answeraa to hodder review questions

AnswersReview questions ow fast? – rates

Answers to Topic 1 Review questions

1 a) Labelled axes (1); accurate plot (1); smooth curve (1).b) Three half-lives = 40 103 s (2); half-life of first order reaction independent of

initial concentration (1).c) Method of calculating gradient (1); values for gradients (2); units: mol dm−3 s−1

(1).d) Axes (1); accurate plot with straight line (1).

k = 1.8 10−5 (1); s−1 (1).2 a) Axes (1); plot and curve (1).

Half-life from 20 10−3 mol dm−3: 3.6 103 s (1)Half-life from 12 10−3 mol dm3: 7.6 103 s (1)Lower starting concentration – high half-life (1).

b) Tangents drawn (1); gradients calculated (1); concentration units included (1).c) Options:

Plot a concentration–time graph and take a series of readings from the graph to show that it takes the form y = x2

plot 1/[A]t against time to get a straight line if the reaction is second order, where [A]t is the concentration at time t

plot log(rate) against log(concentration) to get a graph with gradient 2.Choice of method (1); plot (2); explanation (2).

3 a) H2O2(aq) + 2H+(aq) + 2I−(aq) 2H2O(l) + I2(aq) (1).b) Rate = k[H2O2(aq)][I−(aq)] (2)c) Second order (1)d) The first step (1); which includes the two species which feature in the rate

equation (1).4 a) i) Changing [X] does not affect the rate (1); zero order (1).

ii) Rate quadruples if [Y] doubles (1); second order (1).b) Rate = k[Y]2 (2)

c) k =

= 1 10−2 (1) dm3 mol−1 s−1 (1)d) Rate-determining step involves two Y molecules (1).

Y + Y Y2 slow (1)Y2 + X XY2 fast (1)

e) Range of answers possible (1 mark for each point made up to 5). Such as:

rate equations help to deduce reaction mechanisms; this information can inform methods of chemical synthesis, understanding enzyme action and other biochemical processes, and hence guide drug design and development

rate measurements are important in controlling chemical changes in laboratory and industrial processes; chemical engineers use rate data when designing chemical plants.

5 a) The clock procedure measures the initial rate – in this case by finding the time taken for a small, fixed amount of iodine to form (1). At first the thiosulfate ions immediately turn the iodine back to iodide ions (1); but once all the thiosulfate is

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© G. Hill and A. Hunt 2009 Edexcel Chemistry for A2


used up the iodine produces a deep blue-black colour with the starch indicator (1).

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b) Completion of table (3):

Temperature, T/K 288 292.5 299 308 315

Time, t, for the blue colour to appear/s

10.0 7.0 5.0 3.5 2.5

ln (1/t) 2.30 1.95 1.61 1.25 0.92

1/T /K1 0.00347 0.00342 0.00334 0.00325 0.00318

Plot of graph of ln (1/t) against 1/T with appropriate scales and accurate linear plot (3).

Gradient = 4180 K (1) =

Ea = 35 kJ mol −1 (1) c) The catalyst provides an alternative reaction pathway with a lower activation

energy (1). One possible alternative route is that when the catalyst is present the iron(III) ions oxidise iodide ions to iodine (1) and that peroxodisulfate(VI) ions then oxidise the resulting iron(II) ions back to iron(III) thus regenerating the catalyst. (1). Perhaps the negative peroxodisulfate(VI) ions react more readily with positive iron(II) ions than with negative iodide ions (1).

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1 a) At 0 K water is solid and the atoms are highly ordered. (1); The crystals are not disrupted by molecular movement (1).

b) The rises in entropy at T1 and T2 correspond to the changes of state: melting and vaporisation (1).There is a greater increase in disorder when molecules separate from a liquid and turn to gas than when molecules melt (1); because of the much greater freedom of movement of molecules in the gas phase (1).

c) Water and steam are in equilibrium (1);at 373 K and 1 atmosphere pressure so there is no tendency for the change to go in either direction (1).

d) Stotal = Ssystem + Ssurroundings (1)Since ΔStotal = 0 kJ mol−1 andSsurroundings = −H/T (1)it follows that Ssystem = −H/T (1)

Ssystem = (1)

= 0.110 kJ mol−1 K−1

= 110 J mol−1 K−1 (1)2 a) Al2O3(s) + 3C(s) 2Al(s) + 3CO(g)

(1)b) Total enthalpy change of formation of

products = 3 (−111 kJ mol−1) = −333 kJ mol−1 (1)Total enthalpy change of formation of reactants = −1669 kJ mol−1 (1)Overall enthalpy change for the reaction = (−333 kJ mol−1) − (−1669 kJ mol−1)= +1336 kJ mol−1 (1)

c) Total entropy of products = (2 28.3 J mol−1 K−1) + (3 198 J mol−1 K−1)= 651 J mol−1 K−1 (1)Total entropy of reactants

= 50.9 J mol−1 K−1 + (3 5.7 J mol−1 K−1)= 68 J mol−1 K−1 (1)Overall entropy change of the system = 651 J mol−1 K−1 − 68 J mol−1 K−1 (1)= 583 J mol−1 K−1

d) i) Entropy change of the surroundings

=

= −4483 J mol−1 K−1 (1)The total entropy change = 583 J mol−1 K−1 − 4483 J mol−1 K−1 (1)= −3900 J mol−1 K−1

ii) The reaction is not feasible at 298 K because the total entropy energy change is negative (1).

e) The reaction becomes feasible when:

Ssystem − > 0

TSsystem > Hreaction (1)This just happens when T =

= 2292 K (1)f) Reactions between solids are slow.

The reaction has to be very high for it to proceed at a measurable rate (1).

3 a) i) Ssurroundings =

= +1742 J mol−1 K−1 (1)Stotal = + 82 J mol−1 K−1 + 1742 J mol−1 K−1

= +1824 J mol−1 K−1 (1)ii) The total entropy change is

positive so the reaction is feasible (1).

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iii) At 298 K the activation energy of the reaction is too high (the bonds in the molecules are too strong) (1) and the energy of collisions between molecules too low for the reaction to happen at a measurable rate (1).

b) In the reaction forming CO and steam, 2.5 mol gases react to form 3 mol of gas (1). This increases disorder so that the entropy change is substantially positive (1).In the reaction forming soot, 2 mol gases react to form an ordered solid and 2 mol gas (1). Overall there is little change in entropy (1).

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4 a)

b) Hsolution[CaCl2(s)]= −(−2258 kJ mol−1) − 1657 kJ mol−1

+ (2 −340 kJ mol−1) (1)= −79 kJ mol−1

(1) for 79; (1) for sign and unitsc) The Ca2+ ion has a higher charge and a larger radius than Li+. The higher charge

will tend to make its electron density greater than Li+ (1); but its larger radius will tend to make its electron density less than Li+ (1).

d) Water molecules are polar with a + charge between the H atoms and a − charge on the O atom (1). The + charge attracts anions and the − attracts cations (1); so hydration is an exothermic process and the hydration enthalpies of both anions and cations are negative (1).

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1 a) Kc =

(1)= 49 (1); no units (1)

b) i) The quantity of iodine that has reacted = 0.003 mol dm−3 This reacts with 0.003 mol dm−3 hydrogen and produces 0.006 mol dm−3 HI.So the new concentrations are: [H2(g)] = 0.017 mol dm−3 (1); and[HI(g)] = 0.076 mol dm−3 (1).

ii) Substituting in the expression for Kc gives

(1)= 48.5This is very close to the value for Kc calculated in a). The system is again at equilibrium (1).

c) Doubling the hydrogen concentration causes a shift in the position of equilibrium to the right giving more HI and less iodine (1); some, but not all, of the added hydrogen reacts (1).

2 a) Concentration = 0.50 mol dm−3

(1)= 0.85 mol dm−3 (1)

b) Concentration = 0.01 mol dm−3

(1)= 0.006 mol dm−3 (1)

c) Kc = = 140 (1); no

units (1)

3 a) i) Kc = (2)

ii) Kc has no units (1); the concentration terms cancel – (mol dm−3)3 appears on the top and the bottom of the expression for Kc (1).

b) i) The reciprocal of the value given in the question (1): Kc = 1 10−10 (1).

ii) The square root of the value given in the question (1): Kc = 1 105 (1).

4 a) At equilibrium the forward and backward reactions happen at the same rate so that overall there is no change (1). Nitrogen and hydrogen react to form ammonia as fast as ammonia decomposes back to nitrogen and hydrogen (1).

b) i) Mole fractions: NH3, 0.137 (1);H2, 0.655 (1); N2, 0.209 (1)

ii) Partial pressures: NH3, 1.37 atm;H2, 6.55 atm; N2, 2.09 atm (2)

iii) Kp = (1)

= 3.20 10−3 (1); atm−2 (1)

c) Stotal =Ssystem – (1)

H is negative and so the total entropy change gets smaller as T rises (1).Stotal =R ln K (1)This means that Kp gets smaller as T rises (1).As a result, the mole fraction of ammonia decreases and the mole fractions of nitrogen and hydrogen increase as T rises (1).

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1 a) Kc = (1)

b) Kc = = 125 (1) dm3

mol−1

c) [PCl3(g)] = [Cl2(g)] = x (1)

= 8 × 10−3 mol dm−3

(1)x = 2 × 10−2 mol dm−3 (1)

d) i) Some, but not all, of the added PCl5 decomposes (1) and increases the concentrations of PCl3 and chlorine (1).

ii) Increasing the pressure causes the equilibrium to shift in the direction that gives fewer moles of gas (1); so the concentration of PCl5 increases while the concentrations of PCl3 and chlorine decrease (1).

iii) Raising the temperature causes the equilibrium to shift in the direction that is endothermic (1); so the concentration of PCl5 decreases while the concentrations of PCl3 and chlorine increase (1).

e) i) Changing concentrations does not affect the value of Kc (1).

ii) Changing the pressure does not affect the value of Kc (1).

iii) Kc increases (1). For an endothermic reaction the value of Kc gets larger as the temperature rises (1).

2 a) Chemical equilibrium is dynamic (1); on a molecular scale there is constant change with opposing reactions cancelling each other out (1). However, the overall macroscopic effect is that nothing appears to be changing when the system is at equilibrium (1).

b) Kc does not vary with concentration (1). However, it is true that adding more of a reactant to a system at equilibrium means that the mixture is no longer at equilibrium (1); some of the added chemical reacts and more products form until the system is again at equilibrium (1).

c) Adding a catalyst does not affect the value of Kc (1). At a constant temperature, the yield of product at equilibrium is not changed by adding a catalyst (1); however, many processes are so slow that they never reach equilibrium in the absence of a catalyst, so in that sense an added catalyst increases the yield by enabling the reaction mixture to reach equilibrium (1).

d) This is true if the reaction to make products is endothermic (1). If the reaction is exothermic, then the

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(1)


equilibrium yield of product falls as the temperature rises (1). However, as with catalysts, raising the temperature can increase the yield of a product by allowing a reaction mixture to reach equilibrium that reacts too slowly at room temperature (1).

e) A reaction is feasible in the thermodynamic sense if the total entropy change for the process is positive (1). Adding a catalyst speeds up the rate of reaction but cannot change the feasibility of a reaction at a particular temperature (1). Only if the reaction is not going because it is kinetically inert can a catalyst enable a reaction to happen at a particular temperature (1).

3 a) Kp = (2)

b) i) Kp does not change (1).ii) Kp increases (1).iii) Kp does not change (1).

c) i) Concentration of products decreases and the concentration of reactants increases (1).

ii) Concentration of products increases and the concentration of reactants decreases (1).

iii) Composition of the equilibrium mixture does not change (1).

4 a) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) (2)

b) In theory the yield of NO at equilibrium is high if the temperature is low (1) and the pressure low (1); adding excess oxygen also favours the conversion of ammonia to NO (1).

c) There is an excess of oxygen (even though only a fifth of the air is oxygen) (1). The temperature has to be high because reaction only happens at a practicable rate if the catalyst is hot (1). Raising the pressure compresses the gases and this increases the mass of material flowing through

the reactor and hence the mass of product produced in a given time from a given piece of plant (1).(The chosen pressure also depends on the price of platinum. Working at higher pressures increases the wear on the catalyst and shortens the life of the catalyst gauzes.)

d) The gases flow through the solid catalyst gauzes and so the catalyst is used continuously and does not have to be recovered from the reaction mixture (1).

e) The gases flow through the catalyst gauzes so the mixture of product and unchanged reactants is only in contact with the catalyst for a short time (1); as soon as the gases flow away from the catalyst reaction stops – the mixture has time to approach

equilibrium but not reach equilibrium (1).

f) The hot gases flow though a heat exchanger where they turn water to hot steam (1); that can be used in other processes or to generate electricity (1).

5 a) In a simple addition reaction two chemical combine to form a single product (1); so all the reactant molecules end up in the product – the atom economy is 100% (1); example (1). In any substitution reaction there is at least one other chemical formed in addition to the desired product (1); example (1).

b) The yield of a reaction measures the amount of product made compared

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with the amount expected according to the balanced chemical equation (1); the calculation of yield takes no account of the other products of the reaction. (1); it is possible to have a very high yield from a reaction with a low atom economy (1); example used to distinguish yield from atom economy (2).

c) Changes in temperature affect both the rate of reaction (1); and the

composition of the equilibrium mixture (1); reactions have to go fast enough to produce product from a chemical plant at an economic rate (1); in practice this can mean that the temperature has to be higher than would otherwise be chosen to maximise the percentage of product in the equilibrium mixture. (1); example explained (1).


1 a) HX(aq) H+(aq) + X−(aq) (1)

b) (1)

c) i) As the temperature rises the equilibrium shifts in the direction that is endothermic. So the acid ionises more as the temperature rises and this is because the value of Ka increases (1).

ii) Since the acid ionises more as the temperature rises, the hydrogen ion concentrations rises and the pH decreases (1).

iii) Ka will not change. Ka is a constant and only varies if the temperature changes (1).

2 a) A buffer solution evens out the large swings in pH which can happen without a buffer (1).

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A typical buffer mixture consists of a solution of a weak acid and one of its salts, for example a mixture of ethanoic acid and sodium ethanoate (1); there must be plenty of both the acid and its salt in the buffer mixture (1).Adding a small amount of strong acid temporarily increases the concentration of H+(aq) so the equilibrium shifts to the left to counteract the change (1); adding a small amount of strong alkali temporarily decreases the concentration of H+(aq) so the equilibrium shifts to the right to counteract the change (1).The pH of blood, for example, is closely controlled by buffers within the narrow range 7.35 to 7.45. Chemists use buffers when they want to investigate chemical reactions at a fixed pH (1).

b) i) [H+(aq)] =

1.7 10−5 mol dm−3 ×

= 3.4 10−5 mol dm−3

pH = 4.5 (1)ii) [H+(aq)] = 1.7 10−5 mol dm−3

× (1) = 1.7 10−5 mol dm−3

pH = 4.8 (1)

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(1)

Answers6 Stereochemistry

iii) [H+(aq)] =

= 5.0 10−12 mol dm−3 (1)pH = 11.3 (1)

iv) The change of pH is very much less when adding the alkali to a buffer solution than when adding it to water (1).

