MEMORANDUM

QUESTION 1
1.1 D (2)
1.2 C (2)
1.3 A (2)
1.4 A (2)
1.5 C (2)
1.6 B (2)
1.7 B (2)
1.8 D (2)
1.9 D (2)
1.10 B (2)          [20]

QUESTION 2
2.1

• A series of organic compounds that can be described by the same general formula
  OR
  A series of organic compounds in which members differ by the number of – CH₂ units  (2)

2.2 A compound that contains carbon and hydrogen (atoms) only  (2)
2.3.1 C_nH2_n+2 (1)
2.3.2 4-ethyl -2,2-dimethyl hexane
Marking criteria

• Hexane
• Methyl/Metiel and
• whole name correct 3/3
  Deduct 1 mark for any error, hyphens omitted, incorrect sequence etc  (3)

2.4.1   (2)

Marking criteria

• Hexane
• Methyl and ethyl 
• whole name correct  3/3
  Deduct 1 mark for any error, hyphens omitted, incorrect sequence etc

2.4.2 Butan-2-one OR 2-butanone (2)

Marking criteria

• Functional group and correct position g pent-2-one (1/2)
• Whole name correct (2/2)

2.4.3

• The functional group can only be in position 1
  OR
  Side chain / branch / methyl group can only be in position 2  (1)

2.5.1 Polymerisation (1)
2.5.2 Contains single bonds only (1 or 0) (1)
2.5.3 Use in plastics (toy cars) (1)
[16]

QUESTION 3
3.1 The temperature at which the vapour pressure of a liquid equals the atmospheric/external pressure.  (2)
3.2 To ensure a fair test /To ensure there is one independent variable  (1)
3.3

• Hexane has London forces (only)
  Pentanal has dipole-dipole forces (and London forces)
• Dipole-dipole forces are stronger ( than the london forces in hexane)
  OR London forces are weaker (than the Dipole-dipole forces in pentanal)
• More energy is required to overcome the intermolecular /Dipole-dipole forces in Pentanal  (4)

3.4 Higher than  (1)
3.5

• Chain isomer of pentan-2-ol (2-methylbutan-2-ol) has a shorter chain length/ smaller surface area than pentan-2-ol
• London forces in the isomer of pentan-2-ol (2-methylbutan-2-ol) will be weaker (than that of pentan-2-ol).
  OR
• Pentan-2-ol has a larger chain length/ surface area than its chain isomer
• London forces in butan-2-ol is stronger (than that of the isomer of pentan-2-ol (2-methylbutan-2-ol)  (3)

3.6 2 C₆H₁₄ + 19 O₂ → 12 CO₂ + 14 H₂O
Marking criteria

• Reactants
• Products 
• Balancing     (3)

[14]

QUESTION 4
4.1.1 Substitution /Hydrolysis  (1)
4.1.2 Butan-2-ol OR/OF 2-butanol (2)
4.2   (6)

Balancing
Marking criteria
For organic reagents

• Whole structure correct(2/2)
• Functional group correct (1/2)

4.3
4.3.1 Hydrogenation (1)
4.3.2 Platinum/Palladium/Nickel (1)
4.3.3

Marking criteria
For organic reagents

• Whole structure correct (2/2)
• Functional group correct (1/2)  (2)

4.4
4.4.1 Esterification/Condensation  (1)
4.4.2 Heat/Add a catalyst/H₂SO₄  (1)
4.4.3 Water/H₂O (1)
4.4.4


Marking criteria

• Whole structure correct (2/2)
• Functional group correct (1/2)
• Name correct (2/2)

Propyl butanoate  (4)
[20]

QUESTION 5
5.1 The change in concentration of reactants or products per unit time.(2)
5.2 Carbon dioxide/CO₂ (1)
5.3 Stopwatch  (1)
5.4

• Rate = -Δ m/Δt
  Rate= –(200 -184,80) / (5 – 0)
  Rate = 3,04 (g∙min^-1)
  Accept:
  Rate = (184,80 –200) / (5 – 0)        Rate = (200 –184,80) /(5 – 0)
          = - 3,04 (g∙min-1)                    Rate = 3,04 (g∙min^-1)   (3)

5.5

• m_CO2 produced = 200- 184,80
  = 15.2 g
  nCO₂ = m/M
  = 15,2 /44
  = 0,35 mol
  nCaCO₃ reacting = 0.35 mol
  = 0.35 x 100
  = 35 g

Marking guidelines

• Mass of CO₂
• Formula n = m/M
• Substitution of 44 into n = m/M
• Use of ratio CaCO₃: CO₂ (1:1)
• Substitution of 100 into  n = m/M
• Final answer
  (Ratio) mCaCO₃ reacting= n.M                (6)

5.6.1 Temperature (1)
5.6.2 B (1)
5.6.3

• Higher temperature increases average kinetic energy of the particles increases.
• More particles have sufficient kinetic energy (to collide effectively) / More particles have Ek greater or equal to Ea
• More effective collisions per unit time  (3)

5.7.1

• What effect will the concentration (of a substance) have on the rate of reaction?
  OR
• What is the relationship between reaction rate and concentration?
  OR
• How does concentration affect reaction rate?

