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3.1.4.2 Calorimetry

exam questions for chemistry aqa

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0% found this document useful (0 votes)

27 views99 pages

3.1.4.2 Calorimetry

exam questions for chemistry aqa

Uploaded by

haasina85

We take content rights seriously. If you suspect this is your content, claim it here.

Available Formats

Download as PDF, TXT or read online on Scribd

You are on page 1/ 99

Name: ________________________

3.1.4.2 Calorimetry
Class: ________________________

Date: ________________________

Time: 365 min.

Marks: 322 marks

Comments:

Page 1 of 99
Q1.
This question is about enthalpy changes.

(a) When ethanoic acid reacts with sodium hydroxide, the enthalpy change, ∆H, is
–56.1 kJ mol–1

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

Calculate the temperature rise when 25 cm3 of 2.0 mol dm–3 aqueous ethanoic acid
react with 25 cm3 of 2.0 mol dm–3 aqueous sodium hydroxide.

Assume that both solutions have the same initial temperature, have a density of 1.0
g cm–3 and a specific heat capacity of 4.18 J K –1 g–1

Temperature rise ____________________ °C

(4)

(b) A student recorded the temperature of aqueous ethanoic acid in a polystyrene cup
for three minutes.

At the fourth minute, the student added sodium hydrogencarbonate.

The student stirred the mixture and carried on recording the temperature every
minute for several minutes.

The student’s measurements are shown in the graph.

A best-fit line showing the temperature before mixing has been drawn.

Draw an appropriate best-fit line on the graph and use it to find the temperature
change at the time of mixing.

Page 2 of 99
Temperature change at time of mixing ____________________ °C
(2)
(Total 6 marks)

Q2.
The oxidation of propan-1-ol can form propanal and propanoic acid.
The boiling points of these compounds are shown in the table.

Compound Boiling point / °C

propan-1-ol 97

propanal 49

propanoic acid 141

In a preparation of propanal, propan-1-ol is added dropwise to the oxidising agent and the
aldehyde is separated from the reaction mixture by distillation.

(a) Explain, with reference to intermolecular forces, why distillation allows propanal to
be separated from the other organic compounds in this reaction mixture.

___________________________________________________________________

Page 3 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(b) Give two ways of maximising the yield of propanal obtained by distillation of the
reaction mixture.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(c) Describe how you would carry out a simple test-tube reaction to confirm that the
sample of propanal obtained by distillation does not contain any propanoic acid.

___________________________________________________________________

___________________________________________________________________
(2)

Page 4 of 99
(d) A student carried out an experiment to determine the enthalpy of combustion of
ethanol.
Combustion of 457 mg of ethanol increased the temperature of 150 g of water from
25.1 °C to 40.2 °C

Calculate a value, in kJ mol–1, for the enthalpy of combustion of ethanol in this

experiment.
Give your answer to the appropriate number of significant figures.

(The specific heat capacity of water is 4.18 J K –1 g–1)

Enthalpy of combustion ____________________ kJ mol –1

(3)

(e) A mixture of isomeric alkenes is produced when pentan-2-ol is dehydrated in the

presence of hot concentrated sulfuric acid. Pent-1-ene is one of the isomers
produced.

Name and outline a mechanism for the reaction producing pent-1-ene.

Name of mechanism _________________________________________________

Mechanism

(4)

(f) A pair of stereoisomers is also formed in the reaction in part (e).

Name the less polar stereoisomer formed.

Explain how this type of stereoisomerism arises.

Name _____________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

Page 5 of 99
___________________________________________________________________

___________________________________________________________________
(2)
(Total 16 marks)

Q3.
A student carried out an experiment to find the temperature rise for a reaction between
hydrochloric acid and sodium hydroxide solution.

• The student used a measuring cylinder to place 50 cm 3 of 0.400 mol dm–3

hydrochloric acid into a glass beaker.
• The student recorded the temperature at one-minute intervals for three minutes.
• At the fourth minute the student added 50 cm3 of 0.400 mol dm–3 sodium hydroxide
solution and stirred to mix the solutions, but did not record the temperature.
• The student recorded the temperature at one-minute intervals for a further eight
minutes.

The results are shown in the table.

(a) Plot a graph of temperature against time on the grid below.

Page 6 of 99
Use your graph to find the temperature rise, ∆T, at the fourth minute.
Show your working on the graph by drawing suitable lines of best fit.

Page 7 of 99
∆T ____________________ °C
(5)

(b) The uncertainty in each of the temperature readings from the thermometer used in
this experiment was ±0.1° C

Calculate the percentage uncertainty in the value for the temperature rise.

Percentage uncertainty ____________________

(1)

___________________________________________________________________

___________________________________________________________________
(1)

(d) Suggest and explain another change to the experiment that would decrease the
percentage uncertainty in the use of the same thermometer.

___________________________________________________________________

___________________________________________________________________
(2)

(e) A second student completed an experiment to determine the enthalpy of

neutralisation for the reaction between ethanedioic acid solution (HOOCCOOH) and
potassium hydroxide solution.

The student added 25 cm3 of 0.80 mol dm–3 ethanedioic acid solution to 75 cm3 of
0.60 mol dm–3 potassium hydroxide solution.
The temperature increased by 3.2 °C

Give an equation for the reaction between ethanedioic acid solution and potassium
hydroxide solution.
Calculate the enthalpy change (∆H) per mole of water formed in this reaction.
Assume that the specific heat capacity of the reaction mixture is 4.2 J K –1 g–1
Assume that the density of the reaction mixture is 1.00 g cm –3

Equation ___________________________________________________________

Page 8 of 99
∆H ____________________ kJ mol–1
(5)

(f) In a similar experiment to that in part (e), the enthalpy of neutralisation for the
reaction between sulfuric acid and potassium hydroxide solution was found to be
–57.0 kJ mol–1 per mole of water formed.

Suggest an explanation for the difference between this value and your answer to
part (e).

(If you were unable to obtain an answer to part (e) you should assume a value of
–28.5 kJ mol–1. This is not the correct answer.)

___________________________________________________________________

___________________________________________________________________
(2)
(Total 16 marks)

Q4.
A student planned and carried out an experiment to determine the enthalpy of reaction
when magnesium metal displaces zinc from aqueous zinc sulfate.

Mg(s) + Zn2+(aq) ⟶ Mg2+(aq) + Zn(s)

The student used this method:

• A measuring cylinder was used to transfer 50 cm3 of a 1.00 mol dm−3 aqueous
solution of zinc sulfate into a glass beaker.

• A thermometer was placed in the beaker.

• 2.08 g of magnesium metal powder were added to the beaker.

• The mixture was stirred and the maximum temperature recorded.

The student recorded a starting temperature of 23.9 °C and a maximum temperature of

61.2 °C.

(a) Show by calculation which reactant was in excess.

Use the data to calculate the experimental value for enthalpy of reaction in kJ
mol−1(Assume that the specific heat capacity of the solution is 4.18 J K −1g−1and the
density of the solution is 1.00 g cm−3).

Page 9 of 99
Reactant in excess ____________________

Enthalpy of reaction ____________________ kJ mol −1

(6)

(b) Another student used the same method and obtained a value for the enthalpy of
reaction of −142 kJ mol−1

A data book value for the enthalpy of reaction is −310 kJ mol−1

Suggest the most likely reason for the large difference between the student’s
experimental value and the data book value.

___________________________________________________________________

___________________________________________________________________
(1)

(c) Suggest how the students’ method, and the analysis of the results, could be
improved in order to determine a more accurate value for the enthalpy of reaction.

Justify your suggestions.

Do not refer to the precision of the measuring equipment. Do not change the
amounts or the concentration of the chemicals.

___________________________________________________________________

___________________________________________________________________
(6)
(Total 13 marks)

Q5.
Anhydrous magnesium chloride, MgCl2, can absorb water to form the hydrated salt
MgCl2.4H2O

Page 10 of 99
MgCl2(s) + 4H2O(l) ⟶ MgCl2.4H2O(s)

(a) Suggest one reason why the enthalpy change for this reaction cannot be
determined directly by calorimetry.

___________________________________________________________________

___________________________________________________________________
(1)

(b) Some enthalpies of solution are shown in Table 1.

Table 1

Enthalpy of
Salt
solution / kJ mol−1

MgCl2(s) −155

MgCl2.4H2O(s) −39

Calculate the enthalpy change for the absorption of water by MgCl 2(s) to form
MgCl2.4H2O(s).

Enthalpy change _________________________________________ kJ mol−1

(2)

(c) Describe how you would carry out an experiment to determine the enthalpy of
solution of anhydrous magnesium chloride.

You should use about 0.8 g of anhydrous magnesium chloride.

Explain how your results could be used to calculate the enthalpy of solution.

___________________________________________________________________

___________________________________________________________________
(6)

Page 11 of 99
(d) Anhydrous magnesium chloride can be formed by direct reaction between its
elements.

Mg(s) + Cl2(g) ⟶ MgCl2(s)

The free-energy change, ΔG, for this reaction varies with temperature as
shown in Table 2.

Table 2

T/K ΔG / kJ mol−1
298 −592.5

288 −594.2

273 −596.7

260 −598.8

240 −602.2

Use these data to plot a graph of free-energy change against temperature on the
grid below.

Calculate the gradient of the line on your graph and hence calculate the entropy
change, ΔS, in J K−1 mol−1, for the formation of anhydrous magnesium chloride from
its elements.

Show your working.

Page 12 of 99
ΔS _________________________________________ J K−1 mol−1

Page 13 of 99
(5)
(Total 14 marks)

Q6.
A student carried out a reaction between magnesium ribbon and aqueous
trichloroethanoic acid in order to determine the enthalpy change. The equation for the
reaction is shown:

Mg(s) + 2H+(aq) ⟶ Mg2+(aq) + H2(g)

The student measured the initial temperature of the trichloroethanoic acid and again every
minute for 3 minutes before adding the magnesium ribbon at the fourth minute.
The student continued to measure the temperature every minute for a further 10 minutes.
The graph for these measurements is shown below.

The student used 240 mg of magnesium and 10.0 cm3 of aqueous trichloroethanoic acid
(an excess).

Use these data and information determined from the graph above to calculate the
enthalpy change, in kJ mol−1, for this reaction.
Show your working.
Give your answer to an appropriate precision.
(The specific heat capacity of water = 4.18 J K−1 g−1)

Page 14 of 99
Enthalpy change = ___________ kJ mol−1
(Total 7 marks)

Q7.
What is the temperature rise, in K, when 504 J of heat energy are absorbed by 0.110 kg of
solid iron?
Specific heat capacity of iron = 0.448 J K −1 g−1

A 9.78 × 10−2

B 1.02 × 101

C 2.83 × 102

D 1.02 × 104
(Total 1 mark)

Q8.
The figure below shows apparatus used in an experiment to determine the enthalpy of
combustion of leaf alcohol.

The alcohol is placed in a spirit burner and weighed. The burner is lit and the alcohol
allowed to burn for a few minutes. The flame is extinguished and the burner is re-weighed.
The temperature of the water is recorded before and after heating.

The following table shows the results obtained.

Initial mass of spirit burner and alcohol / g 56.38

Final mass of spirit burner and alcohol / g 55.84

Initial temperature of water / °C 20.7

Final temperature of water / °C 40.8

Page 15 of 99
(a) Write an equation for the complete combustion of leaf alcohol
(CH3CH2CH=CHCH2CH2OH).

___________________________________________________________________
(1)

(b) Use the results from the table above to calculate a value for the enthalpy of
combustion of leaf alcohol. Give units in your answer.
(The specific heat capacity of water is 4.18 J K −1 g−1)

Enthalpy of combustion = Units =

(4)

(c) State how your answer to part (b) is likely to differ from the value quoted in
reference sources.
Give one reason for your answer.

___________________________________________________________________

___________________________________________________________________
(2)

(d) A 50.0 g sample of water was used in this experiment.

Explain how you could measure out this mass of water without using a balance.

___________________________________________________________________

___________________________________________________________________
(2)

Page 16 of 99
(Total 9 marks)

Q9.
A 5.00 g sample of potassium chloride was added to 50.0 g of water initially at 20.0 °C.
The mixture was stirred and as the potassium chloride dissolved, the temperature of the
solution decreased.

(a) Describe the steps you would take to determine an accurate minimum temperature
that is not influenced by heat from the surroundings.

___________________________________________________________________

___________________________________________________________________
(4)

(b) The temperature of the water decreased to 14.6 °C.

Calculate a value, in kJ mol−1, for the enthalpy of solution of potassium chloride.

You should assume that only the 50.0 g of water changes in temperature and that
the specific heat capacity of water is 4.18 J K −1 g−1.
Give your answer to the appropriate number of significant figures.

Enthalpy of solution = _______________ kJ mol −1

(4)

(c) The enthalpy of solution of calcium chloride is −82.9 kJ mol−1.

The enthalpies of hydration for calcium ions and chloride ions are −1650 and
−364 kJ mol−1, respectively.

Use these values to calculate a value for the lattice enthalpy of dissociation of
calcium chloride.

Page 17 of 99
Lattice enthalpy of dissociation = _______________ kJ mol −1
(2)

(d) Explain why your answer to part (c) is different from the lattice enthalpy of
dissociation for magnesium chloride.

___________________________________________________________________

___________________________________________________________________
(2)
(Total 12 marks)

Q10.
An engineer was trying to develop a new fuel for a motorboat by blending mixtures of
different alcohols in order to find out which mixture released the most energy when used
in the engine.

The engineer had a number of alcohols in unlabelled bottles. It was decided to identify the
alcohols by determining their enthalpies of combustion and comparing these values with
those from a data book.

(a) Outline a simple practical experiment that the engineer could use, including the
measurements to be taken, in order to determine the enthalpy of combustion for one
of the unknown alcohols. You do not need to include details of any calculations.

___________________________________________________________________

___________________________________________________________________
(5)

Page 18 of 99
(b) Other than heat loss to the surroundings, identify two major sources of error in the
experiment. Do not refer to the precision of the equipment.

___________________________________________________________________

___________________________________________________________________
(2)

(c) The engineer found that the experimental values for the enthalpies of combustion of
butan-1-ol and methylpropan-2-ol were very similar and so these values could not
be used to distinguish between the two alcohols.

Identify a reagent that the engineer could use to distinguish between these two
alcohols.
Give the observation in each case.

Reagent ___________________________________________________________

___________________________________________________________________

Butan-1-ol __________________________________________________________

___________________________________________________________________

Methylpropan-2-ol ____________________________________________________

___________________________________________________________________
(3)

(d) The filter in the air intake for the engine in the motorboat may become partially
blocked by dust and debris.

Explain with the aid of an equation why combustion of methylpropan-2-ol under

these circumstances would be of economic and environmental concern to the
engineer.

___________________________________________________________________

___________________________________________________________________
(3)
(Total 13 marks)

Q11.

Page 19 of 99
Alcohols such as methanol (CH3OH), ethanol (CH3CH2OH) and propan-1-ol
(CH3CH2CH2OH) are good fuels.

(a) A student carried out an experiment to determine the enthalpy of combustion of

methanol.

Methanol was placed in a spirit burner and the mass of the spirit burner measured.
The student placed 100 g of water in a copper calorimeter and clamped it above the
spirit burner. The burner was lit and allowed to burn for a few minutes. The flame
was then extinguished and the new mass of the spirit burner found.

The measured temperature rise was 38.0 °C. The specific heat capacity of water is
4.18 J K−1 g−1.

A diagram of the apparatus is shown alongside a table which shows the

measurements the student recorded.

Use the student’s data to calculate an experimental value for the enthalpy of
combustion of methanol in kJ mol−1.

___________________________________________________________________

___________________________________________________________________
(4)

(b) Suggest one reason, other than incomplete combustion or heat transfer to the
atmosphere, why the student’s value for the enthalpy of combustion of
methanol is different from that in a Data Book.

Page 20 of 99
___________________________________________________________________

___________________________________________________________________
(1)

(c) The uncertainty in each of the temperature readings from the thermometer in
this experiment was ±0.25 °C. This gave an overall uncertainty in the
temperature rise of ±0.5 °C.

Calculate the percentage uncertainty for the use of the thermometer in this
experiment.

___________________________________________________________________

___________________________________________________________________
(1)

(d) The student said correctly that using a thermometer with an overall
uncertainty for the rise in temperature of ±0.5 °C was adequate for this
experiment.

Explain why this thermometer was adequate for this experiment.

___________________________________________________________________

___________________________________________________________________
(1)

(e) The enthalpy of combustion of ethanol is −1371 kJ mol−1. The density of

ethanol is 0.789 g cm−3.

