Physical Sciences Grade 12 P2 Eastern Cape June 2025 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2025 PHYSICAL SCIENCES: CHEMISTRY P2/ FISIESE WETENSKAPPE: CHEMIE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye.

2 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) QUESTION/VRAAG 1 1.1 B (2) 1.2 D (2) 1.3 D (2) 1.4 B (2) 1.5 C (2) 1.6 C (2) 1.7 B (2) 1.8 D (2) 1.9 C (2) 1.10 B (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 3 QUESTION/VRAAG 2 2.1 A bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds. Binding of atoom of groep atome wat die fisiese en chemiese eienskappe van groep organiese verbindings bepaal. (2) 2.2 2.2.1 A (1) 2.2.2 C (1) 2.2.3 B (1) 2.3 2.3.1 3-methylbutanal Marking criteria/Nasienkriteria: 3-metielbutanaal Butanal / butanaal Whole name correct / Hele naam korrek (2) 2.3.2 4-methylpent-2-yne Marking criteria/Nasienkriteria: 4-metielpent-2-yn Pent-2-yne / pent-2-yn Whole name correct / Hele naam korrek OR/OF 4-methyl-2-pentyne 4-metiel-2-pentyn (2) 2.3.3 2,3-dibromo-4-ethylhexane 2,3-dibromo-4-etielheksaan Marking criteria/Nasienkriteria: Hexane / heksaan dibromo and ethyl / dibromo en etiel Whole name correct / Hele naam korrek (3) 2.4 2.4.1 Markingcriteria H H O H H /Nasienkriteria Correct functional group / H C C C C C H Funksionele groep korrek 1/2 Whole structure correct/ H H H Hele struktuur korrek 2/2 H C H H (2) 2.4.2 Esterification / Condensation / Esterifikasie / Kondensasie (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) 2.4.3 Marking criteria/Nasienkriteria H H H O Correct functional group / Funksionele groep korrek H C C C C O H 1/2 Whole structure correct/ Hele H H H struktuur korrek 2/2 (2) 2.5 2.5.1 M 43,24 = x 100 74 M = 31,9976 M = 32 g·mol-1 M(O) = 16 g·mol-1 CxHyO2 Organic compound is either carboxylic acid or ester. Organiese verbinding is karboksiesuur of ester. General formula / Algemene formule: CnH2nO2 12 x n + 2 x n + 32 = 74 n=3 C3H6O2 (3) 2.5.2 H O H O H H H C C O C H OR/ H C O C C H OF H H H H AND/EN H H O H C C C O H H H Marking criteria/Nasienkriteria Correct functional group / Funksionele groep korrek 1/2 Whole structure correct/ Hele struktuur korrek 2/2 (4) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 5 QUESTION/VRAAG 3 Marking criteria/ Nasienkriteria If any of the underlined key words/ phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/ frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase. 3.1 Boiling point is the temperature at which the vapour pressure of a substance /liquid equal the atmospheric pressure Kookpunt is die temperatuur waarteen die dampdruk van stof/vloeistof gelyk aan die atmosferiesedruk is (2) 3.2 Same functional group / Same homologous series Dieselfde funksionele groep / Dieselfde homoloëreeks Accept same type of intermolecular forces/Aanvaar dieselfde tipe van intermolekulêre kragte (1) 3.3 Marking criteria The boiling point increases down the table Chain length / molecular size increases down the table Increase in the strength of the London forces/Dispersion forces/Induced dipole forces / down the table Relate the strength of London forces /dispersion forces/induced dipole to energy involved Nasienkriteria Die kookpunt styg teen die tafel af Kettinglengte / molekulêre grootte neem teen die tabel af Toename in die sterkte van die Londonkragte /Dispersiekragte /Geïnduseerde dipoolkragte / teen die tabel af Verband hou met die sterkte van die Londonkragte/ verspreidingskrag/ kragte met die energie betrokke The trend shows an increase in the boiling point down the table Increase in the surface area/chain length/molecular size down the table Strength of London forces/dispersion forces/induced dipole forces / increases More energy is needed to overcome intermolecular forces Die tendens toon toename in die kookpunt teen die tabel af Toename in die oppervlakte/ketting lengte/molekulêre grootte teen die tabel af Sterkte van die Londonkragte/v dipoolkragte neem toe Meer energie word benodig om die intermolekulêre kragte te oorkom (4) 3.4 Decreases. The boiling point increases. Afneem. Die kookpunt neem toe. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) 3.5 2-methylpropan-1-ol / 2-methyl-1-propanol Marking criteria/Nasienkriteria: 2-metielpropan-1-ol/ 2-metiel-1-propanol Propan-1-ol Whole name correct / Hele naam korrek (2) 3.6 Marking criteria Compare the boiling point of the alcohol to the alcohol with 3-carbon atom alcohol Compare the surface area/chain length/molecular size of the alcohol to the alcohol with 3-carbon atoms Compare the boiling point of the alcohol to the alcohol with 4-carbon atoms Compare the surface area/chain length of the alcohol to the alcohol with 4-carbon atoms Nasienkriteria Vergelyk die kookpunt van die alkohol met die 3-koolstof atoom alkohol Vergelyk die oppervlakte/kettinglengte/molekulêre grootte van die alkohol met die 3-koolstof atoom alkohol Vergelyk die kookpunt van die alkohol met die 4-koolstof atoom alkohol Vergelyk die oppervlakte/kettinglengte van die alkohol met die 4-koolstof atoom alkohol The boiling point of the unknown alcohol (2-methylpropan-1-ol) is higher than that of the alcohol with 3-carbon atoms (propan-1-ol) Surface area / chain length / molecular size is greater than the alcohol with 3-carbon atoms (propan-1-ol) The boiling point of the unknown alcohol (2-methylpropan-1-ol) is smaller than that of the straight chain alcohol with 4-carbon atoms (butan-1-ol) Surface area / chain length is smaller than the straight chain alcohol with 4-carbon atoms (butan-1-ol) Die kookpunt van die onbekende alkohol (2-metielpropan-1-ol) is hoër as dié alkohol met 3-koolstofatome (propan-1-ol) Oppervlakte / kettinglengte / molekulêre grootte is grootter as die alkohol met 3-koolstof atome (propan-1-ol) Die kookpunt van die onbekende alkohol (2-metielpropan-1-ol) is laer as die reguitketting alkohol met 4-koolstofatome (butan-1-ol) Oppervlakte / kettinglengte is kleiner as die reguitketting alkohol met 4-koolstof atome (butan-1-ol) (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 7 QUESTION/VRAAG 4 4.1 4.1.1 Breaking down of long chain hydrocarbon molecules into more useful shorter chains (2 or 0) Die afbreek van lang kettingkoolwaterstof-molekules in meer bruikbare, molekule korter kettings (2 of 0) (2) 4.1.2 C2H4 (2) 4.1.3 Used as a fuel / Word as brandstof gebruik (1) 4.1.4 C5H12 + 8 O2 5 CO2 + 6 H2O ( bal.) Reactants, products and balancing Reaktanse, produkte en balansering (3) 4.2 4.2.1 2-bromopentane / 2-bromopentaan (2) 4.2.2 Substitution/ Hydrolysis of haloalkanes / Substitusie / Hydrolise van haloalkane (1) 4.2.3 Diluted strong base / diluted NaOH Verdunde sterk basis / verdunde NaOH (1) 4.2.4 H Marking criteria/Nasienkriteria Correct functional group / Funksionele H H H O H groep korrek 1/2 Whole structure correct/ Hele struktuur H C C C C C H korrek 2/2 H H H H H (2) 4.2.5 Dehydration / Dehidrasie / Dihidratering (1) Marking criteria/Nasienkriteria 4.2.6 H H H H H Correct functional group / Funksionele groep korrek 1/2 H C C C C C H Whole structure correct/ Hele struktuur korrek 2/2 H H H (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) QUESTION/VRAAG 5 5.1 EXOTHERMIC / EKSOTERMIES. 0 (2) 5.2 Carbon dioxide (CO2) gas escapes / Koolstof dioksied (CO2) gas ontsnap (2) 5.3 189 g Accept/Anvaar (188,5 189) (1) 5.4 5.4.1 11 (minutes / minute) accept/ aanvaar 10 (minutes / minute) (1) 5.4.2 EQUAL TO The amount of CaCO3 consumed is equal to the amount of CO2 produced / the mol ratio of CaCO 3 to CO2 is 1 : 1. The time taken for consumption and production is the same. GELYK AAN Die hoeveelheid CaCO3 verbruik is gelyk aan die hoeveelheid CO2 vervaardig / die molverhouding van CaCO3 tot CO2 is 1 : 1. Die tyd geneem vir die verbuik en vervaardiging is dieselfde. (3) 5.4.3 Marking criteria / Nasienkriteria (a) Subtract / Aftrek 200 185 (b) Subst. into / Vervanging in n = m/M (c) Formula / Formule V = nVm (d) Subst. into/ Vervanging in V = nVm (e) Final answer/ Finale antwoord M n = m 200 185 (a) n = (b) 44 n = 0,34 mol V = nVm (c) V = (0,34)(24000) (d) V = 8 160 (cm3) (e) Range = 8160 8181,82 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 9 5.4.4 Marking criteria/Nasienkriteria (a) Use of mol ratio / Gebruik van molverhouding CO2 : CaCO3 (b) Subst. into/ Vervanging in n = m/M (c) Subst. into rate equation/ Vervanging in tempo vergelyking (d) Final answer / Finale antwoord Positive marking from/Positiewe nasien van 5.4.3 n (CO2) = n (CaCO3) = 0,34 mol (a) m = nM m = (0,34)(100) (b) m = 34 g rate/ = tempo rate/ 0 34 = (c) tempo 11 m = 3,09 g·min-1 (d) (4) 5.5 5.5.1 Marking criteria for hypothesis Statement regarding the correct independent and dependent variable Statement is measurable Nasienkriteria vir hipotese Stelling om die onafhanklike en afhanklike veranderlike is korrek Stelling is meetbaar A(n) decrease / increase in temperature will decrease / increase the reaction rate. Afname/ toename in temperatuur sal die reaksietempo laat afneem/ toeneem. (2) 5.5.2 INCREASES / TOENEEM (1) 5.5.3 Higher temperature means the average kinetic energy of the particles increases More particles will have sufficient kinetic energy / more particles will have kinetic energy equal to or higher than the activation energy More effective collisions per unit time / Frequency of the effective collisions increases Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) Hoër temperatuur beteken die gemiddelde kinetiese energie van die deeltjies neem toe Meer deeltjies het genoeg kinetiese energie / meer deeltjies het kinetiese energie gelyk of hoër as die aktiveringsenergie Meer effektiewe botsings per tydseenheid / Frekwensie van die effektiewe botsings neem toe (3) [24] QUESTION/VRAAG 6 6.1.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Indien enige van die sleutelwoorde/frases in die korrekte konteks weggelaat word: - 1 punt per woord/frase. When the equilibrium in a closed system is disturbed, the system will re- instate a new equilibrium by favouring the reaction that will oppose/cancel the disturbance. Wanneer die ewewig in geslote sisteem versteur word, sal die sisteem nuwe ewewig deur die reaksie wat die versteuring teenwerk/kanseleer, te bevoordeel. (2) 6.1.2 (a) NO EFFECT / GEEN EFFEK (1) (b) INCREASES / TOENEEM (1) (c) DECREASES / AFNEEM An increase in pressure favours the reaction that produces the least number of gaseous moles. Mole ratio of reactants : products 1 : 2 The reverse reaction is favoured / the equilibrium position shifted towards left. Toename in druk bevoordeel die reaksie wat die minste gasmol produseer. Die molverhouding reaktanse : produkte 1 : 2 Die terugwaartse reaksie word bevoordeel / die ewewigsposisie verskuif links. (4) 6.2 6.2.1 2 NO2 (g) N2O4 (g) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 11 Marking criteria/ 6.2.2 Nasienkriteria Both graphs rates increases vertically and reached new equilibrium Increase in the reverse reaction rate is higher than the forward rate Beide grafieke se tempo to t1 t2 time/tyd (min) neem vertikaal toe en bereik nuwe ewewig Toename in die tempo van die terugwaartse reaksie is hoër as die tempo van die voortwaartse reaksie (2) 6.3 6.3.1 [NO2]2 Kc = [N2O4] (0,0457)2 Kc = (0,448) Kc = 4,66 x 10-3 (3) 6.3.2 LOW YIELD. Kc is less than 1. LAE OPBRENGS. Kc is kleiner as 1. (2) 6.3.3 POSITIVE MARKING FROM / POSITIEWE NASIEN VANAF 6.3.1 MARKING CRITERIA a. Subst. Kc value and correct N2O4 concentration into the correct Kc expression b. Determine the change in concentration of NO2 c. Correct ratio N2O4 and NO2 d. Determine the initial concentration N2O4 e. Subst. change in concentration and initial concentration of N2O4 in % change formula f. Final answer NASIENKRITERIA a. Vervang die Kc-waarde en korrekte N2O4 konsentrasie in die korrekte Kc-uitdrukking