1.
The reaction between sulphur dioxide and oxygen is a dynamic equilibrium.
2SO2 + O2
(a)
2SO3
H = –196 kJmol–1
Explain what is meant by dynamic equilibrium.
..................................................................................................................................
..................................................................................................................................
..................................................................................................................................
(2)
(b)
In the table below state the effect on this reaction of increasing the temperature and of
increasing the pressure.
Effect on the rate of the
reaction
Increasing the
temperature
Effect on the position of
equilibrium
Increases
Increasing the pressure
(3)
(c)
This reaction is one of the steps in the industrial production of sulphuric acid. The
normal operating conditions are a temperature of 450 °C, a pressure of 2 atmospheres
and the use of a catalyst.
Justify the use of these conditions.
(i)
A temperature of 450 °C:
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
Maltby Academy
1
(ii)
A pressure of 2 atmospheres:
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(iii)
A catalyst:
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
(d)
Give the name of the catalyst used.
..................................................................................................................................
(1)
(e)
Give one large scale use of sulphuric acid.
..................................................................................................................................
(1)
(Total 13 marks)
2.
In the vapour phase sulphur trioxide dissociates:
2SO3(g)
(a)
(i)
2SO2(g) + O2(g)
Write an expression for Kp for this dissociation.
(1)
Maltby Academy
2
(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.
............................................................................................................................
............................................................................................................................
(1)
(Total 8 marks)
Maltby Academy
3
3.
Consider the following equation:
2SO2 + O2
2SO3
2.0 moles of SO2 and 1.0 mole of O2 were allowed to react in a vessel of volume 60 dm3.
At equilibrium 1.8 moles of SO3 had formed and the pressure in the flask was 2 atm.
(a)
(i)
Write the expression for Kc for this reaction between SO2 and O2.
(1)
(ii)
Calculate the value of Kc, with units.
(3)
(b)
The reaction between SO2 and O2 is exothermic. State the effect on the following, if the
experiment is repeated at a higher temperature:
(i)
Kc ……………………………………………………………………………..
(1)
(ii)
the equilibrium position ....................................................................................
(1)
(c)
State the effect of a catalyst on:
(i)
Kc ……………………………………………………………………………..
(1)
Maltby Academy
4
(ii)
the equilibrium position ....................................................................................
(1)
(d)
(i)
Write the expression for Kp for the reaction between SO2 and O2.
(1)
(ii)
Calculate the mole fractions of SO2, O2 and SO3 at equilibrium.
(2)
(iii)
Calculate the partial pressures of SO2, O2 and SO3 at equilibrium.
(1)
(iv)
Calculate the value of Kp, with units.
(2)
(Total 14 marks)
Maltby Academy
5
4.
The equation below shows a possible reaction for producing methanol.
CO(g) + 2H2(g)  CH3OH(l)
(a)
ΔHο = 129 kJ mol–1
The entropy of one mole of each substance in the equation, measured at 298 K, is shown
below.
(i)
Substance
Sο
/J mol1 K1
CO(g)
197.6
H2(g)
130.6
CH3OH(l)
239.7
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)
Maltby Academy
6
(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)
(i)
CH3OH(g)
Suggest reasons for the choice of temperature and pressure.
Temperature ........................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
Pressure ...............................................................................................................
.............................................................................................................................
.............................................................................................................................
(3)
Maltby Academy
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(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)
Maltby Academy
8
(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)
5.
The reaction between nitrogen and hydrogen can be used to produce ammonia.
N2(g) + 3H2(g)
2NH3(g)
ΔHο = – 92.2 kJ mol–1
Standard entropies are given below
Sο [N2(g)]
= +191.6 J mol–1 K–1
Sο [H2(g)]
= +130.6 J mol–1 K–1
Sο [NH3(g)] = +192.3 J mol–1 K–1
Maltby Academy
9
(a)
Calculate the entropy change of the system, ΔSοsystem, for this reaction. Include a sign and
units in your answer.
(2)
(b)
Calculate the entropy change of the surroundings, ΔSοsurroundings, at 298 K. Include a sign
and units in your answer.
(2)
(c)
(i)
Calculate the total entropy change, ΔSοtotal, at 298 K. Include a sign and units in
your answer.
(1)
Maltby Academy
10
(ii)
Is this reaction feasible at 298 K? Justify your answer.