Calculate the heat energy released in kJ when 0.500 dm 3 of ethanol is burned.

Give your answer to an appropriate number of significant figures.

___________________________________________________________________

___________________________________________________________________
(3)
(Total 10 marks)

Q12.
(a) Propanone can be formed when glucose comes into contact with bacteria in the
absence of air.

(i) Balance the following equation for this reaction of glucose to form propanone,
carbon dioxide and water.

.......C6H12O6 .......CH3COCH3 + .......CO2 + .......H2O

(1)

Page 21 of 99
(ii) Deduce the role of the bacteria in this reaction.

______________________________________________________________
(1)

(b) Propanone is also formed by the oxidation of propan−2−ol.

(i) Write an equation for this reaction using [O] to represent the oxidising agent.

______________________________________________________________
(1)

(ii) State the class of alcohols to which propan−2−ol belongs.

______________________________________________________________
(1)

(c) A student determined a value for the enthalpy change when a sample of propanone
was burned. The heat produced was used to warm some water in a copper
calorimeter.
The student found that the temperature of 150 g of water increased by 8.0 °C when
4.50 × 10−3 mol of pure propanone was burned in air.

Use the student’s results to calculate a value, in kJ mol−1, for the enthalpy change
when one mole of propanone is burned.
(The specific heat capacity of water is 4.18 J K −1 g−1)

___________________________________________________________________

___________________________________________________________________
(3)

(d) Define the term standard enthalpy of combustion.

___________________________________________________________________

Page 22 of 99
___________________________________________________________________

___________________________________________________________________
(3)

(e) Use the mean bond enthalpy data in the table and the equation given below the
table to calculate a value for the standard enthalpy change when gaseous
propanone is burned.

C−H C−C C−O O−H C=O O=O

Mean bond 412 348 360 463 805 496

enthalpy / kJ mol−1

CH3COCH3(g) + 4O2(g) 3CO2(g) + 3H2O(g)

___________________________________________________________________

___________________________________________________________________
(3)

(f) Suggest two reasons why the value obtained by the student in part (c) is different
from the value calculated in part (e).

Reason 1 ___________________________________________________________

___________________________________________________________________

Reason 2 ___________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)
(Total 15 marks)

Q13.
A value for the enthalpy of combustion of an alcohol can be determined using the
apparatus shown in the diagram. The calorimeter is held in position by a clamp.

Page 23 of 99
This experiment can be repeated by using a different volume of water that would result in
a more accurate value for the enthalpy of combustion because there would be a reduction
in the heat lost.

State a change in the volume of water that would cause a reduction in heat loss and
explain your answer.

Change in volume: _______________________________________________________

Explanation: ____________________________________________________________

_______________________________________________________________________
(Total 2 marks)

Q14.
Ethanol is an important fuel.

(a) A dilute aqueous solution of ethanol can be produced by the fermentation of an

aqueous solution of glucose.
It is claimed that the ethanol obtained from this solution is a carbon-neutral biofuel.

Write an equation for this fermentation reaction.

Give two other essential conditions for this reaction to produce a good yield of
ethanol.

Name a process used to produce a much more concentrated solution of ethanol

from a dilute aqueous solution.

State the meaning of the term carbon-neutral in the context of this biofuel.

___________________________________________________________________

Page 24 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(5)

(b) A student carried out a laboratory experiment to determine the enthalpy change
when a sample of ethanol was burned. The heat produced was used to warm some
water in a copper calorimeter. The student found that the temperature of 75.0 g of
water increased by 5.50 °C when 2.40 × 10 –3 mol of pure ethanol was burned in air.

Use the student’s results to calculate a value, in kJ mol–1, for the enthalpy change
when one mole of ethanol is burned.
(The specific heat capacity of water is 4.18 J K –1 g–1)

Deduce two reasons why the student’s value for the standard enthalpy of
combustion of ethanol is different from a Data Book value of –1279 kJ mol–1.

___________________________________________________________________

___________________________________________________________________
(5)

(c) Mean bond enthalpies can be used to calculate enthalpies of reaction.

(i) Give the meaning of the term mean bond enthalpy.

______________________________________________________________

______________________________________________________________
(2)

(ii) Consider the mean bond enthalpy data in the following table.

Page 25 of 99
C—H C—C C—O O=O C=O O—H

Mean bond enthalpy / to be

412 348 360 805 463
kJ mol–1 calculated

Use the data in the table above and the equation shown to calculate a value
for the bond enthalpy for the O=O double bond in an oxygen molecule.

CH3CH2OH(g) + 3O2(g) 2CO2(g) + 3H2O(g) ΔH = –1279 kJ mol–1

______________________________________________________________

______________________________________________________________
(3)
(Total 15 marks)

Q15.
The alcohol 2-methylpropan-2-ol, (CH3)3COH, reacts to form esters that are used as
flavourings by the food industry. The alcohol can be oxidised to produce carbon dioxide
and water.

A student carried out an experiment on a pure sample of 2-methylpropan-2-ol to

determine its enthalpy of combustion. A sample of the alcohol was placed into a spirit
burner and positioned under a beaker containing 50 cm 3 of water. The spirit burner was
ignited and allowed to burn for several minutes before it was extinguished.

The results for the experiment are shown in Table 1.

Table 1

Initial temperature of the water / °C 18.1

Final temperature of the water / °C 45.4

Initial mass of spirit burner and alcohol / g 208.80

Final mass of spirit burner and alcohol / g 208.58

(a) Use the results from Table 1 to calculate a value for the heat energy released from
the combustion of this sample of 2-methylpropan-2-ol.
The specific heat capacity of water is 4.18 J K –1 g–1.
Show your working.

Page 26 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(2)

(b) Calculate the amount, in moles, of 2-methylpropan-2-ol burned in the experiment.

Hence calculate a value, in kJ mol–1, for the enthalpy of combustion of
2-methylpropan-2-ol.
Show your working.

(If you were unable to calculate an answer to part (a), you should assume that the
heat energy released was 5580 J. This is not the correct value.)

___________________________________________________________________

___________________________________________________________________
(3)

(c) An equation for the combustion of 2-methylpropan-2-ol is

(CH3)3COH(I) + 6O2(g) 4CO2(g) + 5H2O(I)

Table 2 contains some standard enthalpy of formation data.

Table 2

(CH3)3COH(I) O2(g) CO2(g) H2O(I)

∆Hf / kJ mol–1 –360 0 –393 –286

Use the data from Table 2 to calculate a value for the standard enthalpy of
combustion of 2-methylpropan-2-ol. Show your working.

___________________________________________________________________

Page 27 of 99
___________________________________________________________________

___________________________________________________________________
(3)

(d) An accurate value for the enthalpy of combustion of 2-methylpropan-2-ol in which

water is formed as a gas is –2422 kJ mol–1.

Use this value and your answer from part (b) to calculate the overall percentage
error in the student’s experimental value for the enthalpy of combustion of
2-methylpropan-2-ol.

___________________________________________________________________

___________________________________________________________________
(1)

(e) Suggest one improvement that would reduce errors due to heat loss in the
student’s experiment.

___________________________________________________________________

___________________________________________________________________
(1)

(f) Suggest one other source of error in the student’s experiment. Do not include heat
loss, apparatus error or student error.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 11 marks)

Q16.
The enthalpy of hydration for the chloride ion is −364 kJ mol−1 and that for the bromide ion
is −335 kJ mol−1.

(a) By describing the nature of the attractive forces involved, explain why the value for
the enthalpy of hydration for the chloride ion is more negative than that for the
bromide ion.

___________________________________________________________________

Page 28 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(b) The enthalpy of hydration for the potassium ion is −322 kJ mol−1. The lattice
enthalpy of dissociation for potassium bromide is +670 kJ mol−1.

Calculate the enthalpy of solution for potassium bromide.

___________________________________________________________________

___________________________________________________________________
(2)

(i) Explain why the free-energy change for the dissolving of potassium chloride in
water is negative, even though the enthalpy change is positive.

______________________________________________________________

______________________________________________________________
(3)

(ii) A solution is formed when 5.00 g of potassium chloride are dissolved in 20.0 g
of water. The initial temperature of the water is 298 K.

Calculate the final temperature of the solution.

In your calculation, assume that only the 20.0 g of water changes in

temperature and that the specific heat capacity of water is 4.18 J K−1 g−1.

______________________________________________________________

Page 29 of 99
______________________________________________________________

______________________________________________________________

______________________________________________________________
(5)
(Total 13 marks)

Q17.
A student used Hess’s Law to determine a value for the enthalpy change that occurs
when anhydrous copper(II) sulfate is hydrated. This enthalpy change was labelled ΔHexp
by the student in a scheme of reactions.

(a) State Hess’s Law.

___________________________________________________________________

___________________________________________________________________
(1)

(b) Write a mathematical expression to show how ΔHexp, ΔH1 and ΔH2 are related to
each other by Hess’s Law.

___________________________________________________________________
(1)

(c) Use the mathematical expression that you have written in part (b), and the data
book values for the two enthalpy changes ΔH1 and ΔH2 shown, to calculate a value
for ΔHexp

Page 30 of 99
ΔH1 = −156 kJ mol−1
ΔH2 = +12 kJ mol−1

___________________________________________________________________

___________________________________________________________________
(1)

(d) The student added 0.0210 mol of pure anhydrous copper(II) sulfate to 25.0 cm 3 of
deionised water in an open polystyrene cup. An exothermic reaction occurred and
the temperature of the water increased by 14.0 °C.

(i) Use these data to calculate the enthalpy change, in kJ mol −1, for this reaction
of copper(II) sulfate. This is the student value for ΔH1

In this experiment, you should assume that all of the heat released is used to
raise the temperature of the 25.0 g of water. The specific heat capacity of
water is 4.18 J K−1 g−1.

______________________________________________________________

______________________________________________________________
(3)

(ii) Suggest one reason why the student value for ΔH1 calculated in part (d)(i) is
less accurate than the data book value given in part (c).

______________________________________________________________

______________________________________________________________
(1)

(e) Suggest one reason why the value for ΔHexp cannot be measured directly.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 8 marks)

Q18.
Methanol (CH3OH) is an important fuel that can be synthesised from carbon dioxide.

Page 31 of 99
(a) The table shows some standard enthalpies of formation.

CO2(g) H2(g) CH3OH(g) H2O(g)

∆HfƟ/kJ mol–1 – 394 0 – 201 – 242

(i) Use these standard enthalpies of formation to calculate a value for the
standard enthalpy change of this synthesis.

CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

______________________________________________________________

______________________________________________________________
(3)

(ii) State why the standard enthalpy of formation for hydrogen gas is zero.

______________________________________________________________

______________________________________________________________
(1)

(b) State and explain what happens to the yield of methanol when the total pressure is
increased in this synthesis.

CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

Effect on yield _______________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

Page 32 of 99
(c) The hydrogen required for this synthesis is formed from methane and steam in a
reversible reaction. The equation for this reaction is shown below.

CH4(g) + H2O(g) C0(g) + 3H2(g) ∆H = +206 kJ mol–1

State and explain what happens to the yield of hydrogen in this reaction when the
temperature is increased.

Effect on yield _______________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)

(d) The methanol produced by this synthesis has been described as a carbon-neutral
fuel.

(i) State the meaning of the term carbon-neutral.

______________________________________________________________

______________________________________________________________
(1)

(ii) Write an equation for the complete combustion of methanol.

______________________________________________________________
(1)

(iii) The equation for the synthesis of methanol is shown below.

CO2(g) + 3H2(g) CH3OH(g) + H2O(g)

Use this equation and your answer to part (d)(ii) to deduce an equation to
represent the overall chemical change that occurs when methanol behaves as
a carbon-neutral fuel.

Page 33 of 99
Equation ___________________________________________________
(1)

(e) A student carried out an experiment to determine the enthalpy change when a
sample of methanol was burned.

The student found that the temperature of 140 g of water increased by 7.5 °C when
0.011 mol of methanol was burned in air and the heat produced was used to warm
the water.

Use the student’s results to calculate a value, in kJ mol–1, for the enthalpy change
when one mole of methanol was burned.
(The specific heat capacity of water is 4.18 J K –1 g–1).

___________________________________________________________________

___________________________________________________________________
(3)
(Total 16 marks)

Q19.
(a) Anhydrous calcium chloride is not used as a commercial de-icer because it reacts
with water. The reaction with water is exothermic and causes handling problems.

A student weighed out 1.00 g of anhydrous calcium chloride. Using a pipette, 25.0
cm3 of water were measured out and transferred to a plastic cup. The cup was
placed in a beaker to provide insulation. A thermometer was mounted in the cup
using a clamp and stand. The bulb of the thermometer was fully immersed in the
water.

The student recorded the temperature of the water in the cup every minute, stirring
the water before reading the temperature. At the fourth minute the anhydrous
calcium chloride was added, but the temperature was not recorded. The mixture
was stirred, then the temperature was recorded at the fifth minute. The student
continued stirring and recording the temperature at minute intervals for seven more
minutes.

The student’s results are shown in the table below.

Time / minutes 0 1 2 3 4

Page 34 of 99
Temperature / °C 19.6 19.5 19.5 19.5

Time / minutes 4 5 6 7 8 9 10 11 12

Temperature / °C 24.6 25.0 25.2 24.7 24.6 23.9 23.4 23.0

Plot a graph of temperature (y-axis) against time on the grid below.

Draw a line of best fit for the points before the fourth minute.
Draw a second line of best fit for the appropriate points after the fourth minute.
Extrapolate both lines to the fourth minute.

Page 35 of 99
(5)

(b) Use your graph to determine an accurate value for the temperature of the water at
the fourth minute (before mixing).

Temperature before mixing ____________________________________________

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(1)

(c) Use your graph to determine an accurate value for the temperature of the reaction
mixture at the fourth minute (after mixing).

Temperature after mixing ______________________________________________

(1)

(d) Use your answers from parts (b) and (c) to determine an accurate value for the
temperature rise at the fourth minute.
Give your answer to the appropriate precision.

Temperature rise ____________________________________________________

(1)

(e) Use your answer from part (d) to calculate the heat given out during this
experiment. Assume that the water has a density of 1.00 g cm –3 and a specific heat
capacity of 4.18 JK–1 g–1. Assume that all of the heat given out is used to heat the
water.
Show your working.

___________________________________________________________________

___________________________________________________________________
(2)

(f) Calculate the amount, in moles, of CaCl2 in 1.00 g of anhydrous calcium chloride
(Mr = 111.0).

___________________________________________________________________
(1)

(g) Use your answers from parts (e) and (f) to calculate a value for the enthalpy
change,
in kJ mol–1, for the reaction that occurs when anhydrous calcium chloride dissolves
in water.

CaCl2(s) + aq CaCl2(aq)

___________________________________________________________________

___________________________________________________________________
(2)

(h) Explain why it is important that the reaction mixture is stirred before recording each
temperature.

___________________________________________________________________

Page 37 of 99
___________________________________________________________________
(1)

(i) Anhydrous calcium chloride can be prepared by passing chlorine over heated
calcium.
To prevent unreacted chlorine escaping into the atmosphere, a student suggested
the diagram of the apparatus for this experiment shown below.

(i) Suggest one reason why the student wished to prevent unreacted chlorine
escaping into the atmosphere.

______________________________________________________________

______________________________________________________________
(1)

(ii) Suggest one hazard of using the apparatus as suggested by the student for
this experiment.

______________________________________________________________

______________________________________________________________
(1)
(Total 16 marks)

Q20.
A student calculated that a value for the enthalpy change of neutralisation
is –51.2 kJ mol–1.

The design of a possible hand-warmer using hydrochloric acid and sodium hydroxide was
discussed. It was proposed that 500 cm3 of hydrochloric acid should be used in a flexible,
sealed plastic container with a breakable tube of solid sodium hydroxide also in the
container. On breaking the tube, the sodium hydroxide would be released, react with the
acid and produce heat.
A 40 °C temperature rise was thought to be suitable.

(a) Calculate the heat energy, in J, required to raise the temperature of the reaction
mixture by 40 °C. Assume that the reaction mixture has a density of 1.00 g cm–3 and
a specific heat capacity of 4.18 J K –1 g–1.
Assume that all of the heat energy given out is used to heat the reaction mixture.

___________________________________________________________________

Page 38 of 99
(2)

(b) Use your answer from part (a) and the value for the enthalpy change of
neutralisation of –51.2 kJ mol–1 to calculate the minimum amount, in moles, and
hence the minimum mass of sodium hydroxide required in the breakable tube.
(If you could not complete the calculation in part (a) assume that the heat energy
required was 77 400 J. This is not the correct answer).