b. Bepaal die verandering in die konsentrasie van NO2 c. Korrekte verhouding tussen N2O4 en NO2 d. Bepaal die aanvanklike konsentrasie van N2O4 e. Vervanging in verandering in konsentrasie en die aanvanklike konsentrasie van N2O4 in die % verandering formule f. Finale antwoord Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) [NO2]2 Kc = [N2O4] [NO2]2 4,66 x 10-3 = (a) (0,491) [NO2] = 0,0478 mol·dm-3 N2O4 2 NO2 (g) Initial conc/ 0,5 (d) 0,03 Aanvangskonsentrasie Change in conc./ (c) ratio/ Verandering in 8,9 x 10-3 0,0178 (b) verhouding konsentrasie Equilibrium conc./ 0,491 0,0478 Ewewigskonsentrasie 8,9 x 10-3 % [N2O4] = X 100 % (e) 0,5 % [N2O4]= 1,78 % (f) (6) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 13 QUESTION/VRAAG 7 7.1 7.1.1 An acid is a proton (H+) donor (2 or 0) Suur is proton (H+) skenker (2 of 0) (2) 7.1.2 CH3COOH + H2O CH3COO + H3O+ Ignore phases / Ignoreer fases (2) 7.1.3 Weak acid / Swaksuur Ka value is less than 1 / Ka-waarde is kleiner as 1 (2) 7.1.4 Methanoic acid It has the larger Ka-value / Ionises to a greater extend / It is the stronger acid. The higher the concentration of the H3O+ the lower the pH-value. Metanoësuur Dit is het die groter Ka-waarde / Ioniseer tot groter mate / Dit is die sterker suur. Hoe hoër die konsentrasie van H3O+ hoe kleiner is is die pH-waarde (3) 7.2 7.2.1 Burette / Buret (1) 7.2.2 To ensure reliable results / accurate results Om betroubare resultate / akkurate resultate te verseker (1) 7.2.3 8,3 10 (1) 7.2.4 C2O42 + H2O HC2O4 + OH (reactants / reaktanse and / en products / produkte) Excess OH is produced / Oormaat OH is geproduseer (3) 7.3 7.3.1 n = cV n = (0,1)(25 x 10-3) n = 2,5 x 10-3 mol (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 (EC/JUNE/JUNIE 2025) 7.3.2 OPTION 1 OPTION 2 Marking criteria Marking criteria (a) pH formula (a) Formula pH + pOH = 14 (b) Subst. into pH-formula (b) Subst. into pH + pOH =14 (c) Subst. into Kw = [H3O+][OH ] (c) Subst. into pOH formula (d) Final answer (d) Final answer OPSIE 1 OPSIE 2 Nasienkriteria Nasienkriteria (a) pH-formule (a) Formule pH +pOH = 14 (b) Vervang in pH-formule (b) Vervang in pH + pOH =14 (c) Vervang in Kw = [H3O+][OH ] (c) Vervang in pOH-formule (d) Finale antwoord (d) Finale antwoord pH = - log[H3O+] (a) pH + pOH = 14 (a) 12,52 = -log[H3O+] (b) 12,52 + pOH = 14 (b) [H3O+] = 3,02 x 10-13 mol·dm-3 pOH = 1,48 Kw = [H3O+][OH ] pOH = -log[OH ] 1 x 10-14 = (3,02 x 10-13)[OH ] (c) 1,48 = -log[OH ] (c) [OH ] = 0,033 mol·dm-3 (d) [OH ] = 0,033 mol·dm-3 (d) (4) 7.3.3 POSITIVE MARKING FROM QUESTION 7.3.1 AND 7.3.2 POSITIEWE NASIEN VAN VRAAG 7.3.1 EN 7.3.2 Marking criteria (a) Subst. 0,033 and V into n = cV (b) Subst. 0,033V, 0,1V and n equivalence point in nexcess = nTotaal nep (c) Subst. into V (total volume) = V (volume titrated) + V (acid) (d) Subst. eq (2) into (1) (e) Final answer Nasienkritria (a) Vervang 0,033 en V in n = cV (b) Vervang 0,033V , 0,1V en ekwivalente punt in noormaat = ntotaal - nekw (c) Vervang in V (totale volume) = V (volume getitreer) + V (suur) (d) Vervang vergelyking (2) in (1) (e) Finale antwoord n (excess) = n (total) n (equivalence point) n (oormaat) = n (totaal) n (ekwivalente punt) cV (total volume) = cV(volume titrated) n (equivalence point) cV (totaal volume) = cV (volume getitreer) n (ekwivalente punt) 0,033V (total volume) (a) = 0,1V (volume titrated) 2,5 x 10-3 (b) V (total volume) = V (volume titrated) + V (acid) V (totale volume) = V(getitreer) + V (suur) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/JUNE/JUNIE 2025) PHYSICAL SCIENCES P2/FISIESE WETENSKAPPE V2 15 V (total volume) = V (volume titrated) + 25 x 10-3 (c) Subst. equation (2) into (1) / Vervang vergelyking (2) in (1) 0,033 [Volume titrated + 25 x 10-3] (d) = 0,1 Volume titrated 2,5 x 10-3 Volume titrated / volume getitreer = 0,050 dm3 Volume titrated / volume getitreer = 50 cm3 (e) (5) [27] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief