Physical Chemistry_Chemical Equilibria Marks Scheme

1. Ammonia is manufactured from hydrogen and nitrogen in the Haber Process:

N2(g) + 3H2(g) 2NH3(g) H = –92 kJ mol-1

(a) What is meant by the term dynamic equilibrium? reacts both directions (12) at equal rates (1) must mention rates for second mark

(2)

(b) State the conditions employed industrially in the manufacture of ammonia, and justify them on physico-chemical grounds.

70 - 1000 atm (1) high P moves equilibrium to right (1) 350 - 550 C (1) low temperature moves equilibrium to right (1) but rate too slow (1) ( use) iron or iron oxide catalyst (1) if no other marks awarded high pressure and low temperature (1) if both given

(5)

(c) What effect does a catalyst have on the rate of achievement of the equilibrium and the composition of the equilibrium mixture?

Faster give mark if mentioned in (b) (1) no effect on equilibrium position (1)

(2)

(d) A significant proportion of the ammonia made is oxidised to nitrogen monoxide and steam, using oxygen and a platinum / rhodium catalyst.

(i) Give the equation for the reaction. 4NH3 + 5O2 4NO + 6H2O (1)

(ii) Use the enthalpies of formation, Hf, given below to evaluate H for the oxidation of ammonia.

Hf / kJ mol-1 NO(g) +90 H2O(g) -242 NH3(g) -46

H = 4Hf(NO) + 6Hf(NH3) = 4(90) + 6(-242) - 4(-46) (-908) kJ mol-1 (3) look for 4 6 4 gives (3) if signs wrong (-I) on each occasion correct answer with working (3) if equation wrong mark consequentially.

Khalid Mazhar Qureshi www.edexcel-cie.com

(iii) Ammonia does not spontaneously catch fire in air or oxygen; use this fact and your

result from part (ii) to explain the difference between kinetic and thermodynamic stability. thermodynamic stability: products have higher energy than reactants OR reverse for instability (1) kinetic stability: (products more stable / lower energy than reactants) / ammonia thermodynamically unstable but (1) (reaction slow due to) high E2 (1)

(7) (Total 16 marks)

2. Ammonia is manufactured from hydrogen and nitrogen in the Haber Process:

N2(g) + 3H2(g) 2NH3(g) H = –92 kJ mol-1

What is meant by the term dynamic equilibrium? reacts both directions (12) at equal rates (1) must mention rates for second mark

(2)

3. (a) Ammonia is manufactured by direct synthesis in the Harber Process:

N2(g) + 3H2(g) = 2NH3(G) H = 92Kj mol–1

(i) Write an expression for the equilibrium constant, Kc, for this reaction and give its units.

KNH

N Hmust be square bracketsc

[ ]

[ ][ ]( )3

2

2 23

1

mol–2 dm6 (1) units marked consequentially on above Kc ( or Kp if used) the unit mark can be obtained if the correct units are mentioned in (ii)

(ii) When 3 mol of hydrogen and 1 mol of nitrogen were allowed to reach equilibrium in a vessel of 1 dm3 capacity at 500C and 1000 atm pressure, the equilibrium mixture contained 0.27 mol of N2, 0.81 mol of H2 and 1.46 mol of NH3.

Calculate Kc at this temperature.

Kc ( . )

( . )( . )( )

146

0 27 0811

2

3

= 14.86 or 14.9 or 15 (1)

penalise more than 4 sig figs (lose (1))

(4)

(b) Predict and explain the effect of an increase in temperature on:

(i) the value of Kc: Kc decreases (1) since reaction exothermic / reverse reaction endothermic (1) (increase in temperature) moves equilibrium to left / causes endothermic reaction to occur (1)

(ii) the rate of forward reaction.


increases (1) energies / speed of molecules increased (1) more have energy > EAct (1) or sufficient energy to react instead of EAct reference to EAct or sufficient energy to react essential.


4. Benzene and methylbenzene may be separated by fractional distillation. Sketch the general form of the boiling point/composition diagram for such a mixture and use it to explain the basis on which fractional distillation rests.

0 1

b.p. of benzene

temperature/ºC

b.p. of methylbenzene

mole fraction of methylbenzene vapour line (1) liquid line (1) tie line (1) vapour richer in more volatile component (1) must be a verbal statement repeat tie lines (1) distillate is benzene (½) and residue is methylbenzene (½) must be a verbal statement

(6) (Total 6 marks)

5. (a) State Hess’s Law enthalpy/heat/heat energy change (1) independent of route (1) allow a clear diagram

(2)

(b) (i) Write an equation the enthalpy change for which would be the enthalpy of formation of zinc sulphide, ZnS. Zn(s) + S(s) ZnS(s) (1) allow 1/8S8

must have correct state symbols

(1)


(ii) In the smelting of zinc ores, the following reaction occurs:

ZnS(s) + 1½O2(g) ZnO(s) H = –441 kJ mol–1

Use this, together with the data below, to calculate a value for the enthalpy of formation of ZnS.