Show your working.

Moles of NaOH ______________________________________________________

___________________________________________________________________

Mass of NaOH ______________________________________________________

___________________________________________________________________
(3)

(c) Use the amount, in moles, of sodium hydroxide from part (b) to calculate the
minimum concentration, in mol dm–3, of hydrochloric acid required in the 500 cm3 of
solution used in the sealed container.

___________________________________________________________________

___________________________________________________________________
(1)

(d) Suggest one possible risk to a person who uses a hand-warmer containing sodium
hydroxide and hydrochloric acid.

___________________________________________________________________

___________________________________________________________________
(1)

(e) A commercial hand-warmer uses powdered iron sealed in a plastic container.

A valve allows air to enter the container, and oxygen in the air reacts slowly with the
iron to form solid iron(lll) oxide. The heat released warms the container.

(i) Write an equation for this reaction between iron and oxygen to form iron(lll)
oxide.

______________________________________________________________
(1)

(ii) One version of an iron-oxygen hand-warmer advertises that it is designed to

stay warm for up to four hours.
Other than by increasing the amount of iron in the container, state one change
to the iron in the hand-warmer that would increase this time.
Explain why this change to the iron might not be an advantage.

Change to the iron ______________________________________________

______________________________________________________________

Page 39 of 99
Explanation ____________________________________________________

______________________________________________________________

______________________________________________________________
(3)

(f) Another type of hand-warmer uses sodium thiosulfate. Sodium thiosulfate is very
soluble in water at 80 °C but is much less soluble at room temperature.
When a hot, concentrated solution of sodium thiosulfate is cooled it does not
immediately crystallise. The sodium thiosulfate stays dissolved as a stable
’super-saturated’ solution until crystallisation is triggered.
Heat energy is then released when the sodium thiosulfate crystallises.

(i) This type of hand-warmer is re-usable.

Suggest one environmental advantage that a sodium thiosulfate hand-warmer
has over the other two types.

______________________________________________________________
(1)

(ii) Describe the two steps that you would take to make the sodium thiosulfate
hand-warmer ready for re-use.

Step 1 ________________________________________________________

______________________________________________________________

Step 2 ________________________________________________________

______________________________________________________________
(2)
(Total 14 marks)

Q21.
Glucose, produced during photosynthesis in green plants, is a renewable source from
which ethanol can be made. Ethanol is a liquid fuel used as a substitute for petrol.
The processes involved can be summarised as follows.

Process 1 Photosynthesis in green plants

6CO2 + 6H2O → C6H12O6 + 6O2

Process 2 Fermentation of glucose to form ethanol

Process 3 Complete combustion of ethanol

CH3CH2OH + 3O2 → 2CO2 + 3H2O

(a) State three essential conditions for the fermentation of aqueous glucose in Process
2.

Write an equation for the reaction that takes place during this fermentation.

___________________________________________________________________

Page 40 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(4)

(b) It has been claimed that there is no net carbon (greenhouse gas) emission to the
atmosphere when ethanol made by Process 2 is used as a fuel.

State the term that is used to describe fuels of this type.

Use the equations for Processes 1, 2 and 3 to show why it can be claimed that there
is no net emission of carbon-containing greenhouse gases.

___________________________________________________________________

___________________________________________________________________
(3)

(c) Use the information from the equation for Process 3 above and the mean bond
enthalpies from the table below to calculate a value for the enthalpy change for this
process.

C–H C–C C–O O–H C=O O=O

Mean bond
+412 +348 +360 +463 +743 +496
enthalpy / kJ mol–1

Give one reason why the value calculated from mean bond enthalpies is different
from the value given in a data book.

___________________________________________________________________

Page 41 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(4)

(d) A student carried out a simple laboratory experiment to measure the enthalpy
change for Process 3. The student showed that the temperature of 200 g of water
increased by 8.0 °C when 0.46 g of pure ethanol was burned in air and the heat
produced was used to warm the water.

Use these results to calculate the value, in kJ mol –1, obtained by the student for this
enthalpy change. (The specific heat capacity of water is 4.18 J K –1 g–1)

Give one reason, other than heat loss, why the value obtained from the student’s
results is less exothermic than a data book value.

___________________________________________________________________

___________________________________________________________________
(4)
(Total 15 marks)

Q22.
A group of students devised an experiment which they believed would enable them to
investigate the strength of the intermolecular forces between ethyl ethanoate molecules
(CH3COOCH2CH3) and trichloromethane molecules (CHCl3).

They mixed exactly 0.10 mol of each of the two liquids in a copper calorimeter and
recorded the following results. The starting temperature of both liquids was the same.

Mass of 0.10 mol of ethyl ethanoate / g 8.80

Mass of 0.10 mol of trichloromethane / g 11.95

Increase in temperature (∆T) on mixing / K 9.5

(a) (i) Write an expression for the heat change (q) which relates mass (m), specific
heat capacity (c) and change in temperature (∆T).

______________________________________________________________
(1)

Page 42 of 99
(ii) Calculate the amount of heat required to increase the temperature of 8.80 g of
ethyl ethanoate by 9.5 K during the mixing process. (You should assume that
c for ethyl ethanoate = 1.92 J g–1 K–1)

______________________________________________________________
(1)

(iii) Calculate the amount of heat required to increase the temperature of 11.95 g
of trichloromethane by 9.5 K during the mixing process. (You should assume
that c for trichloromethane = 0.96 J g –1 K–1)

______________________________________________________________
(1)

(iv) Using the values from parts (a) (ii) and (a) (iii), calculate the molar enthalpy
change in kJ mol–1 for the mixing process.

______________________________________________________________

______________________________________________________________
(2)

(b) The students deduced that the heat change was due only to the formation of
intermolecular forces between ethyl ethanoate molecules and trichloromethane
molecules.

Ignoring all experimental errors, give one reason why the students may have made
an incorrect deduction.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 6 marks)

Q23.
(a) Define the term standard enthalpy of formation, ∆Hfο

___________________________________________________________________

___________________________________________________________________
(3)

(b) Use the data in the table to calculate the standard enthalpy of formation of liquid
methylbenzene, C7H8

Substance C(s) H2(g) C7H8(l)

Standard enthalpy of combustion, ∆Hcο /kJ mol–1 –394 –286 –3909

7C(s) + 4H2(g) → C7H8(l)

___________________________________________________________________

Page 43 of 99
___________________________________________________________________

___________________________________________________________________
(3)

(c) An experiment was carried out to determine a value for the enthalpy of combustion
of liquid methylbenzene using the apparatus shown in the diagram.

Burning 2.5 g of methylbenzene caused the temperature of 250 g of water to rise by

60°C. Use this information to calculate a value for the enthalpy of combustion of
methylbenzene, C7H8

(The specific heat capacity of water is 4.18 J K –1 g–1. Ignore the heat capacity of the
container.)

___________________________________________________________________

___________________________________________________________________
(4)

(d) A 25.0 cm3 sample of 2.00 mol dm–3 hydrochloric acid was mixed with 50.0 cm3 of a
1.00 mol dm–3 solution of sodium hydroxide. Both solutions were initially at 18.0 °C.

After mixing, the temperature of the final solution was 26.5°C.

Use this information to calculate a value for the standard enthalpy change for the
following reaction.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

In your calculation, assume that the density of the final solution is 1.00 g cm –3 and
that its specific heat capacity is the same as that of water. (Ignore the heat capacity
of the container.)

___________________________________________________________________

___________________________________________________________________
(4)

(e) Give one reason why your answer to part (d) has a much smaller experimental

Page 44 of 99
error than your answer to part (c).

___________________________________________________________________

___________________________________________________________________
(1)
(Total 15 marks)

Q24.
The sketch graph below shows how the entropy of a sample of water varies with
temperature.

(a) Suggest why the entropy of water is zero at 0 K.

___________________________________________________________________
(1)

(b) What change of state occurs at temperature T1?

___________________________________________________________________
(1)

___________________________________________________________________

___________________________________________________________________
(2)

(d) It requires 3.49 kJ of heat energy to convert 1.53 g of liquid water into steam at
373 K and 100 kPa.

(i) Use these data to calculate the enthalpy change, ∆H, when 1.00 mol of
liquid water forms 1.00 mol of steam at 373 K and 100 kPa.

______________________________________________________________

Page 45 of 99
______________________________________________________________

______________________________________________________________

(ii) Write an expression showing the relationship between free-energy change,

∆G, enthalpy change, ∆H, and entropy change, ∆S.

______________________________________________________________

(iii) For the conversion of liquid water into steam at 373 K and 100 kPa,
∆G = 0 kJ mol–1

Calculate the value of ∆S for the conversion of one mole of water into steam
under these conditions. State the units.

(If you have been unable to complete part (d)(i) you should assume
that ∆H = 45.0 kJ mol–1. This is not the correct answer.)

Calculation _____________________________________________________

______________________________________________________________

Units _________________________________________________________
(6)
(Total 10 marks)

Q25.
Methanol, CH3OH, is a convenient liquid fuel.

(a) An experiment was conducted to determine the enthalpy of combustion of liquid

methanol. The energy obtained from burning 2.12 g of methanol was used to heat
150 g of water. The temperature of the water rose from 298 K to 362 K. (The
specific heat capacity of water is 4.18 J K–1 g–1)

(i) Define the term standard enthalpy of combustion.

______________________________________________________________

______________________________________________________________
(3)

(ii) Use the data above to calculate a value for the enthalpy of combustion of one
mole of liquid methanol.

Page 46 of 99
______________________________________________________________

______________________________________________________________

______________________________________________________________
(4)

(b) Methanol can be synthesised from methane and steam by a process that occurs in
two stages.

Stage 1 CH4(g) + H2O(g) 3H2(g) + CO(g) ΔHO = +206 kJ mol–1

Stage 2 CO(g) + 2H2(g) CH3OH(g) ΔHO = –91 kJ mol–1

(i) Explain why, in Stage 1, a higher yield of hydrogen and carbon monoxide is
not obtained if the pressure is increased.

______________________________________________________________

______________________________________________________________
(2)

(ii) Stage 2 is carried out at a compromise temperature of 500K. By considering

what would happen at higher and lower temperatures, explain why 500 K is
considered to be a compromise for Stage 2.

______________________________________________________________

______________________________________________________________
(3)

(c) The standard enthalpies of combustion of carbon monoxide and of hydrogen are
–283 kJ mol–1 and –286 kJ mol–1, respectively. Use these data and the enthalpy
change for Stage 2 to calculate a value for the standard enthalpy of combustion of
gaseous methanol.

Page 47 of 99
___________________________________________________________________

___________________________________________________________________

___________________________________________________________________
(3)
(Total 15 marks)

Q26.
(a) Define the term standard enthalpy of formation.

___________________________________________________________________

___________________________________________________________________
(3)

(b) State Hess’s Law and use it, together with the data given in the table below, to
calculate the standard enthalpy change for the following reaction.

MgO(s) + 2HCl(g) → MgCl2(s) + H2O(l)

MgO HCl(g) MgCl2 H2O

Page 48 of 99
ΔHf /kJ mol–1 –602 –92 –642 –286

___________________________________________________________________

___________________________________________________________________
(4)

(c) In an experiment, an excess of solid magnesium oxide was added to 50 cm 3 of

3.0 mol dm–3 hydrochloric acid. The initial temperature of the solution was 21 °C.
After reaction, the temperature had risen to 53 °C. (The specific heat capacity of
water is 4.2 J K–1 g–1)

Use this information to calculate the enthalpy change for the reaction of one mole of
magnesium oxide with hydrochloric acid. For your calculation you should assume
that all the heat from the reaction is used to raise the temperature of 50 g of water.

___________________________________________________________________

Page 49 of 99
___________________________________________________________________
(8)
(Total 15 marks)

Q27.
(a) Write an equation for the complete combustion of propanone, C 3H6O, to form
carbon dioxide and water.

___________________________________________________________________
(1)

(b) In a laboratory experiment, 1.45 g of propanone were burned completely in oxygen.

The heat from this combustion was used to raise the temperature of 100 g of water
from 293.1 K to 351.2 K.

(i) Calculate the number of moles of propanone in the 1.45 g.

______________________________________________________________

(ii) Calculate the heat energy required to raise the temperature of 100 g of water
from 293.1 K to 351.2 K.
(The specific heat capacity of water is 4.18 J K –1 g–1)

______________________________________________________________

(iii) Hence, calculate a value, in kJ mol–1, for the enthalpy of combustion of

propanone.

______________________________________________________________

______________________________________________________________
(5)

(c) In a similar experiment, the enthalpy of combustion of butanone, C 4H8O, was found
to be –1290 kJ mol–1. A data book value for the same reaction is ΔHc = –2430 kJ
mol–1.

(i) Suggest one reason why the experimental value is very different from the data
book value.

______________________________________________________________

(ii) This data book value of ΔHc for butanone (–2430 kJ mol–1) refers to the
formation of carbon dioxide gas and water in the gaseous state. How would
this value differ if it referred to the formation of water in the liquid state?
Explain your answer.

Difference _____________________________________________________

Page 50 of 99
Explanation ____________________________________________________

______________________________________________________________
(3)

(d) Calculate a value for the standard enthalpy of formation for liquid
ethanethiol,C2H5SH. Use the equation given below and enthalpy of combustion data
from the following table.

Substance C2H5SH(l) C(s) H2(g) S(s)

ΔHc / kJ mol–1 –1170 –394 –286 –297

2C(s) + 3H2(g) + S(s) → C2H5SH(l)

___________________________________________________________________

___________________________________________________________________
(3)
(Total 12 marks)

Q28.
A 50.0 cm3 sample of a 0.200 mol dm–3 solution of silver nitrate was placed in a
polystyrene beaker. An excess of powdered zinc was added to this solution and the
mixture stirred. Zinc nitrate, Zn(NO3)2, and silver were formed and a rise in temperature of
3.20 °C was recorded.

(a) Write an equation for the reaction between silver nitrate and zinc.

___________________________________________________________________
(1)

(b) Calculate the number of moles of silver nitrate used in the experiment.

___________________________________________________________________

___________________________________________________________________
(2)

(c) Calculate the heat energy evolved by the reaction in this experiment assuming that
all the energy evolved is used to heat only the 50.0 g of water in the mixture.
(Specific heat capacity of water is 4.18 J g –1 K–1)

___________________________________________________________________

___________________________________________________________________
(2)

(d) Calculate the heat energy change for the reaction per mole of zinc reacted.

___________________________________________________________________

Page 51 of 99
___________________________________________________________________
(2)

(e) Explain why the experimental value for the heat energy evolved in this experiment
is less than the correct value.

