chemistry mark scheme, Schemes and Mind Maps of Chemistry

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Mark schemes

(a) (Enthalpy change = lattice dissociation energy + hydration energies of ions) M1 Enthalpy change = + 2237 − ( 2 x 364) − 1650 M2 = − 141 kJ mol− 2

(b) Temperature goes up/increases Allow answer consequential on (a) 1 (c) M1 fluoride ions/F−^ (ions) are smaller OR M1 F−^ has a higher charge density Do not accept fluorine atoms/ions are smaller M2 stronger attraction (of fluoride ion) to δ+ on H/ electron deficient H (in water) M2 do not accept ionic bonds Do not accept more energy to break bonds Do not accept stronger attraction to H+ Ignore electronegativity and shielding 2 (d) 2 (e) −795 = 193 + 590 + 2nd^ IE + (121 x 2 ) + (−364 x 2 ) − 2237 = (+) 1145 (kJ mol-1) M1: Allow −795 = −1940 + 2nd^ IE 2

Stage 1 comparing values from perfect ionic model 1a Value for CaCl 2 is larger OR Values for KCl and AgCl are similar OR Values for CaCl 2 > AgCl > KCl 1b Ca2+^ has a larger charge/ is a smaller ion OR Ag+^ and K+^ have smaller charge or larger ions 1c CaCl 2 has stronger ionic bonds or stronger attraction between + and - ions (Ca2+and Cl-^ ) OR AgCl and KCl have weaker ionic bonds or weaker attraction between + and - ions (Ag+/ K+ and Cl-) Stage 2 similarities in the perfect ionic model and Born-Haber cycle values 2a CaCl 2 has similar values (between the perfect ionic model and Born-Haber cycle) 2b KCl has similar values (between the perfect ionic model and Born-Haber cycle) 2c CaCl 2 and KCl have (almost) perfect ionic bonding or + ions are point charges/(perfectly) spherical Stage 3 difference in the perfect ionic model and Born-Haber cycle values 3a AgCl has larger difference in values (between the perfect ionic model and Born-Haber cycle) 3b AgCl contains (some) covalent character 3c Ag+^ more polarising/distorts electron cloud more [16] (a) toxic/poisonous/too much chlorine causes death

(b) Cl 2 + H 2 O → HCl + HClO allow Cl 2 + H 2 O → 2 H+^ + Cl−^ + ClO− 1 chlorine/Cl/Cl 2 gains electron(s) (to form Cl−) and loses electron(s) (to form ClO−) 1 ignore chlorine is oxidised and reduced ignore disproportionation ignore oxidation numbers unless incorrect

(c) brown solution or black solid (forms) do not accept purple 1 Cl 2 + 2I−^ → 2Cl−^ + I 2 allow multiples ignore state symbols 1 (d) H 2 SO 4 + 2H+^ + 2I–^ → SO 2 + 2H 2 O + I 2 allow SO 4 2–^ + 4H+^ + 2I–^ → SO 2 + 2H 2 O + I 2 1 H 2 SO 4 + 8H+^ + 8I–^ → H 2 S + 4H 2 O + 4I 2 allow SO 4 2–^ + 10H+^ + 8I–^ → H 2 S + 4H 2 O + 4I 2 1 oxidising agent 1 equations can be in either order allow alternative correct balanced equations starting from NaI to form SO 2 and H 2 S eg 2 H 2 SO 4 + 2 NaI → Na 2 SO 4 + SO 2 + 2 H 2 O + I 2 3 H 2 SO 4 + 2 NaI → 2 NaHSO 4 + SO 2 + 2 H 2 O + I 2 5 H 2 SO 4 + 8 NaI → 4 Na 2 SO 4 + H 2 S + 4 H 2 O + 4I 2 9 H 2 SO 4 + 8 NaI → 8 NaHSO 4 + H 2 S + 4 H 2 O + 4I 2 (e) NaF or sodium fluoride 1 CO 2 or carbon dioxide 1 CO 3 2−^ + 2H+^ → CO 2 + H 2 O allow multiples 1

(c) M1 (Ions hit a detector/electron multiplier and) each ion gains an electron (generating a current) M2 current is proportional to abundance 2 (d) M1 Abundance 87 Sr = 2 × 18 ÷ 3 = 12(%) M2 A r = M3 = 87. Answer to 1 decimal place 3 (e) the protein (ion) does not break up/fragment 1 [12] (a) Thymol blue

(b) Square brackets essential 1 (c) M1 K a = 10-p K a^ = 1.35 × 10- M2 OR [H+]^2 = K a [CH 3 CH 2 COOH] M2 Square brackets or with numbers M3 [H+] = √(1.35 × 10-5^ x 0.1) = 1.16 x 10-3^ mol dm- M4 pH = -log 10 (1.16 x 10-3) = 2. M4 = -log 10 M Answer to 2 decimal places Allow 2. 4