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
(1)
(d)
In industry the reaction is carried out at about 700 K using an iron catalyst and high
pressures.
(i)
The yield of ammonia produced at equilibrium is less at 700 K than at 298 K, if the
pressure remains constant. In terms of entropy, explain why this happens.
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
(1)
(ii)
Higher pressures increase the yield of ammonia at equilibrium. Suggest a reason
why pressures greater than 300 atmospheres are not routinely used.
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
(1)
(iii)
Iron is a heterogeneous catalyst. Explain what is meant by heterogeneous.
...............................................................................................................................
...............................................................................................................................
(1)
(Total 9 marks)
Maltby Academy
11
6.
(a)
Define the term standard enthalpy of formation.
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
.....................................................................................................................................
(3)
(b)
In the Haber process, ammonia is manufactured from nitrogen and hydrogen as shown in
the equation.
N2(g) + 3H2(g)
(i)
2NH3(g)
Use the bond enthalpies below to calculate the standard enthalpy of formation of
ammonia.
Bond
Bond enthalpy / kJ mol–1
N≡N in N2
+945
H–H in H2
+436
N–H in NH3
+391
(4)
Maltby Academy
12
(ii)
Draw a labelled enthalpy level diagram for the formation of ammonia in the Haber
process.
Enthalpy
(2)
(iii)
State the temperature used in the Haber process and explain in terms of the rate of
reaction and position of equilibrium, why this temperature is chosen.
Temperature .......................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(3)
Maltby Academy
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(iv)
Identify the catalyst used in the Haber process and state what effect, if any, it has
on the equilibrium yield of ammonia.
Catalyst ................................................................
Effect on yield ...................................................................................................
(2)
(v)
Explain why it is necessary to use a catalyst in this process.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(1)
(c)
The pressure used in the Haber process is 250 atmospheres.
(i)
State and explain an advantage of increasing the pressure to 1000 atmospheres.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(ii)
Suggest a disadvantage of using a pressure of 1000 atmospheres.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(1)
(Total 18 marks)
Maltby Academy
14
7.
One stage in the manufacture of sulphuric acid is
2SO2(g) + O2(g)
The equilibrium constant Kp =
(a)
2SO3(g)
2
p SO
3
2
p SO
 pO2
2
10.0 mol of SO2 and 5.00 mol of O2 were allowed to react. At equilibrium, 90.0% of the
SO2 was converted into SO3.
(i)
Calculate the number of moles of SO2, O2 and SO3 present in the equilibrium
mixture.
(2)
(ii)
Calculate the mole fractions of SO2, O2 and SO3 at equilibrium.
(1)
Maltby Academy
15
(iii)
Assuming that the total pressure of the equilibrium mixture was 2.00 atm, calculate
the partial pressures of SO2, O2 and SO3 at equilibrium.
(1)
(iv)
Calculate the value of Kp.
(2)
(b)
The reaction between sulphur dioxide and oxygen is exothermic.
(i)
State the effect, if any, on Kp of increasing the temperature at constant pressure.
...........................................................................................................................
(1)
Maltby Academy
16
(ii)
Use your answer to (i), and the expression Kp =
2
p SO
3
2
p SO
 pO2
2
to explain the effect
on the position of equilibrium of increasing the temperature at constant pressure.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(2)
(c)
The reaction was repeated at a higher pressure whilst maintaining a constant temperature.
(i)
State the effect, if any, of an increase in the total pressure on the value of Kp.
...............................................................................
(1)
(ii)
State the effect, if any, of this increase in pressure on the amount of sulphur
trioxide in the equilibrium mixture.
...............................................................................
(1)
(d)
State the effect, if any, of a catalyst on:
(i)
Kp
...............................................................................
(1)
(ii)
the equilibrium position.
...............................................................................
(1)
(Total 13 marks)
Maltby Academy
17
8.
Methanoic acid and ethanol react together to form ethyl methanoate, HCOOC2H5, and water.
This reaction is reversible and can be allowed to reach equilibrium.
HCOOH(l) + C2H5OH(l)
(a)
HCOOC2H5 (l) + H2O(l)
ΔH = +45 kJ mol–1
Draw the full structural formula of ethyl methanoate, showing all bonds.
(1)
(b)
What type of organic compound is ethyl methanoate?