___________________________________________________________________

___________________________________________________________________
(1)
(Total 8 marks)

Page 52 of 99
Mark schemes

Q1.
(a) M1 moles ( = × 2.0) = 0.050
1

M2 heat released = 0.050 × 56.1 (= 2.805 kJ or 2805 J)

M3 ∆T =
1

M4 ∆T = or = 13(.4) (°C)
Correct answer (to at least 2 sig fig) scores 4 marks
27 or 26.8°C (from moles of two reagents being added
together for M2, or use of 25 cm3 in M4) scores 3 marks
0.013(.4)°C (from not converting kJ to J) scores 3 marks
(loses M4) [0.027 or 0.0268°C would score 2 marks (loses
M2 and M4)
M1 moles can be shown for either substance or without
specifying the substance; if it is shown for both substances,
must be correct for both for M1
Allow ECF from M1 to M2
Allow ECF from M2 to M4 (providing an attempt to calculate
q has been made – no ECF if 56100 or 56.1 is used as q)
Correct M4 scores M3. If error made in M4, M3 could score
from substituted values in this expression in M4
M4 final answer to at least 2 sig fig.
Penalise M4 for negative temperature rise
1

(b) M1 draws suitable best fit curve to 4 minutes

M1 line must be a curve and ignore value at 5 minutes

M1 line should not go to times before 4 minutes
1

Page 53 of 99
M2 (17.2 – value read from graph line at 4 minutes) ± 0.2 (°C)
M2 allow use of any curved or straight line that is an attempt
to draw a line through the values after 4 minutes (that may
include the point at 5 minutes)
M2 allow negative values
1
[6]

Q2.
(a) Aldehyde/propanal has dipole-dipole forces (between molecules)
If any ‘covalent bonds broken’ CE=0 for clip.
Ignore Van der Waal forces
M1

Alcohol/propan-1-ol AND Carboxylic acid/ propanoic acid have hydrogen bonding

(between molecules).
Ignore reference to energy
M2

The forces between the molecules in aldehyde are weaker (than those in alcohol
and acid so it will evaporate first.)
M3 only awarded following correct M1 OR M2
Allow converse for M3
M3

(b) Keep the temperature of the reaction mixture below the boiling point of
propan-1-ol/below 97 °C
Allow temperature in range 49-96 inclusive
Allow description of cooling the vessel
M1

Cool the distillate / collecting vessel

Ignore reference to oxidising agents
Penalise lid / sealed container
M2

(c) Add named carbonate/hydrogencarbonate OR magnesium to a sample of the

distillate.
Incorrect chemical CE=0
Allow formula (mark on for incorrect formula)
Allow blue litmus or correct named indicator
M1

Effervescence/fizz/bubbles would confirm presence of acid or converse

Blue litmus turns red confirms acid present or converse
Allow gas/CO2 produced which turns lime water cloudy OR
gas/H2 produced which burns with a squeaky pop
M2

(d) (Temperature difference = 15.1 °C)

If ∆T wrong – AE mark on otherwise can only award M2
If use 457 in M1, can only score M2

Page 54 of 99
q = 150 × 4.18 × 15.1 or 9467.7 J or 9.4677kJ
M1

amount ethanol burned = 0.457/46.0 = 9.93 × 10 -3 mol

If use 457 in M2 can score 2 for - 0.953 kJ mol-1
M2

Heat change per mole = (M1/1000)/M2 = 952.99 kJ mol-1

∆H = – 953 kJ mol–1 must be 3sfs and must be negative

(allow range -953 to -954)

BEWARE if they miss conversion to kJ and also miss
conversion to g, they get answer = - 953 which scores 1
+953 can score M1 and M2
Allow -950 or -960 for rounding to 2sf
M3

(e) Elimination
Penalise base elimination
M1

Mechanism : Either (E1)

M2 for protonation of alcohol, i.e. lp plus arrow to H +

or to H of H–O– in H2SO4 and from H-O bond to O
M3 for protonated alcohol plus arrow showing loss of
water
M4 for arrow showing loss of H+
From correct carbocation (E1)
wrong alcohol used / alkene formed loses M4
3

OR (E2)

Page 55 of 99
M2 for protonation of alcohol, i.e. lp plus arrow to H +
or to H of H–O– in H2SO4 and from H-O bond to O
M3 for protonated alcohol plus arrow showing loss of
water
M4 for arrow showing simultaneous loss of H+
wrong alcohol used / alkene formed loses M4
3

(f) E-pent-2-ene
Allow trans
M1

C=C bond cannot rotate and

Each carbon in the double bond has (2) different groups attached.
Allow (two) different groups on each/either side of the double
bond.
M2
[16]

Q3.
(a) M1 Temperature on the y-axis and uses sensible scales (i.e. minimum 20 little
squares for each °C on y-axis)
Lose mark if temperature scale starts at 0°C
This mark scores if all points fit on the grid. Do not penalise
M1 if extrapolation to 4 mins goes off the grid – this is
penalised in M3.
1

M2 Plots all the points correctly (± half a small square)

Lose mark if the points go off the grid Ignore a plotted point
at 4 mins used to work out ∆T
1

M3 Draws two best fit lines (0-3 mins) and (6-12 mins)
Both lines must be straight and through all points except 5th
minute; lose mark if the lines are kinked/doubled. Any line
through 5th minute loses mark
“S-shaped curve” through points loses M3 and M4
1

M4 Extrapolates both lines to at least the 4th minute

Page 56 of 99
Lose mark if the extrapolation goes off the grid.
1

M5 21.9 – 19.8 = 2.1 (°C)

Allow calculation of ∆T from S-shaped curve as: Value at 4th
minute – 19.8 but not if 0 (°C)
∆T value ecf from incorrect lines/extrapolation
∆T must be to at least 1dp
If value of ∆T = 2.1, then award M5
1

(b) 0.2 / 2.1 × 100 = 9.5 %

Conseq on (a)
Ignore no of sfs.
1

(c) Replace the glass beaker with a polystyrene cup / insulate the glass beaker / use a
lid
Ignore use more dilute solutions
Ignore suggested materials for insulation
Do not allow copper calorimeter / bomb calorimeter
1

(d) Increase magnitude of temperature change

Ignore references to volume changes
1

by increasing the concentration of the acid/alkali

Mark independently
1

(e) M1 HOOCCOOH + 2KOH → K2(OOCCOO) + 2H2O

M1 – equation (allow ionic KOH / 2C2O4)
H2C2O4 + 2KOH → K2C2O4 + 2H2O
ignore state symbols
allow multiples
Mark independently

M2 q ( = mc∆T) = 100 × 4.2 × 3.2 = 1344 J

M2 – process
(ignore sign here)
(allow calculations involving 4.18 which leads to 1338 J)
1

M3 n HOOCCOOH = 25 × 0.800 / 1000 = 0.020

n KOH = 75 × 0.6 / 1000 = 0.045
M3 – calculations of amounts, in moles, of both the
ethanedioic acid and potassium hydroxide (both calculations
needed)
1

M4 Moles of water = 0.040 moles

M4 – answer (stated or used in calculation of ∆H)
1

Page 57 of 99
M5 ∆H = –1.344/0.04

= – 33.6 ( kJ mol–1)
M5 – ecf on M2 and M4
Answer must be negative and to at least 2sf
∆H = – M2 (in kJ) / M4
–32.5 - –34 scores 4/4 (M2-M5 + equation)
+32.5 - +34 scores 3/4 (M2, M3, M4 + equation)
–65 - –68 scores 3/4 (+ equation)
+65 - +68 scores 2/4 (+ equation)
–52 - –54 scores 3/4 (+ equation)
+52 - +54 scores 2/4 (+ equation)
2

(f) HOOCCOOH is a weak acid / not fully dissociated

H2SO4 is a strong(er)acid / fully dissociated / dissociates
more
1

(more) energy needed to break bonds/complete dissociation / dissociation is

endothermic
So less energy is needed for dissociation of sulfuric acid
Ignore references to heat loss
1
[16]

Q4.
(a) M1 Amount ZnSO4 = 1.0 × mol or Amount ZnSO4 = 0.050 mol
Mark M1 and M2 independently
1

M2 Amount Mg = mol or Amount Mg = 0.0856 mol

(Hence Mg in excess)
1
M3 Q = mcΔT
M3 could be scored in M4
1
M4 Q = 50.0 × 4.18 × 37.3
or Q = 7795.7 J
If an error in M4, lose M4 and M5 and only award M6 for
correct use of their incorrect M4 and division by their correct
limiting reagent
1

M5 (Energy released per mole) kJ mol−1 or J mol−1

M5 division by their limiting reagent
1
M6 ΔH = − 156 kJ mol−1
1

(b) Heat loss (from the apparatus would mean the experimental value is smaller /
lower / less exothermic than the data source)
1

Page 58 of 99
(c) Marks awarded for this answer will be determined by the quality of the
communication as well as the standard of the scientific response. Examiners
should apply a ‘best-fit’ approach to the marking.

Level 3 (5 – 6 marks)
Covers 3 Stages with matching justifications
Answer is full and detailed and is supported by an appropriate range of relevant
points such as those given below:
• argument is well structured with minimum repetition or irrelevant points
• accurate and clear expression of ideas with only minor errors in the use
of technical terms, spelling and punctuation and grammar

Level 2 (3 – 4 marks)
Covers 2 Stages with matching justification. OR covers 3 Stages with
incomplete justification
Answer has some omissions but is generally supported by some of the relevant
points below:
• the argument shows some attempt at structure
• the ideas are expressed with reasonable clarity but with a few errors in
the use of technical terms, spelling, punctuation and grammar

Level 1 (1 – 2 marks)
Covers 1 Stage with matching justification. OR covers 2 Stages with
incomplete justification
Answer is largely incomplete. It may contain valid points which are not clearly linked
to an argument structure. Unstructured answer. Errors in the use of technical terms,
spelling, punctuation and grammar or lack of fluency

Level 0 (0 marks)

Insufficient correct chemistry

Indicative Chemistry Content

Stage 1 Improved insulation

1a Insulate the beaker or use a polystyrene cup or a lid
1b To reduce heat loss

Stage 2 Improved temperature recording

2a Record the temperature for a suitable time before adding the metal
2b To establish an accurate initial temperature
OR
2c Record temperature values at regular time intervals
2d To plot the temperature results against time on a graph

Stage 3 Improved analysis of results

3a Extrapolate the cooling back to the point of addition
3b To establish a (theoretical) maximum temperature OR temperature change (e.g.
at the 4th minute) OR adjust for the cooling /apply a cooling correction
3a and 3b could be seen on an extrapolated sketch graph

(Note– IGNORE use of measuring equipment with greater precision)

6
[13]

Q5.

Page 59 of 99
(a) Not possible to prevent some dissolving
ALLOW It is soluble / dissolves / other hydrates may form /
suggestions related to difficulty of measuring T (change) of a
solid
1

(b) (ΔhydH =) –155 – (–39)

OR labelled cycle
Minimum needed for ‘labelled cycle’

–116 (kJ mol−1)

1/2 for (+)116 or for –29 or for seeing –116 that has then be
processed further
1

(c) This question is marked using levels of response. Refer to the Mark Scheme
Instructions for examiners for guidance on how to mark this question

Level 3 (5 – 6 marks)

All stages are covered and the explanation of each stage is correct and virtually
complete.
Stage 2 must include use of a graphical method for Level 3 (i.e. ‘highest T reached’
method is max Level 2)

Answer communicates the whole explanation, including reference to enthalpy,

coherently and shows a logical progression through all three stages.Answer is full
and detailed and is supported by an appropriate range of relevant points such as
those given below:
For the answer to be coherent there must be some indication of how the graph is
used to find ΔT

Level 2 (3 – 4 marks)

All stages are covered (NB ‘covered’ means min 2 from each of stage 1 and 3) but
the explanation of each stage may be incomplete or may contain inaccuracies
OR two stages covered and the explanations are generally correct and virtually
complete

Answer is coherent and shows some progression through all three stages. Some
steps in each stage may be out of order and incomplete

Level 1 (1 – 2 marks)

Two stages are covered but the explanation of each stage may be incomplete or
may contain inaccuracies
OR only one stage is covered but the explanation is generally correct and virtually
complete

Answer shows some progression between two stages

Page 60 of 99
Level 0 (0 marks)

Insufficient correct Chemistry to warrant a mark

Indicative Chemistry Content

Stage 1 Method

(1a) Measures water with named appropriate apparatus

(1b) Suitable volume/mass / volume/mass in range 10 – 200 cm3/g
(1c) Into insulated container / polystyrene cup (NOT just ‘lid’)
(1d) Add known mass of MgCl2(s)
(1e) Use of ‘before and after’ weighing method. NOT ‘added with washings’

Stage 2 Measurements (could mark from diagram)

(2a) Record initial temperature (min 2 measurements)

(2b) Record T at regular timed intervals for 5+ mins / until trend seen
(2c) Plot T vs time

Stage 3 Use of Results (3a and 3b could come from diagram)

(3a) Extrapolate lines to when solid added (to find initial and final T)
(3b) Tfinal – Tinitial = ΔT / idea of finding ΔT from graph at point of addition
(3c) q = mcΔT
(3d) amount = mass/Mr (0.80/95.3 = 8.39 × 10−3 mol)
(3e) ΔHsoln = –q/8.39 × 10−3 or in words

This could all be described in words without showing actual calculations but
describing stages

If method based on ‘combustion’ Max Level 1

(d)

M1 = 5 points correctly plotted

M2 = line drawn correctly (NOT if curved, doubled or kinked)
(Check line of best fit –
if through 250, -600.5 and 280, -595.5 +/- one small square

Page 61 of 99
then award M2, if all crosses on line award M1 as well)
2

Gradient = Δ(ΔG)/ΔT = 0.167 (kJ K−1 mol−1)

(ΔG = ΔH – TΔS so gradient = –ΔS)

ΔS = –167 (J K−1 mol−1)

M4 = unit conversion i.e. M3 × 1000; M5 = –sign (process
marks)
Correct answer with sign gets M3, M4 and M5
ALLOW –163 to –171
1+1
[14]

Q6.
TWO correct extrapolations of best fit straight lines
Max 5/7 if no extrapolations or best fit straight lines
drawn
1

Use of their lines to calculate the temperature change at the 4 th minute

(17 °C)

Q = mcΔT

Q = 10 × 4.18 × [student’s temperature change]

Q = [710.6 J]
Correct numbers inserted into expression.
1

Moles of Mg = 0.24 / 24.3 = 0.00988 mol

ΔH = 710.6 / 0.00988 = 71923.07 J mol−1

ΔH = –72 (kJ mol−1)

Must be negative to score M6
1

Answer to 2 significant figures

1
[7]

Q7.
B
[1]

Page 62 of 99
Q8.
(a) C6H11OH + 6CO2 + 6H2O
1

(b) Temperature rise = 20.1

q = 50.0 × 4.18 × 20.1 = 4201 (J)

Mass of alcohol burned = 0.54 g and Mr alcohol = 100.0

∴ mol of alcohol = n = 0.54 / 100 = 0.0054

Heat change per mole = q / 1000n OR q / n

= 778 kJ mol–1 OR 778 000 J mol–1

ΔH = –778 kJ mol–1 OR –778 000 J mol–1

M4 is for answer with negative sign for exothermic reaction
Units are tied to the final answer and must match
1

(c) Less negative than the reference

Heat loss OR incomplete combustion OR evaporation of alcohol OR heat

transferred to beaker not taken into account
1

(d) Water has a known density (of 1.0 g cm–3)

Therefore, a volume of 50.0 cm3 could be measured out

1
[9]

Q9.
(a) Start a clock when KCl is added to water
1

Record the temperature every subsequent minute for about 5 minutes

Allow record the temperature at regular time intervals until
some time after all the solid has dissolved for M2
1

Plot a graph of temperature vs time

Extrapolate back to time of mixing = 0 and determine the temperature

(b) Heat taken in = m × c × ΔT = 50 × 4.18 × 5.4 = 1128.6 J

Max 2 if 14.6 °C used as ΔT
1

Page 63 of 99
Moles of KCl = 5.00 / 74.6 = 0.0670
1

Enthalpy change per mole = +1128.6 / 0.0670 = 16 839 J mol -1

= +16.8 (kJ mol-1)

Answer must be given to this precision
1

(c) ΔHsolution = ΔHlattice + ΔH(hydration of calcium ions) + 2 × ΔH(hydration of

chloride ions)

ΔHlattice = ΔHsolution – ΔH(hydration of calcium ions) –2 ×ΔH(hydration of chloride

ions)
1

ΔHlattice = –82–9 – (–1650 + 2 × –364) = +2295 (kJ mol–1)

(d) Magnesium ion is smaller than the calcium ion

Therefore, it attracts the chloride ion more strongly / stronger ionic bonding
1
[12]

Q10.
(a) Weigh the spirit burner (alcohol) before and after combustion M1
Do not allow “a known mass of alcohol” owtte
1

Water in a calorimeter / beaker M2

Measure volume of water (or mass) M3

Burn the alcohol to heat the water M4

Measure temperature rise in water M5

(b) Incomplete combustion

Evaporation of alcohol

Heat capacity of / heat absorption by the apparatus

Inadequate stirring
Any two correct
1
1

(c) Acidified potassium dichromate / manganate(VII) (Heat)

Allow sodium in place of potassium with appropriate colour
change)

Page 64 of 99
If reagent incomplete lose M1 but mark on.
If reagent incorrect, CE = 0/3
1

butan-1-ol orange to green / purple to colourless

2-methylpropan-2-ol NVC / orange / purple

(d) C4H9OH + 2O2 → 4C + 5H2O

OR
C4H9OH + 4O2 → 4CO + 5H2O
Allow any correct balanced equations which include
combinations of C, CO and/or CO2 in the products but must
be incomplete combustion.
1

Engine would not run as efficiently / would need to use more fuel / would release
less energy
Allow build-up of carbon in engine costly to remove
1

CO / Particulates of carbon toxic

Allow global dimming if carbon given as product
1
[13]

Q11.
(a) M1 (q = mcΔT = 100 × 4.18 × 38(.0))
= 15 884 / 15 880 / 15 900 / 16 000 (J)
(OR 15.884 / 15.88 / 15.9 / 16 (kJ))
Award full marks for correct answer
Mark is for value not expression (at least 2sf); penalise
incorrect units here only if M1 is the only potential scoring
point in M1-M3
1

M2 Moles (methanol = 1.65 / 32.0) = 0.0516 or 0.052

At least 2sf
1

M3 Heat change per moles = M1/M2

(15 884 / 0.0516 / 1000 = 308 (kJ mol−1)
(allow 305 to 310)
At least 2sf; answer must be in kJ mol−1
1

M4 Answer = −308 (kJ mol−1) (allow −305 to −310)

This mark is for – sign (mark independently)
1

(b) Heating up copper / calorimeter / container / thermometer /

heat capacity of copper / calorimeter / thermometer not taken into account
OR
Evaporation of alcohol/methanol

Page 65 of 99
OR
Experiment not done under standard conditions
Not human errors (e.g. misreading scales)
Not impure methanol
Allow evaporation of water
1

(c) (100 × 0.5 / 38 =) 1.3 or 1.32 or 1.316% (minimum 2 sf)

Allow correct answer to at least 2sf;
Allow 1.31 or 1.315%
1

(d) Idea that heat loss is more significant issue OR

Idea that temperature change/rise is (significantly / much)
bigger than uncertainty
One of these two ideas only and each one must involve a
comparison
1

(e) M1 Mass of ethanol = 500 × 0.789 (= 394.5 or 395 (g))

M2 Moles of ethanol = M1 / 46.0 (= 8.576 or 8.58)

M3 Heat released = M2 × 1371 = 11800 (kJ) must be 3 sf

1
Correct answer to 3sf scores 3; correct value to 2sf or more
than 3sf scores 2
Answers that are a factor of 10x out score 2 if given to 3sf or
1 if given to a different number of sf
M3 ignore units, but penalise incorrect units
M3 ignore sign
M2 and M3 – allow consequential marking
[10]

Q12.
(a) (i) 2C6H12O6 3CH3COCH 3 + 3CO2 + 3H2O
Or multiples
1

(ii) to speed up the reaction

OR
(provide a) catalyst or catalyses the reaction or biological catalyst
OR
release / contain / provides an enzyme
Ignore “fermentation”
Ignore “to break down the glucose”
Not simply “enzyme” on its own
1

(b) (i) CH3CH(OH)CH3 + [O] CH3COCH3 + H2O

Any correct representation for the two organic structures.
Brackets not essential.

