Question

Answer

Indicative Content

Credit

Oxygen spin diagram1

D

⬆️⬇️ | ⬆️ | ⬆️

(i) one electron fills each subshell first

(ii) electrons have an up spin and a down spin in the box diagrams

1 mark

What causes the anomalous properties of water:

• Less dense than liquid as a solid
• Hydrogen bonding
• Ice is in open lattice???no hi t sure
• Weaker intermolecular forces?

B

Hydrogen bonding

1 mark1

• Atom economy of Fe  

C

The atom economy of reaction C is 69% which is the highest.

1 mark

Find the Ea when the gradient of the ln|k| to 1/T kgraph is -16,000

B

+133

-16,000 = -EaRi

Ea = (-16,000 x -8.314)/1000

1 mark

Which is the correct concentration-rate graph for a reaction which is zero-order with respect to reactant C?

k

Blets

D

Yh

Explanation:

The concentration of a zero-order reaction will not have any impact on rate.

1 mark

When is equilibrium reached on the graph?

C

After/as they go horizontal

Concentrations become constant

1 m.ark

Dissolved

 + NaOH → Green precipitate

BaSO4 → White ppt

I’m

D

FeSO4

Explanation:

Barium nitrate reacts to form a white precipitate (BaSO4)

When dissolved and NaOH is added, the complex ion [Fe(H2O)6]2+ reacts to give a green precipitate, Fe(OH)2

1 mark

H2SO4 volume

A

-56

1 mark

Effect of adding catalyst on Ea and proportion of molecules with Ea

 B

Adding a catalyst lowers the Ea of a reaction so that a greater number of particles have energy greater th

an or equal to Ea

1 mark

Which of the following is true about the periodic table?

The periodic table is organised by relative atomic mass

1. Elements in the same group have similar chemical properties
2. Transition metals make good catalysts

C

(2 and 3)

1 is not true because the elements are organised by atomic number.

1 mark    

Incorrect disproportionation

Which of the following is not a disproportionation reaction of Chlorine?

1. Cl2 + H2O → HCl + HClO
2. Was the one for bleach i think

no

D

(1) is a disproportionation reaction as the oxidation state of Cl2 (0) changes to 1+ in HCl and 1- in HClO

1 mark

Which of the following are true about Chlorine? we

1. Gain 1 electron to form 1- ions
2. Lose the 2 3s2 electrons to form 2+ ions
3. Lose the 5 3p5 electrons to form 5+ ions

D (1 only)

1 mark

Which form of Magnesium can be used in antacids?

1. Chloride
2. Hydroxide
3. Sulphate
4. Nitrates

b

Hydroxide.

Mg(OH)2 is used to neutralise stomach acid: H+ + OH- → H2O

1 mark

Which of the following can be added to ethanoic acid to form a buffer solution?

1. CH3COONa
2. CHCOOH
3. HCl
4. NaCl

A

CH3COONa

Buffers are made from a weak acid and its salt (conjugate base pair

1 mark

Why do halide melt/boiling points increase down the group?

C

Due to induced dipole forces (London forces) getting stronger.

1 mark

Question

Answer

Indicative Content

Credit      

Calculate Gibbs Free Energy (ΔG) for the reaction shown.

-117

-98-(298)(0.0625)=-117

Award 3 marks for incorrect sign of ΔH, wrong calculation of ΔG.

4 marks

Whilst the reaction above is feasble, explain why it may stiill not occur naturally.

we

High activation energy

Ignore rate of reaction was too slow to be observed - the question states the reaction does not occur.

1 mark

Complete the born-haber cycle showing stages of the formation of MgO

3 x Missing values

Mn(g) + ½ O2(g)

Mn(g)2+ + O(g)- + e o

MnO(s)

Allow for bottom line:

Mn2+(s) + O2-(s)

I is

3 marks

Why was the T2 value higher than  T

The forward reaction was endothermic

So a higher temperature would increase the yield of the products increasing the value of the guy numerator when calculating Kp

Therefore for Kp to increase, the temperature had to have increased, so

the value of Kp for T2 was at a hi higher temperature

2 marks

Definition of lattice enthalpy

Enthalpy change when 1 mole of an ionic lattice is formed from its constituent ions in their gaseous state 

2 marks

pKa

pKa = 2.85

Indicative content

10-1.95 = [H+] ok

States Ka =

States [H+] = [CH3COO-] (approximation)

Then H+ squared / 0.09 = Ka

Then -log(Ka) = pKa

• 1 mark for calculating H+
• 1 mark for calculating Ka
• 1 mark no for calculating pKa

3 marks

2marks for approximation and dissociation equation

Acid dissociation assumption

[H+] = [A-]

1 mark

Acid dissociation constant

 Ka =

2 marks

Configuration of Fe and Fe2+

Fe:

1s22s22p63s23p63d64s²

Fe2+: 1s22s22p63s23p63d6

Allow 4s2 before 3d6

Explanation for Fe2+:

When the d-orbital is occupied, the 4s orbital is promoted in energy level above 3d, so electrons are first removed from the 4s subshell. 

