Physical Sciences Grade 12 P2 KZN June 2023 Paper & Memo

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Downloaded from Stanmorephysics.com Physical Sciences P2 2 June 2023 Common Test NSC INSTRUCTIONS AND INFORMATION de Write your centre number and examination number in the appropriate spaces on the ANSWER BOOK. 2. This question paper consists of NINE questions. Answer ALL the questions in the ANSWER BOOK. 3: Start EACH question on a NEW page in the ANSWER BOOK. 4, Number the answers correctly according to the numbering system used in this question paper. 5. Leave ONE line between two subquestions, for example between QUESTION 2.1 and QUESTION 2.2. 6. You may use a non-programmable calculator. 7. You may use appropriate mathematical instruments. 8. You are advised to use the attached DATA SHEETS. 9. | Show ALL formulae and substitutions in ALL calculations. 10. Round off your final numerical answers to a minimum of TWO decimal places. 44, Give brief motivations, discussions et cetera where required. 12. Write neatly and legibly. Please turn over Copyright reserved

pryawiieaded from Stanmorephysics.com June Common Test QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question number (1.1-1.10) in the ANSWER BOOK, for example 1.11 D. 1.1 What is the homologous series to which the following compound belongs? CH — C = O | fe} | H A Ketones B_ Alcohols C_ _Haloalkanes D —Carboxyllic acids (2) 1.2 Which ONE of the following is a tertiary alcohol? A _ Pentan-1- ol . B_ Pentan-2-ol C 2-methylbutan - 2 - ol Dd 3 — methylbutan — 2 - ol (2) 1.3 Which of the following statements would apply to organic compounds that belong to the same homologous series? They have the same... . | boiling points. ll ‘functional group. Il molecular formula. Il only | and II only Il and II only 0 0aQ 0 > 1, Wand tl (2) Copyright reserved Please turn over

ownloaded from Stanmorephysics.com Physical Sciences/P2 June Common Test 1.4 Which ONE of the following statements is TRUE for the reaction of an UNSATURATED HYDROCARBON with bromine water? A __ The reaction is slow. B The bromine water decolourises quickly. C _Thereaction is a substitution reaction. 6 (2) Ultraviolet light is needed for this reaction. 1.5 The Maxwell-Boltzmann distribution curve for a reaction mixture is shown below: No. of particles ee Ex The TEMPERATURE of the reaction mixture is now DECREASED. Which one of the following graphs shows the new distribution curve as a dotted line? A No. of particles No. of particles No. of particles (2) Copyright reserved Please turn over

PrBeventoaded from Stanmorephysics.com June Common Test 1.6 The mass of a catalyst was determined at intervals during a reaction. with ti * fa a Which = of the following graphs best represents the mass of the catalyst a oO o a o E E —> > Time Time Cc D . £ Time Time (2) iB i Each of the reactions represented below is at equilibrium in a closed container. In which ONE of these reactions will an INCREASE IN PRESSURE (by decreasing the volume) at constant temperature favour the formation of products? A N204(g) = 2 NO2(g) B PCts(g) = PC&(g) + Clog) C Nag) + 3Ha(g) = 2NHs(g) a in ol D NO2a(g) + CO(g) = NO(g) + COa2(g) faant (2) Copyright reserved Please turn over

June Common Test Physical Basmpe@aded from Stanmofephysics.com 1.8 The expression for the equilibrium constant (Kc) for a hypothetical reaction is given as follows: _IRUIST 5 Seip lf ' the following balanced equations represents the hypothetical A 3P(s) = R@Q) + 286) 3P(t) = R(aq) + 2S(aq) ~ 3P(aq) + Q(s) = Rig) + Sxg) DO. xX io 3P(aq) + Q(s) = R(aq) + 2S(g) (2) 1.9 Which ONE of the following pairs represents the conjugate acid and the conjugate base of HPO?" ? CONJUGATE ACID CONJUGATE BASE POF H2PO;, H2PO; POs H2PO; HsPO, D H2PO7 PO? (2) 1.10 | CHsCOOH is a weak acid. Which ONE of the following statements is correct for 20,01 mol-dm* aqueous solution of CH3COOH? CLO @ os There are more hydronium ions than acetate ions. There are fewer acetate ions than unionized CHsCOOH. There are more hydronium ions than unionized CHsCOOH; ! renin ions and There are equal quantities of unionized CH3COOH, hyd acetate ions. Copyright reserved (2) [20] Please turn over

