(2) - KS5Chemistry · When dinitrogen tetroxide, N 2 O 4, dissociates, the following equilibrium is...
Transcript of (2) - KS5Chemistry · When dinitrogen tetroxide, N 2 O 4, dissociates, the following equilibrium is...
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Maltby Academy 1
1. (a) When solid calcium nitrate is heated, brown fumes of nitrogen dioxide, NO2, are seen and
the solid remaining after decomposition is calcium oxide.
(i) Write a balanced equation for the thermal decomposition of calcium nitrate.
......................................................................................................................... (2)
(ii) Describe the changes you would see when cold water is added drop by drop to cold
calcium oxide and give the chemical equation for the reaction.
.........................................................................................................................
.........................................................................................................................
.........................................................................................................................
......................................................................................................................... (3)
(iii) State whether barium nitrate will decompose more easily or less easily than
calcium nitrate on heating with a Bunsen burner.
......................................................................................................................... (1)
(iv) Account for the trend in the thermal stability of the nitrates of the elements in
group 2.
.........................................................................................................................
.........................................................................................................................
.........................................................................................................................
......................................................................................................................... (3)
(b) The brown fumes in part (a) are not pure NO2 but a mixture of N2O4 and NO2.
N2O4(g) 2NO2(g)
Pale yellow dark brown
A transparent glass syringe was filled with the gaseous mixture of N2O4 and NO2 and its
tip sealed. When the piston of the syringe was rapidly pushed well into the body of the
syringe, thereby compressing the gas mixture considerably, the colour of the gas became
momentarily darker but them became lighter again.
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Maltby Academy 2
(i) Suggest why compressing the gases causes the mixture to darken.
.........................................................................................................................
......................................................................................................................... (1)
(ii) Explain why the mixture turns lighter on standing.
.........................................................................................................................
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......................................................................................................................... (2)
(iii) Write an expression for the equilibrium constant, Kp, for this equilibrium.
(1)
(iv) 1.0 mole of N2O4 was allowed to reach equilibrium at 400K. At equilibrium the
partial pressure of N2O4 was found to be 0.15 atm.
Given that the equilibrium constant Kp for this reaction is 48 atm, calculate the
partial pressure of NO2 in the equilibrium mixture.
(3)
(Total 16 marks)
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Maltby Academy 3
2. (a) (i) Calculate the concentration, in mol dm–3
, of a solution of hydrochloric acid, HCl,
which has a pH of 1.13.
(1)
(ii) Calculate the concentration, in mol dm–3
, of a solution of chloric(l) acid, HOCl,
which has a pH of 4.23.
Chloric(l) acid is a weak acid with Ka = 3.72 × 10–8
mol dm–3
.
(4)
(b) The pH of 0.100 mol dm–3
sulphuric acid is 0.98.
(i) Calculate the concentration of hydrogen ions, H+, in this solution.
(1)
(ii) Write equations to show the two successive ionisations of sulphuric acid, H2SO4, in
water.
......................................................................................................................
...................................................................................................................... (2)
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Maltby Academy 4
(iii) Suggest why the concentration of hydrogen ions is not 0.20 mol dm–3
in 0.100 mol
dm–3
sulphuric acid.
......................................................................................................................
...................................................................................................................... (1)
(c) Many industrial organic reactions produce hydrogen chloride as an additional product.
This can be oxidised to chlorine by the Deacon process:
4HCl(g) + O2(g) 2Cl2(g) + 2H2O(g) H = –115 kJ mol–1
.
0.800 mol of hydrogen chloride was mixed with 0.200 mol of oxygen in a vessel of
volume 10.0 dm3 in the presence of a copper(I) chloride catalyst at 400 ºC. At
equilibrium it was found that the mixture contained 0.200 mol of hydrogen chloride.
(i) Write an expression for the equilibrium constant Kc.
(1)
(ii) Calculate the value of Kc at 400 ºC.
(4)
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Maltby Academy 5
(d) State and explain the effect, if any, on the position of equilibrium in (c) of:
(i) decreasing the temperature;
......................................................................................................................
......................................................................................................................
...................................................................................................................... (2)
(ii) decreasing the volume;
......................................................................................................................
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...................................................................................................................... (2)
(iii) removing the catalyst.
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...................................................................................................................... (2)
(Total 20 marks)
3. In the vapour phase sulphur trioxide dissociates:
2SO3(g) 2SO2(g) + O2(g)
(a) (i) Write an expression for Kp for this dissociation.
(1)
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Maltby Academy 6
(ii) At a particular temperature, 75% of the sulphur trioxide is dissociated, producing a
pressure of 10 atm. Calculate the value of Kp at this temperature paying, attention
to its units.