1  mark

Draw the shape of an s-orbital and a p-orbital

Sphere and Dumbbell Shape on the x,y,z plane

Allow sphere drawn as circle - you can’t draw a sphere on 2D paper

Possibly sx,sy,sz and pz,py,px plane not necess K ary

1 mark

Why is the \catalyst  heterogeneous

The catalyst is in a different physical state/ phase to reactants

IGNORE Catalyst in a different state to products or species oe

This varies from series to series

Mark awarded if reference to liquid nitrogen and gases

1 mark

Type of reaction

Neutralisation

Ignore redox, displacement, precipitation oe

1 mark

Barium + oxygen (equation 2)

2Ba +O2 → 2BaO

Allow Ba +½ O2 → BaO

Allow multiples

Ignore state symbols

1 mark

Barium oxide + water

BaO + H2O → Ba(OH)2

1 mark

SO3(g) ⇌ SO2(g) + ½O2(g)

Calculate Kp for this reaction, which happens at 550K.

Initially, 0.245 moles of SO3(g) are added to the reaction vessel. The reaction is allowed to reach equilibrium and the student finds there are 0.09 moles of SO2(g) in the vessel.

The pressure of the reaction vessel at equilibrium is 2.8 atm.

0.207 at

m

or

0.455atm½

*Total moles of gas at equilibrium: 0.29

5 marks

Kp change when no catalyst used and when pressure is increased

Kp change when the catalyst is removed: no change

Kp change when pressure i  no s increased: no change

2 marks

Melting points of Al Si P4 and S8

*formatted in a table, order of melting points:

• Si
• Al
• S8
• P4

Level of Response

Level 3 (5-6 marks)

Correctly compares melting points with reference to bonding and structure.

Level 2 (3-4 marks)

Level 1 (1-2 marks)

Level 0(0 marks)

Nothing credit-worthy no

Indicative content

Bonding in all species

• Aluminium: electrostatic attraction between Al3+ ions and negative electrons
• Silicone: covalent bonding between atoms to form a giant covalent lattice
• P4 and S8: weak intermolecular forces (London forces) between simple gaseous molecules

Bonding in Aluminium

• Bonding in Aluminium is giant metallic lat
• Aluminium(iii) ions are held together by electrostatic forces of attraction (between the ions and delocalised electrons)

Bonding in Silicon

• Silicon forms a giant covalent no lattice
• The covalent bonds require a high amount of energy to overcome and break

Comparing bonding in P4 and S8

• S8 has more electrons so stronger London forces between molecules, resulting in higher mp than P4

Comparison of bonding required

6 marks

Enthalpy profile diagram  was

l

lol

1. Correctly labelled Ec and Ea
2. Correctly labelled ΔH
3. Correctly labelled reactants and products
4. Correct exothermic graph - the energy of the products is lower than the  energy of the reactants.

3 marks

Mass o f tablet question

Anyone remember the actual question/values/ equation they gave you etc …?

Ratio was 1:5 so had to divide moles by 5 to get moles of KIO3

Think it was 0.056 mol dm3 of smth

84mg

4 marks1

Explain how Oxygen is delivered to body cells via Haemoglobin.

There is a buffer system in the blood which maintains the pH between the healthy range of 7.35 and 7.45.

This is done by the carbonic acid equilibrium:

H2CO3 ⇌ HCO3- + H+

Ka for this equilibrium is 4.27x10-3

A person has a ratio of [HCO3-]:[H2CO3] of 8.5:1

Determine whether this person has healthy blood.

Indicative content  no no

Explanation of Haemoglobin oxygen delivery to body cells (1 mark)

• Oxygen forms coordinate bonds/dative covalent bonds with Fe2+
• Oxygen is a poor ligand so the bond breaks (easily) and oxygen can leave

Reference to ligand substitution (1 mark)

• Oxygen replaces/is replaced by H2O or CO2 ligands in a ligand substitution.

