Correct Answer Reject Mark. Correct Answer Reject Mark [ (2 x 110.5) + (2 x 285.8)] (1)

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1 (a)(i) Correct Answer Reject Mark CH CHCHCH CO H O HOOC(CH ) 4COOH Hʅ f / kj mol Sʅ / J mol K - 4 values correct () marks 3 / values correct mark 0 / values correct (0) marks Question (a)(ii) Correct Answer Reject Mark [ ( x 0.5) + ( x 85.8)] = 3.6 (kj mol ) Allow TE from (a) NOTE If both 0.5 and 85.8 are not doubled, answer CQ = (kj mol ) for mark Ignore SF except SF

2 (a)(iii) Correct Answer 50(.0) [ ( x 97.6) + ( x 69.9)] Rejec t Mark = (J mol K ) Allow TE from (a) NOTE If both 97.6 and 69.9 are not doubled, answer CQ = 96. (J mol K ) for mark Ignore SF except SF Question Correct Answer Reject Mark S surr at 98 K = H/T 3 (a)(iv) = ( 3.6 x 000) / 98 = (+) (J mol K ) Allow TE from (a)(ii) e.g. S surr = (+)375.5(0) (J mol K ) scores () if no doubling in (a)(ii) S tot = S surr + S sys / S tot = / S tot = (+) 48.9 (J mol K ) Allow TE from (a)(ii) and (a)(iii) Allow correct answers given in kj mol K e.g kj mol K Ignore SF except SF If candidates forget to convert H into J mol, then S tot = 56.7 (J mol K ) would score () if correct working is included

3 (a)(v) Correct Answer Reject Mark (Decrease in T) st mark: consideration of S system S system is not (significantly) changed / is unchanged / remains (approximately) constant 3 nd mark: consideration of S surr S surr or H/T is more positive / larger / greater COMMENT ALLOW less negative 3rd mark: consideration of S total (So) increases S tot / makes S tot more positive / makes S tot greater NOTE IF no reference / an incorrect reference made to S system, then only the nd and 3rd marks can be awarded NOTE If candidate states that S surr becomes less +ve, no M But if then states CQ that S tot decreases award M3 as a TE

4 (b) Correct Answer Reject Mark DIMINISHING: (Peak between) (cm ) (due to C=C) (Peak between) (cm ) (due to alkene C-H) INCREASING: (Peak between) (cm ) (due to C=O in carboxylic acid) (Peaks due to alkane C H bonds at) EITHER (cm ) (cm ) ALLOW (Peak between) (cm ) (due to O H in carboxylic acid)

5 Correct Answer Reject Mark (c) (Makes it taste) sour / sharp / tart fruity IGNE acidic / bitter NOTE Contradictory answers (e.g. sharp and sweeter ) score (0) sweet(er) none

6 (d) (i) Correct Answer Reject Mark st mark: 3 (% of oxygen =) 43.9 (%) nd mark: Amount of C = 49.3/ = 4. (mol) Amount of H = 6.8/ = 6.8 (mol) Amount of O = 43.9/6 =.7 (mol) 3rd mark: Ratio.5 C :.5 H : O ( 3 C : 5 H : O) ALLOW for 3rd mark:- Decimal values that round up to these values (e.g..497 C :.478 H : O scores the 3rd mark) ALLOW M r of C 3 H 5 O = 73 (g mol - ) %C = 36 x 00 = 49.3% 73 and %H = 5 x 00 = 6.8% 73 %O = 43.9% ALLOW 43.8%

7 (d) (ii) Correct Answer Reject Mark For Chemical shift column, allow any range or any single value within range 4 Feature of compound Q Chemical shift / ppm Splitting pattern CH Triplet Allow (splits into) three CH.7 3(.0) OH 0(.0) (.0) Quartet Allow quadruplet / (splits into) four singlet

8 (a) (The energy / enthalpy change that accompanies the formation of) energy required / energy needed / energy it takes one mole of a(n ionic) compound ALLOW as alternative for compound: lattice /crystal / substance / solid / product / salt from (its) gaseous ions from one mole of gaseous ions (no nd mark) IGNE References to standard conditions or any incorrect standard conditions from gaseous elements (no nd mark) ALTERNATIVE RESPONSE If no mark(s) already awarded from above, can answer by giving:- energy change / enthalpy change per mole Na + (g) + O (g) Na O(s) NOTE If lattice energy of dissociation is given (e.g. energy required to break down mol of an ionic lattice into its gaseous ions ) max for the nd scoring point gaseous ions

9 (b)(i) Na + (g) + O - (g) 3 Na + (g) + O(g) + e - B A C (or C) Na + (g) + O - (g) + e - Na(g) + O(g) E E G Na(g) + ½O (g) D (or DD) Na(s) + ½O (g) F Na O(s) All seven letters correct Five six letters correct Three four letters correct (3) () ALLOW Either D or D Either C or C ALLOW Correct numerical values (see question paper) may be given as an alternative to the correct letters

10 (b)(ii) FIRST, CHECK THE FINAL ANSWER IF answer = 50 (kj mol ) then award () marks, with or without working Otherwise look for 44 = ( 08) ( 496) + ( 4) ) + H LE H LE = 44 [ ( 08) ( 496) + ( 4) ] A correct expression using letters e.g. F = ()D + E + ()C + A + B + G (= 44 06) = 50 (kj mol ) NOTE 088 (kj mol ) scores (0) overall (as two errors) (+)088 (kj mol ) also scores (0) overall (as several errors) ALLOW for mark: 69 (wrong sign for 44) 96 ( 08 and 496 not used for Na + ) 4 ( 08 not used for Na + ) 04 ( 496 not used for Na + ) +50 (wrong sign for final answer) 80 (sign changed for st electron affinity of oxygen) (atomization of oxygen halved) NOTE Penalise incorrect units (e.g. kj mol) ONCE only NO ECF from incorrect answers to (b)(i)

11 *(c) ALLOW reverse argument where appropriate 4 First mark MgO more exothermic (than MgS) IGNE greater / higher / larger Second mark S larger than O Third mark Charges on O and S same Charges on (all) ions same S smaller charge density than O MgS is larger than MgO S has a larger atomic radius than O NOTE This mark is awarded if both formulae for the ions O and S are mentioned Fourth mark O (forms) stronger (electrostatic) attractions (than S ) IGNE just stronger (ionic) bonds Penalise ONCE ONLY the use of the word atom(s) or molecule(s) / use of formulae such as Mg O O, etc. AND/ Penalise ONCE ONLY use of words such as just magnesium (instead of magnesium ions/mg + ) and/or just oxygen (instead of oxide ions/o ) Mark each point independently (Total for Question = marks)

12 3 (a) xx [:Li] + xx l xx xx Accept all or mixture of dots and crosses Check inner electrons present on lithium If no element symbols but fully correct with Li first give max If no / incorrect charge(s) if the electrons are correct max If arrow drawn from third / outer shell electron on lithium to join electrons in iodine / iodide with correct charges scores max Brackets are not essential Question 3 (b) 3 Li(s) and Li + (g) and I (g) ½I (s) and I(g) (ΔH at )[½I (s)] Notice the square brackets are essential for this mark If wrong state for iodine element ie if ½I (g/l) and consistent (ΔH at )[½I (g/l)] allow third mark If I(s) given for element and (ΔH at ) [I(s)] allow third mark If wrong state with monatomic iodine both the last two marks lost If Li + (g) + e appears ignore electron

13 3 (c) First mark for one of: 70 = electron affinity 759 Or Electron affinity = 70 ( ) Electron affinity = Second mark for: (Electron affinity =) 97 (kj mol ) 97 (kj mol ) alone scores () Wrong unit e.g. NB providing method is recognisable with one transcription error eg 795 for 759 and the final answer is consistent max NB (+) 97 (kj mol ) max

14 3 (d) (Experimental lattice energy is) more negative / exothermic Theoretical lattice energy is less negative / exothermic Greater / less Increase / decrease alone 3 Recognition that more energy released Irrespective of first answer then, any two from: Due to a degree of covalency Deviation from pure ionic model (in experimental value) The theoretical model is pure ionic bonding Polarization / distortion of the iodide / negative ions (by the lithium ion). Can be shown by diagram Iodine/ I / I ion is not acceptable but iodine / I anion is allowed Note I anion is not allowed Question 3 (e) Electron affinities become less negative / less exothermic / more positive (going down Group 7) Greater / less / Increase / decrease alone As (added) electron further from the nucleus More shielding / shielded (from the nucleus) Any indication of ionization/ removing an electron Second mark stands alone Ignore larger (ionic) radius / atom / ion / charge density