Physheioveimesed from Stanmorephysics.com QUESTION 2 (Start on a new page.) The letters A to F in the table below represent six organic compounds. June Common Test wo = | =mI—o—=z | x=IxmI—oO—xI=z | I—0—o—o-2 | ° | pf C | CHsCO2H _ | D | CHsCO2C2Hs E | CsHsO F | 2—bromobutane Use the information in the table to answer the questions that follow. 2,1 2.2 Write down the IUPAC name of: 2.1.1 Compound A. 2.1.2 Compound B. Write down the letter that represents: 2.2.1 Ahaloalkane. Janay ) = 2.2.2 Acompound that belongs to the same homologous series as compound B. Copyright reserved (3) (3) (1) (1) Please turn over

Physical StspMfifoaded from Stanmbrephysics.com sine Commence Tet 2.3. Compound D is produced when compound C is reacted with a primary alcohol. 2.3.1 Define the term functional group. (2) yn the NAME of the FUNCTIONAL GROUP of the homologous which compound C belongs. (1) 2.3.2 Write seri 2.3.3 Write ¢ 1 the structural formula of the functional group of the homologous series to which compound D belongs. (2) 2.3.4 Write a balanced equation using MOLECULAR FORMULAE to represent the reaction that takes place. (5) 2.3.5 Write down the NAME or FORMULA of the substance that can be used to accelerate the above reaction. (1) 2.3.6 Write down the IUPAC name of the primary alcohol used in this reaction. (2) 2.4 Define the term functional isomer. (2) 2.5 Write down the STRUCTURAL FORMULA of the straight chain FUNCTIONAL ISOMER of compound D. (2) [25] QUESTION 3 (Start on a new page.) 3.1. X and Y are two saturated hydrocarbons with the same molecular formula. Both X and Y have 4 carbons each. X has a lower boiling point than Y 3.1.1 Define boiling point. (2) 3.1.2 Fully explain the difference inthe boiling points of these two compounds. (3) 3.1.3. Write down the STRUCTURAL FORMULA for compound X. (2) 3.2 In an experiment the boiling points of alcohols are compared to that of the carboxylic acids. The boiling point of pentan — 1 —ol is found to be 138 °C. 3.2.1 Write down the IUPAC name of the carboxylic acid that must be used in this experiment to ensure that the comparison is fair. — Give a reason for the answer. ? (3) 3.2.2 Will the boiling point of the carboxylic acid be LESS THAN, EQUAL TO or GREATER THAN 138 °C? — (1) 3.2.3. Fully explain the answer to QUESTION 3.2.2. (3) [14] Copyright reserved Please turn over

Phyawuegded from Stanmorephysics.com QUESTION 4 (Start on a new page.) In the flow diagram below, 1, 2, 3, and 4 represent organic reactions. P and Q represent organi¢-compounds. Q is aHYDROCARBON. 4.1 4.2 4.3 44 45. * Copyright reserved June Common Test pentan — 2 —ol 4 1 7 Bra Q P +“ CsHi2 * —$$____—. " + 3 2 (major product) Consider REACTION 1: 4.1.1 Name the type of reaction that takes place. Choose from SUBSTITUTION, ELIMINATION or ADDITION. 4.1.2 | Write down the structural formula of Q. 4.1.3 Apart from heat, state one other reaction condition (2) REACTION 2 is an example of an addition reaction. 4.2.1. Write down the TYPE of addition reaction represented here. (1) 4.2.2 Write down the NAME or FORMULA of the inorganic reactant required for this reaction. . (1) 4.2.3 Name ONE reaction condition for reaction 2. (1) Bromine is used as an inorganic reactant in REACTION 3. 4.3.1. Name the type of reaction that takes place. (1) 4.3.2 State the reaction condition for this reaction. (1) 4.3.3 Write down the structural formula of compound P. (2) Consider REACTION 4: 4.4.1 Apart from heat, state one other reaction condition. (1) 4.4.2 Using condensed structural formula, write a balanced jequation to represent the reaction that takes place. (3) Compound P can be converted directly ta compound Q in an elimination reaction. Write down the TYPE of elimination reaction taking place. O. Please tasocl (1) (2)

