Markscheme November 2016 Chemistry Higher level Paper 2

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1 N6/4/EMI/P/ENG/TZ0/XX/M Markscheme November 06 hemistry igher level Paper 0 pages

2 N6/4/EMI/P/ENG/TZ0/XX/M This markscheme is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of the IB Assessment entre.

3 3 N6/4/EMI/P/ENG/TZ0/XX/M. a i [O O] «K c =» 3 [O] [ ]. a ii Position of equilibrium: moves to right favours product K c : no change is a constant at constant temperature. a iii Bonds broken: O + 3(-) / (077 kj mol - ) + 3(436 kj mol - ) / 346 «kj» Bonds formed: (-O) + (O-) + 4(-) + (-) / (358 kj mol - ) + (463 kj mol - ) + 4(44 kj mol - ) kj mol - / 3644 «kj» «Enthalpy change = bonds broken - bonds formed = 346 kj kj =» -8 «kj». b i «= Σ f products - Σ f reactants = kj mol - - (-0.5 kj mol - ) =» «kj». b ii in (a)(iii) gas is formed and in (b)(i) liquid is formed products are in different states conversion of gas to liquid is exothermic conversion of liquid to gas is endothermic enthalpy of vapourisation needs to be taken into account Award [3] for correct final answer. Award [ max] for «+»8 «kj». Accept product is «now» a liquid. Accept answers referring to bond enthalpies being means/averages. 3

4 4 N6/4/EMI/P/ENG/TZ0/XX/M (Question continued). b iii «S is negative because five mols of» gases becomes «one mol of» liquid increase in complexity of product «compared to reactants» product more ordered «than reactants». b iv S = «kj K -» G = kj - (98 K kj K - ) = «kj». b v increasing T makes G larger/more positive/less negative -T S will increase. c Ethene: - Ethane-,-diol: - Accept fewer moles of gas but not fewer molecules. Award [] for correct final answer. Award [ max] for «+» If kj used, answer is: G = kj - (98 K «kj» Award [] for correct final answer. Do not accept, respectively. kj K - ) =

5 5 N6/4/EMI/P/ENG/TZ0/XX/M (Question continued). d ethane-,-diol can hydrogen bond to other molecules «and ethene cannot» ethane-,-diol has «significantly» greater van der Waals forces Accept converse arguments. Award [0] if answer implies covalent bonds are broken. hydrogen bonding is «significantly» stronger than other intermolecular forces. e acidified «potassium» dichromate«(vi)» / + AND K r O 7 / + AND - r O 7 «acidified potassium» manganate(vii) / «+» KMnO 4 / «+ -» MnO 4. f Number of signals Splitting pattern Ethanedioic acid: AND singlet Ethane-,-diol: Not required Accept SO 4 or 3 PO 4 for +. Accept permanganate for manganate(vii). Accept none/no splitting for singlet.. a Weak acid: partially dissociated/ionized «in solution/water» AND Strong acid: «assumed to be almost» completely/00 % dissociated/ ionized «in solution/water» Accept answers relating to p, conductivity, reactivity if solutions of equal concentrations stated.

6 6 N6/4/EMI/P/ENG/TZ0/XX/M (Question continued). b «log scale» reduces a wide range of numbers to a small range Do not accept easy for calculations. simple/easy to use converts exponential expressions into a linear scale/simple numbers. c i phenolphthalein phenol red. c ii 4.0 «n(nao) = dm mol dm -3 =» «mol» 000. c iii « =» «mol»

7 7 N6/4/EMI/P/ENG/TZ0/XX/M (Question continued). c iv ALTERNATIVE : «mass of pure hydrated ethanedioic acid in each titration = mol 6.08 g mol - =» / «g» Award suitable part marks for alternative methods. Award [3] for correct final answer. mass of sample in each titration = « g =» 0.5 «g» Award [ max] for 50.4 % if anhydrous ethanedioic 000 acid assumed. «% purity = g 00 =» 70.6 «%» 0. 5g ALTERNATIVE : «mol of pure hydrated ethanedioic acid in dm 3 solution = =» «mol» «mass of pure hydrated ethanedioic acid in sample = mol 6.08 g mol - =» 3.53 «g» «% purity = 353. g 00 =» 70.6 «%» 500. g 3 ALTERNATIVE 3: mol of hydrated ethanedioic acid (assuming sample to be pure) = 500. g = «mol» 6. 08gmol - actual amount of hydrated ethanedioic acid = « =» «mol» «% purity = =» 70.6 «%»