.....................................................................................................................................
(1)
(c)
In an experiment, 3.00 mol methanoic acid, HCOOH, and 6.25 mol ethanol, C2H5OH,
were mixed together. A small quantity of catalyst was added. The mixture was left for
several days in a water bath to reach equilibrium at constant temperature.
(i)
Complete the table.
Number of moles in the reaction mixture
HCOOH
C2H5OH
HCOOC2H5
H2O
at start of
experiment
3.00
6.25
0.00
0.00
at equilibrium
0.50
(2)
Maltby Academy
18
(ii)
Write an expression for the equilibrium constant, Kc, for the reaction.
(1)
(iii)
Calculate Kc for the reaction at the temperature of the experiment. The total volume
of the equilibrium mixture was 485 cm3.
(2)
(iv)
State and explain whether Kc for this reaction has units.
...........................................................................................................................
...........................................................................................................................
(1)
Maltby Academy
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(d)
(i)
The temperature of this equilibrium mixture is lowered.
Explain the effect of this on the value of the equilibrium constant and hence on the
yield of ethyl methanoate.
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
...........................................................................................................................
(4)
(ii)
A student added more catalyst to the mixture.
State, giving a reason, what would happen to the composition of the equilibrium
mixture.
...........................................................................................................................
...........................................................................................................................
(1)
(Total 13 marks)
9.
(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)
(b)
2
Temp
(Increases)
Left / to SO2 / to endothermic / lower yield (1)
Press
Increases/faster (1)
Right to SO3 / to smaller number of molecules (1)
3
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20
(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]
PSO2  PO2
2
10.
(a)
(i)
Kp =
PSO3
2
(1)
[ ] no mark
( ) OK
(ii)
Mols at start
mols at equ
1
2SO3
2
0.5

2SO2
0
1.5
+
0
0.75
O2
(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
Maltby Academy
21
i.e.
1 .5
× 10 = 5.46
2.75
0 .7
PO2 =
× 10 = 2.73
2.75
0 .5
PSO3=
× 10 = 1.83
2.75
PSO2 =
n.b. could show mole fraction for all 3 and then × 10 later to
give partial pressure.
(b)
Kp = (5.46)2 × (2.73) / (1.83)2 = 24.5 (1) atm (1)
5
(i)
No effect (1)
1
(ii)
No effect (1)
1
[8]
11.
(a)
(i)
(ii)
(b)
(c)
(d)
Kc = [SO3]2 / [SO2]2 [O2] (1)
1
0 .2
30
= 3.33  103
0 .1
1 .8
60
60
–3
1.67  10
0.03 (1)
2
(0.03)
Kc =
= 4860or 4.86  104 (1)
3 2
(3.33  10 )  1.67  103
mol1 dm3 (1)
3
(i)
Kc decreases (1)
1
(ii)
shifts to left / in reverse (1)
1
(i)
no effect (1)
1
(ii)
no effect (1)
1
(i)
Kp = pSO32 / pSO22  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
Maltby Academy
22
(iii)
Partial pressures: SO2 = 0. 190 (5) atm; O2 = 0.0952 (4) atm;
SO3 = 1.71(4) atm (1) i.e. multiply answer in (ii) by 2
(iv)
1.7142 / 0.19052  0.09524 = 850 (1)
atm1 (1)
1
2
[14]
12.
(a)
(i)
Methanol is the biggest/ most complex molecule / greatest MR /most
atoms/most electrons
1
Ssystem = 239.7 – 197.6 – 2(130.6)
= –219.1/ –219 J mol–lK–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
Stotal = –219.1 + 433 = +213.9 / +213.8 J mol–1 K–1/ +214 J mol–1 K–1/
+0.214 kJ mol –1K–1 (1)
Positive so possible (1)
2
(ii)
(v)
(b)
(c)
(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
Number of molecules / fraction of molecules with energy  EA /number
of molecules which have enough energy to react.
1
Vertical line / mark on axis to show value to the left of line EA
1
(i)
(ii)
[17]
Maltby Academy
23
13.