PhysicalSeammtdaded from Stanmoréphysics.com QUESTION 5 (Start on a new page) June Common Test Calcium carbonate chips are added to excess dilute hydrochloric acid solution in a flask placed on a balance as illustrated below. wool plug Dilute hydrochloric acid Gas bubbles Calcium carbonate chips The cotton wool plug placed in the mouth of the flask prevents spillage of reactants and products, while simultaneously allowing the formed gas to escape. The balanced equation for the reaction that takes place is: CaCOs(s) + 2HC&(aq) — CaCte(aq) + CO2(g) + H20(£) The mass of the flask and its contents is recorded in intervals of 30 seconds. The results obtained are shown in the table below. Time in seconds (s) Mass of the flask and its contents in grams (g) () 217,50 30 215,00 60 212,95 90 211,05 120 209,20 150 207,40 180 xX 210 203,95 240 203,95 270 yY 5.1 Define reaction rate. 5.2 Write down the evidence from the table that suggests that the r: reaction DECREASES with time. ha (2) the above (2) 5.3. Use the collision theory to explain why the rate of the reaction decreases with time. Assume that the temperature remains constant. Copyright reserved (3) Please turn over

Prisdinntetded from Stanmorephysics.com June Common Test 5.4 How will an increase in the volume of the hydrochloric acid used affect the rate of this reaction? Choose from INCREASES, DECREASES or REMAINS THE SAME (1) 5.5 The aver: ate of the reaction for the first 180s is 6,58 x 107 g.s'. Calculate the value 9% in the table. (4) 5.6 Write d @ mass, in grams, represented by Y. Give a reason for the answer. (2) — 5.7 Calculate=the total volume of carbon dioxide gas produced during the above reaction at S.T.P. (6) 5.8 It is observed that when a catalyst is added to the above reaction, the reaction reaches completion in a shorter time. Use the collision theory to explain the observation. (3) 5.9 The CaCOs chips are replaced with an equal mass of CaCO: powder. How will each’ of the following be affected by this change? (Choose from INCREASES, DECREASES or REMAINS THE SAME) 5.9.1 The mass in grams represented by Y. (1) 5.9.2 Time taken to reach the mass represented by X. (1) ‘ [25] |=) | Copyright reserved Please turn over

June Common Test Physical BesWAilbaded from Stanmoréphysics.com QUESTION 6 (Start on a new page) 6.1. An investigation is performed to determine the effect of TEMPERATURE, PRESSURE and CONCENTRATION on equilibrium in the production of ammonia in aisealed container. The volume of the container is kept constant. The balan “equation below represents the reaction that takes place in the sealed container. > Na(g) + 3 Hog) = 2NH3(g) AH<O The graph below represents the results obtained: 4 > E 3 £ N2 5 é -_——_ FS pewewatencnnceneeneen , a E i ’ c %, Pd ° eeeccccecce oO : > 0 th te ts ts Time in minutes 6.1.1 What information about the reaction does the graph provide between O and ti? (2) 6.1.2 Will the Ke value for this reaction between 0 and t; be GREATER THAN 1, EQUAL TO 1 or LESS THAN 1? (1) 6.1.3 Atwhich time, ti, tz or ts was the concentration of a reactant increased? Give a reason for the answer. (2) 6.1.4. State Le Chatelier’s principle. (2) 6.1.5 Which ONE of the factors, TEMPERATURE, PRE SURE or CONCENTRATION was changed at ts? Lonny (1) 6.1.6 Was the factor identified in QUESTION 6.1.5 INCREASED or DECREASED? Explain the answer by referring to Le Chatelier’s principle. (4) Copyright reserved Please turn over