8 8 N6/4/EMI/P/ENG/TZ0/XX/M (Question continued). d O O. e electrons delocalized «across the O O system» resonance occurs O O O O O O Accept single negative charges on two O atoms singly bonded to. Do not accept resonance structures. Allow any combination of dots/crosses or lines to represent electron pairs. Accept delocalized π-bond(s). No EF from (d). «pm» < O < 43 «pm». f coordinate/dative/covalent bond from O to «transition» metal «ion» acts as a Lewis base/nucleophile can occupy two positions provide two electron pairs from different «O» atoms form two «coordinate/dative/covalent» bonds «with the metal ion» chelate «metal/ion» Accept any answer in range 3 «pm» to 4 «pm». Accept bond intermediate between single and double bond or bond order.5.

9 9 N6/4/EMI/P/ENG/TZ0/XX/M 3. a O AND (l) Do not accept O (aq). 3. b SO (g) is an irritant/causes breathing problems SO (g) is poisonous/toxic 3. c n(l) = « dm3.00 mol dm -3 =» / «mol» AND n(na S O 3 ) = « dm mol dm -3 =» / «mol» «mol» > «mol» «mol» > «mol» «mol» > «mol» Accept SO (g) is acidic but do not accept causes acid rain. Accept SO (g) is harmful. Accept SO (g) has a foul/pungent smell. Accept answers based on volume of solutions required for complete reaction. Award [] for second marking point. Do not award M unless factor of (or half) is used.

10 0 N6/4/EMI/P/ENG/TZ0/XX/M (Question 3 continued) 3. d Rate of reaction t / 0-3 s [Na S O 3 ] / moldm -3 five points plotted correctly best fit line drawn with ruler, going through the origin

11 N6/4/EMI/P/ENG/TZ0/XX/M (Question 3 continued) 3. e i first order «because» [Na S O 3 ] is «directly» proportional to rate of reaction Do not accept linear for M. «t» 3. e ii rate = k [Na S O 3 ][l] 3. f Award [] for correct final answer. Accept value based on candidate s graph. t / 0-3 s Rate of reaction «s -» «Time = =» 44.4 «s» mol dm [Na S O 3 ] / mol dm -3 Award M as EF from M. Award [ max] for methods involving taking mean of appropriate pairs of values. t Award [0] for taking mean of pairs of time values. Award [] for answers between 4.4 and 46.4 «s».

12 N6/4/EMI/P/ENG/TZ0/XX/M (Question 3 continued) 3. g i T Accept probability «density» / number of particles / N / fraction on y-axis. Accept kinetic E/KE/E k but not just Energy/E on x-axis. Fraction of particles T Kinetic energy correctly labelled axes peak of T curve lower AND to the right of T curve

13 3 N6/4/EMI/P/ENG/TZ0/XX/M (Question 3 continued) 3. g ii greater proportion of molecules have E E a or E > E Accept more molecules have energy greater a than E a. greater area under curve to the right of the E a Do not accept just particles have greater kinetic energy. greater frequency of collisions «between molecules» more collisions per unit time/second Accept rate/chance/probability/likelihood instead of frequency. Accept suitably shaded/annotated diagram. 3. h shorter reaction time so larger «%» error in timing/seeing when mark disappears Do not accept just more collisions. Accept cooling of reaction mixture during course of reaction.

14 6 4. a Mg 4. b Award [] for correct final answer. «A r =» Do not accept data booklet value (4.3). 00 «= =» c contamination with sodium/other «compounds» 4. d i energy levels are closer together at high energy / high frequency / short wavelength 4. d ii ionisation energy 4. e MgO(s) + O(l) Mg(O) (s) Accept. MgO(s) + O(l) Mg + (aq) + O - (aq) 4. f from basic to acidic through amphoteric Accept alkali/alkaline for basic. Accept oxides of Na and Mg: basic AND oxide of Al: amphoteric for M. Accept oxides of non-metals/si to l acidic for M. Do not accept just become more acidic. 4. g Mg 3 N Accept MgO, Mg (O), Mg (NO x ), MgO h «3-D/giant» regularly repeating arrangement «of ions» lattice «of ions» electrostatic attraction between oppositely charged ions electrostatic attraction between Mg + and O ions 4. i i Anode (positive electrode): l l (g) + e athode (negative electrode): Mg + + e Mg (l) 4 N6/4/EMI/P/ENG/TZ0/XX/M Accept giant for M unless giant covalent stated. Do not accept ionic without description. Penalize missing/incorrect state symbols at l and Mg once only. Award [ max] if equations are at wrong electrodes. Accept Mg (g).