Data: ZnS(s) + 1½O2(g) ZnO(s) H = –441 kJ mol–1

S(s) + O2(g) SO2 (g) H = –297 kJ mol–1 correct answer by any method with full sensible working (5)

if correct answer but no working then (1) only

incorrect answer means max 4 as follows:

(2) for cycle of H statement (penalise (–1) for each error)

(2) for substitution of correct values and signs in the appropriate equation

(penalise (–1) for each error)

Example Zn + S ZnS

–348

ZnO + SO

–297 –441

2

OR

ZnS + 1½O2 ZnO + SO2

Hreact = Hf[ZnO] + Hf[SO2] –Hf[SnS]

–441 = – 348 – 297 – Hf[ZnS]

Hf[ZnS] = –204 ignore units

(5)

(c) One way of utilising he large quantities of sulphur dioxide formed in reaction such as that in (b) is to convert it into sulphur trioxide thus:

2SO2 (g) + O2 (g) 2SO3 (g) H = –98 kJ mol–1

Using your knowledge of Le Chatelier’s principle (which need not be stated), state and explain the effect on the position of equilibrium of:

(i) increasing the temperature at constant pressure; left (1) OWTTE

direction of endothermic reaction/

K decreases because H –ve (1)

(2)

(ii) increasing he total pressure at constant temperature. right (1) OWTTE

number of gas molecules decreases (1)

reduces pressure because gas molecules relate to pressure or volume (1)

marks can be awarded for Kp explanation



6. One way of utilising he large quantities of sulphur dioxide formed in reaction such as that in (b)

is to convert it into sulphur trioxide thus:

2SO2 (g) + O2 (g) 2SO3 (g) H = –98 kJ mol–1

Using your knowledge of Le Chatelier’s principle (which need not be stated), state and explain the effect on the position of equilibrium of:

(i) increasing the temperature at constant pressure; left (1) OWTTE

direction of endothermic reaction/

K decreases because H –ve (1)

(2)

(ii) increasing he total pressure at constant temperature. right (1) OWTTE

number of gas molecules decreases (1)

reduces pressure because gas molecules relate to pressure or volume (1)

marks can be awarded for Kp explanation

(3) (Total 5 marks)

7. Hydrogen and iodine react together to give an equilibrium:

H2(g) + I2(g) 2HI(g)

(a) Write an expression for Kp for this equilibrium, giving consideration to its units. correct expression for Kp (1) no square brackets

must state “no units” or atmatm

2

2 (1)

(2)

(b) When 0.50 mol of I2 and 0.50 mol of H2 were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of I2 were found to be present when equilibrium was established.

(i) Calculate the partial pressures of I2 and 0.50 mol of H2 were mixed in a closed container at 723 K and 2 atm pressure, 0.11 mol of I2 were found to be present when equilibrium was established. moles HI = 0.78 (1)

calculation of the mole fraction(s) (1)

calculation of the partial pressure(s) (1)

(3)

(ii) Hence calculate the value of Kp at 723 K.

Kp = (1.56)

(0.22)

2

2 (consequential on values in (b)(i) and expression in (a)) (1)

= 50.28 or 50.3 or 50 (1) this answer only

(2)

(c) In an experiment to establish the equilibrium concentrations in (b), the reaction was allowed to reach equilibrium at 723 K and then quenched by addition to a known, large volume of water. The concentration of iodine in this solution was then determined by titration with standard sodium thiosulphate solution. (i) Write an equation for the reaction between sodium thiosulphate and iodine.


2Na2S2O3 + I2 Na2S4O6 + 2NaI

species (1)

balance (1) conditional on correct species

ionic equation acceptable

(2)

(ii) What indicator would you use? Give the colour change at the end point. starch (1)

blue/black to colourless (1)

not white or clear instead of colourless

(2)

(iii) In this titration and in titrations involving potassium manganate(VII), a colour change occurs during reaction. Why is an indicator usually added in iodine/thiosulphate titrations but not in titrations involving potassium manganate(VII)? ease of discernibility compared (1)

actual colour change for iodine / thiosulphate

ie yellow colourless (1)


8. (a) (i) Kc = ][

][

42

22

ON

NO (1) mol dm–3 (1)

must be [ ] and unit consequential on expression given for Kc if incorrect expression but correct units for this expression then award the units mark if give correct expression for Kp and correct units then award the units mark 2

(ii) N2O4 2NO2 1 – x 2x

[N2O4] = (1) 10

)1)(5.01(3dm

mol = 0.05 mol dm–3

[NO2] = 10

(1) 0.5 2 x = 0.1

Kc = 3-

2–3

0.05

) 1.0(

dm mol

dm mol (1) = 0.2 (mol dm–3)

for full credit the candidate must show how concentrations have been calculated candidates cannot score full marks if the answer at the end is incorrect even if the process marks are all correct if KC expression is incorrect in (a)(i) then mark consequentially (max 3)

correct answer with no working (1) only 4

(iii) H = 2Hf ( NO2) – Hf(N2O4) = 2 (+33.9) – (+9.7) = +58.1 kJ mol–1 cycle or some working (1) answer (marked consequentially on working) (1) ignore units 2

(iv) amount of NO2 increases / equilibrium moves L R (OWTTE) (1) increasing temp increases Kc / reaction endothermic (1) part (iv) is consequential on answer to part (iii) 2


(b) decrease in pressure/increase in volume/ use bigger vessel (1)

not remove the NO2 formed

to encourage equilibrium to move to the side with more molecules/

L.C.P. lower pressure favours side with the more molecules (1) 2

(c) (i) no change (1) 1

(ii) no change (1) 1

(iii) faster / takes less time (1) 1

(d) bond enthalpy of N2 is very high / high activation energy / the NN triple bond is strong / the bonding in the nitrogen molecule is strong / mixture of N2 + O2 is kinetically stable (1)

reaction very slow at reasonable temperature/ endothermic reaction/ needs high temperature/ or high cost low yield arguments (1) 2

[17]

9. (a) (i) ]][[

][

22c

2

IHHI

K (1) 1

(ii) to freeze reaction at equilibrium (composition) (1) 1

(iii) 2S2O32– + I2 S4O62– + 2I– (1) mol S2O32– = 0.02 × 0.5 = 0.01 (1) mol iodine = 0.02 × 0.5 × 0.5 × 10 (1) = 0.05 in 1dm3 [H2] = 0.05 mol dm–3 also (1)

[Hl]eqm = ]][[ 22c IHK (1)

= 05.005.054 xx = 0.37 mol dm–3 (1) if ratio not 1:2 then (max 4) (if ratio is 1:1 then answer = 0.735) if do not multiply by 10 then (max 5) (answer = 0.037) if 1/Kc in part (a)(i) then mark (iii) consequentially 6

(b) (i) oxidation of I– by atmos O2 (1) is feasible because E is +0.69 V (1) 2

(ii) NaCl + H2SO4 NaHSO4 + HCl (1) not Na2SO4 1

(iii) any acceptable equation (2) to give H2S or SO2 or S ie (1) for S / SO2 / H2S + I2 (1) for balanced equation I– larger than Cl– (1) and therefore more easily oxidised or Hl a better reducing agent than HCl (1) 4

(c) (i) triple [I2] rate × 3 (1) [I2]1 (1) same argument for H2 (1) rate = K [H2][I2] (1) consequential if both I2 and H2 1st order but no argument (max 2) 4


(ii) order of reaction has to be determined experimentally (1)

depends on mechanism not stoichiometry (1) 2

(iii) rds = slowest step identified (1) suitable mechanism (3) 4

[25]

10. (a) lower temperature results in lower energy (1) fewer effective collisions/ not enough collisions with activation energy/ fewer successful collisions (1) rate slower (1) candidates can argue whole case from opposite point of view and explain why it is carried out at the high temperature 3

-

(b) some reference to cost (1) not danger increase in yield does not justify cost of higher pressure (1) 2

-

(c) (i) increased rate (1) provides alternative route of lower activation energy / molecules H2 + N2 held on surface of catalyst and bonding in molecules weakened (1) not “it lowers the activation energy” 2

(ii) no change / no effect / nothing happens (1) catalysts do not influence position of equilibrium / catalyst only affects rate / catalyst affects both reactions equally (1) 2

(d) ammonia liquifies (1) removed from mixture (1) gases recycled (1) 3

[12]

11. (a) (i) H = 436 + 151 - 2(299) = -11 (kJmol–1)

2

(ii) marked consequentially on answer to (a)(i) correct relationship reactants H2 + I2 and products 2HI (1) hump (1) 2


(b) (i) H correctly labelled on diagram in right direction (1) 1

(ii) Ea (f) correctly labelled on diagram (1) 1

(iii) Ea (r) correctly labelled on diagram (1) 1

(c) Cl–Cl bond stronger than I–I (1) EA greater (1) however if candidate makes a sensible chemical statement and makes a correct deduction based on this statement (2) (an example of this might be free radical reaction with chlorine

in sunlight quick so activation energy must be lower) 2

(d) (i) KC = ]][[

][

22

2

IH

HI (1) if ( ) or KP or use HCl (0) 1

(ii) This is consequential on (d)(i)

H2 + I2 2HI a a 0 a–x a–x 2x 0.4 0.4 3.0 (1)

KC = )25.0/4.0)(25.0/4.0(

)25.0/3( 2

(1) to gain this mark there must be some reference

to volume in the calculation and it could be in words

KC = 2

2

)4.0(

)0.3( = 56.3 (1)

answer must be given to 2–4 sig figs 3

(e) (i) one/ first (1) 1

(ii) rate = k[H2][I2]

1.5 × 10–5 = k[0.1/0.5][0.1/0.5] (1)

k = 3.75 × 10–4 (1)

units: 33

13

moldmmoldm

smoldm= mol–1dm3s–1 (1) 3

[17]

12. (a) (i) Kc = OH]HCOOH][C[CH

O]][HHCOOCCH[

523

2523 (1)

Must be [ ] 1

(ii) Kc = )235.0/1.2)(235.0/1.0()235.0/9.0( 2

(1) = 3.86 or 3.9 or 4.0 (1)

NO units (1) To gain both marks for the calculation candidate must show the division by the volume 0.235 dm3 or 235 cm3 or state that volumes cancel, to show that they have used concentrations in the Kc expression 3


(iii) Yield of ester increases or equilibrium moves to right (1)

Some correct justification based on constant value of Kc or on Le Chatelier's principle which must have some elaboration eg ‘oppose change ‘in acid concentration’ (1) 2

(b) No effect on position of equilibrium (1) Catalyst only affects the rate (1) 2

(c) (i) Understanding of the idea that the value of H = difference between energy needed to break bonds and energy given out as new bonds form (1) Use of idea to show that since the same bonds are broken as are formed then H=0 (1) 2

(ii) No effect (1) (conditional on some reasoning) Since H=0 (K value independent of temperature) (1) 2

(iii) Reaction with sodium hydroxide is not an equilibrium reaction (1) so the yield better or the reaction goes to completion. (1) 2

[14]

13. (a) Rates of forward and reverse reactions equal (1) No net reaction or reaction continues in both directions or no change in concentrations (of products or reactants) (1) 2

(b) Kc = 1]O[]SO[

]SO[

22

2

23 mol–1 dm3 (1) or 3–dmmol

1 2

(c) (i) Moves to LHS or endothermic direction (1) justification (1) e.g the forward reaction is exothermic 2

(ii) No effect on equilibrium position (1) only affects rate or does not affect value of K or both forward and reverse reaction rates changed by same amount (1) 2

(d) (i) rate increased (1) greater proportion of molecules have necessary activation energy or more collisions are successful or effective (1) 2

NOTE ‘molecules have greater activation energy’ or ‘collision frequency’ increases.

(ii) rate is increased (1) mechanism altered to one of lower Ea (1) 2

There needs to be some reference to an alternative route for the second mark. The words “lowers the activation energy” do not score.

(e) Read the candidate’s whole answer; if full marks cannot be given then look for the individual points.