PryBeurdaaded from Stanmorepfysics.com June Common Test 6.1.7. A partial graph of reaction rate versus time is drawn below. The dotted line represents the reverse reaction a >, E] =| 0 th te ts ts Time in minutes Copy the above graph in your answer book. Complete the graph for each of the time periods period till ts. Use the solid line for the forward 6) reaction rate and the dotted line for the reverse reaction rate. 6.2 The’ equation below represents the reversible reaction that takes place when sulphur dioxide gas, SO2, and oxygen gas, Oz, are allowed to react in a sealed container. 2 SO2(g) + O2(g) = 2 SOs(g) AH<0 6.2.1. Write down the meaning of the term reversible reaction. (2) 6.2.2 The activation energy for the FORWARD reaction is compared to the activation energy of the REVERSE reaction. Which reaction, FORWARD or REVERSE, will have a higher activation energy? (1) 6.3 A hypothetical reaction is represented by the balanced equation below: 2X(g) + Y(g) = 2g) 2,5 moles of X (g) were mixed with 1,75 moles of Y (g) in a sealed 2.dm* container. The reaction reaches equilibrium at T °C. At equilibrium the concentration of C (g) was 0,5 mol.dm*. ‘Calculate the value of the equilibrium constant, Ke. (7) [27] Copyright reserved Please turn over

Physical eyAtoaded from Stanmoréphysics.com QUESTION 7 (Start on a new page.) 7.1 7.2 7.3 74 Copyright reserved Sulphuric acid, H2SOs, ionizes in water in TWO steps as follows: STEP I: H2SO4(aq) + H20(2) = HSO;(aq) + H30*(aq) STEPH HSO;(aq) + H20(f) = SO (aq) + Hs0*(aq) 7.1.1 Define af acid in terms of the Arrhenius theory. 7.1.2 Writ down the NAME or FORMULA of the substance that acts as an ampholyte in the above equations. Give a reason for the answer. The Ka value for TWO acids are as follows: NAME OF ACID | FORMULA OF ACID | Ka VALUE Oxalic Acid (COOH)2 5,6 x 107 Carbonic Acid H2CO3 4,3 x 102 7.2.1 Oxalic acid ionizes in water according to the following balanced equation: (COOH)2(aq) + 2H20(2) = (COO)> (aq) + 2 H3s0*(aq) Write down the formula of the TWO BASES in this reaction. 7.2.2 Which ONE, oxalic acid or carbonic acid, will have a higher conductivity? Explain the answer by referring to the information in the table. Water is added to a 0,01 mol.dm* solution of hydrochloric acid. How does the addition of the water affect each of the following? (Choose from: INCREASES, DECREASES or REMAINS THE SAME). 7.3.1 The strength of the solution of the hydrochloric acid. 7.3.2 The concentration of the solution of the hydrochloric acid. Give a reason for the answer. A solution of sodium hydrogen carbonate, (NaHCOs) of volume 250 cm}, is prepared by dissolving x g of sodium hydrogen carbonate in sufficient distilled water in a volumetric flask. 20 cm® of the prepared sodium hydrogen carbonate solution neutralises 12,58 cm’ of a sulphuric acid(H2SOs) solution of 0,29 mol.dm*. The balanced equation for the reaction is: H2SO4(aq) + 2NaHCOs(aq) — NazSO4(aq) + 2H20(t) + CO2(g) T 7.4.1 Define the equivalence point of a titration. 7.4.2 Calculate the concentration of the NaHCOs solution that neutralised the H2SOz solution. re 7.4.3. Calculate x, the mass of sodium hydrogen carbonate used to prepare the initial solution of sodium hydrogen carbonate. June Common Test (2) (2) (2) (3) (1) (2) (2) (4) (4) [22] [150] Please turn over

Phyaiinoeaded from Stanmorephysics.com 1 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 2 (CHEMISTRY) TABLE 1: a CONSTANTS June Common Test NAME/NAAM [7-7 SYMBOL/SIMBOOL VALUE/WAARDE pad: p® 1,013 x 10° Pa landaa Molar gas volume at STP Vin 22,4 dm*-mol" Molére gasvolume by STD Standard temperature T 273 K Standaardtemperatuur Charge on electron E -1,6 x 107°C Lading op electron Avogadro's constant Na 6,02 x 1073 mo!" Avogadro-konstante TABLE 2: FORMULAE N m pri neu n Ny '‘V c =4 or/of ga: a TLL pH = -log[H30*] Cag Ny Kw = [H30*][OH"] = 1 x 10°4 at/by 298 K Copyright reserved Please turn over

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Downloaded from Stanmorephysics.com QUESTION 1 11. Dvv (2) 12 Cvv (2) 13. Avv (2) 14. Byv (2) 15 AVY (2) 16 Bvv (2) 17. Cvv (2) 1.8 Dvv (2) 19 Bvv (2) 110 Byv (2) [20] QUESTION 2 24.1 dbs stimethyhexaa 18 =diene co REMOVED 2.1.2 | 2—methylbutan —2-ol OR 2 - methyl -2-butanol Marking criteria: © correct stem i.e. butanol” e substituent methyl correctly identifiedv © IUPAC name completely correct including numbering, sequence and hyphen and commasY (3) 221 FY (1) 2.2.2 Ev (1)

Downloaded from Stanmorephysics.com 2.3.1. A_bond/an_atom/a_ group of atoms that determine(s) the (physical and) chemical properties of a group of organic compounds. Marking criteria: If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark Rion (2) i J 2.3.2 carboxyb<— (1) 2.3.3 fe) | | Marking criteria: —~Cc—o~—c~— YY Whole structure correct 2 or zero (2) 2.3.4 C2HsO2Vv + C2HsOv — CaHsO2¥ + H20Vv balancing v Marking criteria: e 1 for each correct reactant and product e 1 for correct balancing (5) e If structural formulae are used Max 4/5 e Any additional reactant or product: -1 mark 2.3.5. Sulphuric acid/H2SO4 ¥ (1) 2.3.6 ethanolvv (2) 2.4 Compounds with the same molecular formula ~ but different functional groups/homologous series. ¥ Marking criteria: If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark (2) H H H ~ O H L i i | l O—H ) — — ~~ \7 _— _ } oor Marking criteria: Ce e Only Functional group correctly drawnY % e Whole structure correct ¥ % (2) [25]

Downloaded from Stanmorephysics.com QUESTION 3 3.1.1 The temperature _at_which the vapour pressure of a substance equals atmospheric pressure VV Marking: :eriteria: “of the underlined key phrases in the correct context is omitted, deduct (2) 3.1.2 Compound X / methylpropane Is branched / has a smaller surface area / is less spherical while compound Y / butane has a straight chain / is unbranched / has a bigger surface area / is more spherical” The intermolecular forces in compound Y are stronger than the intermolecular forces in comnpound X. ” More energy is required to overcome the intermolecular forces in compound Yiv (3) | Marking criteria: e 3-carbons in the longest chain’ e substituent methyl on the second carbon and everything else correctY =x | xmim—o-=z | (2) 3.2.1 Butanoic acid vv The compounds must be of comparable molecular mass OR butanoic acid has the same molecular mass as pentan — 1 - olV OR Pentanoic acid vv The compounds must be of comparable chain length OR butanoic acid has the same chain length as pentan — 1 - olv (3) 3.2.2 GREATER THANY (1) 3.2.3. The carboxylic acid has 2 sites for hydrogen bonding while the alcohol has only 1 site for hydrogen bonding. “ The intermolecular forces will therefore be stronger between)thi ‘molecules of the carboxylic acid. ¥ More energy will therefore be required to overcome the intérmolecular forces between the molecules of the acid. v (3) [14]