Page 66 of 99
Not “sticks” for the structures in this case
1

(ii) Secondary (alcohol) OR 2° (alcohol)

OR q =150 × 4.18 × 8.0

Award full marks for correct answer
In M1, do not penalise incorrect cases in the formula

M2 = (±) 5016 (J) OR 5.016 (kJ) OR 5.02 (kJ)

(also scores M1)

M3 This mark is for dividing correctly the number of kJ by the number

of moles and arriving at a final answer in the range shown.
Using 0.00450 mol

therefore ΔH = − 1115 (kJ mol−1 )

OR − 1114.6 to − 1120 (kJ mol−1 )

Range (+)1114.6 to (+)1120 gains 2 marks

BUT − 1110 gains 3 marks and +1110 gains 2 marks

AND − 1100 gains 3 marks and +1100 gains 2 marks

Award full marks for correct answer
In M1, do not penalise incorrect cases in the formula
Penalise M3 ONLY if correct numerical answer but sign is
incorrect; (+)1114.6 to (+)1120 gains 2 marks
Penalise M2 for arithmetic error and mark on
If ΔT = 281; score q = m c ΔT only
If c = 4.81 (leads to 5772) penalise M2 ONLY and mark on
for M3 = − 1283
Ignore incorrect units in M2
If units are given in M3 they must be either kJ or kJ mol−1 in
this case
3

(d) M1 The enthalpy change / heat change at constant pressure when

1 mol of a compound / substance / element

M2 is burned / combusts / reacts completely in oxygen

OR
burned / combusted / reacted in excess oxygen

M3 with (all) reactants and products / (all) substances in standard /

specified states

OR
(all) reactants and products / (all) substances in normal states under standard
conditions / 100 kPa / 1 bar and specified T / 298 K
For M3
Ignore reference to 1 atmosphere

Page 67 of 99
3

(e) M1
Σ B (reactants) − Σ B (products) = ΔH
OR
Sum of bonds broken − Sum of bonds formed = ΔH
OR
2B(C−C) + B(C=O) + 6B(C−H) + 4B(O=O) (LHS)

− 6B(C=O) − 6B(O−H) (RHS) = ΔH

M2 (also scores M1)

2(348)+805+6(412)+4(496) [LHS = 5957]

(696) (2472) (1984)

− 6(805) − 6(463) [RHS = (−) 7608] = ΔH

(4830) (2778)

OR using only bonds broken and formed (5152 − 6803)

M3
ΔH= − 1651 (kJ mol−1)

Candidates may use a cycle and gain full marks.

Correct answer gains full marks
Credit 1 mark for (+) 1651 (kJ mol−1)
For other incorrect or incomplete answers, proceed as
follows
• check for an arithmetic error (AE), which is either a
transposition error or an incorrect multiplication / addition
error; this would score 2 marks (M1 and M2)
• If no AE, check for a correct method; this requires
either a correct cycle with 4O2, 3CO2 and 3H2O OR a clear statement of
M1 which could be in words and scores only M1
Allow a maximum of one mark if the only scoring point is
LHS = 5957 (or 5152) OR RHS = 7608 (or 6803)
Award 1 mark for + 1651
3

(f) For the two marks M1 and M2, any two from
• heat loss or not all heat transferred to the apparatus or heat absorbed by
the apparatus or (specific) heat capacity of the apparatus not considered
• incomplete combustion / not completely burned / reaction is not
complete
• The idea that the water may end up in the gaseous state (rather than
liquid)
• reactants and / or products may not be in standard states.
• MBE data refers to gaseous species but the enthalpy of combustion
refers to liquids in their standard states / liquid propanone and liquid
water in standard states
• MBE do not refer to specific compounds OR MBE values vary with
different compounds / molecules OR are average / mean values taken
from a range of compounds / molecules
Apply the list principle but ignore incomplete reasons that

Page 68 of 99
contain correct chemistry
Ignore “evaporation”
Ignore “faulty equipment”
Ignore “human error”
Not enough simply to state that “MBE are mean / average
values”
2
[15]

Q13.
Increase in volume
If a volume is quoted it must be less than 300
1

Smaller increase in T above room temperature

Or increased contact between calorimeter and water
Or smaller heat loss by evaporation / from the surface
1
[2]

Q14.
(a)
M1 C6H12O6 2CH3CH2OH + 2CO2
(2C2H5OH)
Penalise C2H6O for ethanol in M1.

M2 and M3
Mark M2 and M3 independently.

Any two conditions in any order for M2 and M3 from

• (enzymes from) yeast or zymase

• 25 °C ≤ T ≤ 42 °C OR 298 K ≤ T ≤ 315 K

• anaerobic / no oxygen / no air OR neutral pH

A lack of oxygen can mean either without oxygen or not
having enough oxygen and does not ensure no oxygen,
therefore only credit “lack of oxygen” if it is qualified.
Penalise ‘bacteria’, ‘phosphoric acid’, ‘high pressure’ using
the list principle.

M4 (fractional) distillation or GLC

Ignore reference to ‘aqueous’ or ‘water’ (ie not part of the list
principle).

M5 Carbon-neutral in this context means

There is no net / overall (annual) carbon dioxide / CO2 emission to the

atmosphere

Page 69 of 99
There is no change in the total amount / level of carbon dioxide / CO2 present
in the atmosphere
For M5 – must be about CO2 and the atmosphere.
The idea that the carbon dioxide / CO2 given out equals the
carbon dioxide / CO2 that was taken in from the atmosphere.
5

(b) M1 q = m c ∆T (this mark for correct mathematical formula)

Full marks for M1, M2 and M3 for the correct answer.
In M1, do not penalise incorrect cases in the formula.

M2 = (75 × 4.18 × 5.5)

1724 (J) OR 1.724 (kJ) OR 1.72 (kJ) OR 1.7 (kJ)

(also scores M1)

Ignore incorrect units in M2.

M3 Using 0.0024 mol

therefore ∆H = − 718 (kJ mol−1)

(Accept a range from −708 to −719 but do not penalise more than 3 significant
figures)
Penalise M3 ONLY if correct numerical answer but sign is
incorrect. Therefore +718 gains two marks.
If units are quoted in M3 they must be correct.
If ∆T = 278.5, CE for the calculation and penalise M2 and
M3.

M4 and M5 in any order

Any two from

• incomplete combustion

• heat loss

• heat capacity of Cu not included

• some ethanol lost by evaporation

• not all of the (2.40 × 10−3 mol) ethanol is burned / reaction is incomplete
If c = 4.81 (leads to 1984) penalise M2 ONLY and mark on
for M3 = − 827
5

(c) (i) M1 enthalpy / heat / energy change (at constant pressure) or enthalpy /
heat / energy needed in breaking / dissociating (a) covalent bond(s)
Ignore bond making.

M2 averaged for that type of bond over different / a range of molecules /

compounds
Ignore reference to moles.
2

Page 70 of 99
(ii) M1

∑ B(reactants) − ∑ B(products) = ∆H

Sum of bonds broken − Sum of bonds formed = ∆H

B(C-C) + B(C-O) + B(O-H) + 5B(C-H) + 3B(O=O)

– 4B(C=O) – 6B(O–H) = ∆H = −1279
Correct answer gains full marks.
Credit 1 mark for − 496 (kJ mol−1)
For other incorrect or incomplete answers, proceed as
follows
• check for an arithmetic error (AE), which is either
a transposition error or an incorrect multiplication;
this would score 2 marks (M1 and M2).
If no AE, check for a correct method; this requires either a
correct cycle with 2CO2 and 3H2O OR a clear statement of
M1 which could be in words and scores only M1.

M2 (also scores M1)

348+360+463+5(412)+ 3B(O=O)

(3231) (or 2768 if O–H cancelled)

− 4(805) − 6(463) = ∆H = − 1279

(5998) (or 5535 if O–H cancelled)

3B(O=O) = 1488 (kJ mol−1)

Credit a maximum of one mark if the only scoring point is
bonds formed adds up to 5998 (or 5535) OR bonds broken
includes the calculated value of 3231 (or 2768).

B(O=O) = 496 (kJ mol−1)

Award 1 mark for −496

Students may use a cycle and gain full marks

3
[15]

Q15.
(a) (Q = mcΔT)

= 50 × 4.18 × 27.3
If incorrect (eg mass = 0.22 or 50.22 g) CE = 0 / 2
1

= 5706 J (accept 5700 and 5710)

Accept 5.7 kJ with correct unit. Ignore sign.

Page 71 of 99
1

(b) Mr of 2-methylpropan-2-ol = 74(.0)

For incorrect Mr, lose M1 but mark on.
1

Moles = mass / Mr

= 0.22 / 74(.0)

= 0.00297 moles
1

ΔH = –5706 / (0.002970 × 1000)

= –1921 (kJ mol–1)

If 0.22 is used in part (a), answer = –8.45 kJ mol–1 scores 3

(Allow –1920, –1919)

If uses the value given (5580 J), answer = –1879 kJ mol–1
scores 3
Answer without working scores M3 only.
Do not penalise precision.
Lack of negative sign loses M3
1

(c) ΔH = ΣΔH products – ΣΔH reactants

OR a correct cycle
Correct answer with no working scores 1 mark only.
1

ΔH = −(−360) + (4 × −393) + (5 × −286)

M2 also implies M1 scored.
1

ΔH = –2642 (kJ mol–1) This answer only.

Allow 1 mark out of 3 for correct value with incorrect sign.
1

(d) (–2422 – part (b)) × 100 / –2422

Ignore negative sign.

Expect answers in region of 20.7

If error carried forward, 0.22 allow 99.7
If 5580 J used earlier, then allow 22.4
1

(e) Reduce the distance between the flame and the beaker / put a sleeve around
the flame to protect from drafts / add a lid / use a copper calorimeter rather
than a pyrex beaker / use a food calorimeter
Any reference to insulating material around the beaker must
be on top.
Accept calibrate the equipment using an alcohol of known
enthalpy of combustion.
1

Page 72 of 99
(f) Incomplete combustion
1
[11]

Q16.
(a) Chloride (ions) are smaller (than bromide ions)
Must state or imply ions.
Allow chloride has greater charge density (than bromide).
Penalise chlorine ions once only (max 2 / 3).
1

So the force of attraction between chloride ions and water is stronger

This can be implied from M1 and M3 but do not allow
intermolecular forces.
1

Chloride ions attract the δ+ on H of water / electron deficient H on water

Allow attraction between ions and polar / dipole water.
Penalise H+ (ions) and mention of hydrogen bonding for M3
Ignore any reference to electronegativity.
Note: If water not mentioned can score M1 only.
1

(b) ΔHsolution = ΔHL + ΔHhyd K+ ions + ΔHhyd Br − ions / = 670 − 322 − 335

Allow ΔHsolution= ΔHL + ΣΔHhyd

= (+)13 (kJ mol−1)

Ignore units even if incorrect.
+13 scores M1 and M2
−13 scores 0
−16 scores M2 only (transcription error).
1

(c) (i) The entropy change is positive / entropy increases

ΔS is negative loses M1 and M3
1

Because 1 mol (solid) → 2 mol (aqueous ions) / no of particles increases

Allow the aqueous ions are more disordered (than the solid).
Mention of atoms / molecules loses M2
1

Therefore TΔS > ΔH

(ii) Amount of KCl = 5/Mr = 5/74.6 = 0.067(0) mol

If moles of KCl not worked out can score M3, M4 only
(answer to M4 likely to be 205.7 K)
1

Heat absorbed = 17.2 × 0.0670 = 1.153 kJ

Page 73 of 99
Process mark for M1 × 17.2
1

Heat absorbed = mass × sp ht × ΔT

(1.153 × 1000) = 20 × 4.18 × ΔT

If calculation uses 25 g not 20, lose M3 only (M4 = 11.04, M5
= 287)
1

ΔT = 1.153 × 1000 / (20 × 4.18) = 13.8 K

If 1000 not used, can only score M1, M2, M3
M4 is for a correct ΔT
Note that 311.8 K scores 4 (M1, M2, M3, M4).
1

T = 298 − 13.8 = 284(.2) K

If final temperature is negative, M5 = 0
Allow no units for final temp, penalise wrong units.
1
[13]

Q17.
(a) The enthalpy change / heat (energy) change (at constant pressure) in a reaction is
independent of the route / path taken (and depends only on the initial and final
states)
Ignore the use of ΔH for enthalpy
1

(b) ΔHexp + ΔH2 – ΔH1 = 0

Any correct mathematical statement that uses all three terms

ΔHexp + ΔH2 = ΔH1 OR ΔH1 = ΔHexp + ΔH2

ΔHexp= ΔH1 – ΔH2 OR ΔHexp = ΔH1 +( – ΔH2 )

(c) ΔHexp = ΔH1 – ΔH2

ΔHexp = −156 −12 = −168 (kJ mol−1)

Ignore units

Award the mark for the correct answer without any working
1

(d) (i) M1 q = m c ΔT OR calculation (25.0 x 4.18 x 14.0)

Award full marks for correct answer

M2 = 1463J OR 1.46 kJ (This also scores M1)

In M1, do not penalise incorrect cases in the formula

Page 74 of 99
M3 must have both the correct value within the range specified and the
minus sign
Penalise M3 ONLY if correct numerical value but sign is
incorrect; e.g. +69.5 to +69.7 gains 2 marks (ignore +70
after correct answer)

For 0.0210 mol, therefore

ΔH1 = − 69.67 to − 69.52 (kJ mol-1)

OR ΔH1 = − 69.7 to − 69.5 (kJ mol−1)

Penalise M2 for arithmetic error but mark on

Accept answers to 3sf or 4sf in the range − 69.7 to − 69.5

ΔT = 287, score q = m c ΔT only

Ignore -70 after correct answer

If c = 4.81 (leads to 1684J ) penalise M2 ONLY and mark on
for M3 = − 80.17 (range − 80.0 to − 80.2)
Ignore incorrect units
3

(ii) The idea of heat loss

NOT impurity

Incomplete reaction (of the copper sulfate)

NOT incompetence

Not all the copper sulfate has dissolved

NOT incomplete combustion
1

(e) Impossible to add / react the exact / precise amount of water

Not just “the reaction is incomplete”

Very difficult to measure the temperature rise of a solid

Difficult to prevent solid dissolving

(Copper sulfate) solution will form

1
[8]