Award 1st mark for correct explanations of thermodynamic stability (or instability) Award 2nd mark for correct explanation of kinetic stability. Award the 3rd mark for applying these definitions to THIS system. thermodynamic: products at lower energy level than reactants thus thermodynamically unstable(1) can be argued the other way

Kinetic: large Ea or means very few molecules have enough energy to react (1) so reaction is very slow or is kinetically stable (1) 3


(f) Time taken to reach equilibrium yield is not compensated by a proportionate increase in

yield or product is removed or not a closed system (1) 1 [16]

14. (a) (i) Reagent: potassium dichromate (VI)/potassium manganate (VII) (1) or formulae sulphuric acid or hydrochloric acid (1) or formulae

If potassium manganate(VII) chosen not HCl or conc H2SO4 for second mark

‘Acidfied dichromate’ or H+ / Cr2O72–(1) 2

(ii) amount of propanol = 5.67/60 = 0.0945 mol (1) amount of propanoic acid produced = 0.64 × 0.0945 = 0.06048 mol (1) yield of propanoic acid = 74 × 0.06048 = 4.5 / 4.48 / 4.476 g (1)

OR by mass ratio: ratio acid/alcohol = 74/60 = 1.23 (1) 100% yield = 1.23 × 5.67g = 6.99 g (1) 64% yield = 6.99 g × 0.64 = 4.5 / 4.48 / 4.476 g (1) 3

(b) (i) increase in temperature:

(position of ) equilibrium goes to the right (1) as endothermic left to right (1) on the addition of sodium propanaoate the position of equilibrium goes to left (1) higher concentration of / more propanoate ions or sodium propanoate produces propanoate ions (1) 4

(ii) pH rises (consequential on above) (1) 1

(c) (i)

14

12

9

6

3

10 20 30 40 50

start pH 2 to 4 ( )

correct general shape ( )

vertical 6/7.6 to 10/12 ( )

at 25 cm ( )

1

1

1

13

4

(ii) indicator : thymol blue (1) consequential on vertical part of graph

reason: pH change sharp around pKin value / its colour changes around end point pH / band pH8 to 10 shown on graph (1) 2


(d) (i) fully dissociated and reactions identical

OR H+ + OH– H2O (1) 1

(ii) HCN weak acid / partially dissociated (1) Hionisation of HCN endothermic (1) 2

[19]

15. (a) Still reacting / rate of forward reaction and backward reaction equal / implication that forward and backward reactions are still taking place (1)

But concentrations constant / no macroscopic changes (1) 2

(b) Temp (Increases) Left / to SO2 / to endothermic / lower yield (1)

Press Increases/faster (1) Right to SO3 / to smaller number of molecules (1) 3

(c) (i) Increases rate / or suitable comment on rate (1)

Moves position of equilibrium to endothermic side / or suitable comment on equilibrium such as reasonable yield / less SO3 (1)

Either compromise in which the rate is more important than the position of equilibrium or optimum temperature for catalyst to operate or valid economic argument (1) 3

(ii) Increases rate / more SO3 / only needs small pressure to ensure gas passes through plant / high or reasonable yield obtained at 1 atms or at low pressure anyway (1) and references to economic cost against yield benefit e.g increase in pressure would increase yield of product but the increase in yield would not offset the cost of increasing the pressure (1) 2

(iii) Catalyst speeds up reaction (1) 1

(d) Vanadium (V) oxide / vanadium pentoxide / V2O5 (1) 1

(e) Any one use production of fertilizers, detergents, dyes, paints, pharmaceuticals (in) car batteries, pickling metal 1

[13]

16. (a) (i) Kp = p(NO2)²/p(N2O4) (1) units are atm or any SI unit of pressure (1) (2)

(ii) Kp = p(CO2) (1) units are atm or any SI unit of pressure (1) (2) 4

(b) (i) N2O4 NO2

moles at start 1 0 change –x + 2x total moles = 1 + x but x = 0.81 equilibrium moles 1 –0.81 = 0.19 1.62 total moles = 1.81 mole fraction 0.19 /1.81 = 0.105 1.62 /1.81 = 0.895 partial pressure /atm 0.105 × 1.2 = 0.126 0.895 × 1.2 = 1.07

marking points


For correct equilibrium moles (1) for consequential total moles (1) for consequential mole fractions (1) for partial pressure N2O4(1) for partial pressure NO2 (1) (5)

(ii) Kp = (1.07)² /0.126 = 9.1 (atm)(accept 9.2) (1) 6

(c) (i) Kc =initial2initial2

initial3

][O]²[SO

]²[SO (1)

(ii) Catalyst has no effect on the position of equilibrium (1)

(iii) .(0.010/2)(0.40/2)

(0.20/2)

][O][SO

][SO2

2

22

2

23 (1) = 50

this is the value of Kc (1.7 × 106) or this is not = Kc (1) so reaction will move to the right to increase value of [SO3] /or to get value up to 1.7 × 106 (1) (3) 5

[15]

17. (a) (i) Equilibrium shifts to left / lower yield (1) must be clear not implied more (gas) molecules on right (1) 2

(ii) Equilibrium shifts to left / lower yield (1) must be clear not implied forward reaction exothermic / shift to endothermic direction / moves to absorb heat / rate of reverse reaction increases more than rate of forward reaction (1) 2

(i) increased rate (1) Molecules closer together / less space between molecules / higher concentration / same number of molecules in less space (1) more collisions/ collide more frequently (1) 3

(ii) Increased rate (1) More molecules /collisions have (at least) the activation energy (1) More of the collisions successful / more of the collisions lead to reaction (1) 3

(c) (i) Platinum (plus rhodium) / Pt (Rh) (1) 1

(ii) Alternative route (1) With lower activation energy (1) this mark is consequential on previous mark Increase in rate because there are more successful collisions (1) 3

(iii) Increased surface area / more active sites (1) 1 [15]


18. (a) (i) Dynamic:

reaction occurring in both directions / rate of forward reaction and reverse reactions equal (1) Equilibrium: constant concentrations / no change in macroscopic properties (1) 2

(ii) all substances in same phase / are all in the gaseous state (1) 1

(b) (i) Higher yield of ammonia / (equilibrium position) moves to. r.h.s (1) Fewer product molecules (1) 2

(ii) Lower yield of ammonia / (equilibrium position) moves to 1.h.s. (1) since this absorbs heat/ shift in endothermic direction / the reaction is exothermic (1) 2

(c) (i) 350 – 500C /623 – 773 K (1) 1

(ii) High temp favours high rate (1) (*) Low temp favours good yield (1) (*) (*) Or reverse argument Temperature used compromise / balance between yield and rate (1) 3 consequential on first two points correct

(d) (i) Iron (not Fe) (1) 1 ignore references to oxides

(ii) Provides alternative pathway / route or Explanation of what happens at the surface (1)

of lower activation energy (1) 2 Second mark consequential on the first

[14]

19. (a) (i) fraction of the total pressure generated by a gas or or pressure gas would generate if it alone occupied the volume or Ptotal × mol fraction (1) 1

(ii) O)p(H)p(CH)p(Hp(CO)

K24

32

p

(1) not [ ] 1

(iii) Increase in total pressure will result in less product molecules in the equilibrium mixture / equilibrium moves to left (1) because more molecules on product side of the equilibrium than on left (1) 2

(b) (i) No change (1) 1

(ii) KP increase (1) 1

(iii) No change (1) 1

(c) (i) )p(CH1K

4p (1) 1

(ii) 9.87 × 10–3 kPa–1/ 9.87 × 106 Pa1 consequential on (i) (1) 1 Allow 3 – 5 sig fig

(iii) equilibrium has moved left in favour of gas (1) exothermic going left to right/in the forward direction / as


written (1) Stand alone 2

(iv) Answer yes or no with some sensible justification (1) e.g. No the costs would not justify the amount produced 1

[12]

20. (a) The marks are for:

writing the expression for K

substituting correctly

calculating p(SO3)

correct generation of the ratio

calculation of the ratio to give answer which rounds to 95 t

Kp = pSO32/ pSO2

2 pO2 (= 3.00 × 104) (1)

3.00 × 104 = pSO32 / 0.1 0.1 0.5 (1) if no expression for Kp is

given this correct substitution can score 2 marks

pSO32 = 150

pSO3 = 12.25 (1)

Ratio of SO3 = 0.5) 0.1 (12.25 100% 12.25

(1) = 95% (1) 5

(b) (i) The marks are for

Recognizing the existence of hydrogen bonds ( between molecules) (1)

That each molecule can form more than one hydrogen bond because of the two OH (and two S=O groups) / or a description of hydrogen bonds in this case / or a diagram showing the hydrogen bonds (1)

That hydrogen bonds make for strong intermolecular forces (and hence high boiling temperature) which requires higher energy to break / separate molecules (1) 3

(ii) If water is added to acid heat generated boils and liquid spits out (1) if acid added to water the large volume of water absorbs the heat generated (and the mixture does not boil) (1) 2

(c) (i) pH = –log10 (0.200) = 0.70 (1) allow 0.7 or 0.699 1

(ii) realising that the first ionisation / dissociation of sulphuric and that of HCl are both complete (1)

that the second ionisation of sulphuric is suppressed by the H+ from the first (1)

little contribution from 2nd ionisation so reduces the pH very little / increases the [H+] very little (1) 3


(d) (i) Lead equations 1 mark

Pb + H2SO4 PbSO4 + 2H+ + 2e– (1) or

Pb + SO42 PbSO4 + 2e–

Lead(IV) oxide equations 2 marks PbO2 + H2SO4+ 2H+ + 2e– PbSO4 + 2H2O or

PbO2 +SO42– + 4H+ + 2e PbSO4 + 2H2O+

Species (1) balancing (1) 3

(ii) PbO2 + Pb + 2H2SO4 2PbSO4 + 2H2O (1) 1 [18]

21. (a) (i) Rates of forward and back reactions the same (1). 1

(ii) Moves equilibrium to left hand side (1) being the side with the smaller number of molecules/moles (1) 2

(b) (i) 2NO + O2 2NO2(1) or N2O4 1

(ii) 4NO2 + 2H2O + O2 4HNO3 (2) (1) species (1) balance. 2 Answer could be in terms of a reactions that give HNO2 and HNO3 as the first stage. i.e. 2NO2+ H2O HNO2 + HNO3 (2HNO2 + O2 2HNO3) or 3NO2+ H2O 2HNO3 + NO

(c) (i)

Fractionofmolecules

E cat

E uncat

Energy

a

a

Axis labels (1) Starts at or near origin (not on y or x axis) skewed distribution

that is reasonably asymptotic to the x-axis (1)


(ii) Ea for the uncatalysed reaction shown well to the right of the

peak and Ea for catalysed reaction to the left of this, still to the right of the peak (1) Some comment concerning the areas under the curve to right of

Ea lines or labelled shading (1) Greater number of collisions (or particles) have energy greater

than the activation energy / enough energy to react (1) Therefore greater number of successful collisions (1). 4

[12]

22. (a) (i) Kc = [SO3]2 / [SO2]2 [O2] (1) 1

(ii) 30

2.0

60

1.0

60

8.1

= 3.33 103 1.67 10–3 0.03 (1)