QUESTION 4 4.1.1 Elimination. v 4.1.2 H H H H | | | | H — — T —- C= C — 7 H H H Marking criteria: e Double bond on second carbonY WA 2 e Whole structure correct ¥ % 4.1.3 4.2.1 Hydrogenationv 4.2.2 Hydrogen/Hav 4.2.3. Platinum/Pt catalystv’ 4.3.1 Halogenation/bromination/substitutionv 4.3.2 Require uv light. Or heatv 4.3.3 Downloaded from Stanmorephysics.com concentrated sulphuric acid v VY (If concentrated is not mentioned, then 1/2) HH H oH H | | | | | nT TP pe HH H Br H Marking criteria: e Bron second carbonY A e Whole structure correct Vv x [a (1) (2) (2) (1) (1) (1) (1) (1) (2)

Downloaded from Stanmorephysics.com 4.41 Dilute strong base/KOH/NaOH/H20v (1) 4.4.2 | CHsCH2CH2CHBrCHs + KOH -— CHsCH2CH2CHOHCHs + KBr OR CHsCH2CH2CHBrCHs + NaOH — CHsCH2CH2CHOHCH3 + NaBr OR CH3sCH2GH2CHBrCH3 + H20 -> CHsCH2zCH2CHOHCH3 + HBr Marking criteria: e | Correct reactantsY e Correct productsY e BalancingY (3) 4.5 Dehydrohalogenation/dehydrobrominationv (1) [17] QUESTION 5 5.1 Marking criteria: Give the mark for per unit time only if in context of reaction rate. ANY ONE e Change in concentration V of products/reactants per (unit) time. VW e Change in amount/number of moles/volume/mass” of products or reactants per (unit) time. Vv e Amount/number of moles/volume/mass of products formed/reactants used per (unit) time. vv e Rate of change in concentration/amount of moles/number of moles/volume/mass. v ¥ (2 or 0) (2) 5.2 Change in the mass of the flask and contentsY decreases per unit time / per second / every 30 seconds. v (2) 5.3 Mass of CaCOs decreases or is used up. ¥ A decrease in the exposed surface area. ¥ The number of effective collisions per unit time decreases. v OR NOTE: If no_mention made of CaCOs or HCilused up, award HCl is used up. ” 2 marks tg'second bullet. Concentration of the HCI decreases. v ont The number of effective collisions per unit time decreases. vm mi (3) 5.4 REMAINS THE SAME. v (1)

Downloaded from Stanmorephysics.com 5.5 Average rate -_ Am Marking criteria: Bs Substitute ratev © _X - 217,50 )¥% e Substitute Amv 180 - 0 e Atv e 205,66 g ¥ Final answerv 6,58 x 107 SN (4) 5.6 203,95 Gx Reactidn ha is reached completion. ” (2) 5.7 Marking criteria: ¢ Calculate mass of pure CO2Y m e Formula: n =— M e Correct substitution of 44 ( ue in the above ¥ Either equation formula e Formula n(CO2)produced == — e Substitute n = 0,31 v e Substitute Vm = 22,4 v e Final answer = 6.944 dm3 v m(COz) = 217,50 - 203,95 V = 13,559 n(CO2) = _ 13,55 44 Vv = 0,31 mols ¥ Either equation n(COz)produced /= ~ _ Vv 22,4 ¥ V_ = 6,944 dm? (6,90)v (6) sheray / lowers L rs (3) 5.9.1. REMAINS THE SAMEY (1) 0,31v 5.8 A catalyst provides an alternate pathway of lower activation the activation energyY More molecules will therefore have sufficient energyY The number of effective collisions per unit time increases Vv 5.9.2 DECREASESV (1) [25]