Penalise units only once in this question
(a)
(b)
(c)
(d)
(2×192.3)–[191.6 + (130.6 × 3)] (1)
= –198.8/199 J mol–1 K –1 (1)
2
– – 92.2  1000/ – –92.2 /
–H / T (1)
298
298
= + 309(.4) J mol–1 K–1 / + 0.309(4) kJ mol–1K–1 (1)
2
–198.8 + 309 = + 110 J mol–1 K–1 (3 SF)
OR
– 198.8 + 309.4 = + 111 J mol–1 K–1 (3 SF)
[Do not penalise missing + sign if penalised already in (b)]
NOT 4SF. Penatise SF only once on paper
1
(ii)
Yes, as Stotal is positive / total entropy change
1
(i)
Higher T makes Ssurroundings decrease (so Stotal is less positive)
1
(ii)
Cost (of energy) to provide compression/ cost of equipment
to withstand high P/ maintenance costs.
NOT safety considerations alone
1
Different phase/state (to the reactants)
1
(i)
(iii)
[9]
14.
(a)
Enthalpy/heat/energy change for one mole of a
compound/substance/ a product (1)
NOT solid/molecule/species/element
Reject “heat released or heat required” unless both mentioned
to be formed from its elements in their standard states (1)
ALLOW normal physical state if linked to standard conditions
Reject “natural state” / “most stable state”
standard conditions of 1 atm pressure and a stated temperature (298 K) (1)
3
Reject “room temperature and pressure”
Reject “under standard conditions”
Maltby Academy
24
(b)
(i)
Bonds broken
Bonds made
N≡N (+)945
6N–H (–)2346 (1)
and
()1308
3H–H
(1)
()2253
∆H = 945 +1308 – 2346
= –93 sign and value (1)
∆Hο = –93 = –46.5 (kJ mol–1)
sign and value q on 3rd mark (1)
2
4
Accept –46.5 (kJ mol–1) with working (4)
Accept + 46.5 with working max (3)
Accept +93 with working max (2)
(ii)
N2 + (3)H 2
(Enthalpy)
H
OR
–93
(2)NH 3
Accept –46.5
Correct labelled levels (1)
Reject “Reactants” and “Products” as labels
∆H labelled (1)
direction of arrow must agree with thermicity
Accept double headed arrow
Diagram marks cq on sign and value of ∆H in (b)(i)
IGNORE activation energy humps
Maltby Academy
2
25
(iii)
350–500 °C (1)
Accept any temperature or range within this range
higher temperature gives higher rate (1)
but a lower yield because reaction is exothermic (1)
Accept favours endothermic reaction more than exothermic so
lower yield
OR
Lower temperature give higher yield because reaction is exothermic (1)
but rate is slower (1)
3
Accept cq on sign of ∆Hf in (b)(i) or levels in (ii)
Reject lower temp favours exothermic reaction
(iv)
(v)
Iron / Fe (1) IGNORE any promoters
no effect on yield (1)
2
temp would have to be much higher for a reasonable rate then
yield would be too low
“lower activation energy” implies reasonable rate
OR
Allows reaction at a lower temp at a reasonable/fast rate giving
a reasonable yield.
1
Accept rate too slow without catalyst at a temp giving a
reasonable yield
Reject to lower activation energy of reaction
(c)
(i)
advantage
higher (equilibrium) yield/more NH3 in equilibrium
mixture/equilibrium shifts to right (1)
because smaller number of (gaseous) moles/molecules on rhs (1)
IGNORE any reference to change in rate
2
Reject just “more ammonia”
Maltby Academy
26
(ii)
disadvantage
(plant more) expensive because thicker pipes would be needed
OR
cost (of energy) for compressing the gases/cost of pump
OR
Cost of equipment/pressure not justified by higher yield
1
Accept stronger or withstand high pressure for thicker
Accept vessel/container/plant /equipment/reaction vessels for
pipes
Reject “just more expensive”
Reject “just thicker pipes etc”
Reject apparatus
[18]
15.
(a)
IGNORE s.f. throughout this question
(i)
moles SO2 (10.0 – 9.00) = 1.00 (mol)
moles O2 (5.00 – 4.50) = 0.500 (mol)
moles SO3 9.00 (mol)
all 3 correct → (2)
2 correct → (1)
2
Reject multiples of the stated moles
(ii)
All three ÷ total number of moles (1)
i.e.