Kc = 323

2

1067.1)1033.3(

)03.0( = 4860or 4.86 104 (1)

mol1 dm3 (1) 3

(b) (i) Kc decreases (1) 1

(ii) shifts to left / in reverse (1) 1

(c) (i) no effect (1) 1

(ii) no effect (1) 1

(d) (i) Kp = pSO32 / pSO2

2 pO2 (1) penalise square brackets 1

(ii) Total number of moles (1) consequential on a (ii) SO2 = 0.0952(4); O2 = 0.0476 (2); SO3 = 0.857 (1) (1) 2

(iii) Partial pressures: SO2 = 0. 190 (5) atm; O2 = 0.0952 (4) atm; 1 SO3 = 1.71(4) atm (1) i.e. multiply answer in (ii) by 2

(iv) 1.7142 / 0.19052 0.09524 = 850 (1)

atm1 (1) 2 [14]

23. (a) A series of compounds with the same functional group / same or similar chemical properties (1)

NOT same (physical) properties (0)

NOT same physical and chemical properties (0)

with the same general formula (1)

where one member differs from the next by CH2 (1) 3

(b) (i) It will move to the right (1)

Which is the exothermic direction (1) 2 2nd mark conditional upon first statement


(ii) The rate of reaction would be too slow if cooled (1)

yield too small / too little reactants converted

OR converse arguments (1) 2 NOT equilibrium yield

(c) (i)

H H

C C C CH H

H H 1

(ii) No, as both the end carbons have two H atoms 1

(iii) H H H H

C C C CH H

OH OH OH OH

Penalise C-HO once

Or CH2(OH)CH(OH)CH(OH)CH2OH (1) for adding two OH groups across one double bond, and (1) for adding it across both. Consequential on structure in (c)(i) 2

1 mark per double bond max 2 [11]

24. (a) Kc = (1)

224

422

OHCH

]H[]CO[

Starting amounts:

(1)

mol0.3molg18/g54OH

mol625.0molg16/g10CH1

2

14

Equilibrium amounts: CH4 = 0.625 – ¼ × 2.0 = 0. 1 25 mol (1)

H2O = 3.0 – ½ 2.0 = 2.0 mol (1) CO2 = 0.0 + ¼ × 2.0 = 0.500 mol (1)

H2 = 2.0 mol (given) Equilibrium concentrations:

above values ÷ 4 dm3 (1) (mark consequently)

[CH4] = 0.03125 mol dm-3

[H2O] = 0.500 mol dm-3

[CO2] = 0.125 mol dm-3

[H2] = 0.50 mol dm-3

2

4

2

eq2eq4

4eq2eq2

c500.003125.0

50.0125.0

]OH[]CH[

]H[]CO[K

= 1.0 (or 1 or 1.00) (1) mol2 dm-6 (1) 8

(b) H = (-394) – [(-76) + (2 × – 242)] (2) 1 mark for × 2, 1 mark for signs and values

= + 166 (1) kJ mol–1 3 166 scores zero

(c) A catalyst (of nickel) is used because the reaction, even at 750°C, is too slow / to speed up the reaction (1)


Then any six of the following eight points: a temperature of 750°C is used: as the reaction is endothermic (1) a high temperature increases the value of the equilibrium constant (1) and so increases the equilibrium yield (1) a high temperature also favours a fast rate (1) but a temperature > 750 would be too expensive / cause engineering problems (1) Temperature could score up to 5 but max 6 for T and P combined.

If their calculation in (b) gives an exothermic answer, mark consequentially [exothermic (1), decreases Kp (1), decreases yield (1) but faster rate(1), so 750 is compromise of fast rate and lower yield (1)] A pressure of 30 atm is used even though the reaction goes from 3 to 5 gas moles / more gas moles of

right of equation (1) causing a decrease in equilibrium yield (1) but a moderately high pressure is needed to push the gases through the 7

plant (1) Ignore any reference to rate Pressure could score up to 3 but max 6 for T and P combined. Do not give all 7 marks unless the candidate has expressed their ideas clearly

[18]

25. (a) (i) The heat (energy) / enthalpy change when 1 mol of gaseous atoms (1) NOT “energy” on its own is formed from the element (1) 2

(ii) ½ Cl2 (g) Cl (g) 1

(iii)

Sr (g) 2Cl (g)

2Cl(g)

Sr (g)

Sr(g)

Sr(s) + Cl (g) SrCl (s)2 2

+

2+ –

Hf

Cycle marks: 1 mark for cycle showing all balanced species 1 mark for all state symbols correct Allow Sr(g) Sr2+(g) as a single stage


Calculation marks:

Hf = sum of other terms and / or – 829 = +164 +550 +1064 + 2 x ( +122) + 2 x ( – 349) + LE or LE= –829–164–550–1064–2 x(+122) – 2x(–349) 2 ie 1 mark for multiplying Ha of Cl by 2 1 mark for multiplying EA of Cl(g) by 2

LE = – 2153 (kJ mol–1) (1) The answer mark is consequential on whether they multiply neither or just one of Ha and EA by 2 , but not consequential if they have wrong signs. 5

(b) (i) + 737 = Hf SrCl + 122 – ( – 829) OR correct cycle (1)

Hf SrCl = – 214 (kJ mol–1) (1) 2

(ii) As the reaction (left to right) is endothermic (1) a decrease in temperature will decrease the value of K (1) thus the equilibrium position will shift to the left (1) 3

(iii) Kp = psrCl × pCI (1) MUST be stated moles of SrCl = moles of Cl (1) – This can be stated or implied pSrCl = pCl = ½ × 4.2 = 2.1 (atm) (1)

Kp = 2.1 atm × 2.1 atm = 4.4(l) (1) atm2 (1) Units are NOT consequential on wrong Kp 5

[18]

26. (a) (i) Al2Cl6 2 AlCl3

Initial: 2 0 Equil: 0.5 3 (1) Equil conc: 0.05 0.3 (1)