Downloaded from Stanmorephysics.com QUESTION 6 6.1.1. The reaction is in a state of (dynamic) equilibrium. vv OR The rate of forward reaction equals the rate of reverse reaction. OR The concentrations of the reactants and products remain constant. Note > IF: Forwa Steaction equals reverse reaction. VY, (2) 6.1.2 Less than (1) 6.1.3 tiv The concentration of Nz was increased./graph shows an increase in concentration of Nav (2) 6.1.4 Marking criteria: If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark. When the equilibrium in _a closed system is disturbed, the system will re- instate a new equilibrium by favouring the reaction that will cancel/oppose the disturbance. Vv (2) 6.1.5 temperature’ (1) 6.1.6 INCREASEDY According to the graph the concentration of the reactants increases OR the concentration of the products decreases. ¥ Therefore the reverse reaction / endothermic reaction is favoured. V (According to Le Chatelier’s Principle) an increase in temperature favours the endothermic reaction. ¥ (4) 6.1.7 MARKING CRITERIA At t1, forward reaction is favoured (vertical upward climb of solid line) ¥ At te, forward reaction is favoured (vertical upward climb of solid line) ¥ At ts, reverse reaction is favoured (vertical upward climb of broken line) v Vertical upward increase in rates at tz, ts and ts. ¥ Equilibrium rate at ts .> equilibrium rate at ts > equilibrium rate at ta” oar

Downloaded from Stanmorephysics.com 6.2.1 6.2.2 6.3 REACTION RATE 0 t te ts Products can be converted back to reactants. vv REVERSEY ta Time (minutes) Marking criteria: Change in no. of mol of C =1v Using the correct ratiow Subtraction: initial mol X — change in mol of X AND initial mol Y — change in mol of Y. e Equilibrium mol of X and Y; divide by 2v e Correct Kc expression (formulae in square brackets) “ e Substitution of equilibrium concentrations into Kc expression. “ e Final answer 0,71v” Y Ratio 2 1 2 Initial quantity (mol) 2,5 1,75 0 Change (mol) 1 - 0,6 Iv Using ratio v Quantity at equilibrium (mol) C45 | 1,250) 1 v Equilibrium concentration (mol-dm) 0,75 | 0,625 | 0,5 | ¥ Divide by 2 TeV ce iat = v ° [XP] No K- expression, correct subst den. % . . __ 105) . 2 [0,75]" [0,625] Wrong K- expression 4, = 0,71” (5) (2) (1) (7) [27]

Downloaded from Stanmorephysics.com QUESTION 7 7.1.1 7.2.1 7.2.2 7.3.1 7.3.2 7.4.1 acids produce hydrogen ions (H*/H3O*/hydronium ions) in aqueous solutions. vv Hydrogen sulphate ion/ HSO; v Acts as an acid and a basev H20v (COO)? v Oxalic acid. ~ Has a higher Ka value. ¥ lonises to a greater extent / more completely. OR has a higher ion concentration. Vv REMAINS THE SAMEY DECREASESY The volume increases while number of moles is constant / The number of moles of acid decreases in proportion to the volume of water v The point at which the acid has completely reacted with the base / The point at which the base has completely reacted with the acid Vv (2) (2) (2) (3) (1) (2) (2)

742 Downloaded from Stanmorephysics.com Marking criteria: VA on ¢ Formula CaVa _ —A/n=cVv¥ Ve Ng Substitute for Ca, Va and Ve in the above formula/cVY Ratio na : nev Final answer: 0,365 mol.dm3v Mal = ™” vy CsVB~ ne — V 0,29 x 12,58 = 1 yy Cp x 20 ~ 2 Ce = 0,365 mol.dm? v OR n(NaHCO3) = 2n(H2SOs) = 2cWw = 2(0,29 )(0,01258) = 7,2964 x 10% mols oN c(NaHCOs) Vv v v _ 7,2964 x109 oo 0,02 = 0,365 mol.dm? v 7.4.3. POSITIVE MARKING FROM QUESTION 7.4.2 Marking criteria: e Any Formula: n= CV/m= nM/m=CMV Formula: m= nMv e Substitute 0,365 from 7.3.1 and 0,25v e Substitute M = 84v in any of the above formulae e Final answer: 7,665 g OPTION1 OPTION2 n(NaHCOs) in 250cm? = cVY m = CMVY _ = v sia] 0,365)(0.25 = (0,365) (0,25) ¥i(84) v = 0,09125 mols m(NaHCOs) in 250cm = nM = 767qv Adal = 0,09125 x 84v = 7,67 g¥ TOTAL: (4) (4) [22] 150