (i) TherewouldbemoreCH3COOC2H5 and H2O TherewouldbelessCH3COOH

Theequilibriumshiftstotherighttocounteractthe effect of increasing the ethanol concentration.

(ii) Kc stays the same. (e) Thecompositionstaysthesamebecausethecatalyst

does not shift equilibrium position. Theratesoftheforwardandreversereactionsare

increased by the same amount. (f) (i) A decrease in Kc results in more reactants and

fewerproducts.TheequilibriumthereforeshiftstotheleftinresponsetothenewvalueofKc.

(ii) Increasingthetemperaturemovestheequilibriumto the left to counteract the effect of the increase inenergy.Theforwardreactionisexothermicandthereversereactionisendothermic.

4 (a) Kc = [Hl]2

______ [H2] [l2]

(b) (i) H2,0.14mol;I2,0.04mol;HI,0.32mol

(ii) Kc = 0.322 ___________

0.14 0.04 = 18.285 714 29

= 18 (to 2 sig. figs), no units (c) Kcisconstantbecauseallvolumescancelandthe

temperature is constant. Thecompositionisthesameastherearethesame

number of gas moles on either side of the equation. (d) Theforwardreactionisexothermic.Theequilibrium

movestothelefttocompensatefortheincreaseintemperature.

(e) (i) I2(aq) + H2S(g) 2HI(aq)+S(s) (ii) amountofI2 reacted = 1.89 mol, ThereforeamountofHIformed=3.44mol TheoreticalamountofHIproduced=3.78mol

% yield = 3.44 100 __________ 3.78

= 91.0%


(iii) [HI]=3.44 1000 ___________ 750

= 4.59 mol dm3

pH = log 4.59 = 0.665 (a) (i) A BrnstedLowry acid is a proton donor. (ii) A weak acid is partially dissociated. (iii) pH = log[H+] (iv) A buffer solution minimises changes in pH after

addition of small amounts of both acids and alkalis.

(b) IntheequilibriumH2CO3(aq) Y H+(aq) + HCO3(aq) Hydrogencarbonate, HCO3, reacts with any added

acid: HCO3 + H+ H2CO3 Theequilibriummovestothelefttocounteract

increased [H+]. H+ reacts with any added alkali: H+ + OH H2O Theequilibriummovestotherighttocounteracta

decrease in H+.

(c) Ka = [H+][HCO3(aq)] _____________

[H2CO3(aq)]

[H+] = 10pH = 107.4 = 3.98 108 mol dm3

[HCO3(aq)] __________ [H2CO3(aq)]

= Ka ____

[H+] = 4.17 10

7

__________ 3.98 108

= 10.5

6 (a) A weak acid partially dissociates: e.g. HCOOH Y H+ + HCOO

(b) (i) pH = log (1.55 103) = 2.81 [H+]dealswithnegativeindicesoveraverywide

range.

(ii) Ka = [H+(aq)][HCOO(aq)]

_________________ [HCOOH(aq)]

(iv) Ka = [H+(aq)]2 ____________

[HCOOH(aq)] =

(1.55 103)2 ____________

0.015

= 1.60 104 mol dm3

pKa = log Ka = log (1.60 104) = 3.80

(v) Percentage dissociating = (1.55 103) 100

_________________ 0.015

= 10.3% (c) (i) HCOOH + NaOH HCOONa + H2O (ii) n(HCOOH) = 0.0150 25.00/1000

= 3.75 104 mol From the graph, the rapid rise in pH takes place

after addition of 30 cm3ofNaOH(aq).Therefore30cm3 of NaOH(aq) are required for neutralisation.

Therefore[NaOH]=3.75104 1000/30 = 0.0125 mol dm3

(iii) Kw = [H+(aq)][OH(aq)] pH = log(1 1014/0.0125) = 12.10

(iv) metacresol purple TheindicatorspHrangecoincideswiththepH

change during the sharp increase in pH in the titrationcurve.

7 (a) Theextentofdissociationoftheacid (b) (i) H2SO3(aq) + CH3COOH(aq) Y

HSO3(aq) + CH3COOH2+(aq) acid 1 base 2 base 1 acid 2 (ii) CH3COOH is a stronger acid than C6H5OH.