(5)
(b) Solid vanadium(V) oxide, V2O5, is an effective catalyst for this reaction. State the effect
of using double the mass of catalyst on:
(i) the position of the equilibrium;
............................................................................................................................
............................................................................................................................ (1)
(ii) the value of Kp.
............................................................................................................................
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(Total 8 marks)
4. (a) Methane reacts with steam in a reversible reaction. In industry this reaction, carried out at
a pressure of 30 atm, is used to produce hydrogen for the manufacture of ammonia
CH4(g) + H2O(g) CO(g) + 3H2(g) H = +210 kJ mol–1
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Maltby Academy 7
(i) Define the term partial pressure as applied to a gas mixture.
............................................................................................................................
............................................................................................................................ (1)
(ii) Write an expression for the equilibrium constant, Kp, for this reaction.
(1)
(iii) State and explain the effect of increasing the total pressure on the position of this
equilibrium;
............................................................................................................................
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............................................................................................................................ (2)
(b) State the effect on the value of Kp for this equilibrium of the following.
(i) Increasing the total pressure.
............................................................................................................................ (1)
(ii) Increasing the temperature.
............................................................................................................................ (1)
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Maltby Academy 8
(iii) Adding a catalyst.
............................................................................................................................ (1)
(c) There is a theory that methane, CH4, constantly leaks from the earth’s crust. This is not
noticeable on land but at the bottom of a cold sea, such as off the Canadian coast, the
methane is trapped in a solid cage of water molecules.
CH4(g) + 6H2O(s) [CH4(H2O)6](s)
methane hydrate
At –29 °C the equilibrium pressure of the methane is 101.3 kPa.
(i) Write an expression for Kp for this equilibrium.
(1)
(ii) Deduce the value of Kp at –29 °C, stating its units.
(1)
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Maltby Academy 9
(iii) At 0 °C the equilibrium pressure of methane rises to 2600 kPa. What does this tell
you about the effect of temperature change on the position of equilibrium and
about the enthalpy change for this reaction?
............................................................................................................................
............................................................................................................................
............................................................................................................................ (2)
(iv) Some people have suggested collecting the methane hydrate from the bottom of the
sea and allowing it to warm up to 0 °C on board a ship. Comment on whether this
would be a useful method for collecting methane.
............................................................................................................................
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............................................................................................................................ (1)
(Total 12 marks)
5. (a) Define the term partial pressure.
...............................................................................................................................
...............................................................................................................................
............................................................................................................................... (1)
(b) If phosphorus pentachloride is heated in a sealed tube the following equilibrium is set up
PCl5(g) PCl3(g) + Cl2(g)
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Maltby Academy 10
(i) Write an expression for the equilibrium constant, Kp, for the above reaction.
(1)
(ii) At a given temperature T a sample of phosphorus pentachloride is 40.0%
dissociated, the total equilibrium pressure being 2.00 atm. Calculate the partial
pressures of each of the components of the equilibrium mixture, to three significant
figures. Hence calculate the value of Kp under these conditions.
(5)
(c) At a higher temperature, T + t, the equilibrium mixture in (b) contains a greater
proportion of phosphorus trichloride and chlorine.
(i) State, with a reason, whether the dissociation of phosphorus pentachloride is
exothermic or endothermic.
......................................................................................................................
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...................................................................................................................... (2)
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Maltby Academy 11
(ii) How does Kp change when the temperature is increased?
...................................................................................................................... (1)
(d) If calcium carbonate is heated in a sealed vessel in the absence of air the following
equilibrium is set up:
CaCO3(s) CaO(s) + CO2(g)
(i) Write the expression for Kp for this equilibrium.
...................................................................................................................... (1)
(ii) At 1030 °C the total pressure in the vessel is 16 atm. What is the value of Kp?
...................................................................................................................... (1)
(Total 12 marks)
6. When dinitrogen tetroxide, N2O4, dissociates, the following equilibrium is established.
N2O4(g) 2NO2(g)
(a) State a property which could be measured to follow the progress of this reversible
reaction.
.................................................................................................................................... (1)
(b) Write an expression for the equilibrium constant, Kc, for this reaction.
(1)
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Maltby Academy 12
(c) When a sample of 0.0370 moles of gaseous dinitrogen tetroxide is allowed to dissociate at
25 °C in a container of volume 1 dm3, 0.0310 moles of N2O4(g) remain in the equilibrium
mixture.
Complete the table below, and use the data to calculate Kc for the reaction. Include a unit
in your answer.
N2O4 NO2
Number of moles at start 0.0370 0
Number of moles in 1 dm3
at equilibrium 0.0310
Kc calculation:
(3)
(d) The reaction was repeated at a higher pressure, maintaining the temperature at 25 °C.
(i) How does this increase in pressure affect the amount of nitrogen dioxide, NO2(g),
in the equilibrium mixture?
.......................................................................................................................... (1)
(ii) How does this increase in pressure affect the value of Kc?
.......................................................................................................................... (1)
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Maltby Academy 13
(e) The reaction was repeated at the original pressure, but the temperature was increased to
75 °C. The value of Kc was approximately twenty times greater.
How does this information show that the reaction is endothermic?
....................................................................................................................................
....................................................................................................................................
.................................................................................................................................... (1)
(f) Predict the sign of ΔSsystem for the reaction, giving a reason for your answer.
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.................................................................................................................................... (2)
(g) Write the equation for the relationship between ΔSsurroundings and ΔH for the reaction.
(1)
(h) The magnitude of ΔSsystem for the reaction is greater than the magnitude of ΔSsurroundings.
Explain why this must be the case.
....................................................................................................................................
....................................................................................................................................
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.................................................................................................................................... (2)
(Total 13 marks)
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Maltby Academy 14
7. This question concerns the equilibrium
2NO(g) N2(g) + O2(g) H = –180 kJ mol–1
(a) Define the term partial pressure.
............................................................................................................................................
............................................................................................................................................ (1)
(b) (i) Write the expression for Kp for the above reaction.
(1)
(ii) At 1600 C and 1.5 atm pressure NO is 99 % dissociated at equilibrium. Calculate
the value of Kp under these conditions.
(4)
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Maltby Academy 15
(c) State and explain the effect on Kp and hence on the position of equilibrium of decreasing
the temperature at constant pressure.
............................................................................................................................................
............................................................................................................................................
............................................................................................................................................
............................................................................................................................................
............................................................................................................................................ (3)
(d) The reaction
Ni(s) + 4CO(g) Ni(CO)4(g)
is used to purify nickel.
(i) Write the expression for Kp for this system.
(1)
(ii) In order to achieve a high equilibrium yield of Ni(CO)4 should a low or a high
partial pressure of carbon monoxide be used? Explain your answer in terms of Kp.
.....................................................................................................................................
.....................................................................................................................................
..................................................................................................................................... (2)
(Total 12 marks)
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Maltby Academy 16
8. The equation below shows a possible reaction for producing methanol.
CO(g) + 2H2(g) CH3OH(l) ΔHο = 129 kJ mol
–1
(a) The entropy of one mole of each substance in the equation, measured at 298 K, is shown
below.
Substance
Sο
/J mol1
K1
CO(g) 197.6
H2(g) 130.6
CH3OH(l) 239.7
(i) Suggest why methanol has the highest entropy value of the three substances.
.............................................................................................................................
............................................................................................................................. (1)
(ii) Calculate the entropy change of the system, ΔSοsystem, for this reaction.
(2)
(iii) Is the sign of ΔSοsystem as expected? Give a reason for your answer.
.............................................................................................................................
.............................................................................................................................
............................................................................................................................. (1)
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Maltby Academy 17
(iv) Calculate the entropy change of the surroundings ΔSοsurroundings, at 298 K.
(2)
(v) Show, by calculation, whether it is possible for this reaction to occur spontaneously
at 298 K.
(2)
(b) When methanol is produced in industry, this reaction is carried out at 400 ºC and 200
atmospheres pressure, in the presence of a catalyst of chromium oxide mixed with zinc
oxide. Under these conditions methanol vapour forms and the reaction reaches
equilibrium. Assume that the reaction is still exothermic under these conditions.
CO(g) + 2H2(g) CH3OH(g)
(i) Suggest reasons for the choice of temperature and pressure.
Temperature ........................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
Pressure ...............................................................................................................
.............................................................................................................................
............................................................................................................................. (3)
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Maltby Academy 18
(ii) The catalyst used in this reaction is heterogeneous. Explain this term.
.............................................................................................................................
............................................................................................................................. (1)
(iii) Write an expression for the equilibrium constant in terms of pressure, Kp, for this
reaction.
CO(g) + 2H2(g) CH3OH(g)
(1)
(iv) In the equilibrium mixture at 200 atmospheres pressure, the partial pressure of
carbon monoxide is 55 atmospheres and the partial pressure of hydrogen is 20
atmospheres.
Calculate the partial pressure of methanol in the mixture and hence the value of the
equilibrium constant, Kp. Include a unit in your answer.
