A-LEVEL PAPER 1 PP6 MS

1.      (a)     (i)      ΔH atomisation/sublimation of magnesium
1
(ii)     Bond/dissociation enthalpy of Cl-Cl
OR 2 × H atomisation of chlorine
1
(iii)     Second ionisation enthalpy of magnesium
1
(iv)    2 × electron affinity of chlorine
1
(v)     Lattice formation enthalpy of MgCl2
1
(b)     Equation     2MgCl(s)  →  MgCl2(s) + Mg(s)
State symbols not required but penalise if incorrect
1
          Calculation    ΔH reaction = ΣΔHf products – ΣΔHf reactants
1
= – 653 – (2 × –133)
1
= – 427 (kJmol–1)
Allow +427 to score (1) mark
Other answers; award (1) for a correct D H reaction expression
1
(c)     ΔH soln MgCl2 = – ΔH Lat.form. + ΔH hyd.Mg2+ + 2ΔH hyd.Cl–
1
          or cycle
= 2502 – 1920 – (2 × 364)
1
= – 146 (kJmol–1)
Allow + 146 to score (1) mark
Other answers; award (1) for a correct ΔH soln MgCl2 expression/cycle
1
[12]
 
2.    (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
1
= (+)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
1
(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
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]

3.    (a)     (i)      (At 0 K) particles are stationary / not moving / not vibrating
Allow have zero energy.
Ignore atoms / ions.
1
No disorder / perfect order / maximum order
Mark independently.
1
(ii)     As T increases, particles start to move / vibrate
Ignore atoms / ions.
Allow have more energy.
If change in state, CE = 0
1
Disorder / randomness increases / order decreases
1
(iii)    Mark on temperature axis vertically below second ‘step’
Must be marked as a line, an ‘x’ , Tb or ‘boiling point’ on the temperature axis.
1
(iv)    L2 corresponds to boiling / evaporating / condensing / l → g / g → l
And L1 corresponds to melting / freezing / s → l / l → s
There must be a clear link between L1, L2 and the change in state.
1
Bigger change in disorder for L2 / boiling compared with L1 / melting
M2 answer must be in terms of changes in state and not absolute states eg must refer to change from liquid to gas not just gas.
Ignore reference to atoms even if incorrect.
1
(b)    (i)      ΔG = ΔH − TΔS
1
ΔH = c and (−)ΔS = m / ΔH and ΔS are constants (approx)
Allow ΔH is the intercept, and (−)ΔS is the slope / gradient.
Can only score M2 if M1 is correct.
1
(ii)     Because the entropy change / ΔS is positive / TΔS gets bigger
Allow -TΔS gets more negative
1
(iii)    Not feasible / unfeasible / not spontaneous
1
(c)    (i)      + 44.5 J K−1 mol−1
Allow answer without units but if units given they must be correct (including mol−1)
1
      (ii)      At 5440 ΔH = TΔS
= 5440 × 44.5 = 242 080
1
(OR using given value = 5440 × 98 = 533 120)
Mark is for answer to (c)(i) × 5440
1
ΔH = 242 kJ mol−1
(OR using given value ΔH = 533 kJ mol−1)
Mark is for correct answer to M2 with correct units (J mol−1 or kJ mol−1) linked to answer.
If answer consequentially correct based on (c)(i) except for incorrect sign (eg −242), max 1 / 3 provided units are correct.
1
[15]

4.     (a)     (i)      - log[H+]
penalise missing [ ] here and not elsewhere
1
(ii)     [H+][OH–]
Allow ( ) brackets, but must have charges
1
(iii)     Mark independently from a(ii)
[H+] = 10-13.72 = 1.905 × 10–14
If wrong no further mark
1
Kw = 1.905 × 10–14 × 0.154 = = (2.93 – 2.94) × 10–15
1
(b)     (i)      Ka = 
Must have charges and all brackets, allow ( )
Acid/salt shown must be CH3COOH not HA
and correct formulae needed
1
(ii)     In pH values penalise fewer than 3 sig figs each time
but allow more than 2 dp
For values above 10, allow 3sfs - do not insist on 2 dp
Ka = 
Allow HA
1
([H+]2 = 1.75 × 10–5 × 0.154 = 2.695 × 10–6 = 2.70 × 10–6)
If √ shown but not done gets pH = 5.57 (scores 2)
[H+] = 1.64 × 10–3
Allow mark for pH conseq to their [H+] here only
1
pH = 2.78 or 2.79
1

(c)     (i)     In pH values penalise fewer than 3 sig figs each time but allow more than 2 dp. For values above 10, allow 3sfs - do not insist on 2 dp
M1 Initially
mol OH– = (10 × 10–3) × 0.154 and
mol HA = (20 × 10–3) × 0.154
or mol OH– = 1.54 × 10–3 and mol HA = 3.08 × 10–3
1
M2 [H+] = Ka 
or with numbers
Allow Henderson Hasselbach
pH = pKa + log 
M3 mol ethanoic acid left = (mol ethanoate ions) = 1.54 × 10–3
Ka = [H+] or pH = pKa scores M1, M2 and M3
1 If either mol acid in mixture or mol salt wrong
- max 2 for M1 and M2
Any mention of [H+]2 - max 2 for M1 and M3
1
M4 pH (= - log 1.75 × 10–5) = 4.76 or 4.757
Not 4.75
1
If no subtraction (so mol ethanoic acid in buffer = original mol)
pH = 4.46 scores 2 for M1 and M2
If [H+]2 used, pH = 3.02 scores 2 for M1 and M3
(ii)     In pH values penalise fewer than 3 sig figs each time but
allow more than 2 dp
For values above 10, allow 3sfs - do not insist on 2 dp
M1 XS mol KOH (= (20 × 10–3) × 0.154) = 3.08 × 10–3
If no subtraction: max 1 for correct use of volume
No subtraction and no use of volume scores zero
If wrong subtraction or wrong moles
Can only score M2 and M3 for process
1
M2 [OH-] = 3.08 × 10–3 ×  = 0.0513(3)
Mark for dividing their answer to M1 by correct volume (method mark)
If no volume or wrong volume or multiplied by volume, max 2 for M1 and M3 process
1
M3 [H+] =  (= 1.948 × 10–13 to 1.95 × 10–13)
or pOH = 1.29
Mark for Kw divided by their answer to M2
If pOH route, give one mark for 14 – pOH
1
M4 pH = 12.7(1) 
Allow 3sf but not 12.70
1
If no subtraction and no use of volume (pH = 11.79 scores zero)
If no subtraction, max 1 for correct use of volume, (60cm3)
(pH = 13.01 scores 1)
If volume not used, pH = 11.49 (gets 2); If multiplied by vol, pH = 10.27 (gets 2)
[16]
 5.   (a)    Diagram of an Fe3+ / Fe2+ electrode that includes the following parts labelled:
Solution containing Fe2+ and Fe3+ ions
1
Platinum electrode connected to one terminal of a voltmeter
Must be in the solution of iron ions (one type will suffice)
1
Salt bridge
Do not allow incorrect material for salt bridge and salt bridge must be in the solution (ie it must be shown crossing a meniscus)
1
298 K and 100 kPa / 1 bar
1
all solutions unit / 1 mol dm−3 concentration
Allow zero current / high resistance voltmeter as alternative to M4 or M5
Ignore hydrogen electrode even if incorrect
1
(b)     Cu2+ + Fe → Cu + Fe2+
Ignore state symbols
1
Fe|Fe2+||Cu2+|Cu   correct order
Allow Cu|Cu2+||Fe2+|Fe
1
Phase boundaries and salt bridge correct, no Pt
Allow single / double dashed line for salt bridge
Penalise phase boundary at either electrode end
Can only score M3 if M2 correct
1
Copper electrode
Allow any reference to copper
1
(c)     Eϴ Au+( / Au) > Eϴ O2 ( / H2O)
Allow E cell / e.m.f. = 0.45 V
Allow 1.68 > 1.23
1
So Au+ ions will oxidise water / water reduces Au+
QoL
1
2Au+ + H2O → 2Au + O2 + 2H+
Allow multiples
1
(d)     Eϴ Ag+( / Ag) > Eϴ Fe2+( / Fe)
Allow E cell / e.m.f. = 1.24
Allow 0.80 > −0.44
1
And Eϴ Ag+( / Ag) > Eϴ Fe3+( / Fe2+)
Allow E cell / e.m.f. = 0.03
Allow 0.80 > 0.77
1
So silver ions will oxidise iron (to iron(II) ions) and then oxidise Fe(II) ions (further to Fe(III) ions producing silver metal)
Allow Ag+ ions will oxidise iron to iron(III)
1
[15]

6.      (a)     Equation:
e.g. [Cu(H2O)6]2+ + 4Cl– → [CuCl4]2– + 6H2O
Species
1
Balance
1
Colours:
e.g [Cu(H2O)6]2+ blue
1
e.g. [CuCl4]2– yellow/green
1
(b)     (i)      ΔE:        The energy absorbed
1
         h:           Planck’s constant
1
(ii)     Factor 1   Change of ligand
1
Factor 2   Change in oxidation state
1
Factor 3   Change in co-ordination number
1
[9]

7.(a)    Cobalt has variable oxidation states
Allow exists as Co(II) and Co(III)
1
(It can act as an intermediate that) lowers the activation energy
Allow (alternative route with) lower Ea
1
CH3CHO + 2Co3+ + H2O → CH3COOH + 2Co2+ + 2H+
Allow multiples; allow molecular formulae
Allow equations with H3O+
1
O2 + 2Co2+ + 2H+ → 2Co3+ + H2O
1
(b)    (i)      [Co(H2O)6]2+ + 3H2NCH2CH2NH2 → [Co(H2NCH2CH 2NH2)3]2+ + 6H2O
Do not allow en in equation, allow C2H8N2
1
The number of particles increases / changes from 4 to 7
Can score M2 and M3 even if equation incorrect or missing provided number of particles increases
1
So the entropy change is positive / disorder increases / entropy increases
1
(ii)     Minimum for M1 is 3 bidentate ligands bonded to Co
Ignore all charges for M1 and M3 but penalise charges on any ligand in M2
1
Ligands need not have any atoms shown but diagram must show 6 bonds from ligands to Co, 2 from each ligand
Minimum for M2 is one ligand identified as H2N-----NH2
Allow linkage as −C−C− or just a line.
1
Minimum for M3 is one bidentate ligand showing two arrows from separate nitrogens to cobalt
1

(c)     Moles of cobalt = (50 × 0.203) / 1000 = 0.01015 mol
Allow 0.0101 to 0.0102
1
Moles of AgCl = 4.22/143.4 = 0.0294
Allow 0.029
If not AgCl (eg AgCl2 or AgNO3), lose this mark and can only score M1, M4 and M5
1
Ratio = Cl− to Co = 2.9 : 1
Do not allow 3 : 1 if this is the only answer but if 2.9:1 seen somewhere in answer credit this as M3
1
[Co(NH3)6]Cl3 (square brackets not essential)
1
Difference due to incomplete oxidation in the preparation
Allow incomplete reaction.
Allow formation [Co(NH3)5Cl]Cl2 etc.
Some chloride ions act as ligands / replace NH3 in complex.
Do not allow 'impure sample' or reference to practical deficiencies
1
[15]

8.	MnO4- will oxidise the chloride ion / reaction of MnO4- and Cl- feasible
Accept converse argument with Cr2O72-
Accept calculations of overall E° values.
1
Larger volume needed
1
[2]

9.