Calculation of person’s pH of blood (2 marks)

or

H

Calculation of healthy  Ik[HCO3-]:[H2CO3] ratio (2 marks)

it was 8.5:1 in

and

States the pH 7.30 is unhealthy or the ratio falls outside the calculated range of healthy ratio (if this no method  was used) oe (1 mark)

5 m

arks

Rate-determining step

Step 1 by

F2 + NO2 ->

 F + FNO2 only

Rate equation

Rate = k[F2][NO2]

2 marks

Acidic/alkaline hydrogen hi

Fuel cells and lithium equation

1.23V

Li + CoCO3

2H2 + O2 → 2H2O

*systems 3

and 6, 2 and 4

1 lithium ion cell

3 marks of comparing

Suggest two different oxidation states of Manganese in Mn3O4, hausmannite.

3+ and 2+ only it

Disallow 6+ and 1+, the 1+ oxidation state does not happen, and confirm - the given molecule in the exam (hausmannite) is 3+ and 2+.

See https://en.wikipedia.org/wiki/Manganese(II,III)_oxide

Do not allow 4+ - this was stated in the question

Explanation

For 2+ and 3+

D

Mn2+ and O2- and 2Mn3+ and 3O2-, these pairs cancel out.

1 mark

Half-life calculation

Level of Response Q

Please refer to the marking instructions on page 5 of the mark scheme for guidance on marking this question.

Level 3 (5-6 marks)

A comprehensive conclusion using quantitative data from the graph to correctly determine initial rate AND half lives / gradient with 1st order conclusion for CV AND determination of k.

There is a well-developed line of reasoning which is clear and logically structured.

Clear working relationships for initial rate, half life / gradient and order and k.

Units are mostly correct throughout.

Level 2 (3-4 marks)

Attempts to describe all three scientific points but explanations may be incomplete.

OR

Explains two scientific points thoroughly with few omissions.

There is a line of reasoning with some structure and supported by some evidence. The scientific points are supported by evidence from the graph.

Level 1 (1-2 marks)

Reaches a simple conclusion using at least one piece of quantitative data from the graph.

Attempts to calculate initial rate OR half life.

There is an attempt at a logical structure with a reasoned conclusion from the evidence.

0 marks - nothing credit-worthy um

Indicative content:

I’m

Calculation of the rate constant, k

• k 
• or using a gradient drawn at 3 minutes.
• Half-life 2.5 minutes (150s) must be shown on a graph by two or more half-lives correctly measured (to show constancy).
• k=0.277 min^-1

Determination of the order with respect to [CV]

• The reaction is first order with respect to [CV].

• Rate equation + rate at 3 min

rate= k[CV]

Units: s-1

( if in minutes units must relate eg min-1)

Determination of the rate of reaction at 3 minutes

• Rate = ‘k value obtained’ *  [CV]  at 3 minutes
• =3.73*10^-8 mol. dm^-3 min^-1
• OR
• Draw tangent at 3.0 minutes
• Rate = (1.12*10^-7.- 0)/3.0
• = 3.73*10^-8 mol dm^-3 min^-1

6 marks

benefits of Cl in water treatment

Kills harmful bacteria/microorganisms (to make water safe to drink)b

Harmful may be required

ALLOW kills germs OR kills microorganisms OR kills pathogens OR sterilises/disinfects OR makes water potable/ safe to drink OR purifies water

1 mark

Ionic equation for chlorine and bromine

Cl2 + 2Br- → 2Cl- + Br2

1 mark correct reactant and product species AND correctly balanced ionic equation

Part 1 of 2 marks

Chlorine ionising bromide but not iodine

• Chlorine is more reactive/a stronger oxidising than Bromine
• Allow reference to electron structure.

Part 2 of 2 marks

Titration curve graph

0.14 if volume of Ba(OH)2 taken to be 12.50 cm-3 

subject to a range of answers

Allowing between 12.4 and 12.8 likely

5 marks

Indicator for titraion

Phenol red only

• Equivalence point of titration
• i’mat ~ ph 8, and phenol red has range 6.9-8.4

1 mark

Why SO3 has ~120° bond angles

The S atom of SO3 has 3 bonding regions of electron density and no lone pairs of electrons

1 mark

Why SO3 and SO2 have polar bonds, but only SO2 is a polar molecule

Indicative content

Explanation of polar bonding

• There is a difference in electronegativity of Oxygen and Sulphur
• Oxygen is more electronegative so it attracts bonding electrons more than sulphur
• So there is an uneven distribution of electrons in the (covalent) bonds.

Explanation of polar SO2

• There is one lone pair in SO2 whereas there is not in SO3.
• So SO2 is asymmetrical/non-linear/bent
• There is a dipole so SO2 is polar.

And in

Debate: must be reminded that sulphur and oxygen  undoubtedly have differences in electronegativity in so3 and so2 so explanation of bonding needs work!

2 marks