A-Level Chemistry Question and Answers 2013 (A-level).pdfThe equation for this reaction is shown below: 2HgO(s) = 2Hg(l) + O 2(g) a) What volume of O 2(g) would be obtained if 1.08g

A-Level Chemistry Question and Answers 2013

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Table of Contents Atomic Structure (Answers) ................................................................................................................ 7

Atoms, Molecules and Stoichiometry (Questions) ........................................................................... 9

Atoms, Molecules and Stoichiometry (Answers) ............................................................................ 10

Chemical Equilibria (Questions) ........................................................................................................ 12

Chemical Equilibria (Answers) .......................................................................................................... 13

Electrochemistry (Questions) * ........................................................................................................ 14

Electrochemistry (Answers)............................................................................................................... 15

Group II and Group IV (Questions) ................................................................................................. 16

Group VII (Questions)........................................................................................................................ 18

Group VII (Answers) .......................................................................................................................... 19

Reaction Kinetics (Questions) ........................................................................................................... 24

States of Matter (Questions) ............................................................................................................. 26

States of Matter (Answers) ............................................................................................................... 27

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Aliphatic Compounds

Aromatic and Plastics

Chemical Energetics

General Principles

Ionic Equilibria

Periodicity

Transition Metals

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5

Atomic Structure (Questions) *

1. Chemists use a model of an atom that consists of three types of sub-atomic particle. These are protons,

neutrons and electrons.

a) Complete the table below, which shows the properties of these particles.

Particle Relative Charge Relative Mass

Electron

Proton

1

Neutron 0

(4 marks)

b) Types of atoms and ions vary from each other by the number of sub-atomic

particle found in them. What differences are there between 12C and

14C?

(1 mark)

c) Define the term 'isotope'.

(1 mark)

d) Write down the electronic configuration of Cr3+.

(1 mark)

(Marks available: 7)

2.

a) The first ionisation energy for lithium is 519 kJmol-1.

Write an equation to show what is meant by this statement.

(1 mark)

6

b) The second ionisation energy for lithium is 7300 kJmol-1.

Explain the difference in value for the first and second ionisation energy for

lithium.

(1 mark)


7

Atomic Structure (Answers)

Answer outline and marking scheme for question: 1

a)

Particle Relative Charge Relative Mass

Electron -1 1/1820 or negligible

Proton +1 1

Neutron 0 1

(4 marks)

b) 12C has 6 protons, 6 neutrons and 6 electrons. 14C

has 6 protons, 6 electrons and 8 neutrons. The difference between this pair

is that 14C has two more neutrons than 12C.

(1 mark)

c) An isotope of an element has he same number of protons but a different

number of neutrons.

(1 mark)

d) 1s2 2s2 2p6 3s2 3p6

3d3 4s0

(1 mark)


8


a) Li(g) - e- = Li+(g)

The electronic configuration of lithium is 1s2 2s1. It

is relatively easy to remove the electron from the 2s orbital as it is shielded

from the nuclear charge by the 1s electrons.

(1 mark)

b) To remove the second electron for the second ionisation energy requires

considerably more energy as this electron comes from the 1s shell which is closer

to the nucleus and not subject to any shielding.

(1 mark)


9

Atoms, Molecules and Stoichiometry (Questions)

1. Calculate the mean relative atomic mass of nickel from the mass spectrum.


2. In 1774 Joseph Priestly conducted one of his most famous experiments which

lead to a method for the preparation of oxygen. The experiment involved heating

a sample of mercury II oxide with a large lens.

The equation for this reaction is shown below:

2HgO(s) = 2Hg(l) + O2(g)

a) What volume of O2(g) would be obtained if 1.08g of mercury

II oxide were completely decomposed? (Given that 1 mole of a gas occupies 24

dm3 under the experimental conditions)


3. An organic compound is found to have the following composition by mass:

C 72.7%; H 6.1%; N 21.2%.

a) Calculate the empirical formula of the compound.

(2 marks)

b) The molar mass of the compound is 198 gmol-1. Calculate

the molecular formula of the compound.

(3 marks)


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Atoms, Molecules and Stoichiometry (Answers)


Intensities of peaks:

58Ni = 10.2 = 69.4% of total (abundance)



(Sum of intensities = 14.7)

Mean relative atomic mass = (58 x 69.4) + (59 x 26.5) + (62 x 4.1) =

58.4

100



Molar mass of HgO = 201 + 16 = 217 gmol-1

1.08g of HgO will contain 1.08 / 217 mols = 0.005mol

From the equation, 1 mole of O2(g) is produced from 2 moles of HgO

This means that 0.005 mol of HgO will produce 0.005 / 2 mol of O2 = 0.0025 mol

0.0025 mol of O2 will occupy 0.0025 x 24dm3 = 0.06dm3


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Ratio of elements in compound C : H : N

Ratio C : N : H = 72.7 / 12 : 6.1 / 1 : 21.2 / 14 = 6.1 : 6.1 : 1.5

Whole number ratio = 4 : 4 : 1

Empirical formula of compound is C4H4N


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Chemical Equilibria (Questions)

1.

Esters are a useful group of compounds due to their distinctive smells. One example

of an ester is ethyl ethanoate, its formation is shown below.

CH3COOH(aq) + C2H5OH(aq)λCH3COO C2H5(aq)

+ H2O(l)

a) Systems like this are described as being a 'dynamic equilibrium'. Explain

the term 'dynamic equilibrium'

b) Write down the expression for the equilibrium constant, Kc, for this reaction.

c) Calculate the value of Kc for this reaction given the equilibrium concentrations below.

[CH3COOH] = 0.08 moldm-3

[C2H5OH] = 0.08 moldm-3

[CH3COO C2H5] = 0.25 moldm-3

[H2O] = 0.1 moldm-3

d) Concentrated sulphuric acid is added to the reaction mixture as it removes water molecules. What effect

would this have on the equilibrium position of this system?


2. The Haber-Bosch process is used for the large-scale production of ammonia from

nitrogen and hydrogen gas. The reaction is shown below:

N2(g) + 3H2(g)λ2NH3(g)

a) Write an expression for Kp for this reaction.

b) What are the units of Kp for this reaction? (Assume pressure is measured in kPa)

c) When the temperature is raised for this process the proportion of NH3(g) in the mixture decreases. Explain this

observation.

d) What effect will an increase in the total pressure have on the equilibrium position?


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Chemical Equilibria (Answers)


a)

- concentrations of the species are constant

- forward and backward reactions are continually taking place but at the same rate.

b) Kc =[CH3COO C2H5] [H2O]

[CH3COOH]

[C2H5OH]

c) Kc = [CH3COO C2H5] [H2O] = (0.25) (0.1) = 3.91 (no units)

[CH3COOH] [C2H5OH] (0.08)(0.08)

(1 mark numerical answer, 1 mark stating no units)

d) Concentrated sulphuric acid is added to the reaction mixture as it removes water molecules. What effect

would this have on the equilibrium position of this system?



a) Kp = (p NH3)2 (1)

(p N2)(pH2)2

b) Units = kPa2 =

1 / kPa2 = kPa-2

kPa x kPa3

c) Increase in temperature leading to decrease in NH3 means reaction

towards right is exothermic.

d) An increase in pressure will lead to an increase in the proportion of NH3(g) found in the mixture.


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Electrochemistry (Questions) *

1. Dichromate (VI) ions are powerful oxidising agents and are reduced to chromium III ions. This colour change

was once used in 'breath test' apparatus to determine if a driver had consumed excessive alcohol.

Explain the term 'redox reaction'


2. Aluminium metal is extracted from molten bauxite (Al2O3.2H2O)

using electrolysis. Cryolite (AlF3) added to the ore in order to

lower the melting point required and thus the energy required by the process.

a) Write an half equation to show how aluminium metal is produced from

the ore.

(1 mark)

b) What mass of aluminium metal would be produced if a current of 30,000A

is applied to a cell for 1 hour.

(4 marks)

c) In the molten mixture there is a mixture of anions which mostly consists

of O2- and F-. Write an equation to show which

of these anions will be oxidised in the cell?

(2 marks)


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Electrochemistry (Answers)


A redox reaction is one where electrons are transferred from one species to another.



a) Al3+(l) + 3e- = Al(l)

(1 mark)

b) Charge passed; E = It

E = 30,000 x 3600 = 1.08 x 108C (1)

1 mole of electron has a charge of 96,500C

Mols electrons = (1.08 x 108) / 96,500 = 1119 mols (1)

To reduce 1 mol of Al3+, 3 mols of electrons are required

In the cell 1119 / 3 = 373 mols of Al produced (1)

Ar Al = 27.0

Mass Al produced = 373 x 27.0 = 10071g = 10.071kg of Al (1)

(4 marks)

c) 2O2-(l) - 4e- = O2(g)

(1 for correct anion, 1 for correct equation)

(2 marks)


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Group II and Group IV (Questions)


a) Ba(s) + 2H2O(aq) = Ba(OH)2(aq) +

H2(g)

(1 mark)

b) 2Ba(NO3)2(s) = 2BaO(s) + 4NO2(g)

+ O2(g)

(1 mark)

c) There are two factors that affect the solubility in water; lattice

energy and enthalpy of hydration.

As both salts are sulphates, there is little difference in lattice enthalpy

(1)

The enthalpy of hydration is much higher for barium as it has a higher ionic

radius (lower charge density) which makes it difficult for water to hydrate.

(1) (less of an attraction between cation and polar water molecules)

(2 marks)



a) Covalent

(1 mark)

b) SiCl4(l) + 2H2O(l) = SiO2(aq)

+ 4HCl(g)

(1 mark)

c) The silicon has 3d electron orbitals available (1)

Which can accept a lone pair of electrons from the oxygen atom of water. (1)

Carbon does not have any d electron orbitals available.

(2 marks)

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d) The 6s electrons in lead are drawn in to the nucleus of the atom

due to poor shielding by d and f orbital electrons. This means that they behave

like an 'inert pair' of inner electrons rather than an 'active pair' of outer

electrons. (1)

Silicon does not have any d or f electron orbitals filled therefore the 3s electron

orbitals become hybridised with the 3p electrons giving silicon a valency of

4. (1)

(2 marks)


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Group VII (Questions)

1. Astatine (As) is a member of Group VII of the periodic table.

(You may consult a data book if necessary for this question)

a) What is the colour and state of Astatine at room temperature?

(2 marks)

b) What is the oxidation number of As in NaAsO3?

(1 mark)

c) Write an equation to show the reaction between astatine and cold

aqueous sodium hydroxide.

(1 mark)

d) Predict the bond energy of H-As bond.

(1 mark)

e) Chlorine gas is passed through an aqueous solution of potassium astatide,

KAt. Write an equation to show what reaction will take place.

(1 mark)

f) Under certain conditions it is possible for the following reaction

to take place:

3AsO- = 2As- + AsO3-

What is the name given to this type of reaction?

(1 mark)

g) Give a method for testing for the presence of astatide

ions in aqueous solution.

(2 marks)

(Marks available: 9 marks)

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Group VII (Answers)


a) Astatine will be a black(1) solid(1)

(2 marks)

b) Oxidation number = +5

(1 mark)

c) Predict the bond energy of H-As bond.

(1 mark)

d) Bond energies / kJmol-1

H-F = 562

H-Cl = 431

H-Br = 366

H-I = 299

H-As = ?

From pattern between H-Cl and H-I.: Bond energy of H-As = 232 kJmol-1 (1)

(1 mark)

e) Cl2(g) + 2As-(aq) = As2(s)

+ 2Cl-(aq)

(1 mark)

f) Disproportionation

(1 mark)

g) Add aqueous silver nitrate (1)

Presence of astatide ions would be indicated by a dark yellow / orange precipitate

of AgAs (1)

(2 marks)


20

Ionic Equilibria (Questions)

1. Benzoic acid, C6H5COOH is a weak acid. When dissolved in water the following

reaction takes place:

C6H5COOH(aq) + H2O(aq)λC6H5COO-(aq) + H3O

+(aq)

a) Explain what is meant by the term 'weak acid'.