(2)
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Maltby Academy 19
(c) The diagram below shows the distribution of energy in a sample of gas molecules in a
reaction when no catalyst is present. The activation energy for the reaction is EA.
(i) What does the shaded area on the graph represent?
............................................................................................................................. (1)
(ii) Draw a line on the graph, labelled EC, to show the activation energy of the
catalysed reaction. (1)
(Total 17 marks)
9. (a) (i) 2Ca(NO3)2 2CaO + 4NO2 + O2
formulae correct (1) balance (1). Ignore any state symbols.
The balance mark is not stand–alone. 2
(ii) steam / fizzing sound / crumbles (1)
solid swells up / milky liquid produced / comment about sparingly soluble
substance (1)
CaO + H2O Ca(OH)2 (1) ignore any state symbols 3
(iii) less (1) 1
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Maltby Academy 20
(iv) (Cat)ion size increases down the Group / charge density decreases (1)
(not atom size)
The polarizing power of the cation decreases down the Group (1).
The less polarized the anion is by the cation the more difficult the nitrate is to
decompose (1).
Polarisation mark could come from ‘the less the electron cloud is distorted…’
or
trend in cation size (1)
comparison of the lattice energies of the nitrate and the oxide (1)
balance in favour of oxide at top of group
and the nitrate at the bottom (1) 3
(b) (i) same number of particles in a smaller volume / gas density increased (1) 1
(ii) comment related to the number of molecules on each side to explain a shift to
l.h.s. (1) (not just ‘due to Le Chatelier…’)
so at higher pressure equilibrium moves to favour N2O4 (1) 2
(iii) Kp = p(NO2)2
p(N2O4)
There must be some symbolism for pressure, and no [ ] 1
(iv) (Kp = p(NO2)2 = 48)
p(N2O4)
p(NO2)2 = 48 × 0.15 = 7.2 (1)
p(NO2) = 2.7 (1) atm (1) accept 2.683 / 2.68 / 2.7
Answer and units conditional on (iii). 3 [16]
10. (a) (i) HCl: pH = 1.13
[H+] = 0.074 mol dm
–3
[HCl] = 0.074 mol dm–3
[0.074 to 0.07413] 1
(ii) HOCl: pH = 4.23
[H+] = 5.89 × 10
–5 mol dm
–3 (1)
Ka = [H+] [OCl
–] (1)
[HOCl]
[H+] = [OCl
–] (1) or implied later in calculation
[HOCl] = [H+]2/ Ka = 0.0932 mol dm
–3 (1) 4
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Maltby Academy 21
(b) (i) [H+] = 0.10 / 0.1047 / 0.105 (1)
(ii) H2SO4 H+ + HSO4
–(1) or
H2SO4 + H2O H3+O + HSO4
– ignore state symbols
HSO4– H
+ + SO4
2– (1) Must be
H2SO4 + H2O H3O+ + SO4
2– ignore state symbols
(iii) second ionisation suppressed by the first ionisation (1) 4
(c) (i) ][O[HCl]
O][H][ClK
24
22
22
c
1
(ii) 4HCl + O2 2Cl2 + 2H2O
equilibrium mols 0.20 0.050 (1) 0.30 and 0.30 (1)
[ ] eq ÷ 10 (1) 0.020 0.0050 0.030 0.030
Kc = [0.030]2 x [0.030]
2 = 1010 or 1012 or 1013 or 1012.5
[0.020]4 x [0.005] (mol
–1 dm
3) (1) 4
(d) (i) As reaction (left to right) is exothermic (1)
Decrease in temperature drives equilibrium to from left to right (1) 2
(ii) As more (gas) molecules on the left (1), equilibrium is driven from left to right
(1)
2
(iii) A catalyst has no effect (1)
As it only alters the rate of the reaction not the position of equilibrium / it alters the
rate of the forward and reverse reactions equally (1) 2 [20]
11. (a) (i) Kp = 2
SO3
O2
2
SO2
P
PP (1)
[ ] no mark
( ) OK 1
(ii) 2SO3 2SO2 + O2
Mols at start 2 0 0
mols at equ 0.5 1.5 0.75 (1)
Mark by process
1 mark for working out mole fraction
1 mark for × 10
1 mark for correct substitution in Kp and answer
1 mark for unit
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Maltby Academy 22
i.e. PSO2 = 75.2
5.1 × 10 = 5.46
PO2 = 75.2
7.0 × 10 = 2.73
PSO3= 75.2
5.0 × 10 = 1.83
n.b. could show mole fraction for all 3 and then × 10 later to
give partial pressure.
Kp = (5.46)2 × (2.73) / (1.83)
2 = 24.5 (1) atm (1) 5
(b) (i) No effect (1) 1
(ii) No effect (1) 1 [8]
12. (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)K
24
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(CH
1K4
p (1) 1
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Maltby Academy 23
(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]
13. (a) The pressure which the gas exerts if it alone occupies the
same volume at the same temperature (1)
or
partial pressure of a gas in a mixture = mole fraction of gas ×
total pressure (1) 1
(b) (i) Kp = p(PCl3)p(Cl2)/p(PCl5) (1) no square brackets
Or
Kp = PCl5
Cl2 PCl3
P
PP
1
(ii) mark consequentially on (i)
PCl5 PCl3 + Cl2
mols at start 1 0 0
mols at eqm 0.60 0.40 0.40 (1)
mol fraction 0.60 0.40 0.40 (1)
1.40 1.40 1.40
partial pressures 0.60 ×2 0.40 ×2 0.40 ×2
1.40 1.40 1.40
= 0.857 = 0.571 = 0.571 (1)
Kp = (0.571)2/0.857 = 0.38 or 0.381 (1) atm (1) but
2-4 sf acceptable
i.e.
mols at eqm (1)
mols fraction (1)
partial pressures (ie. multiply x2) (1)
substitute in Kp + answer (1)
units (1) 5
(c) (i) Endothermic (1)
conditional on increase of temperature moves equilibrium in
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Maltby Academy 24
direction that absorbs heat (1) 2
(ii) Kp increases (1) 1
(d) (i) Kp = P(CO2) (1)
or
Kp = PCO2 1
(ii) 16 (atm) (1) ignore units. Consequential on (d)(i) 1 [12]
14. (a) Pressure NOT partial pressure )
intensity or change of colour ) Any one
volume ) 1
(b) Kc = [NO2(g)]2 / [N2O4(g)]
State symbols required 1
(c) Mol NO2 at equilibrium = 0.0120 / 1.20 × 10–2
(1)
Kc = (0.0120)2 ÷ (0.0310)
= 4.6 / 4.65 × 10–3
(1) mol dm–3
(1) 3
(d) (i) Amount of NO2 reduced 1
(ii) No effect 1
(e) As Kc is bigger, more NO2 is produced so heat helps forward reaction /
by Le Chatelier’s principle reaction goes forward to use up heat /
as temperature increases Stotal must be more positive so Ssurroundings
(= –H/T must be less negative 1
(f) Positive / + with some attempt at explanation (1)
1 mol / molecule gas 2 mol / molecule gas / products more disordered
than reactants (1) 2
(g) Ssurroundings = –T
H OR
T
H 1000– 1
(h) Stotal is positive as reaction occurs (1)
So Ssystem must be more positive than Ssurroundings is negative (1) 2
[13]
15. (a) Pressure exerted by the gas if it alone occupied the same volume at the
same temperature/mole fraction × total pressure 1
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Maltby Academy 25
(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]
16. (a) (i) Methanol is the biggest/ most complex molecule / greatest MR /most
atoms/most electrons 1
(ii) Ssystem = 239.7 – 197.6 – 2(130.6)
= –219.1/ –219 J mol–l
K–1
Method (1)
answer + units (1) 2
(iii) yes as 3 molecules 1 OR yes as (2) gases a liquid 1
(iv) Ssurr = –H/T (stated or used) (1)
= –(–129/ 298) = +0.433 kJ mol–1
K–1
/ +433 J mol–1
K–1
/+ 432.9 (1)
–1 for wrong units/ no units / more than 4 SF
–1 for wrong sign/ no sign 2
(v) Stotal = –219.1 + 433 = +213.9 / +213.8 J mol–1
K–1
/ +214 J mol–1
K–1
/
+0.214 kJ mol –1
K–1
(1)
Positive so possible (1) 2
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Maltby Academy 26
(b) (i) Temperature
Faster at 400°C (1)
even though yield is lower (1)
Pressure
Higher pressure improves yield of methanol (1)
Higher pressure increases rate (1)
Maximum 3 3
(ii) Not in same phase as reactants. ALLOW state instead of phase 1
(iii) Kp = p(CH3OH)/p(CO)×p(H2)2 1
(iv) Partial pressure of methanol = 200 – 55 – 20 = 125 atm (1)
Kp = (125)/55×202
= 5.68 × 10–3
/ 5.7 × 10–3
atm–2
(1) 2
(c) (i) Number of molecules / fraction of molecules with energy EA /number
of molecules which have enough energy to react. 1
(ii) Vertical line / mark on axis to show value to the left of line EA 1
[17]