15 5 N6/4/EMI/P/ENG/TZ0/XX/M (Question 4 continued) 4. i ii reduction 4. i iii Anode (positive electrode): oxygen/o hydogen ion/proton/ + AND oxygen/o athode (negative electrode): hydrogen/ hydroxide «ion»/o - AND hydrogen/ 4. j Any two of: «inert» Pt electrode platinum black conductor mol dm 3 + (aq) (g) at 00 kpa Award [ max] if correct products given at wrong electrodes. Accept atm (g). Accept bar (g) Accept a labelled diagram. Ignore temperature if it is specified. 4. k i Mg (s) + u + (aq) Mg + (aq) + u (s) 4. k ii «+0.34 V - (-.37 V) = +».7 «V» 4. k iii cell potential increases Accept correct answers based on the Nernst equation. reaction «in Q4(k)(i)» moves to the right potential of the copper half-cell increases/becomes more positive max

16 6 N6/4/EMI/P/ENG/TZ0/XX/M 5. a Propane: AND Propene: 5. b i Sigma (σ): Pi (π): 5. b ii Number of sigma (σ) Number of pi (π) bonds bonds Propane 0 0 Propene 8 Award [] for two or three correct answers. Award [] for all four correct.

17 7 N6/4/EMI/P/ENG/TZ0/XX/M (Question 5 continued) 5. c i Br 3 7 Br + Br «sun»light/uv/hv high temperature Do not accept reflux for M. 5. c ii Br 3 6 Br 5. c iii Propane: «free radical» substitution / S R AND Propene: «electrophilic» addition / A E 5. d 3 3 Award [ max] for formation of -bromopropane. Br Br 3 3 curly arrow going from = to of Br and curly arrow showing Br leaving representation of carbocation curly arrow going from lone pair/negative charge on Br - to + Br

18 8 N6/4/EMI/P/ENG/TZ0/XX/M 6. a Br Br correct isomer mirror image shown clearly 5 6. b S N would give inversion of configuration «almost 00 %» S N would give «approximately» 50 % of each so mechanism is a mixture of both mechanisms 6. c I bond «longer, so» weaker «than Br bond» I is a better leaving group than Br

19 7. a alculation: ALTERNATIVE : [ + ] = (K a [A]) / / ( ) / / «mol dm 3» p = «-log 0 [ + ]».9 ALTERNATIVE : p = 0.5(pK a - log 0 [A]) p =.9 9 N6/4/EMI/P/ENG/TZ0/XX/M Award [] for correct final answer. Assumption: ionisation is << so [A - ] [A] eqm = [A] initial all + ions in the solution come from the acid «and not from the self-ionisation of water» [ + ] = [OO ] Do not accept partial dissociation. 3

20 0 N6/4/EMI/P/ENG/TZ0/XX/M (Question 7 continued) b i M: must show buffer region at p < 7 and equivalence at p > 7. p 7 pk a V n Volume V n at neutralization Accept graph starting from where two axes meet as p scale is not specified. Volume of strong base added correct shape of graph p at half neutralization/equivalence b ii ALTERNATIVE : OO OO ions consumed in reaction with O - are produced again as equilibrium moves to the right «so [ + ] remains almost unchanged» ALTERNATIVE : OO + O - OO - + O added O - are neutralized by OO strong base replaced by weak base Accept A or any other weak acid in equations. Equilibrium sign must be included in equation for M.

Markscheme November 2015 Chemistry Standard level Paper 2

N15/4/CHEMI/SP2/ENG/TZ0/XX/M Markscheme November 2015 Chemistry Standard level Paper 2 14 pages 2 N15/4/CHEMI/SP2/ENG/TZ0/XX/M This markscheme is the property of the International Baccalaureate and must