(1 mark)

b) A solution is made of benzoic acid at 25°C. The acid dissociation

constant for benzoic acid is 6.3 x 10-5 moldm-3. Write

down an expression for Ka.

(1 mark)

c) Calculate the pH of a 0.005M solution of benzoic acid.

(1 mark)


2. Calculate the pH of the following solutions.

a) 0.1M HCl(aq).

(2 marks)

b) 0.1M Be(OH)2(aq)

(2 marks)


3. Precipitation reactions are an important method of identifying the inorganic

ions in a solution.

For example silver nitrate is used to identify halides in solution.

a) Write down the expression for the solubility product, Ksp, for silver

chloride, AgCl.

(1 mark)

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b) Ksp for AgCl is 1.8 x 10-10 mol2dm-6. Calculate

the concentration of Ag+(aq) in a saturated solution of

AgCl.

(1 mark)

c) What effect will the addition of HCl have on the solubility of AgCl?

(2 marks)


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Ionic Equilibria (Answers)


a) A weak acid is one that does not completely dissociate when dissolved

in aqueous solution.

(1 mark)

b) Ka = [C6H5COO-][ H3O+]

[C6H5COOH]

(1 mark)

c) From equation we know that [C6H5COO-]

= [ H3O+]. We can replace these concentrations in the

expression of Ka with x2.

Take [C6H5COOH] = 0.005M (as degree of dissociation

is small there will be a negligible effect on this concentration).

Ka = x2 / 0.005 = 6.3 x 10-5 (1)

x2 = 0.005 x (6.3 x 10-5) = 3.2 x 10-7

x = 5.7 x10-4 (1)

[H+] = 5.7 x 10-4 moldm-3

pH = -lg[H+]

pH of solution = -lg(5.7 x 10-4) = 3.2

(1 mark)



a) Acid is fully dissociated - [H+] = 0.1 moldm-3

pH = -lg[H+]

pH = -lg[0.1] = 1

(2 marks)

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b) Base is fully dissociated - [OH-] = 2 x 0.1 = 0.2 moldm-3+

pOH = -lg[OH-] = 0.7

pH + pOH = 14

pH = 14 -0.7 = 13.3

(2 marks)



a) Ksp = [Ag+(aq) ][Cl-(aq)]

(1 mark)

b) From equation [Ag+(aq) ] = [Cl-(aq)]

Ksp = [Ag+(aq) ]2 = 1.8 x 10-10

[Ag+(aq) ] = 1.3 x 10-5 moldm-3

(1 mark)

c) Addition of HCl leads to and increase in [Cl-(aq)]

This means that [Ag+(aq) ][Cl-(aq)]

is greater than Ksp

And AgCl will precipitate from the solution i.e. solubility has decreased.

(2 marks)


24

Reaction Kinetics (Questions)

1. Nitrogen (II) oxide reacts with oxygen in the following way:

2NO(g) + O2(g) = 2NO2(g)

Experiments were carried out to determine the orders of reaction with respect to NO and O2. The results of these

experiments are shown below.

Experiment [NO(g)] / moldm-3 [O2(g)] / moldm-3 Rate / moldm-3 s-1

1 1.5 x 10-5 0.5 x 10-5 2.1 x10-7

2 4.5 x 10-5 0.5 x 10-5 1.9 x10-6

3 1.5 x 10-5 2.0 x 10-5 8.4 x10-7

a) Using the data, deduce the orders of reaction with respect to NO and O2.

b) Write down the rate equation for this reaction and use it to calculate k.

c) Explain the meaning of the term 'rate determining step'?