Marking criteria
For organic reagents

• Independent variable and dependent variable correct
• In the form of a question  (2)

5.7.2 LOWER THAN (1)
5.7.3 EQUAL TO(1)
5.8 The same amount of CaCO3 (the limiting reagent) is used in both experiments  (2)
[24]

QUESTION 6
6.1 Reaction in which products can be converted back to reactants   (2)
6.2 FORWARD REACTION(1)
6.3 No

• The rate of forward reaction is equal to the rate of reverse reaction /Reaction reached equilibrium   (3)

6.4.1 Increases  (1)
6.4.2 Remains the same  (1)
6.4.3 Increases (1)
6.5

• The amount of HI remains constant.
• The volume decreases
• The concentration increases according to c = n/V (2)

6.6.1 High yield                   Kc is large            OR             Kc >1 (2)
6.6.2 EXOTHERMIC 

• The value Kc decreases with an increase in temperature.
• As temperature increases, the [prdoducts] decreases 
• Reverse reaction is favoured by an increase in temperature
  OR
• The value Kc increases with a decrease in temperature.
• As temperature decreases the [products] increases
• Forward reaction is favoured by a decrease in temperature  (4)

6.7.1

• Kc = [HI]²/[H₂].[I₂]
  50,3 = [HI]²/(0.46)(0,39)
  [HI] = 3 mol.dm^-3            (4)

6.7.2 POSITIVE MARKING FROM 6.7.1 
OPTION 1 : CALCULATIONS USING NUMBER OF MOLES
Marking Criteria for Mole Option

• Multiplication/ van c _equilibrium by/ met 0,5 dm³ for/ vir I₂ ,H₂ and HI
• Calculations of moles of HI reacting
• Using mole ratio 2 n(HI) = n(H₂) = n (I₂) reacting
• Calculation of (H₂) initial and n (I₂) initial (Δn + n_equilibrium)
• Multiplication /  van n (I₂) by 2 to find  n (HI) theoretical 
• Substitution into Yield = n_produced x 100
• Final answer  (RANGE : 78,95% – 79,37%)

CALCULATION USING MOLES

I₂ is the limiting reagent 
n (HI) theoretical  = 2 (n I₂ initial) = 2 x (0,945) = 1,89 mol
% Yield  = n_produced X 100 = 1,5/1,89 x 100 = 79,37 %
H₂ used as limiting reagent Max 3/7

OPTION 2: CALCULATION USING CONCENTRATIONS
Marking Criteria for concentration Option 

• Calculations of  cHI reacting
• Using mole ratio 2 c(HI) = c(H₂) = c(N₂) reacting 
• Calculation of cH₂ initial  and cI₂ initial (Δc + cequilibrium )
• Multiplication of  cI₂ by 2 to find vind nHI theoretical 
• Substitution into Yield  = n
  HI_produced x 100
• Final answer 
  (RANGE: 78,95% -79,37%)

I₂ is the limiting reagent 
C (HI) theoretical = 2 x 1,89 = 3,78 mol.dm^-3
% Yield  = cproduced x 100 =3/3,78 x 100 =79,37%          (7)

H₂ used as limiting reagent Max 3/7
[28]

QUESTION 7 
7.1.1 An acid is a substance that donates protons /H+ions  (2)
7.1.2 H₂PO₄ (1)
7.1.3

• Reaction I : Reverse reaction it accepts a proton (H+) / acts as a base
• Reaction II: Foward reaction donates a proton (H+)/ act as an acid.  (2)

7.1.4 HPO₄²

• The conjugate base of a weak acid
• lower K_a value is the stronger base   (3)

7.2.1 Reaction of a salt with water (2)
7.2.2 C₂HO₄ (2)
7.2.3 (Excess) OH- ions/hydroxide ions are produced   (2)
7.3.1 A strong base undergoes complete ionisation /disociation  (2)

7.3.3 OPTION 1 

• caVa  = n_a
  cbVb     n_b
     ca(25)     = 1 
  (0,5)(24)        2
  ca = 0,24 mol∙dm^-3
  c = m 
       MV
  0,24 =         7,56        
            (90+18x)(0,25)
  x = 2

OPTION 2 :
Marking guideline 

• Substitution of  0,5 and  24/100 in n = cV
• Use of mole ratio  in Acid: Base 1:2
• Calculating number of moles of acid in original solution
• Use of  90 in m = nM
• Calculation of mass of water of crystallization in original solution
• Calculating ration of nWater/nAcid 
• Final answer 

n NaOH reacting  = cV= 0.5 x 24/1000 = 0.012 mol
n_{oxalic acid reacting} = ½ x 0.012 mol = 0,006 mol
n_{oxalic acid in original solution}  = 250/25 x 0,006
= 0,06 mol
m_{oxalic acid in original solution}  = nM = 0,06 x 90
= 5,4 g
m H₂O of crystallisation in original solution  = 7,56 - 5,4
= 2.16 g
n H₂O crystalllisation  = 2,16 /18 = 0,12 mol
x = 0,12/0,06 = 2                (7)

[28]
TOTAL  150