Q18.
(a) (i) M1 (could be scored by a correct mathematical expression which must

Page 75 of 99
have
all ∆Hsymbols and the ∑ or SUM)

M1 ΔHr = ΣΔHf (products) - ΣΔHf (reactants)

OR a correct cycle of balanced equations with 1C, 3H2 and 1O2

M2 ΔHr = – 201 + (– 242) – (– 394)

ΔHr = – 201 – 242 + 394
ΔHr = – 443 + 394
(This also scores M1)

M3 = – 49 (kJ mol–1)
(Award 1 mark ONLY for + 49)
Correct answer gains full marks
Credit 1 mark ONLY for + 49 (kJ mol–1)
For other incorrect or incomplete answers, proceed as
follows
• check for an arithmetic error (AE), which is either
a transposition error or an incorrect multiplication;
this would score 2 marks (M1 and M2)
• If no AE, check for a correct method; this requires
either
correct cycle of balanced equations with 1C, 3H2 and
1O2
OR a clear statement of M1 which could be in words
and
scores only M1
3

(ii) It is an element / elemental

Ignore reference to “standard state”

By definition
1

(b) M1 (The yield) increases / goes up / gets more

If M1 is given as “decreases” / “no effect” / “no change” then
CE= 0 for clip, but mark on only M2 and M3 from a blank M1

M2 There are more moles / molecules (of gas) on the left / of reactants
OR fewer moles / molecules (of gas) on the right
/ products
OR there are 4 moles /molecules (of gas) on the left and 2 moles / molecules on
the right.
OR (equilibrium) shifts / moves to the side with less moles / molecules
Ignore “volumes”, “particles” “atoms” and “species” for M2

M3: Can only score M3 if M2 is correct

The (position of) equilibrium shifts / moves (from left to right) to oppose the increase
in pressure
For M3, not simply “to oppose the change”

Page 76 of 99
For M3 credit the equilibrium shifts / moves (to right) to lower
/ decrease the pressure
(There must be a specific reference to the change that is
opposed)
3

(c) M1 Yield increases goes up

M2 The (forward) reaction / to the right is endothermic OR takes in/ absorbs

heat

The reverse reaction / to the left is exothermic OR gives out / releases heat
If M1 is given as “decrease” / “no effect” / “no change” then
CE= 0 for clip, but mark on only M2 and M3 from a blank M1

Can only score M3 if M2 is correct

M3 The (position of) equilibrium shifts / moves (from left to right) to oppose the
increase
in temperature (QoL)
For M3, not simply “to oppose the change”
For M3, credit the (position of) equilibrium shifts / moves
(QoL)
to absorb the heat OR
to cool the reaction OR
to lower the temperature
(There must be a specific reference to the change that is
opposed)
3

(d) (i) An activity which has no net / overall (annual) carbon emissions to the
atmosphere
OR
An activity which has no net / overall (annual) greenhouse gas emissions
to the atmosphere.
OR
There is no change in the total amount / level of carbon dioxide /CO2 carbon
/greenhouse gas present in the atmosphere.
The idea that the carbon /CO2 given out equals the carbon
/CO2 that was taken in from the atmosphere
1

(ii) CH3OH + 1½ O2 CO2 + 2H2O

Ignore state symbols
Accept multiples
1

(iii) 3H2 + 1½ O2 3H2O

Ignore state symbols

OR
Accept multiples

Page 77 of 99
2H2 + O2 2H2O
Extra species must be crossed through
1

(e) M1 q = m c ∆T
Award full marks for correct answer
Ignore the case for each letter

OR q = 140 × 4.18 × 7.5

M2 = 4389 (J) OR 4.389 (kJ) OR 4.39 (kJ) OR 4.4 (kJ)(also scores M1)

M3 Using 0.0110 mol

therefore ∆H = – 399 (kJmol–1 )
OR – 400
Penalise M3 ONLY if correct numerical answer but sign is
incorrect; +399 gains 2 marks
Penalise M2 for arithmetic error and mark on
In M1, do not penalise incorrect cases in the formula
If ∆T = 280.5; score q = m c ∆T only
If c = 4.81 (leads to 5050.5) penalise M2 ONLY and mark on
for M3 = – 459

+399 or +400 gains 2 marks

Ignore incorrect units
3
[16]

Q19.
(a) Temperature on y-axis
If axes unlabelled use data to decide that temperature is on
y-axis.
1

Uses sensible scales

Lose this mark if the plotted points do not cover half of the
paper.
Lose this mark if the temperature axis starts at 0 °C.
1

Plots all of the points correctly ± one square

Lose this mark if the graph plot goes off the squared paper.
1

Draws two best-fit lines

Candidate must draw two correct lines.
Lose this mark if the candidate’s line is doubled or kinked.
1

Both extrapolations are correct to the 4th minute

Award this mark if the candidate’s extrapolations are within
one square of your extrapolations of the candidate’s best-fit
lines at the 4th minute.

Page 78 of 99
1

(b) 19.5 (°C)

Accept this answer only.
1

(c) 26.5 ± 0.2 (°C)

Do not penalise precision.
1

(d) (c) – (b)

Only award this mark if temperature rise is recorded to 1 d.p.
1

(e) Uses mcΔT equation

Allow use of this equation with symbols or values for M1
even if the mass is wrong.
1

Correct value using 25 × 4.18 × (d)

7.0 gives 732 J.
Correct answer with no working scores one mark only.
Do not penalise precision.
Allow answer in J or kJ.
Ignore sign of enthalpy change.
1

(f) 9.0(1) × 10–3

Do not allow 0.01
Allow 9 × 10–3 or 0.009 in this case.
1

(g) If answer to (e) in J, then (e) / (1000 × (f))

If answer to (e) in kJ, then (e) / (f)

7.0 and 9.01 × 10–3 gives 81.2 kJ mol–1
If answer to (e) is in J must convert to kJ mol–1 correctly to
score mark.
1

Enthalpy change has negative sign

Award this mark independently, whatever the calculated
value of the enthalpy change.
1

(h) The idea that this ensures that all of the solution is at the same temperature
Do not allow ‘to get an accurate reading’ without
qualification.
1

(i) (i) Chlorine is toxic / poisonous / corrosive

Do not allow ‘harmful’.
1

Page 79 of 99
(ii) Explosion risk / apparatus will fly apart / stopper will come out
Ignore ‘gas can’t escape’ or ‘gas can’t enter the tube’.
1
[16]

Q20.
(a) q = 500 × 4.18 × 40
Do not penalise precision.
1

= 83600 J
Accept this answer only.
Ignore conversion to 83.6 kJ if 83600 J shown.
Unit not required but penalise if wrong unit given.
Ignore the sign of the heat change.
An answer of 83.6 with no working scores one mark only.
An answer of 83600 with no working scores both marks.
1

(b) Moles (= 83.6 / 51.2) = 1.63

Using 77400 alternative gives 1.51 mol
Allow (a) in kJ / 51.2
Do not penalise precision.
1

Mass = 1.63 × 40(.0) = 65.2 (g)

Allow 65.3 (g)
Using 77400 alternative gives 60.4 to 60.5
Allow consequential answer on M1.
1 mark for Mr (shown, not implied) and 1 for calculation.
Do not penalise precision.
2

(c) Molarity = 1.63 / 0.500 = 3.26 mol dm–3

Allow (b) M1 × 2
Using 1.51 gives 3.02
1

(d) Container splitting and releasing irritant / corrosive chemicals

Must have reference to both aspects; splitting or leaking (can
be implied such as contact with body / hands) and
hazardous chemicals.
Allow ‘burns skin / hands’ as covering both points
Ignore any reference to ‘harmful’.
Do not allow ‘toxic’.
1

(e) (i) 4Fe + 3O2 → 2Fe2O3

Allow fractions / multiples in equation.
Ignore state symbols.
1

Page 80 of 99
(ii) Iron powder particle size could be increased / surface area lessened
Decrease in particle size, chemical error = 0 / 3
Change in oxygen, chemical error = 0 / 3
1

Not all the iron reacts / less reaction / not all energy released / slower
release of energy / lower rate of reaction
Mark points M2 and M3 independently.
1

Correct consequence of M2
An appropriate consequence, for example
• too slow to warm the pouch effectively
• lower temperature reached
• waste of materials
1

(f) (i) Conserves resources / fewer disposal problems / less use of landfill / fewer
waste products
Must give a specific point.
Do not allow ‘does not need to be thrown away’ without
qualification.
Do not accept ‘no waste’.
1

(ii) Heat to / or above 80 °C (to allow thiosulfate to redissolve)

Accept ‘heat in boiling water’.
If steps are transposed, max 1 mark.
1

Allow to cool before using again

Reference to crystallisation here loses this mark.
1
[14]

Q21.
(a) Three conditions in any order for M1 to M3

M1 yeast or zymase

M2 30 °C ≥ T ≤ 42 °C

M3 anaerobic/no oxygen/no air OR neutral pH

M4 C6H12O6 2C2H5OH + 2CO2

OR
2C6H12O6 4C2H5OH + 4CO2
Mark independently
Penalise “bacteria” and “phosphoric acid” using the list
principle
Ignore reference to “aqueous” or “water” (i.e. not part of the
list principle)
Or other multiples

Page 81 of 99
4

(b) M1 Carbon-neutral
Ignore “biofuel”
1

M2 6 (mol/molecules) CO2/carbon dioxide taken in/used/used

up (to form glucose or in photosynthesis)
1

M3 6 (mol/molecules) CO2/carbon dioxide given out due to

2 (mol/molecules) CO2/carbon dioxide from fermentation/
Process 2 and 4 (mol/molecules) CO2/carbon dioxide from
combustion/Process 3
It is NOT sufficient in M2 and M3 for equations alone without
commentary or annotation or calculation
1

(c) M1 (could be scored by a correct mathematical expression)

(Sum of) bonds broken – (Sum of) bonds made/formed = ΔH

(Σ) Breactants – (Σ) Bproducts = ΔH

(where B = bond enthalpy/bond energy)

For M1 there must be a correct mathematical expression
using ΔH or “enthalpy change”

M2 Reactants = (+) 4719

OR
Products = (–) 5750

M3 Overall + 4719 – 5750 = –1031 (kJ mol–1) (This is worth 3 marks)

Award full marks for correct answer.
Ignore units.
M2 is for either value underlined
M3 is NOT consequential on M2
3

Award 1 mark ONLY for +1031

Candidates may use a cycle and gain full marks.

M4 Mean bond enthalpies are not specific for this reaction

OR they are average values from many different
compounds/molecules
Do not forget to award this mark
1

(d) M1 q = m c ΔT (this mark for correct mathematical formula)

M2 = 6688 (J) OR 6.688 (kJ) OR 6.69 (kJ) OR 6.7 (kJ)

M3 0.46g is 0.01 mol

therefore ΔH = – 669 kJ mol–1 OR – 670 kJmol–1

Page 82 of 99
OR –668.8 kJ mol–1
Award M1, M2 and M3 for correct answer to the calculation
Penalise M3 ONLY if correct answer but sign is incorrect
In M1, do not penalise incorrect cases in the formula
If m = 0.46 or m = 200.46 OR if ΔT = 281, CE and penalise
M2 and M3
If c = 4.81 (leads to 7696) penalise M2 ONLY and mark on
for M3 = –769.6 OR –770
Ignore incorrect units in M2

M4 Incomplete combustion
Do not forget to award this mark. Mark independently
4
[15]

Q22.
(a) (i) q = mc ΔT
Ignore case except T
1

(ii) 8.80 × 1.92 × 9.5 = 161 (J) to 160.5(12) (J)

Credit 0.161 provided it is clear that it is kJ.
Penalise wrong units
1

(iii) 11.95 × 0.96 × 9.5 = 109 (J) to 108.98(4) (J)

Credit 0.109 provided it is clear that it is kJ.
Penalise wrong units.
1

(iv) M1 Addition of (a)(ii) and (a)(iii)

M2 Multiply by 10 and convert to kJ (divide by 1000)

leading to an answer
Consequential on (a)(ii) and (a)(iii)
Penalise wrong units
Ignore the sign

Therefore ΔH = (–) 2.69 OR (–) 2.7(0) (kJ mol–1)

Ignore greater numbers of significant figures (2.69496)
Subtraction in M1 is CE
2

(b) One from:

• No account has been taken of the intermolecular forces initially

in the two liquids OR each liquid has its own intermolecular
forces in operation before mixing.

• The liquids may react or reference to reaction or reference

to bonds broken or formed
Any statement which shows that there are other
intermolecular forces to consider.
Ignore heat loss and ignore poor mixing.

Page 83 of 99
1
[6]

Q23.
(a) Enthalpy change when 1 mol of compound (1)

Is formed from it’s elements (1)

All substances in their standard state (1)

(b) ΔH = ΣΔHοc (reactants) – ΣΔHοc (products) (1)

= (7x – 394) + (4 x – 286) – (– 3909) (1)

= + 7 kJmol–1 (1)
3

(c) Heat change = m c ΔT (1)

= 250 × 4.18 × 60 = 62700J = 62.7kJ (1)

Moles C7H8 = 2.5 /92 = 0.0272 (1)

ΔH = 62.7 / 0.0272 = – 2307 kJ mol–1 (1)

(allow –2300 to –2323)
4

(d) Mass of water heated = 25 + 50 = 75g

Temp rise = 26.5 – 18 = 8.5 °C
both for (1) mark

Heat change = 75 × 4.18 × 8.5 = 2665 J = 2.665 kJ (1)

Moles HCl = 0.05 (1)

ΔH = – 2.665 / 0.05 = –53.3 kJmol–1 (1)

(allow –53 to –54)
4

(e) Less heat loss (1)

1
[15]

Q24.
(a) Particles are in maximum state of order
(or perfect order or completely ordered or perfect crystal or
minimum disorder or no disorder)
(entropy is zero at 0 k by definition)
1

(b) (Ice) melts

(or freezes or changes from solid to liquid or from liquid to
solid)
1

Page 84 of 99
(c) Increase in disorder
1

Bigger (at T2)

1
Second mark only given if first mark has been awarded

(d) (i) Moles of water = 1.53/18 (= 0.085)

Heat change per mole = 3.49/0.085 = 41.1 (kJ mol –1)

(allow 41 to 41.1, two sig. figs.)
(penalise –41 (negative value), also penalise wrong units but
allow kJ only)
1

(ii) ΔG = ΔH – TΔS
1

(iii) ΔH = TΔS or ΔS = ΔH/T

(penalise if contradiction)
1

ΔS = 41.1/373 = 0.110 kJ K–1 (mol–1) (or 110 (J K–1 (mol–1))

(allow 2 sig. figs.)
(if use value given of 45, answer is 0.12 (or 120 to 121)
(if ΔH is negative in (d) (i), allow negative answer)
(if ΔH is negative in (d) (i), allow positive answer)
(if ΔH is positive in (d) (i), penalise negative answer)
1

Correct units as above (mol–1 not essential)

1
[10]

Q25.
(a) (i) enthalpy change when 1 mol of a substance
(or compound) (QL mark)
1

is (completely) burned in oxygen (or reacted in excess oxygen)

at 298 K and 100 kPa (or under standard conditions)

(ii) heat produced = mass of water × Sp heat capacity

xΔT (or mcΔT)
1

= 150 × 4.18 × 64 (note if mass = 2.12 lose first 2 marks

then conseq) = 40100 J or = 40.1 kJ (allow 39.9 - 40.2
must have correct units)
1

moles methanol = mass/Mr = 2.12/32 (1)

= 0.0663

Page 85 of 99
1

ΔH = – 40.1/0.0663 = – 605 kJ (mol–1)

1
(allow –602 to –608 or answer in J)
(note allow conseq marking after all mistakes but note use of
2.12 g loses 2 marks

(b) (i) equilibrium shifts to left at high pressure

because position of equilibrium moves to favour

fewer moles (of gas)
1

(ii) at high temperature reaction yield is low (or at low T yield is high)
1

at low temperature reaction is slow (or at high T reaction is fast)

therefore use a balance (or compromise) between rate and yield

(c) ΔH = ΣΔHcο(reactants) – ΣΔHcο (products) (or correct cycle)

ΔHcο (CH3OH) = ΔHcο(CO) + 2 × ΔHcο(H2) – ΔH

= (–283) + (2 × –286) – (–91) (mark for previous equation or this)

= –764 (kJ mol–1) ( units not essential but lose mark if units wrong)
(note + 764 scores 1/3)
1
[15]

Q26.
(a) (i) enthalpy (or heat or heat energy) change when
1 mol of a substance (1) (QL mark) is formed from its elements (1)
all substances in their standard states (1) (or normal states at 298K,
100 kPa or std condits)
not STP, NTP
3

(b) enthalpy change (or enthalpy of reaction) is independent of route (1)

ΔH = ΣΔHf prods - ΣΔHf reactants (or cycle) (1)

minimum correct cycle is:

ΔH = -642 – 286 – (–602 + 2 × –92) (1)

= –142 (kJ mol–1) (1)
penalise this mark for wrong units

Page 86 of 99
+142 scores 1 mark out of the last three
4

(c) ΔH = mcT (1) (or mcΔT)

= 50 × 4.2 × 32 = 6720 J = 6.72J (1)
mark is for 6720 J or 6.72 kJ

moles HCl = × conc = × 3 (1)

= 0.15 (1)
if error here mark on conseq.