2. Carbon dating using 14C is a highly important tool in archaeology. Using this technique it is possible

to determine the length of time since an organism has died. The half-life of 14C is 5600 years.

a) Define the term 'half-life'.

b) A sample of wood taken from an archaeological dig is analysed and found to have a radioactive count of 4

counts per minute per gram of carbon. A new sample of the same type of wood has a count of 16 c.p.m.g-1.

What is the approximate age of the archaeological sample?

c) Which order of kinetics does radioactive decay follow?


25

Reaction Kinetics (Answers)


a) For NO - examine experiments 1 and 2. A 3-fold increase in the concentration of NO leads to a 9-fold increase

in rate. (1)

This means that the reaction must be second order with respect to NO. (1)

For O2 - examine experiments 1 and 3. A 4-fold increase in O2 concentration leads to a 4-fold increase in rate.

(1)

This means that the reaction must be first order with respect to O2. (1)

b) Rate = k[NO]2[O2] (1)

2.1 x10-7 = k (1.5 x10-5)2 (0.5 x10-5)

k = (2.1 x10-7) / (1.5 x10-5)2 (0.5 x10-5)

k = 1.9 x 108 mol-2dm6s-2

c) The rate-determining step is the slowest section of the mechanism of the reaction when compared with the

other steps.


a) The half-life of a substance is the time taken for its concentration to decrease by 50%.

b) After 5600 years the count form the piece of new wood would drop to 8 c.p.m, after a further 5600 years it

will drop to 4 c.p.m.

The old wood is therefore approximately two half-lives old = 11200 years.

c) First order


26

States of Matter (Questions)

1. Hydrogen generated from the reaction between magnesium and nitric acid was

collected. At a temperature of 30°C and pressure

of 1.2 x 105 Nm-2, 340cm3 of gas

was collected.

What would the volume of this gas be at s.t.p.?


2. The ideal gas equation can be written as 'PV = nRT'

a) Use this equation to calculate the volume of 0.5 moles of an ideal

gas at 350K and 1x 105Nm-2 pressure.

(2 marks)

b) A compound was vaporised and it was found that 0.15g of the gas occupied

120cm3 at 323K and 100 kPa.

Calculate the molecular mass of the compound.

(2 marks)


3. Explain in terms of structure and bonding:

a) Why graphite conducts electricity but diamond does not.

(3 marks)

b) Why water is a liquid at room temperature but hydrogen sulphide is

a gas.

(2 marks)


27

States of Matter (Answers)


Reaction conditions: P1 = 1.2 x 105 Nm-2 ; T1 = 303K ; V1 = 340cm3

S.t.p.

P2 = 1.01 x105 Nm-2 ; T2 = 273K (1)

P1V1 = P2V2 (1.2

x 105)(340) = (1.01 x105)V2

T1 T2 303

273

V2 = 364.0 cm3



a) Rearranging the equation: V = nRT/P

V = 0.5 (8.314)(350) / (1 x 105)

V = 0.015 m3 (numerical = 1, units = 1)

(2 marks)

b) Using PV = nRT and n = m / Mr

n = PV / RT = (1 x 105)( 0.00012) / (8.314)(323)

n = 0.004

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Mr = 0.15 / 0.004 = 37.5

(2 marks)



a) The carbon atoms in graphite are covalently bonded to three other

carbon atoms and have delocalised electrons in between the sheets of atoms.

(1 mark)

These delocalised electrons are able to move when a p.d. is applied

(1 mark)

The carbons in diamond are covalently bonded to four other carbon

atoms so there are no electrons able to move freely.

(1 mark)

(max 3 marks)

b) Water (H2O) has a higher boiling point due to there being

a lot of hydrogen bonding between the molecules.

Hydrogen sulphide (H2S) has a much lower boiling point because there

is a much lower degree of hydrogen bonding.

(2 marks)


Documents

A-Level Chemistry Question and Answers 2013 (A-level).pdfThe equation for this reaction is shown below: 2HgO(s) = 2Hg(l) + O 2(g) a) What volume of O 2(g) would be obtained if 1.08g