1.00
( 0.0952) or 2 21
10.5
0.500
X O2 
( 0.0476) or 1 21
10.5
9.00
X SO3 
(0.857) or 18 21 or 6 7
10.5
X SO2 
Reject rounding to 1 sig fig
Mark consequential on (a)(i)
Maltby Academy
1
27
(iii)
All three × total pressure (1)
i.e.
1.00
 2.00 or 4 21
10.5
= 0.190 (atm)
0.500
 2.00 or 2 21
pO2 =
10.5
= 0.0952 (atm)
9.00
 2.00 or 36 21 or 12 7
pSO3 =
10.5
= 1.71 (atm)
pSO2 =
Mark consequential on (a)(ii)
(iv)
Kp 
1
(1.71) 2
(0.190) 2  (0.0952)
Kp = 851 (1) atm–1 (1)
Mark consequential on (a)(iii) and (a)(iv)
2
Accept answer with units and no working (2)
Accept “correct answers” between 845 and 855 as this covers
rounding up etc
Reject wrong units e.g. mol–1 dm3
(b)
(i)
(Kp) decreases
(ii)
(Kp decreases so)
1
Reject any Le Chatelier argument (this prevents access to 1st
mark)
fraction/quotient
p 2 SO3
p 2 SO2  pO2
has to decrease (to equal new kp) (1)
so shifts to left hand side (1) – this mark only
available if (b)(i) answer was kp decreases.
Reject shifts to right, even if answer to (b)(i) was kp increases
(c)
(i)
Maltby Academy
(as p SO3 decreases whereas p SO 2 and p O 2 increase)
2
No effect/none/zero (effect)
1
28
(ii)
(d)
Increases
OR
more SO3/more sulphur trioxide
OR
increases amount of SO3/sulphur trioxide
1
(i)
No effect/none/zero (effect)
1
(ii)
No effect/none/zero (effect)
1
[13]
16.
(a)
O
H
C
O
(b)
ester
(c)
(i)
H
H
C
C
H
H
H
1
Moles: C2H5OH: 3.75 (1)
Moles: HCOOC2H5 : 2.50 and moles H2O : 2.50 (1) for both
(ii)
1
Kc 
[HCOOC 2 H 5 ][H 2 O]
[HCOOH][C2 H 5 OH]
2
1
Reject obviously round brackets “( )”
Maltby Academy
29
 2.50
0
.
485
0.485 (1)
Kc 
0.50
 3.75
0.485
0.485
Must have clearly divided moles of each component by
0.485 for 1st mark e.g.
[HCOOC2H5] = [H2O] = 5.16 (mol dm–3)
and [HCOOH] = 1.03 (mol dm–3)
and [C2H5OH] = 7.73 (mol dm–3)
2.50
(iii)
= 3.33 (1) stand alone mark
IGNORE sig.figs.
2
(2.50)2
= 3.33 only scores (2) if it is stated
0.50 3.75
that V cancels either here or in (iv)
Accept K c 
If [H2O] omitted in (ii), then answer
Kc = 0.647 mol–1 dm3
(2) but this will give
Kc = 1.33 mol–1 dm3 with V omitted from calculation (1)
Reject 1st mark if 485 used as V in expression
(iv)
No, (as) equal numbers of moles on both sides
OR volumes cancel
OR mol dm–3 cancel
OR units cancel
OR crossing out units to show they cancel
1
Accept “equal powers/moles on both sides”
OR “powers cancel”
Mark CQ on Kc expression in (ii)
Reject “concentrations cancel”
(d)
(i)
(as reaction) endothermic (1)
Accept exothermic in backward direction (or words to that
effect)
Kc decreases (1)
If state exothermic in forward direction, 1 mark only (out of 4)
for CQ “increase in Kc”
numerator in quotient (has to) decrease
OR denominator in quotient (has to) increase
OR fraction (has to) decrease (1)
yield of HCOOC2H5 decreases (1)
Maltby Academy
4
30
(ii)
no effect as catalysts do not affect (the value of) K
OR
no effect as catalysts do not affect the position of equilibrium
OR
no effect as catalysts do not affect the yield
OR
No effect as catalysts increase the rate of the forward and
backward reactions equally/to the same extent
OR
no effect as catalysts only increase the rate
OR
no effect as catalysts only alter the rate
“no effect” can be stated or implied
IGNORE any references to activation energy
1
Reject just “catalysts increase rate”
[13]
Maltby Academy
31