AS-LEVEL PAPER 2 PP8 MARK SCHEME

1.          (penalty for sig fig error =1mark per question)
          (a)     (i)      moles KNO3 = 1.00/101.1 = 9.89 × 10-3 (mol)
1
(ii)     pV = nRT or n = pV/RT 
1
moles O2 = n =  = (1)  (1)
2
                                  = 4.93 × 10-3 (mol)
1
(mark answer first – check back if wrong)
(transcription error lose M3, mark M4 conseq on error)
(if ‘untraceable’ figures used M3=M4=0)
(if wrong temp conversion – lose M3 – conseq M4)
(if n = RT/pV CE, lose M3 and M4)

(b)     (i)      simplest/lowest ratio of atoms of each / element/s in a compound /
substance / species / entity / molecule
1
(ii)     K                 N              O
                 (1)
                                               (1)
1.17          1.18          2.35
1               1               2               KNO2      (1)
(M3 tied to M2), (M3 can be transferred from equation if ratio correct but EF not given) (if calc inverted, lose M2 and M3), (if used At N 1 / wrong No for Ar then CE, lose M2 and M3) (if % of O missing, award M2 only)
3
(c)     2KNO3 → 2KNO2 + O2 or fractions/multiples 
(accept 2KNO3 → K2N2O4 + O2)
(do NOT accept ‘Y’ in equation) 
1
[10]

2.      (a)     M1    (Free-) radical substitution
Both words needed
1
M2    Cl2 → 2Cl•
1
M3    Cl• + CH4 → •CH3 + HCl
1
M4    Cl2 + •CH3 → CH3Cl + Cl•
1
M5    CH4 + 3Cl2 → CHCl3 + 3HCl
Penalise the absence of a radical dot once only
Ignore termination steps except, if and only if both M3 and M4 do not score, then accept for one mark
Cl• + •CH3 → CH3Cl
1

(b)     M1    UV (light)/ sunlight / light / UV radiation
M2    C–Cl or carbon-chlorine bond breakage
OR
homolysis of C–Cl
OR
equation to show a chlorine-containing organic
compound forming two radicals
For M1 and M2, ignore use of Cl2, but credit UV and C–Cl bond breakage if seen
1
M3    Cl• + O3 → ClO• + O2
1
M4    ClO• + O3 → Cl• + 2O2
Ignore other equations
Penalise the absence of a radical dot once only
Accept radical dot anywhere on either radical.
1
M5    Any one from
•        Combination 2O3 → 3O2
•        Stated that Cl• / chlorine atom is regenerated / not used up
•        Stated that the Cl• / chlorine atom is unaffected
by the process.
1
For M5 accept Cl• on both sides of the equation
M6    Stated that the role of the Cl• / chlorine atom is to find an
alternative route OR lower Ea / activation energy
1
(c)     M1    Halothane contains C–Cl / Cl
OR
Desflurane does not contain C–Cl bonds / Cl
OR
Desflurane contains C–F / F as the only halogen
Mark independently.
For M1, credit the idea that desflurane contains C–F bonds that are difficult to break OR that halothane contains C–Cl bonds which are easy to break.
1
M2    Desflurane / molecules that have fluorine as the only
halogen, cause no damage / do not deplete / do not react
with the ozone (layer)
OR
Halothane / chlorine-containing molecules, damage /
deplete / react with the ozone (layer)
1
[13]
 
3.      (a)     Equation     1/2N2 + 3/2H2 → NH3
1
          ΔHf = [(945 × 0.5) + (426 × 1.5)] – (391 × 3)
1
= –46.5 kJ mol–1
1

Mark
Range	The marking scheme for this part of the question includes an overall assessment for the Quality of Written Communication (QWC). There are no discrete marks for the assessment of QWC but the candidates’ QWC in this answer will be one of the criteria used to assign a level and award the marks for this part of the question
Descriptor
an answer will be expected to meet most of the criteria in the level descriptor	
4-5	–    claims supported by an appropriate range of evidence
–    good use of information or ideas about chemistry, going beyond those given in the question
–    argument well structured with minimal repetition or irrelevant points
–    accurate and clear expression of ideas with only minor errors of grammar, punctuation and spelling 	
2-3	–    claims partially supported by evidence
–    good use of information or ideas about chemistry given in the question but limited beyond this
–    the argument shows some attempt at structure
–    the ideas are expressed with reasonable clarity but with a few errors of grammar, punctuation and spelling 	
0-1	–    valid points but not clearly linked to an argument structure
–    limited use of information or ideas about chemistry
–    unstructured
–    errors in spelling, punctuation and grammar or lack of fluency	

(b)     The higher the temperature the faster the reaction QWC
1
          but, since the reaction is exothermic 
1
the equilibrium yield is lower QWC
1
          The higher the pressure the greater the equilibrium yield QWC
1
because there is a reduction in the number of moles of gas
in the reaction
1
but higher pressure is expensive to produce or plant is more
expensive to build QWC
1
          A better catalyst would lessen the time to reach equilibrium
1
and allow more ammonia to be produced in a given time QWC
1
[11]

4.     (a)      (i)     M1   (could be scored by a correct mathematical expression which must 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”
OR
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”
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
OR
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
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]
 
5.    (a)     = (0.0745) × (0.0813) / (0.0424) × (0.0525)
= 2.72
Allow answer only without working if correct.
Lose this mark if the wrong Kc expression is used.
1
Answer, whether or not correct, given to three significant figs
Do not expect conversion from moles to concentration but allow if shown.
1
(b)     Less acid is present (so less NaOH needed)
1
Equilibrium would shift to right (side with more ester / less acid)
1
[4]

6.      (a)     Alcohol: Reaction = Substitution (/ hydrolysis) (1)
Ignore reference to nucleophilic, but electrophilic give zero
          Alcohol: Role = nucleophile (/ lone pair donor) (1)
          Alkene: reaction = elimination (1)
Ignore ref to nucleophilic or electrophilic
          Alkene: base (/ proton acceptor) (1)
If no indication of order in (a) assume as in question.
If order is wrong can still score 'role' mark.
4
(b)     Alcohol: Role = butan-2-ol (1)
Not 2-hydroxybutane or but-2-ol
          Appropriate structure for CH3CH(OH) CH2CH3 (1)
Brackets not essential
 
SN2 version	SN1 version	
– bond is polar	C–Br bond is polar (1)	
Lone pair of OH–	C–Br bond breaks (1)	
Attacks the Cδ+	forming carbocation / carbonium ion (1)	

M1 can be scored from a diagram, M2 and M3 from written explanation only
5

(c)     

If but-2-ene not given here it may be obtained from cis / trans isomer
H lost from different carbon atoms (1)
H removes from C1 and C3 to give two isomers (1)
Draws clear Cis and trans isomers for but-2-ene
Can score these marks from a diagram

8
[17]

7.	C

8.	B

9.	B

10.	B

11.	D

12.	D

13.	D

14.	B

15.	D