(d) M1 Initial amount of butanoic acid = 25 × 0.1 ×10-3^ = 2.5 ×10-3^ mol M2 Initial amount of NaOH = 20 × 0.1 × 10-3^ = 2.0 × 10-3^ mol M3 Final amount of acid = 2.5 × 10-3^ - 2.0 × 10-3^ = 5.0 × 10-4^ mol M3 = M1-M M4 [H+] = Or [H+] = M4 allow volumes cancelled out M4 Allow [H+] = 3.775 × 10-6^ mol dm- Alternative method for M pH = pKa + log = 4.82 + log M5 pH = 5. M5 = dependent on a correct expression for [H+] in M 5 (e) M1 This is a weak acid and weak base/alkali titration M2 pH change is too gradual/not sharp (at the equivalence point so colour change of indicator is difficult to judge) M2 Allow no vertical/steep section on pH curve 2 [13] (a) The enthalpy change when one mole of a substance is formed from its (constituent) elements. Allow heat energy change All substances in their standard states (at 100 kPa and a stated temperature). 2

(b) (Δ H = ΣΔf H (products) - ΣΔf H (reactants)) M1: +56.2 = (-123+90.4) - 2x 2x = -88.8 kJ mol- M2: x = -44(.4) kJmol- M2 = their M1/ 2

(c) universal indicator 1 SO 2 (aq) orange-red 1 SO 3 (aq) red 1 allow correct comparison of acidic colours (red, orange, yellow) or pH meter SO 2 (aq) pH 2- SO 3 (aq) pH 0- allow correct comparison of acidic pH ignoring values or any named metal carbonate ( or formula) or Mg or Ca or Zn SO 2 (aq) slower effervescence SO 3 (aq) faster effervescence if reagent is incomplete lose M1 and mark on allow observation allow correct comparison allow named oxidising agent eg (acidified) KMnO 4 or (acidified) K 2 Cr 2 O 7 SO 2 (aq) correct colour acidified change SO 3 (aq) no visible change or NVC allow (acidified) barium chloride solution or allow (acidified) barium chloride solution SO 2 (aq) no visible change or NVC SO 3 (aq) white precipitate (d) 31 P 4 + Allow P 4 +^ = 1 mark Allow 31 P = 1 mark 2 (e) P 4 O 10 + 12 NaOH → 4 Na 3 PO 4 + 6 H 2 O allow formation of acid salts P 4 O 10 + 4 NaOH + 2 H 2 O → 4 NaH 2 PO 4 P 4 O 10 + 8 NaOH → 4 Na 2 HPO 4 + 2 H 2 O 1

(f) must show all bonds 1 (g) This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question. Level 3 5-6 marks All stages are covered and the description of each stage is generally correct and virtually complete. Answer is communicated coherently and shows a logical progression from stage 1 to stage 2 and stage 3. Level 2 3-4 marks All stages are covered but the description of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows progression from stage 1 to stage 2 and/or stage 3. Level 1 1-2 marks Two stages are covered but stage(s) may be incomplete or may contain inaccuracies OR only one stage is covered but is generally correct and virtually complete. Answer includes isolated statements and these are presented in a logical order. 0 marks Insufficient correct chemistry to gain a mark. indicative chemistry content contradictions negate statements Stage 1 structure 1a NaCl ionic lattice or giant ionic 1b Cl 2 and HCl molecular (covalent) or Cl 2 and HCl (simple) molecules Stage 2 forces responsible for melting point 2a NaCl attractions between + and – ions 2b Cl 2 vdw forces 2c HCl dipole dipole forces

(c) Correct answer (286) with working scores 3 M1 Δ H (or −3388) = ΣB(reactants) − ΣB(products) or Δ H (or −3388) = ΣBonds broken - ΣBonds formed or Δ H (or −3388) = 4 B(C−C) + 11 B(C−H) + B(C−O) + B(O−H) + 7½ B(O=O) − 10 B(C=O) − 12 B(O−H) If no other marks scored, allow 1 mark for 13606 for bonds formed or 9075 for bonds broken (apart from C−C) M2 −3388 = 4 B(C−C) + 11(412) + 360 + 463 + 7½(496) − 10(805) − 12(463) or −3388 = 4 B(C−C) + 9075 − 13606 or −3388 = 4 B(C−C) − 4531 or 4 B(C−C) = 1143 M2 also scores M M3 B(C−C) = = (+)286 (kJ mol−1) (allow ECF from M2 to M3 ) 229 scores 2 marks (for assuming there are 5 not 4 C−C bonds) −1979.75 scores 2 marks (rearrangement error) M3 ignore units; allow 285.75, 285.8, 286 (not 290) 3 (d) Correct answer scores 2 M1 amount of butan-1-ol in 1 dm^3 = (= 10.95 mol) M2 energy from 1 dm^3 = M1 × 2676 = 29300 (kJ dm-3) Allow correct answers to at least 2sf (29291.351..) Ignore sign of final answer Overall calculation is: 29.3 scores 1 mark Allow ECF from M1 to M2 if M1 is an attempt using 74(.0) Alternative M1 energy per gram of fuel = (= 36.1 kJ g−1) M2 energy per dm^3 = M1 × 810 = 29300 (kJ dm−3) 2 [10]

M1: SF 6 is octahedral (either in words or as a structure) M2: SF 6 bond angle is 90° M3: SF 6 all the bond pairs repel equally M4: SF 3 +^ is (trigonal) pyramidal (either in words or as a structure) (Allow tetrahedral) M5: SF 3 +^ bond angle is 103-107° M6: SF 3 +^ lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion (so bond angle is reduced) [6]

A

The boiling point of HX increases. [1]

B

A polyamide [1]

A

The concentration of hydroxide ions in water at 18 °C is 8.00 × 10-^8 mol dm- [1]

D

SiCl 4 [1]

A

[1]

C

Cyclohexane [1]

C

[1]