3 a) Strong acid ionises completely in aqueous solution (1). An example is

hydrochloric acid: HCl(aq) H+(aq) + Cl−(aq) (1)A weak acid only ionises to a slight extent in solution (1). An example is ethanoic acid:CH3COOH(aq) H+(aq) + CH3CO2

−(aq) (1)b) i) 2 (1)

ii) [OH−(aq)] = 0.01 mol dm−3

[H+(aq)] =

= 1 1012 mol dm−3 (1)pH = 12 (1)

c) i) (1)pH = 3.3 (1)

ii) Line of sketch graph rises at first from pH = 3.3; partly levels off through buffer region (1); then rises to give a steeply rising curve when the titre is 25 cm3 (1) ; half-way up the steeply rising portion is abovepH 7 (1); the curve then levels off at aboutpH 12 (1).

iii) Thymolphthalein (1) full colour change must take place within the steeply rising part of the curve (1).

4 a) i) HClO(aq) H+(aq) + ClO−(aq) (1)

ii) Ka = (1)

b) [H+(aq)] = 10−7.43 (1) = 3.7 10−8 mol dm−3 (1)

Ka = (1)

= 3.7 × 10−8 mol dm−3 (1); pKa = 7.4 (1)

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1 a) Two structural isomers with the same functional groups (1); structures of 1-chloropropane and2-chloropropane (1).

b) Six structural isomers/E–Z isomers: one with F on the C atom with H, one with the Cl on that atom, and one with the Br on that atom. Then each of these can have E–Z isomers (6).

c) Two E–Z isomers: the E isomer has the Br and Cl on opposite sides of the double bond (1), while the Z isomer has them on the same side of the double bond (1).

2 a) A chiral centre is typically a carbon atom with four different groups around it (1); in general a molecule is chiral if it is asymmetric.

b) The mirror image forms or a chiral molecule cannot be superimposed on one another (1); they have opposite effects as they rotate the plane of plane-polarised light (1).

c) Only CH3CHOHCOOH is chiral and can exist as two optical isomers (1); the other two compounds both have two hydrogen atoms on the central carbon atom .

d) A racemic mixture contains equal amounts of the two optical isomers (1); the two isomers rotate the plane of plane-polarised light equally but in opposite directions so their effects cancel (1).

e) i) C13H21O3N (1) ii) There is one chiral carbon atom:

(1)

3 Stereoisomerism involves molecules with the same molecular formula (1); and the same structural formula (1); but different three-dimensional shapes (1); in which their atoms occupy different positions in space (1).There are two forms of stereoisomerism.E–Z (cis/trans) isomerism (1); which occurs principally in alkenes and other compounds with C=C double bonds (1).In E–Z isomerism, the Z-isomer has atoms, attached to the C atoms of the double bond, with higher atomic number (1) on the same side of the double bond (1).In the E-isomer, atoms attached to the C atoms of the double bond with higher atomic number are on opposite sides of the double bond (1). An example of E–Z isomerism is 1,2-dibromoethene.Named structures of the E and Z isomers (2).Optical isomerism (1) involves non-superimposable mirror images (1). In most cases, this requires the molecules to possess a C atom with four different atoms or groups attached to it (1).These mirror images are labelled (+) and () forms (1).The (+) form rotates plane-polarised light clockwise (1) and the () form rotates plane-polarised light anticlockwise (1).A named example of optical isomerism (1); e.g. alanine or bromochlorofluoromethane.

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Diagram showing structures of the optical isomers of the example suggested as mirror images (1).

[Any 15 of these or other legitimate points]

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Answers7 Carbonyl compounds


Reactant Reagent Organic product

Name Displayed formula

CH3CH2CHOTollens’ reagent

propanoic acid (1)

(1)

CH3CH2CHO Cr2O72−/H+ (1) propanoic acid

(1)

CH3CH2CHO LiAlH4 propan-1-ol (1)

(1)

2 a) C10H16O (1)b) i) Carbon–carbon double bond as in an alkene (1); aldehyde group (1).

ii) Ketone group (1).c) i) The blue reagent gives an orange-brown precipitate (1) with citral; but not

with -ionone (1).ii) The reagent gives a bright orange precipitate (1); with citral and with -ionone

(1).iii) The reagent gives a pale yellow precipitate (1) with -ionone; but not with

citral (1).d) i) The structure stays the same except that the ketone group is reduced to a

secondary alcohol; the side chain attached to the ring becomes –CH=CHCHOHCH3 (1).

ii) The structure stays the same except that the aldehyde group is oxidised to a carboxylic acid group –COOH (1).

e) i) Both compounds have E–Z isomers (1).ii)

Z-citral (1)iii) Neither compound has a chiral centre and so neither has optical isomers (1).

3 a) W: empirical formula CH2, molecular formula C4H8 (1); could be but-1-ene or but-2-ene (1); because both give 2-bromobutane when they add HBr.X: 2-bromobutane (1); because it hydrolyses to butan-2-ol (1).Y: butan-2-ol (1); a secondary alcohol because it oxidises to a ketone (1).Z: butanone (1); a ketone because it does not react with Benedict’s reagent (1).

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1

(1)

Answers8 Carboxylic acids and their derivatives

b) CH3CH=CHCH3(l) + HBr(g) CH3CH2CHBrCH3(l) (1).CH3CH2CHBrCH3(l) + NaOH(aq) CH3CH2CHOHCH3(l) (1).CH3CH2CHOHCH3(l) + [O] CH3CH2COCH3 (l) (1).

4) Equations to describe the mechanism of addition to C=C and C=O (2 2)

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Propene with bromine HCN with propanone

Similarities

Addition reaction

Heterolytic bond breaking

Two-step process (any 2)

Differences

Electrophilic reagent

Attacks electron-rich bond

Intermediate is a positive ion (any 2)

Nucleophilic reagent

Attacks + end of polar C=O bond

Intermediate is a negative ion (any 2)



1 Butane is non-polar. The only intermolecular forces are weak London forces. It has the lowest boiling temperature: 273 K (1).Propanal has a polar C=O bond and so there are attractions between permanent dipoles as well as London forces. It boils at a higher temperature than butane: 322 K (1).Hydrogen bonding is the strongest type of intermolecular force. Hydrogen bonding is possible between the molecules of ethanol and between the molecules of ethanoic acid (1).There is greater scope for hydrogen bonding in ethanoic acid with the OH group and O atom in the carboxylic acid group. Hence both compounds boil at a higher temperature than propanal, but ethanoic acid boils at a higher temperature (391 K) than ethanol (371 K) (1).

2 a) Propanoyl chloride, CH3CH2COCl (1)b) Butan-1-ol, CH3CH2CH2CH2OH (1)c) Pentan-1-ol, CH3CH2CH2CH2CH2OH

(1); and sodium ethanoate CH3COONa (1)

d) Calcium ethanoate, (CH3COO)2Ca (1) e) Methylethyl ethanoate,

CH3COOCH(CH3)2 (1)3 a) Step 1: Heat under reflux with dilute

hydrochloric acid (1).Step 2: React with PCl5 at room temperature (1).

b) CH3CH2CH2COCl (1), butanoyl chloride (1)

c) Butanamide (1)

(1)4 a) Possible answers:

reaction with magnesium (1); colourless flammable gas given off as

metal disappears (1); reaction much faster with hydrochloric acid (1).orreaction with sodium carbonate solution (1); colourless gas given off which turns limewater milky (1); effervescence much more vigorous with hydrochloric acid (1).

b) Acid solutions react in a similar way because they contain H+(aq) ions (1).Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) (1)CO3

2−(s) + 2H+(aq) CO2(g) + H2O(l) (1)HCl(aq) is a strong acid and so fully ionised (1); while ethanoic acid is a partially ionised weak acid (1).The reactions with HCl(aq) are much faster because of the higher concentration of hydrogen ions (1).

5 a) C13H18O2 (1).b) The carboxylic acid group can form

hydrogen bonds with water (1). The hydrocarbon part of the molecule is non-polar and cannot easily break into the hydrogen-bonded structure of water (1). The large hydrocarbon part of the molecule, including a benzene ring, means that ibuprofen is only likely to be very sparingly soluble in water (1).

c) i) The same as the structure shown in the question but with the carboxylic acid group turned into a sodium salt: COO−Na+ (1).

ii) The same as the structure shown in the question but with the carboxylic acid group turned into an ethyl ester: COOC2H5 (1).

6 a) The saturated fatty acid molecules have a more regular shape (1); so they pack together more easily with a large area in contact for London forces (1). The double bond in the

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unsaturated compound puts a kink in the molecules, which means that they do not pack so well and melt at a lower temperature (1).

b) A solution of bromine (in water or preferably an organic solvent) (1); turns from orange to colourless (1); when shaken with unsaturated cis-octadec-9-enoic acid (1); there is no change with the saturated acid.

c) It is an ester of propane-1,2,3-triol (1); with three fatty acids (1).

d) i) Hydrogen adds across the double bond to form octadecanoic acid (1).

ii) In the cis compound the two hydrogen atoms in the CH=CH are on the same side of the double bond with both hydrocarbon chains on the other side (1); in the trans compound the two hydrogen atoms are on opposite sides of the molecule (1).

iii) During partial hydrogenation some of the cis unsaturated acids turn to the trans form (1). There is some evidence that trans acids raise the level of the harmful form of cholesterol in the blood; the form that can increase the risk of heart disease. (1)

7 a) A condensation reaction is one in which molecules join together by splitting off a small molecule, such as water, from two functional groups, such as a carboxylic acid group and an alcohol group (1); condensation polymers are produced by a series of condensation reactions between the functional groups of the monomers, where each monomer has at least two of the reactive functional groups (1).

b) A molecule with two carboxylic acid groups

(1)A molecule with two alcohol functional groups.

(1)

c) Polyester is used to make fibres and fabrics for clothing and other uses (1).Polyesters can be melted and spun into fibres which are strong. The fibres in polyester fabrics do not stretch or shrink (1).

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1Type of chromatography

Method of separation: adsorption or relative solubility

Example of an application

Advantages Limitations

HPLC Adsorption (1) Studying the metabolism of drugs in the body

(1)

Can be used to analyse ionic salts and also large biological molecules

Can give quantitative as well as qualitative results (1)

Relatively expensive

Columns cannot be too long or the pressure needed to force the liquid through them is too high and this can limit the degree of separation achieved

(1)

GC Either adsorption or relative solubility, depending on the type of column (1)

Investigation arson, detecting pollutants of water, air and food (1)

Very versatile technique with a range of types of column for different applications

Can only be used to analyse volatile compounds that do not decompose in the hot column

Similar compounds have

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Very sensitive − can analyse very small samples

Can give quantitative as well as qualitative results (1)

similar retention times making it hard to identify components unless combined with MS

Unknowns cannot be identified because of the lack of reference data (1)

2 a) i) Three environments (1)ii) From the integration trace data: 3

: 3 : 2 (1)b) Peak at = 1.0: RCH3 (1); peaks at

= 2.1 and 2.5: consistent with H3CCO and RCH2CO (1).

c) Peak at = 1.0: triplet – two protons on an adjacent carbon atom (1).Peak at = 2.5: quadruplet – three protons on an adjacent carbon atom (1).

d) The compound is a carbonyl compound (1). It is a ketone and not an aldehyde (1). There is no OH group in the molecule so it is not an alcohol or acid (1).

e) Butan-2-one

(1)3 a) Pentan-2-one pentan-3-

one

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answeraa to hodder review questions

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