Kc = 62

23

ClAl

AlCl (1)

= 0.32 / 0.05 mol dm–3 = 1.8 mol dm–3 (1) 4

(ii)

Al Al

Cl ClCl

Cl ClCl

(1) MUST be 3 – D around Al atoms Covalent bonds shown (1) Dative bonds shown (1) 3


(b) (i)

+ C H Cl + HCl

C H

2

2

5

5

1

(ii)

C H

C H2 3

2

2

5

5

3

3

+

+

–

+

CH CH H ( :ClAlCl )

(1)

(1)

(1)

+ HCl + (AlCl ) + H

4

(iii) Ethene has available electrons no delocalisation Benzene has (six) delocalised electrons (1) By substitution benzene regains its delocalisation resonance energy (1) but with ethene more energy is released by / single bond formation (than with reforming bonds) (1) 3

(c) (i) Hydrolysed by water of crystallisation / deprotonated by Cl– and not sufficient water present to dissolve HCl (1) Aluminium oxide (or hydroxide) (1) 2

(ii) Al(H2O)6 3+ deprotonated (1) by solvent water (1) giving H3O+ (1) Last two could be given as an equation. 3

[20]

27. (a) Pressure exerted by the gas if it alone occupied the same volume at the same temperature/mole fraction × total pressure 1

(b) (i) Kp = 2

22

)NO(

)O()N(

p

pp 1

(ii) Correct number of moles (1) Correct mole fractions (1) Correct partial pressures (1) 2.45 × 103 (1) ACCEPT 2–4 SF 4


(c) Kp increases (1)

Equilibrium moves to r.h.s. (1) which is the exothermic direction (1) 3

(d) (i) Kp = p (Ni(CO)4) / p(CO)4 1

(ii) High partial pressure with some reason (1) so the pressure Ni(CO)4 increases to keep Kp constant. (1) 2

[12]

28. (a) (i) rate forward = rate back (1) no change in concentration/partial pressure/amount (1) 2

(ii) (%) increases (1) 1

(iii) None (1) 1

(b) If temperature too low, rate too slow/high temperature gives fast rate(1), but if too high, yield too small/high yield needs low temperature (1) therefore a compromise temperature (of 450 °C) (1) and catalyst for quick rate (at temperature 450 °C)/catalyst not effective if temperature below 400 °C (1) 4

(c) In car batteries (1) ) manufacture of: ) fertilisers / ammonium sulphate/phosphates (1) ) paints (1) ) detergents (1) ) Any one car batteries (1) ) man–made fibres (1) ) dyes (1) ) explosives (1) ) 1

[9]

29. (a) (i) Kc = OH]H)COOH][CCH(NH[CH

O]][HH)COOCCH(NH[CH

5223

25223 IGNORE minor slip in formulae 1

(ii) Bonds broken: O−H and C−O Bonds made: C−O and O−H (1) Notes that there is no change (and therefore ∆H is zero) (1) OR Bonds broken + (464 + 358) = (+) 822 And bonds made − (358 + 464) = (−) 822 (1) therefore ∆H = 0 OR correct signs (1) 2

(iii) No effect (1) Increases (1) 2


(b) (i)

H C C C

H N

HHH

+

H HO

O–

Do not accept +ve charge on covalently bonded H inNH but OK if dative covalent bond to H+ +

3

OR

O

C

O

–

ALLOW CH3CH(NH3+)COO– / CH3CH(NH3

+)CO2– brackets can be

omitted 1

(ii) Attraction between (ionic) charges on different ions/zwitterions (is strong) 1

(iii)

H C C COO + H H C C COOH

H C C COO + OH H C C COO + H O

H

H

H

H

3

3

3

3

3

2

3

3

3

2

3

3 3

3

2

+

+

+

+

+

+–

– ––

NH

NH

NH

NH

(H CCH(NH )COOH) H CCH(NH )COOH

ACCEPT variations starting with H C C COOH

H

NH

MUST be balanced equations Can use NaOH or HCl etc 2

(c) (i) Non–superimposable on its mirror image OR has no plane of symmetry / it has an asymmetric carbon atom NOT “4 different groups on a C atom” on its own 1

(ii) One diagram correct and 3D (1) Mirror image (1) – can be awarded if 1st mark not given because of a nearly correct structure eg ester 2

(iii) Rotation of the plane of (plane)–polarised light in opposite directions 1 [13]

30. (a) vanadium(V) oxide/V2O5 / divanadium pentoxide / vanadium pentoxide 1 NOT vanadium oxide


(b) (i) 400-500 °C / 673-773 K

[any temperature or range of temperatures within these ranges] 1

QWC(ii) rate increases (1) molecules/particles (NOT atoms) have higher (kinetic) energy (1) more molecules / particles / collisions have activation energy / enough (1) energy to react Greater proportion / more collisions are successful / results in a (1) reaction / higher frequency of effective collisions 4th mark not stand alone and must be linked to 3rd mark If no reference to Ea max 2 4

If just talk about increase in number of collisions max 2

(iii) yield decreases (1) because reaction is exothermic /equilibrium shifts to endothermic (1) 2 direction / moves to absorb heat / reverse reaction is endothermic ALLOW K decreases with increase in temperature

(c) (i) 2-5 atm (any number or range within this range) / just above atmospheric 1

(ii) Pushing it through the system (1) (1) Higher pressure would increase yield (1) But yield is high even at this pressure (1) Max 3 (1) Higher pressure too expensive (1) (1)

Increased cost of the extra pressure is not justified by the extra SO3 produced (2) 3 IGNORE reference to rate

(d) (SO3) dissolved / absorbed in conc. H2SO4 OR dissolved in H2SO4 to form oleum (1) if % acid given, must be 95 or above (1) water added – not stand alone

H2SO4 + SO3 H2S2O7 and H2S2O7 + H2O 2H2SO4 for both marks OR 2 SO3 reacts with the water in conc. H2SO4 for both marks

(e) 2NH3 + H2SO4 (NH4)2SO4 1 Allow correct equation based on NH4OH

(f) any one use: making detergent / soap / paint / pigment inc TiO2 / dyestuffs / fibres / plastics / pharmaceuticals (in) car batteries, pickling metal / anodising Al / electrolytic refining of copper 1

[16]

31. (a) (i) KP = p (CO2) allow without brackets, IGNORE p [ ] 1

(ii) 1.48 (atm)

Penalise wrong unit

Answer is consequential on (a) (i) e.g. 48.1

1 must have atm–1 1

(b) (i) Kp = 2

22

(p(NOCl))

p(NO))p(Cl allow without brackets, penalise [ ] 1

(ii) 2NOCl 2NO + Cl2


Start 1 0 0 –0.22 +0.22 +0.11 eq moles 0.78 0.22 0.11 (1) total moles of gas 1.11 mole fractions above values ÷1.11 (1) 0.7027 0.1982 0.09910 partial pressure / atm above values × 5.00 (1) 3.51 0.991 0.495

Kp = 2

2

atm) 51.3(

atm) (0.991 atm 495.0 (1)

= 0.0395/ 0.0394 atm (1) range of answers 0.0408 / 0.041 0.039 / 0.0392 NOT 0.04 ACCEPT 2 S. F Correct answer plus some recognisable working (5)

Marks are for processes

Equilibrium moles

Dividing by total moles

Multiplying by total pressure

Substituting equilibrium values into expression for KP 5

Calculating the value of KP with correct consequential unit.

(iii) As the reaction is endothermic – stand alone (1)

the value of KP will increase (as the temperature is increased) - (1)

consequential on 1st answer (if exothermic (0) then KP decreases (1)) For effect on KP mark, must have addressed whether reaction is endothermic or exothermic 2

(iv) Because (as the value of KP goes up), the value of

pCl2 × (pNO)2 / (pNOCl)2 (the quotient) must also go up (1) and so the position of equilibrium moves to the right – stand alone (1)

But mark consequentially on change in K in (iii) If “position of equilibrium moves to right so Kp increases” (max 1) 2 IGNORE references to Le Chatelier’s Principle

[12]


32. (a) KC =

2322

22

COOHCHOHCHHOCH

OHformulaitsor ester

(1)

both molar masses (1) HOCH2CH2OH + 2CH3COOH Ester +2H2O Moles at start 24.8 / 62 = 0.400 66.0 / 60 = 1.10 (1) Moles at equilibrium 0.400 – 0.320 = 0.080 1.10 – 0.640 = 0.460 0.320 0.640 (1)

Concentration at equilibrium divide above by 0.0900 dm3 (1)

0.080 / 0.0900 = 0.889 5.11 3.56 7.11 OR explain why volume cancels in this case

KC = 2

2

11.5889.0

11.756.3

=

2.23

180

= 7.76 / 7.8 / 7.74 etc. (1) There are no units for K (1) 7

(b) Amount of ethan-1, 2-diol = 1054 / 62 = 17 mol

Amount of ethene = 28

560 = 20 mol or

1240

1054 × 100 (1)

Yield = 17 × 100 20 = 85% (1) 2

(c) (i) Any acid with two COOH groups or its acid dichloride or its dimethyl ester (1) Accept HOOCCOOH

(Where R = the hydrocarbon part of their diacid). for correct ester linkage drawn out (1) for remainder with continuation (1) 3

(ii) No, because the acid would hydrolyse / is a catalyst for the hydrolysis of the ester. OR Yes, not hydrolysed at low temperature / only hydrolysed at high temperature 1

(d) The ester cannot form (intermolecular) hydrogen bonds but the acid can (1) The ester does not have a +hydrogen atom OR the acid has +hydrogen / polar OH / O and H have a large difference in electronegativities (1) thus less energy / heat is required to separate molecules of the ester (1) but as ethanoic acid has fewer electrons than propanoic acid, (1) it has weaker intermolecular instantaneous induced dipole / induced dipole forces / van der Waals / dispersion / London forces (1) Allow vdW 5

[18]


33. (a) (i)

(1)

cycle with state symbols or as energy level diagram. (1) labels (in symbols, words or numbers) (1) Hsoln = – Hlat + Hhyd Ca2+ + 2 × Hhyd OH– OR values Hlat = –1650 + 2 × (–460) – (–16.2) (1) 4 = – 2553.8

(ii) Solubility increases down the group (1) (if this is wrong, no marks available in this part)

QWC

Hhyd of cation decreases / less exothermic (1) but Hlatt decreases more (1) therefore Hsol gets more exothermic / increases (1) 4

(b) (i) Calcium hydroxide will be less soluble at the higher temperature, (1) (if this is wrong, no marks available in this part) because the reaction is exothermic (left to right) (1) (an increase in temperature will cause a) decrease in the value of K (and (1) hence drive the equilibrium to the left). 3

(ii) The solubility will decrease, (if this is wrong, no marks available in this part) because the addition of OH– ions will increase [OH–] / concentration of OH– (1) driving the equilibrium to the left. (1) 3

(c) The three gases are: hydrogen bromide ) (1) bromine ) OR formulae (1) sulphur dioxide ) (1)

The hydrogen chloride / CaCl2 / chloride evolved is not a strong enough reducing agent to reduce / cannot reduce the concentrated sulphuric acid (or sulphuric not a strong enough oxidising agent to …) (1) 4

[18]


Physical Chemistry_Chemical Equilibria Marks Scheme

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