Therefore moles of MgO reacted = moles HCl/2 (1)

(mark is for/2, CE if not/2)
= 0.15/2 = 0.075

Therefore ΔH = 6.72/0.075 (1)

= –90 kJ (mol–1)
kJ must be given, allow 89 to 91
value (1)
sign (1); this mark can be given despite CE for /2
8

Note various combinations of answers to part (c) score as follows:

–89 to –91 kJ (8) (or –89000 to 91000J)

no units (7)

+89 to +91 kJ (7) (or + 89000 to +91000J)

no units (6)

–44 to –46 kJ (5) (or -44000 to -46000J)

no units (4) if units after 6.72 or 6720 (5)

+44 to +46 kJ (4) (or +44000 to + 46000)

if no units and
if no units after 6.72 or 6720 (3)
otherwise check, could be (4)
[15]

Q27.
(a) C3H6O + 4O2 → 3CO2 + 3H2O (1) (or multiple) 1

(b) (i) (1) = 0.0250 (1)

allow 0.025
allow conseq on wrong Mr

1.45/100, CE; C.E.

(ii) heat released = mcΔT

= 100 × 4.18 × 58.1 (1)

Page 87 of 99
if 1.45 used in place of 100 CE = 0

= 24300 J (1) (or 24.3kJ)

allow 24200 to 24300
ignore decimal places
units tied to answer
If use 0.1 × 4.18 × 51.8 allow ½ for 24.3 with no units

(iii) = –972 (kJ mol–1) (1)

allow –968 to –973
allow +972
allow conseq
allow no units
penalise wrong units
5

(c) (i) Heat loss (1) or energy loss

do not allow incomplete combustion

(ii) Difference: more negative (1) (or more exothermic)

QoL mark

Explanation: heat (or energy) released when water vapour condenses (1)
or heat/energy required to vaporise water
or water molecules have more energy in the gaseous state
3

(d) ΔH = ΣΔHreactants – ΣΔHproducts (1)

(or cycle )

= (2 × –394) + (3 × –286) + (–297) – (–1170) (1)

= –773 (1)
ignore units even if wrong
Allow 1/3 for +773
3
[12]

Q28.
(a) 2AgNO3 + Zn → Zn(NO3)2 + 2Ag (1)
Accept an ionic equation i.e.2Ag + +Zn → 2Ag + Zn2+
1

(b) Moles = mv / 1000 (1) = 0.20 × 50/1000 = 1.00 × 10–2

(c) Heat energy change = mCΔT (1) = 50 × 418 × 3.2 J

Page 88 of 99
= 669 J (Ignore signs) (1)
Allow 668, 67.0 0.67kJ
Penalise wrong units if given
2

(d) = 134 kJ mol–1

Mark one : 2 × (answer to (c))
Mark two : Dividing by answers to (b)
Allow 133 – 134
Penalise incorrect units
Mark conseq to equation in (a) for full marks, also to that in
(c)
If No working is shown and answer is incorrect zero
2

(e) Incomplete reaction or Heat loss (1)

1
[8]

Page 89 of 99
Examiner reports

Q1.
(a) Many students struggled with this question, though it did discriminate well. The main
problem stemmed from an inability to find the heat released using the enthalpy
change (per mole) and the amount in moles of the acid or alkali reacting. Many
believed that q was the same as the enthalpy change. Some students also
incorrectly thought that amount in moles reacting’ was the amount in moles of acid +
moles of alkali.

(b) Very few students realised that the best-fit line was a curve and not a straight line.
Some students also extrapolated to a time before the time of mixing. Only 3.8% of
students gained both marks.

Q2.
(a) This question discriminated well: 28.3% of students correctly identified the types of
intermolecular forces and noted that hydrogen bonding in the alcohol and the acid
was stronger than dipole-dipole attractions between propanal molecules. Despite
the request in the question to consider intermolecular forces, a few students still
discussed the breaking of bonds in the molecules and so gained no marks, although
fewer did this than is often the case in this type of question.

(b) Very few students (4%) scored both marks in this question and only 39.2% scored
at least one mark, despite the question being based on one of the required practical
activities.

(c) This part was answered well by nearly two-thirds of the students. The main error
was to test for the aldehyde rather than the acid.

(d) This question discriminated well and there was a good spread of marks. Sadly,
some students did not quote their answers to three significant figures and/or
remember the negative sign for their exothermic enthalpy change of combustion.

(e) The major errors in the answers to this acid-catalysed elimination of water were
where students confused this mechanism with the elimination of HX from a
halogenoalkane using a base. However, just under a third of students scored full
marks in a correct mechanism, either via the formation of a carbocation or by
showing the simultaneous loss of water and H+ from a protonated alcohol.

(f) This part was challenging and a surprising proportion of students (8%) made no
attempt. A large number scored no marks as they discussed the formation of
different alkenes, rather than why pent-2-ene shows E-Z isomerism. Sadly, some of
those who did gain the first mark failed to gain the second as this required two
statements, not only that the C=C bond cannot rotate, but also that each carbon in
the double bond has two different groups attached to it.

Q3.
(a) The usual advice is that a scale on a graph is ‘suitable’ if the plotted points take up
more than half of the available space. In this case, the decision about the suitability
of the scale was more ‘relaxed’ due to the need for students to allow space for the
extrapolation to go higher than the highest plotted point. However, a significant
number of students still managed to miss the mark by choosing scales that led to
the plotted points being squeezed in to only a couple of squares on the vertical axis.

Page 90 of 99
There still seems to be a misconception amongst some students that all scales need
to start at zero. Having plotted the points, most students recognised that their lines
of best fit should ignore the point at 5 minutes. Although a lot of ‘S-shaped’ curves
were seen (the reminder clue in the question was that candidates should draw
suitable lines of best fit), some students failed to draw any line between 0 and 3
minutes, and the lines were sometimes not extrapolated to the 4 th minute. A
surprising number of students quoted the ‘final’ temperature at the 4th minute rather
than giving ∆T. Pleasingly, 34.3% of students scored all five marks.

(b) Far more students than expected failed to double the uncertainty given in the
question to reflect the fact that two temperature readings are taken with the
thermometer. As a result, only 19.3% of students were awarded this mark.

(c) As expected, this was the easiest question on the paper with the vast majority of
candidates (90.7%) able to suggest a way to reduce heat loss.

(d) This was another example of a question where some students were let down by not
reading the question, which clearly asked for a change that would reduce the
percentage uncertainty in the use of the same thermometer. Given that percentage
uncertainty is calculated as apparatus uncertainty divided by measurement made, it
was hoped that more students would recognise the need to increase the size of the
measurement made, i.e. the temperature change. Many students who did recognise
the need to generate a bigger temperature change suggested doing so by using
smaller volumes; they thus failed to appreciate that this would reduce the amounts
of reactants as well as the volume, so that less heat would be given out and the
change in temperature would be the same as in the larger scale experiment. 58% of
students scored zero.

(e) Despite the formula and name being given in the question, many students failed to
recognise that ethanedioic acid is diprotic, which affected their equation and
subsequent calculation. In the calculation, most were able to calculate the initial
amounts of each reactant, but many then failed to recognise the idea that one is
present in excess so that the other would be the limiting reagent and so determine
the amount of water formed. Calculations involving adding 0.02 to 0.045, or
subtracting 0.02 from 0.045, were frequently seen, although it was not always clear
what students thought they were calculating when they did so, due to poor layout of
calculations. There were also a lot of variations seen in the mass used for the mcΔT
calculation. In any enthalpy change or calorimetry experiment of this type, it is the
solution that is increasing in temperature, so it is the mass of solution that needs to
be used (100 g here as 25 cm3 of one solution was added to 75 cm3 of another and
all have a density of 1.00 g cm–3). This calculation discriminated extremely well
between students of different abilities; 21.2% scored maximum marks and only 6.4%
failed to score anything.

(f) This was expected to be a high-level discriminator and that certainly proved to be
the case for the second mark (achieved by only 3.3% of students). Although many
gained a mark for referring to the fact that sulfuric acid is a stronger acid than
ethanedioic acid, the second half of their answer often then incorrectly referred to
ideas such as “it therefore reacts with more KOH”, “more water is produced”, and
“bonds between K+ and sulfate are stronger than between K + and ethanedioate”.
Only the best students recognised that the overall enthalpy change is the net result
of the difference between ‘energy supplied’ and ‘energy released’. In this case, the
‘energy released’ will be the same for both acids. Therefore, the difference in the
enthalpies of neutralisation must be due to ethanedioic acid needing more energy to
be supplied, in order to break O−H bonds during the neutralisation reaction
(whereas the sulfuric acid is already fully dissociated).

Page 91 of 99
Q4.
(a) In this extended calculation the correct limiting reagent had to be used to gain full
marks. This proved difficult for many students. The less able students were
expected to be able to calculate the numbers of moles of the two reagents and/or
state that q = mcΔT, but this was not always the case.

(b) 58.1% of students gained the mark for this question about a well-known practical
issue.

(c) This practical question was marked using a levels of response mark scheme. There
was a wide range of answers, showing that most had done this sort of practical
during their course. Nearly all students were aware that insulation is the most
obvious improvement to the technique but many students failed to read the question
carefully. Many answers, even from the best students, described a technique but did
not suggest clearly and then justify how the method given in the question could be
improved.

Q5.
(a) This proved to be a much tougher starter question than had been anticipated, with
only 12% of students earning the mark. Students did not recognise that enthalpy
change could not be measured directly. In this case, the enthalpy change was for
the formation of a hydrated salt from an anhydrous one, with the expected answer
being that it would be impossible to prevent some salt dissolving during the addition
of water to the anhydrous salt. An alternative answer, related to the difficulty in
measuring the temperature of a solid, was also allowed.

Some students’ answers revealed confusion between calorimetry and colorimetry,

with suggested answers including the lack of colour change during the reaction.
Some students also seemed to believe that calorimetry is only possible if something
is being burned.

(b) This calculation was answered reasonably well, with less than 20% of students
failing to score. There was confusion about the ‘direction’ around the Hess Cycle,
and evidence of many students thinking that an equation of the form ΔH=
ΣΔHf(products) - ΣΔHf(reactants) can be used universally, when it only applies to the
specific use of enthalpies of formation.

(c) Most students’ encounter with this ‘extended response’ style of question produced a
good spread of marks, although 37% of students failed to score. This was usually
due to a completely inappropriate method being described, with incorrect answers
such as: “making a solution of magnesium chloride”, “using it to fill a spirit burner
and lighting it under a beaker of water” and “putting a solution of magnesium
chloride in a polystyrene cup and heating it over a Bunsen burner”. Other incorrect
answers included descriptions of the preparation of a standard solution followed by
a titration and the addition of magnesium chloride to acid instead of to water.

This question was marked using a ‘levels of response’ mark scheme. The key to
success was for students to concentrate first on the inclusion of as much correct
chemistry as possible to ensure access to Level 3 (worth 5 or 6 marks). Within a
level, the mark awarded depended on the clarity and coherence of an answer,
together with a clear, logical progression through the description. Appropriate
apparatus and quantities should have been mentioned as necessary. For example,
rather than writing ‘add water to a container’, a good start to the answer would be to
write ‘A measuring cylinder was used to measure 50 cm 3 of water into a polystyrene
cup’.

Page 92 of 99
Despite the fact that many students suggested in part (a), that it is difficult to
measure the temperature of a solid, many then suggested putting the magnesium
chloride into the polystyrene cup first and then recording its initial temperature,
before adding the water.

(d) This was a challenging question, but one for which it proved relatively easy to score
two marks; nearly 20% of students scored full marks here. Most students could
successfully plot the points and draw a best fit line, although the negative scale on
the y-axis confused some. Most students were also able to calculate the gradient of
their line, although, as mentioned previously, it was not always clear what their
suggested answer was. The calculation of ΔS proved trickier, with many trying to
use the relationship ΔG = ΔH − TΔS, and either ignoring their calculated gradient or
substituting it in for ΔH. Relatively few students recognised that this equation can be
taken as y = mx + c (y = c – xm in this case) so that this graph of ΔG (y) vs T (x)
gives a straight line with a gradient of −ΔS.

Q10.
Part (a) - Only the best candidates were able to gain all five marks. Worryingly, many
candidates seemed to attempt to measure the specific heat capacity of the alcohol by
heating it in a water bath. Vague descriptions such as “burn some alcohol to heat some
water” with a failure to state what variables need to be measured were also fairly
common. Some wanted to use a polystyrene cup to hold the water. There were some very
good answers too, though.

Part (b) - Generally well done.

Part (c) - Generally well answered. An incomplete reagent such as missing the condition
“acidified” lost M1 but subsequent marks were available. Tollens’ reagent and Benedict’s
reagent were favourite incorrect answers.

Part (d) - Only the best candidates were able to score full marks. There were many correct
equations although some gave an unbalanced chemical equation forgetting the “O” in the
alcohol and some gave only an equation for complete combustion. A common mistake
was to give two environmental concerns or two economic concerns even though the
question required one of each.

Q11.
Enthalpy of combustion & calorimetry

In the calculation of the enthalpy of combustion from the experimental data, many
students used the mass of the fuel (rather than the water) when using q = mcΔT. Others
incorrectly added 273 to the temperature rise. When finding the moles of methanol
burned, some students rounded this to 1 significant figure which gave inaccurate answers.
Some students failed to include the minus sign on their final answer to show that the
reaction is exothermic. In (b) many students referred to the problems already stated in the
question rather than an additional one. Many students suggested mistakes made during
the experiment rather than design features. Many students scored the mark for (c) but
many others did not know where to start. The calculation of apparatus percentage
uncertainties should be an important and routine part of practical work. Few students
realised that (d) related to the size of the uncertainty in the temperature rise compared to
the actual temperature rise and/or heat loss in the experiment. Many students were
uncertain how to start the calculation in (e), namely by using the density to find the mass
of the ethanol. Others struggled to convert the volume in dm 3 to cm3.

Page 93 of 99
Q12.
This question was answered well. Balancing the equation in part (a)i looks straightforward
when you know the answer, but it is quite demanding if you have to work it out, and
impressively half the students gave the correct answer. In part (b)i, it was necessary to
draw unambiguous structures both for propan-2-ol (not given in the question) and
propanone (given in part (a)i). The calculation in part (c) was well answered with the focus
on correct chemistry rather than on the number of significant figures and the mathematics
of rounding the final answer. The definition in part (d) was expected to be straightforward
and yet still resulted in only 40% gaining all three marks. As many as 45% gained all three
marks and 80% gained at least one mark from the very straightforward calculation in part
(e), with errors occurring because some students did not know the bonding either in
propanone or in carbon dioxide and used the unnecessary C−O that was given in the data
set. It was decided that the very straightforward idea of “heat loss”, in whatever context,
was worth only one mark in part (f) and that students needed to come up with other
reasons why the values in part (c) and part (e) differed; many were able to do so and this
led to 54% scoring both marks.

Q13.
In this question, the first mark (more water) was gained by many, although some
suggested a volume larger than the calorimeter. Few managed to give a clear
explanation, even though some seemed to have the right kind of idea but couldn’t express
it sufficiently clearly.

Q14.
Part (a) discriminated almost as well as the earlier mechanism questions with well over
half of the students scoring at least 3 of the 5 marks. Relatively few recognised that
(fractional) distillation is a process used to produce a much more concentrated solution of
ethanol and many referred to the hydration of ethene, having missed the requirement to
start “from a dilute aqueous solution”. The concept of carbon-neutral “in the context of this
biofuel” required reference both to carbon dioxide and the atmosphere. Part (b) was well
answered with over half of all students scoring 4 out of 5 marks. By comparison, part (c)
proved more demanding. The definition in part (c)(i) most often yielded only 1 of the 2
marks. Too often in part (c)(ii) it proved difficult to decipher the jumble of numbers that
some students recorded on the page and only about one-third of students scored all 3
marks.

Q15.
(a) This part was mostly well answered, with the exception of those who chose to use
the wrong mass.

(b) In this part, failure to give the negative sign was the most common error where
students did not score full marks.

(c) This part was usually done very well.

(d) In contrast, this part was very poorly answered, almost always because –2642 was
used from (c) rather than their answer to (b).

(e) Many vague answers were seen in response to this part. Simply adding insulation
was insufficient and use of a polystyrene cup was wrong in this situation.

(f) The idea of incomplete combustion seemed to be poorly understood in this part, with
quite a few students stating that ’not all the alcohol had burned’. Comments about

Page 94 of 99
heat loss etc were disqualified by the question.

Q16.
This question proved to be difficult. In part (a), most students scored one mark for
mentioning that the ionic radius of the chloride ion is less than that for the bromide ion.
However, many weaker students lost this mark because they did not make reference to a
chloride ion; often, just chlorine was mentioned. Only the best students went on to discuss
how a chloride ion is attracted to the partially positive hydrogen atom in a water molecule.
Others referred incorrectly to intermolecular forces or to hydrogen bonding. About 60% of
students gained both marks for the calculation in part (b). In part (c)(i), most students
stated that the entropy would increase but many did not clearly explain why and also that
the TΔS term would outweigh the positive enthalpy change. Although about a quarter of
all students scored at least 4 marks in part (c)(ii), less than 10% went on to score full
marks by subtracting the temperature change from 298 K. Marks were generally low
because many students did not recognise that the heat absorbed could be calculated by
multiplying the enthalpy of solution by the amount in moles of potassium chloride. Weak
students also used a wrong value for the mass of water heated – usually 25 g or 5.00 g.

Q17.
Almost 85% of students were able to give a correct statement for Hess’s Law and then
attempted to apply it in parts (b) and (c), with over 70% success. No credit was given for a
value calculated in part (c), consequent on an incorrect expression in part (b), since that is
a chemical error.

Part (d) was generally well-answered although the negative sign was often missed in the
answer leading to only 35% gaining all three marks. Only 7% were able to deduce a
correct answer to part (e).

Q18.
The standard enthalpy change calculation in part (a) was straightforward and high
scoring. In parts (b) and (c), good discrimination occurred and a great many
well–articulated responses were seen. It is worth noting that no marks were scored in
either of parts (b) or (c), if the effect on the yield was assigned incorrectly. The meaning of
carbon–neutral in part (d) is spelled out in the specification and needs to refer to net
emissions of carbon dioxide to the atmosphere. Parts (d)(ii) and (d)(iii) were challenging
and only the best students were able to arrive at the equation for the overall reaction
between hydrogen and oxygen to make water. Most students scored at least one mark in
part (e); the commonest error was a failure to give the final answer a negative sign for this
exothermic process.

Q19.
(a) In this part, most students completed the graph successfully. The usual problem
was a mark being wrongly awarded for plotted points which did not cover half the
grid because the student started the y-axis at zero. Some schools and colleges were
very generous when awarding the mark for drawing a line of best fit. In graph
questions where the points contain one or more anomalies, the line of best fit must
ignore these anomalies to score the mark. The line of best fit mark cannot be
awarded when the line itself is poorly drawn or doubled in places. A few schools and
colleges did not allow the mark for extrapolation when the line of best fit was poor.
As long as the line is extended correctly the mark for extrapolation can be awarded.
If the line becomes kinked when extrapolated the mark for extrapolation cannot be
awarded.

Page 95 of 99
(b) A few teachers failed to realise that the marks for part (b) and (c) were essentially
for drawing the graph correctly rather than for reading from a poorly drawn graph.

(c) A few teachers failed to realise that the marks for part (b) and (c) were essentially
for drawing the graph correctly rather than for reading from a poorly drawn graph.

(d) Few students lost the mark in this part for incorrect precision.

(e) In this part proved straightforward but a fair number of students used an incorrect
mass in the equation.

(f) In this part, a good number of students were incorrectly allowed the second mark for
the answer 0.01. This answer, being to one significant figure only, was disallowed in
the Marking Guidelines.

(g) In this part was well answered, although a surprising number of students omitted the
negative sign in their final answer.

(h) Answers to this part were generally good; the common error was to suggest that
stirring ensured that the mixture was homogeneous.

(i) In this part (ii) a number of students thought that heat was the danger to the
equipment and lost the mark.

The above notes are intended for that minority of schools and colleges experiencing
difficulty in meeting the criteria. They must not be allowed to unduly detract from the
very healthy overall picture. Given the pressures on teachers to deliver the teaching
programme, this was once again a very positive and encouraging session. Schools
and colleges are again warmly commended for the trouble taken to assemble a
sample which proved to be easy to moderate. Their efforts continue to be much
appreciated by the moderator team.

Q20.
Section B proved more demanding. In part (a), the calculation had similar faults to
Question 5 but some students failed to read that the answer was to be in Joules - an
answer in kJ was penalized unless there was a clear correct answer in the working. Part
(b) and part (c) were well done though the alternative answer to part (a) was frequently
used in these. In part (d), only a few students failed to highlight the need for the bag to
split or the chemicals to come into contact with skin / hands. Part (e)(i) indicated that a
considerable number of students found difficulties with the equation; common errors were
the wrong formula for iron (III) oxide, the inclusion of ions on the reagent side of the
equation and using Fe2 for the metal. Part (e)(ii) contained many good partial answers but
there was a significant failure to focus finally on why longer times might not be an
advantage. In part (f)(i), the main errors were to re-state the question about being
reusable or to indicate that there would be no waste following use of these handwarmers.
There were some vague answers to part (f)(ii) which referred to ’heating’ the bag or
’keeping it in a warm place’ and, for some, the contents were removed before further
treatment. Nevertheless, sensible answers were frequently seen.

Q21.
Most candidates seem to know something about fermentation and part (a) discriminated
extremely well.

In part (b), only the best candidates were able both to state that such a fuel is
carbon-neutral and also to demonstrate that the six moles of CO2 taken in during

Page 96 of 99
photosynthesis are later released in the fermentation process followed by the combustion
of ethanol. Failure to recognise that two moles of ethanol were burning resulted in only
four moles of CO2 and left candidates confused. A few candidates seemed to think that a
balanced equation with six carbon atoms on each side is carbon-neutral.

The calculation in part (c) proved demanding as candidates failed to count up accurately
how many of each bond were broken and how many were formed. They were able quite
often to get only one of these correct. The idea that mean bond enthalpies are average
values obtained from many different compounds was not well known.

In part (d) the equation used to calculate the heat change was well known but conversion
to enthalpy change proved difficult for all but the best. The idea of incomplete combustion
was often missed.

Q22.
This was generally a well-answered question suggesting that the students were able to
apply their knowledge in a novel context provided they had sufficient information. More
than 70% of candidates scored full marks in all three parts (a)(i), (a)(ii) and (a)(iii) and
almost half the candidates were able to use the calculated data for each liquid to arrive at
some credit in part (a)(iv). Almost a third of the candidates gave plausible reasons for why
the students may have made an incorrect deduction.

Q23.
Part (a) was generally well done by most candidates. The most common error was not
referring to standard states. The calculation in part (b) proved more demanding for
students than some previous enthalpy calculations with many candidates giving the
incorrect answer of –7kJmor\ A large number of candidates did not give the correct mass
or correct temperature rise in part (c) and therefore only scored the mark for the Q = mcΔt
equation. Many of the candidates who did calculate the heat evolved then failed to
calculate molar quantities. The calculation in part (d) proved very difficult for candidates
with very few scoring full marks. Common errors were incorrect mass, incorrect
temperature rise and lack of conversion to molar quantities. Candidates should know that
exothermic reactions are shown by a negative sign e.g. –53.3kJmol–1. In part (e) many
candidates failed to make the comparison regarding heat loss between the two
experiments.

Q24.
This question also produced a wide range of marks. A major misconception was to relate
entropy directly to the energy of particles rather than to their disorder. Most candidates
gave a correct answer to part (b). In part (c) weak candidates often discussed the
breaking of bonds rather than referring to disorder. Others just reiterated the question by
stating that the entropy would increase. Good candidates usually gained one mark by
stating that the disorder of particles increases when a liquid changes to a gas but only the
best candidates answered the question fully by referring also to the smaller change in
order on melting. It was pleasing to note that part (d) was answered correctly by many
candidates. Marks were sometimes lost when giving units in part (d)(iii) that did not tally
with the numerical answer. For example, when the numerical answer should have
corresponded to J K–1 mol–1 the units were frequently given as kJ K –1 mol–1.

Q25.
In this question, good candidates were able to score full marks. Weaker candidates were
less successful though they were usually able to pick up at least five or six marks.

Page 97 of 99
Answers to part (a) usually gained all three marks. Answers to part (b) by weaker
candidates were less successful. One common error in the expression mcΔT was to use
the mass of methanol (2.12 g) instead of the mass of water (100 g). Another common
error was to give an incorrect sign for the final value of the enthalpy of combustion.
Candidates were expected to recognise that this exothermic process should lead to a
negative value for the enthalpy of combustion of methanol. Weaker candidates also lost
marks for incorrect units or for giving no units. It is important for candidates to give correct
units for intermediate values in the calculation as well as for the answer. In this case, if a
candidate gave a wrong final answer, it was still possible to gain an intermediate mark for
the heat released to the 100 g of water. However, if the candidate did not make it clear
whether the intermediate value was expressed in Joules or in kJ, it was not possible to
give any credit.

Answers to part (b) were of a good standard and in part (b) (i) almost all candidates were
awarded the two marks. Answers to part (b) (ii) were not quite so accurate and, in
particular, only the best candidates stated that 500 K achieves the best balance between
rate of reaction and yield.

Good candidates found part (c) straightforward but weaker candidates often omitted to
allow for two moles of hydrogen and also used the wrong sign for the enthalpy of reaction
(–91 kJ mol–1 ) relative to the enthalpies of combustion of carbon monoxide and hydrogen.

Q26.
Most candidates answered this question well though it proved to be difficult to score
maximum marks. The definition of standard enthalpy of formation in part (a) was usually
correct though some candidates lost one mark because they did not make it clear that
reactants and products should be in their standard states. In part (b) good candidates
were able to score maximum marks but weaker candidates sometimes referred to energy
rather than enthalpy when stating Hess’s Law and such candidates often predicted an
endothermic enthalpy change (+142 kJ mol-1) rather than an exothermic one. Part (c)
proved to be the most difficult part of this question. The major error was in the number of
moles of magnesium oxide. Most candidates assumed that the number of moles of MgO
would be the same as that of hydrochloric acid whereas it should have been half of that.
This error led to a value for the enthalpy change that was too small by a factor of two.
Such an answer resulted in the loss of three marks.

Q27.
This question proved accessible to most candidates; even the weaker candidates were
able to score marks approaching half of those available. Answers to part (a) were usually
correct.

Most candidates also gave a correct answer to part (b)(i) though there were a significant
number of errors in the calculation of the relative molecular mass of propanone. Parts
(b)(ii) and (b)(iii) were also answered well although there were some problems with units.
The correct answer to part (c)(i) was known by most candidates but part (c)(ii) proved to
be much more discriminating and only the best candidates were able to explain why the
enthalpy of combustion is more negative when water is formed in the liquid state. A fair
majority of candidates was able to give a correct answer to part (d) but as usual in this
type of question, weaker candidates made an error in signs leading to an answer of +773
kJ mol–1 or made an error by omitting to multiply the relevant enthalpy of combustion by
the number of moles of the substance involved.

Q28.
It was disappointing to find many wrong equations for the reaction between silver nitrate

Page 98 of 99
and zinc in part (a) especially as the use of silver nitrate solution to distinguish between
halides ions forms part of this module and the formula of zinc nitrate was given.

Parts (b) and (c) were well answered and, as part (d) was marked consequentially to
answers given in parts (a), (b) and (c), most candidates were also able to score full marks
in part (d). Almost all candidates stated correctly in part (e) that loss of heat energy was
the reason why the experimental value obtained was less than the correct value.

Page 99 of 99

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3.1.4.2 Calorimetry

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3.1.4.2 Calorimetry

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Name: ________________________

Time: 365 min.

Marks: 322 marks

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

Temperature rise ____________________ °C

At the fourth minute, the student added sodium hydrogencarbonate.

The student’s measurements are shown in the graph.

Compound Boiling point / °C

propanoic acid 141

Calculate a value, in kJ mol–1, for the enthalpy of combustion of ethanol in this

(The specific heat capacity of water is 4.18 J K –1 g–1)

Enthalpy of combustion ____________________ kJ mol –1

(e) A mixture of isomeric alkenes is produced when pentan-2-ol is dehydrated in the

Name and outline a mechanism for the reaction producing pent-1-ene.

Name of mechanism _________________________________________________

(f) A pair of stereoisomers is also formed in the reaction in part (e).

Name the less polar stereoisomer formed.

• The student used a measuring cylinder to place 50 cm 3 of 0.400 mol dm–3

The results are shown in the table.

(a) Plot a graph of temperature against time on the grid below.

Percentage uncertainty ____________________

(e) A second student completed an experiment to determine the enthalpy of

Mg(s) + Zn2+(aq) ⟶ Mg2+(aq) + Zn(s)

The student used this method:

• A thermometer was placed in the beaker.

• 2.08 g of magnesium metal powder were added to the beaker.

• The mixture was stirred and the maximum temperature recorded.

The student recorded a starting temperature of 23.9 °C and a maximum temperature of

(a) Show by calculation which reactant was in excess.

Enthalpy of reaction ____________________ kJ mol −1

A data book value for the enthalpy of reaction is −310 kJ mol−1

Justify your suggestions.

(b) Some enthalpies of solution are shown in Table 1.

Enthalpy change _________________________________________ kJ mol−1

You should use about 0.8 g of anhydrous magnesium chloride.

Mg(s) + Cl2(g) ⟶ MgCl2(s)

Show your working.

Mg(s) + 2H+(aq) ⟶ Mg2+(aq) + H2(g)

The following table shows the results obtained.

Initial mass of spirit burner and alcohol / g 56.38

Final mass of spirit burner and alcohol / g 55.84

Initial temperature of water / °C 20.7

Final temperature of water / °C 40.8

Enthalpy of combustion = ____________ Units = ____________

(d) A 50.0 g sample of water was used in this experiment.

(b) The temperature of the water decreased to 14.6 °C.

Calculate a value, in kJ mol−1, for the enthalpy of solution of potassium chloride.

Enthalpy of solution = _______________ kJ mol −1

(c) The enthalpy of solution of calcium chloride is −82.9 kJ mol−1.

Explain with the aid of an equation why combustion of methylpropan-2-ol under

(a) A student carried out an experiment to determine the enthalpy of combustion of

A diagram of the apparatus is shown alongside a table which shows the

Explain why this thermometer was adequate for this experiment.

(e) The enthalpy of combustion of ethanol is −1371 kJ mol−1. The density of

Calculate the heat energy released in kJ when 0.500 dm 3 of ethanol is burned.

.......C6H12O6 .......CH3COCH3 + .......CO2 + .......H2O

(b) Propanone is also formed by the oxidation of propan−2−ol.

(ii) State the class of alcohols to which propan−2−ol belongs.

(d) Define the term standard enthalpy of combustion.

C−H C−C C−O O−H C=O O=O

Mean bond 412 348 360 463 805 496

CH3COCH3(g) + 4O2(g) 3CO2(g) + 3H2O(g)

Change in volume: _______________________________________________________

(a) A dilute aqueous solution of ethanol can be produced by the fermentation of an

Write an equation for this fermentation reaction.

Name a process used to produce a much more concentrated solution of ethanol

(c) Mean bond enthalpies can be used to calculate enthalpies of reaction.

(i) Give the meaning of the term mean bond enthalpy.

Mean bond enthalpy / to be

CH3CH2OH(g) + 3O2(g) 2CO2(g) + 3H2O(g) ΔH = –1279 kJ mol–1

A student carried out an experiment on a pure sample of 2-methylpropan-2-ol to

The results for the experiment are shown in Table 1.

Initial temperature of the water / °C 18.1

Final temperature of the water / °C 45.4

Initial mass of spirit burner and alcohol / g 208.80

Final mass of spirit burner and alcohol / g 208.58

Enthalpy of combustion = Units =