Therefore,CH3COOH will donate H+ to C6H5OH CH3COOH(aq) + C6H5OH(aq) Y

CH3COO(aq) + C6H5OH2+(aq) (c) For HCl, pH = log[H+] = log 0.045 = 1.35 For CH3COOH, [H+] =

________________ (Ka [CH3COOH])

[H+] = 8.75 104 mol dm3

pH = log 8.75 104 = 3.058 (d) 2 mol of each acid produce 1 mol of H2 Mg + 2HCl MgCl2 + H2 Mg + 2CH3COOH (CH3COO)2Mg + H2 HCl is the stronger acid and in solution [H+] in

HCl > [H+] in CH3COOH. HCl will therefore react quicker as its [H+] is greater.8 (a) (i) Kw is the ionic product of water (ii) Kw = [H+(aq)] [OH(aq)]

(b) amount of HCl = 5.00 103 21.35 __________________

1000

= 1.067 10 4 mol

amount of Ca(OH)2 = 1.067 104 ___________

2

= 5.34 105 mol concentration of Ca(OH)2 = 40 5.34 105

= 2.136 103 mol dm3

(c) [OH] = 2 2.7 103 = 5.4 103 mol dm3

[H+(aq)] = Kw ________

[OH(aq)] = 1.0 10

14

__________ 5.4 103

= 1.85 1012 mol dm3

pH = log (1.85 1012) = 11.73/11.7 (d) [H+] [OH] = 1014 mol2 dm6; If[OH] is 100 (102) greater than [H+], [OH] must be

106 and [H+] must be 108

pH = 10pH, so pH = 89 (a) (i) a weak BrnstedLowry acid is a proton donor

that partially dissociates

(b) Ka = [HCOO][H+]

___________ [HCOOH]

= [H+]2 _________

[HCOOH]

[H+] = _____________________

{(1.58 104) (0.025)} = 1.99 103 mol dm3

pH = log[H+] = log 1.99 103 = 2.70


(c) (i) A buffer solution minimises pH changes. (ii) sodium methanoate, HCOONa HCOONa supplies HCOO as the conjugate

base. (iii) ThepHofthebuffersolutiondependsuponthe

extentofdissociationgivenbytheKavalueandtherelativeconcentrationsoftheweakacidandconjugate base.

(d) Mass of HNO3 = 1400 65 __________

100 / 910 g

amount of HNO3 = 910 ____ 63

= 14.4 mol

pH = log[H+] = log 14.4 = 1.16

(e) 2HNO3 + CaCO3 Ca(NO3)2 + CO2 + H2O (f) A conjugate acidbase pair is two species differing

by H+

acid 1: HNO3; base 2: NO3

acid 2: HCOOH2+; base 2: HCOOH (g) (i) 6HNO3 + S H2SO4 + 6NO2 + 2H2O or 4HNO3 + S H2SO4 + 4NO2 + H2 (ii) TheoxidationnumberofNdecreasesfrom+5in

HNO3 to +4 in NO2.

2.2 Energy 1 (a) (i) Energyisneededtoovercometheforceofattractionbetweenouterelectronsandnucleus (ii) Electronaffinityinvolvesaneutralatomgainingofanelectronbyattractionfromthenucleus (b) (i)

0

Hf

CaO(s)

Ca(s) O2(g)

Ca(g) O2(g)

Ca(g) O(g)

Ca(g) O(g) e

Ca2(g) O(g) 2e

HEA1 798 kJ mol1

HLE 3459 kJ mol1

HEA1 141 kJ mol1

HI2 1150 kJ mol1

HI1 590 kJ mol1

Ha 249 kJ mol1

Ha 178 kJ mol1

Ca2(g) O(g) e

Ca2(g) O2(g)

(ii) Hf = +178 + 249 + 590 + 1150 + (141) + 798 + (3459) = 635 kJ mol1 (iii) LatticeenthalpyofFeOismorenegativeandmore

exothermic than lattice enthalpy of CaO. IonicradiusofFe2+ must be less than that of a Ca2+ion.Thisleadstogreaterattraction between Fe2+ ions and O2 ions than between Ca2+ ions and O2 ions.


2 (a) Lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions.

Mg2+(g) + 2Cl(g) MgCl2(s) (b)

Mg2(g) 2Cl(g) 2e

Mg(g) 2Cl(g) e

Mg(g) 2Cl(g)

Mg(g) Cl2(g)

Mg(s) Cl2(g)

MgCl2(s)

Mg2(g) 2Cl(g)

HLE

2 349HI2 1451 kJ mol

1

HI1 738 kJ mol1

2 Ha 2 123 kJ mol1

Ha 148 kJ mol1

Hf 641 kJ mol1

0

641 = +148 + 2 123 + 738 + 1451 + 2 (349) + HLE HLE = 641 (+148 + 2 123 + 738 + 1451 + 2 (349)) = 2526 kJ mol1 (c) Na+ has a larger radius than Mg2+ and Br has a larger radius than Cl

Na+ has a smaller charge than Mg2+

Br experiences less attraction than Cl

Na+ experiences less attraction than Mg2+

ThereforethereismuchlessattractionbetweenNa+ ions and Br ions than between Mg2+ ions and Cl ions. One mark for correct spelling, punctuation and grammar in at least two sentences.

3 (a) (i) H1 = Enthalpy change of formation of magnesium oxide

H2 = Enthalpy change of atomisation of magnesium

H3 = First ionisation energy of magnesium (ii) Mg2+(g) + O2(g) (iii) Theelectronbeinggainedisrepelledbythe

negativechargeoftheO(g) ion. (b) (i) 602 = 149 + 736 + 1450 + 248 + 650 + HLE HLE = 602 (149 + 736 + 1450 + 248 +

650) = 3835 kJ mol1

(ii) Lattice enthalpy of barium oxide is less exothermicthanthatofmagnesiumoxide.Thisis because Mg2+ has a smaller ionic radius than Ba2+ so there is a stronger attraction between thepositiveandnegativeions.

(c) Magnesium oxide has a high melting point because it has a highly exothermic lattice enthalpy resulting fromstrongattractiveforcesbetweenMg2+ and O2 ions.

4 (a)

NH4(g) Cl(g)

NH4(aq) Cl(g)

NH4(aq) Cl(aq)

NH4Cl(s)

Hydration of Cl

Hydration of NH4

Latticeenthalpy

Enthalpychange ofsolution

(b) Hhyd(NH4+) + Hhyd(Cl) = HLE(NH4Cl) + Hs(NH4Cl) Hhyd(NH4+) + (364) = 689 + 15 Rearranging: Hhyd(NH4+) = (+364) + 689 + 15 Hhyd(NH4+) = 310 kJ mol1 (c) Thereisanincreaseindisorderondissolving. Increaseinentropyoutweighsincreaseinenthalpy. 5 (a) Energy lost = mcT = 108.48 4.18 12.5

= 5668 J = 5.668 kJ

AmountofLiCldissolved=m __ m = 8.48 ___________

(6.9 + 35.5)

= 0.200 mol

Hsolution = 1.60 ______

0.200 = 28.34 kJ mol1


(b) Thereisanincreaseindisorderondissolving. Increaseinentropyoutweighsincreaseinenthalpy. (c) (i) X: Lattice enthalpy of LiCl; Y: Enthalpy change

of hydration of Li+

(ii) Li+ ions are smaller than Na+ ions. Small ions exert more attraction on water

molecules and more energy is released. 6 (a) InEquation1,Sis+vebecauseagasisreleased. InEquation2,Sisvebecausethenumberofgas

molecules decreases. (b) G = H TS. (c) (i) Increaseinentropyfromformationofagas

outweighs increase in enthalpy. Overall,balancebetweenenthalpyandentropy

is such that H TS < 0. (ii) At high temperatures, decrease in entropy from

decrease in the number of gas molecules outweighs decrease in enthalpy because TS becomes more significant.

Overall,balancebetweenenthalpyandentropyis such that H TS > 0.

7 (a) Sispositiveastheforwardreactionresultsinanincrease in the moles of gas molecules.

(b) S = S (products) S (reactants) = (198 + 3 131) (186 + 189) = +216 J K1 mol1

(c) G = H TS At room temperature, G = +210 298 0.216

= +145.632 kJ mol1

For a reaction to take place, G

Documents

RES12 Chem Book Answers