15 4(a) MgCO 3 (s) + HCl(aq) MgCl (aq) + H O(l) + CO (g) ALLOW MgCO 3 (s) + H + (aq) Mg + (aq) + CO (g) +H O(l) All formulae and balancing State symbols mark independently; can be given even if eg MgCl formula incorrect or for H CO 3 (aq) CO 3 - (s) + H + (aq) CO (g) +H O(l) ( mark max) ALLOW missing/incorrect state symbol Question 4(b) Any two from Bubbles (of gas)/ fizzing/ effervescence Solid disappears/ disintegrates /gets smaller /dissolves MgCO 3 disappears (if given as solid in (i)) IGNE clear solution forms Mixture gets warmer/cooler temperature change occurs/ heat change occurs Carbon dioxide /gas given off Precipitate forms (no TE for MgCl (s)) Just exothermic Question 4(c)(i) Moles acid = ((5 x / 000)) = 0.05/0.050 / 5x0 - Ignore units and sf Question 4(c)(ii) Mass Mg CO 3 = ((0.05 x 84.3 )) =.075/.08 /./. (g) ALLOW TE from (c)(i) and (a) /.(g) ALLOW Moles acid x 84.3 for TE(from (i) (4.(5)) if factor of missing for TE from (a)) Ignore sf except sf Ignore units

16 4(c)(iii) To ensure all acid reacts/ all acid is used up / to ensure product is neutral/ it (HCl) is neutralised All reactants used up To ensure reaction is complete (without reference to HCl) To ensure yield is high To ensure magnesium carbonate is in excess Question 4(c)(iv) Filter ALLOW centrifuge/ decant/ pour off / (use) filter paper Ignore comments about heating solution first to concentrate it Sieve Collect MgCl in filter paper Use filter paper to dry crystals Evaporate Question 4(c)(v) 00% yield = (03.3 x 0.05) /5.08(5)g) yield = (3.75 x 00) = 74 % Mol magnesium chloride = ( ) = /0.0845/0.084/0.08 ( ) yield = (00 X ) 0.05 = 74 % Second mark can be given as TE if expected yield or number of moles is wrong. ALLOW 73.8/73.78/73.8 /73.6 /other answers rounding to 74 % from earlier approximations /7 (from 0.08 moles) Allow TE from (a) and or (c)(i) and or (c)(ii) If the ratio HCl to MgCl is : ans 37 % () If moles of HCl in (c)(i) are wrong () If (a) and (c)(i) are correct 37 % scores If moles MgCO 3 = 0.05 allow TE giving 37/ 36.9% Ignore sf except sf

17 4(c)(vi) Some stays in solution / losses on transferring from one container to another/ loss on filtering /crystals left behind/some left on filter paper etc Any one ALLOW correct answers with other comments which are not incorrect eg there may be some spillage and also. Incomplete reaction/side reaction Lost as waste products Lost to environment Lost in manipulation? Hydrolysis Weighing errors Just spillage Question 4(d)(i) Not 00% ionic /almost completely ionic (partial) covalent character/ almost no covalency Discrepancy in BH values indicates polarisation (of ions) Magnesium chloride is covalent Magnesium chloride is partially ionic Just polarity of ions Mark can be given if answer here refers to bond strength and the answer above is included in (ii) Question 4(d)(ii) Question 4(e)(i) QWC I - larger (than Cl - ) so (ion) easier to polarise /distort ALLOW for nd mark increases covalent character / more covalent than MgCl / converse for MgCl / description of polarisation instead of the term If clearly ions, allow reference to iodine instead of iodide ( iodine has a larger ion ) Size of atoms rather than ions I is larger than Cl I molecules are polarised Mg + is polarised Iodine more electronegative than chlorine Read in conjunction with (i). Direct comparison not needed if (i) covers bonding in chloride. (00 x 0) = x 0-3 (g) 0 6 ALLOW 0.00(g) /500 (g) x 0-6 kg IGNE % as unit x 0-3 = 0.000

18 4(e)(ii) (More) soluble (in water)/ (more) soluble in blood stream/ can be given as solution/ won t produce gas in stomach / won t react with stomach acid/ doesn t produce CO Converse answers for MgCO 3 Or other valid answers ALLOW can be given in liquid form MgCl is a liquid MgCO 3 is too reactive

Periodicity & Trends

Periodicity & Trends Mark Scheme 1 Level Subject Exam Board Topic International A Level Chemistry Edexcel The Core Principles of Chemistry Sub Topic Periodicity & Trends Booklet Mark Scheme 1 Time Allowed: