# Physical Sciences Grade 12 Paper 2 Free State June 2025 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Free State June Exam 2025, 13 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Free State June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 13

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## About this document

Physical Sciences · Grade 12 · Free State June Exam · 2025. Memorandum, 13 pages. Read online or download the PDF.

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### Page 1

education

Department of
Education

FREE STATE PROVINCE

GRADE 12 / GRAAD 12

PHYSICAL SCIENCES /
FISIESE WETENSKAPPE

JUNE / JUNIE 2025

MARKS / PUNTE: 150

MARKING GUIDELINES / MERKRIGLYNE.

This MEMO consists of 12 pages. / Hierdie MEMO bestaan uit 12 bladsye.

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### Page 2

QUESTION /VRAAG 1

1414. Cvv
12 Avv
13° Cvv
14 Dvv
15 Bvv
16 Dvv
17 Bvv
18 Dvv
19 Bvv
1.10 Dvv

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FS June/VS Junie 2025

(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)

(2)
[20]

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QUESTION 2/ VRAAG 2

2.1.1 Organic compounds that consist of hydrogen and carbon atoms only YY (2)
Organiese verbindings wat slegs uit waterstof- en koolstofatome bestaan

2.1.2 CaHoneo ¥ (1)

2.1.3. 2,2,5-trimethylheptane
2,2,5-trimetielheptaan

Criteria/ Kriteria Marks/ Punte
Heptane/heptaan v
trimethyl/trimetiel v
Correct numbers, commas and hyphens/ vi
Korrekte getalle, kommas en koppeltekens (3)
2.2.1 Aldehyde ~ (1)
Aldehied
2.2.2 CH3COCHs Yv/ song io, (2)
Criteria/ Kriteria Marks/Punte
Functional group/ Funksionele groep v
Whole structure correct/ Hele struktuur korrek v
2.2.3. Propanone vv accept: Propan-2-one / 2-Propanone (2/0) (2)

Propanoon —aanvaar: Propaan-2-00n / 2-Propanoon

2.3.1 Simplest whole number ratio in which atoms combine ina compoundYvY (2)
Eenvoudigste heelgetalverhouding waarin atome in ‘n verbinding verbind

232
Mol offvan C [ Mol offvan H | _ Mol of/van O
i
5882, . 9.81, ; 31.37 |
12 1 16
4,90 : 9,81 : 1,96
2,5 t 5 i 1
x2v C 5 : 10 ; 2
| Molecular formula/Molekulére formule : CsHi9O2
(5)
2.3.3. POSITIVE MARKING FROM QUESTION 2.3.2/POSITIEF NASIEN VANAF VRAAG 2.3.2
C2HsOH ¥ v / C2HeO (2)
[20]

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QUESTION 3/ VRAAG 3

3.1. The pressure exerted by a vapour at equilibrium with its liquid in closed
system.v ¥ (2)
Die druk wat deur 'n damp in ewewig met sy vioeistof in 'n geslote
sisteem uitgeoefen word

3.2 Criteria/ Kriteria Marks/
Punte
Dependent and independent variables correctly identified v

(intermolecular force/functional group & vapour pressure)/
Afhanklike en onafhanklike veranderlikes korrek geidentifiseer
(intermolekulére_krag/funksionele groep en dampdruk)

There is a relationship between independent and dependent v
variable/ Daar is 'n verband tussen onafhanklike en afhanklike
veranderlike

If:
e No question mark indicated at the end of the investigative question,
then penalise ONE mark. Max. 1/2
e Temperature is identified as one of the variables, then both marks
are forfeited. Max. 0/2
Indien:
e Geen vraagteken word aan die einde van die ondersoekende vraag
aangedui nie, dan word EEN punt gepenaliseer.
e Temperatuur word as een van die veranderlikes geidentifiseer, dan
word beide punte verbeur.

What is the effect of changing intermolecular force on vapour pressure? V ¥
Wat is die effek van die verandering van intermolekulére krag op

dampdruk?

OR
What is the relationship between the intermolecular force and vapour
pressure? (2)

Wat is die verband tussen die intermolekulére krag en dampdruk?
3.3. 68°C vv (Range/ Variansie (67°C - 69°C) (2)

3.4 Yv; because the vapour pressure of W is higher than that of Y.v v
/boiling point of W is lower than the boiling point of Y.v”
OR
vapour pressure of Y is lower than that of W./ boiling point
of Y is higher than the boiling point of W.
Y ; Die dampdruk van Wis hoér as die van Y. / dus is die kookpunt van
W laer as die van Y.

OF
Die dampdruk van Y is laer as die van W. / dus is die kookpunt van Y
hoér as die van W. (3)
3.5 Carboxyl (group)v ¥ (2)
Karboksiel (groep)

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X/aldehyde (in addition to London forces dispersion forces) has
dipole-dipole forcesY¥

Y/alcohol (in addition to London forces dispersion forces,
dipole-dipole forces has one site) of hydrogen bondingY
Hydrogen bonding in Y/alcohol are stronger than dipole-dipole
forces in X/aldehyde ¥

OR

Intermolecular forces in Y/alcohol are stronger than the
intermolecular forces in X/aldehyde

More energy is required to overcome the intermolecular force in
Y/alcohol than in X/aldehydes¥

X/aldehied (benewens die dispersiekragte van Londense kragte) het
dipoo!-dipoot kragte
Y/alkoho! (benewens Londense kragte se verspreidingskragte,
dipoo!-dipoolkragte het een plek) van waterstofbinding
Waterstofbindings in Y/alkohol is sterker as dipool-dipoolbindings
kragte in X/aldehied
OF
Intermolekulére kragte in Y/alkohol is sterker as die
intermolekulére kragte in X/aldehied
Meer energie word benodig om die intermolekulére krag in Y/alkohol as
in X/aldehiede te oorkom.

OR/OF
X/aldehyde in addition to London forces/dispersion forces has
dipole-dipole forcesY
Y/alcohol (in addition to London forces/dispersion forces/dipole-dipole
forces has one site) of hydrogen bondingY
X/aldehyde has dipole-dipole forces which are weaker than hydrogen
bonding in Y/alcohol¥
OR
Intermolecular forces in Y/alcohol are stronger than the intermolecular
forces in X/aldehyde
More energy is required to overcome the intermolecular force in
Y/alcohol than in X/aldehydesY

X/aldehied benewens Londense kragte/dispersiekragte het
dipool-dipool kragte “

Y/alkohol (benewens Londense kragte/dispersiekragte/dipool-
dipoolkragte het een plek) van waterstofbinding ~

X/aldehied het dipool-dipoolkragte wat swakker is as waterstofbinding
in Y/alkohol ~

OF

Intermolekulére kragte in Y/alkohol is sterker as die intermolekulére
kragte in X/aldehied

Meer energie word benodig om die intermolekulére krag in Y/alkohol
as in X/aldehiede te oorkom. “ (4)

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BA
Criteria/ Kriteria Marks/Punte
| Functional group/ Funksionele groep v
Whole structure correct/ Hele struktuur korrek v |
v
H\ 9\H (2)
nd bw
Hy a
H—C—H
i

3.7.2. Tertiary alcohol”, the carbon atom bonded to the hydroxyl group
is bonded to three other carbon atomsY (2)
Tersiére alkohol, die koolstofatoom gebind aan die hidroksielgroep
is gebind aan drie ander koolstofatome
[19]

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QUESTION 4/ VRAAG 4

4.1.1 Saturated v

It contains carbon-to-carbon single bonds only.” (2)
Versadig
Dit bevat slegs koolstof-tot-koolstofbindings, enkelbindings.
4.1.2 Hydrogen bromide “ Waterstofbromied (1)
Li Ltt
rr oe a ee on i iis cin
head hon a
CRITERIA! KRITERIA “TY “Marks/—
Punte

Correct structure of organic reactant/ Korrekte struktuur van | ¥
organiese reaktant
Correct functional group of organic product/ Korrekte v
funksionele groep van organiese produk
Whole structure organic product is correct / Hele struktuur v
organiese produk is korrek
Correct structure of the inorganic product/ Korrekte struktuur | ¥
van die anorganiese produk

Accept HBr/ Aanvaar HBr

If condensed structures are used/ Indien gekondenseerde Ya

strukture gebruik word max/maks

(4)

4.2.1 Combustion/Oxidation (reaction) “ Verbranding/Oksidasie (reaksie) (1)
4.2.2 CsHi2 + 802 ¥— 5CO2 + 6H20 Y (bal) ¥ (3)
4.3.1 High pressure Y Hoé druk (1)
4.3.2 CHsCH2CH2CH2CH2CH3 ¥Y > 3CH2CH2Y + H2Y (3)

(Give 2/3 if structural formulae are used/Gee 2/3 indien strukturele
formules gebruik word)

4.3.3. Addition/Hydrogenation “/ Addisie/Hidrogenering

CaHe ¥ (2)
4.4.1 Elimination/dehydrohalogenation v’/ Eliminasie/dehidrohalogenering (1)
4.4.2 Concentrated “/ Gekonsentreerd (1)
443 CRITERIA KRITERIA Marks/
A. H H Punte
Correct functional group/ v
H—C—C=C—C—-H Korrekte funksionele groep
| | I | Whole structure correct/ Hele | ¥
H H HH struktuur korrek (2)
[21]

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QUESTION 5/ VRAAG 5

5.1. Change in concentration of reactants or products per unit time.v “ (2)
Verandering in konsentrasie van reaktante of produkte per tydseenheid.
Accept/Aanvaar
Rate of change in concentration/Tempo van verandering in konsentrasie

5.2 No,” more than one independent variable v (2)
Nee, meer as een onafhanklike veranderlike

5.3.1 Reaction rate v/ Reaksietempo (1)

5.3.2 Concentration (of the acid)”’/ Konsentrasie (van die suur) (1)

Accept/Aanvaar: Mass of Mg/ Massa van Mg

5.4 Cwill have a higher reaction rate v/ C sal 'n hoér reaksietempo hé
e HC2 concentration in experiment A is less than in experiment C ¥
e Less particle available to react in experiment A than in experiment
C per unit volume (according to c = - ).%
e Less particles with correct orientation/sufficient kinetic energy in
experiment A than in experiment C. v
e Less effective collision per unit time in experiment A. Vv

e¢ HCé konsentrasie in eksperiment A is minder as in eksperiment C

e Minder deeltjie beskikbaar om in eksperiment A te reageer as in
eksperiment C per eenheidsvolume (volgens c =)

e Minder deeltjies met korrekte oriéntasie/voldoende kinetiese energie in
eksperiment A as in eksperiment C.

e Minder effektiewe botsing per tydseenheid in eksperiment A.

OR /OF

e HC2 concentration in experiment C is greater than in experiment A.

e More particle available to react in experiment C than in experiment
A per unit volume (according to c = = ).

e More particles with correct orientation/sufficient kinetic energy in
experiment C than in experiment A.

e More effective collision per unit time in experiment C.

e HC konsentrasie in eksperiment C is groter as in eksperiment A.

e Meer deeltjie beskikbaar om in eksperiment C te reageer as in
eksperimen A per eenheidsvolume (volgens c = =).

e Meer deeltjies met korrekte oriéntasie/voldoende kinetiese energie in
eksperiment C as in eksperiment A.

e Meer effektiewe botsing per tydseenheid in eksperiment C. (5)

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5.6
_m
Niv)= iy
sie tt
24
= 0,208 mol
Mol ratio/ Molverhouding
Mg: HC£
1: 2v
N(HC2):eacted/gereageer =0,415 |
n
=v
ey
0,9= ae
"0,5
n(HCe) initiavaanvanklike = 0,45 mol
n (HC£) unreacted/oormaat = 0,45 — 0,415v
= 0,035 mol ¥ (6)
Range/ Reeks: (0,033 — 0,037)
5.6.1 Cav HCtis in excess “/ HC? is in oormaat (2)
5.6.2 Nature of the reacting substance v/ Aard van die reagerende stof (1)
563 BY (1)
5.6.4 Number of moles of Mg reacting with HC? is greater than number of moles
of Ca reacting with HCE vv (2)
Aantal mol Mg wat reageer met HC¢f is groter as die aantal mol van Ca wat
met HC? reageer
5.7 Cav (1)
5.8 Reactivity increases down the group ¥
Reaktiwiteit neem toe afwaarts in die groep (1)
[25]

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QUESTIONG / VRAAG 6

6.1

6.2.1
62:2

6.2.3

6.3.1

6.3.2

When the equilibrium of an isolated system is disturbed, the system will
reinstate the new equilibrium by favouring the reaction that opposes
disturbance. v ¥ (2)
Wanneer die ewewig van 'n geisoleerde sisteem versteur word, sal die
sisteem die nuwe ewewig herstel deur die reaksie wat die versteuring
teenstaan, te bevoordeel.

Increases V/ Verhoog (1)
Increases V/ Verhoog (1)
System will decrease CQz2(g) by favouring the reaction that uses up

CO2(g).v

Forward reaction favoured.v

Thus causes an increase in the yield of CO. (2)

Stelsel sal CO2(g) verminder deur die reaksie wat opgebruik, te bevoordeel.
CO2g).

Voorwaartse reaksie bevoordeel.

Dus veroorsaak dit 'n toename in die opbrengs van CO.

Increases “7 Verhoog
e Anincrease in pressure favours the reaction which produces the
least number of gas moles.”
e The reverse reaction is favoured. v (3)

e 'n Toename in druk bevoordeel die reaksie wat die minste aantal
gasmol.
e Die omgekeerde reaksie word bevoordeel.

No effect” Temperature is the only factor affecting Kc value (1)
Geen effek nie, Temperatuur is die enigste faktor wat die Kc-waarde
beinvioed

[10]

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QUESTION 7

7.1.1. Asystem that is isolated to its surroundings/ A system where substances
cannot leave/escape the container.v “ (2)
'n Sisteem wat van sy omgewing geisoleer is/'n Sisteem waar stowwe
nie die houer kan verlaat/ontsnap nie.

7.1.2 Rate of the forward reaction is equal to the rate of the reverse reaction.v Y (2)
Die tempo van die voorwaartse reaksie is gelyk aan die tempo van die
terugwaartse reaksie

7.2 Chemical equilibrium.“ Chemiese ewewig. (1)
7.3
Using moles

[Y2] =

Marking criteria

Gebruik van molle
Merkkriteria

0,2197 Vv=

Divide the equilibrium moles by 2 dmv
Correct Ke expression.”

Substitution of Ke of 0,2197 value correctly.”
The correct value of X2 concentration.
Correct substitution of values on the numerator.

Deel die ewewigsmolle deur 2 dm?

Korrekte Ke uitdrukking.

Korrekte vervanging van Ke van 0,2197 waarde.
Die korrekte waarde van X2 konsentrasie.
Korrekte vervanging van waardes op die teller.

[X2][¥2]*

(0,25)? “
10.411Y2]5 ¥

0,89 mol-dm? ¥

(5)

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7.4 POSITIVE MARKING FROM QUESTION 7.3/POSITIEF NASIEN VANAF VRAAG 7.3.

Marking criteria:
* n(¥2) equilibrium x 2 dm*. ¥

moles of Y2.¥

e Formula n(Y2)= mY

e Final answer.v
Merkkriteria:
© n(¥2) ewewig x 2dmo.v

om gereageerde mol Y2 te vind.v
« Aanvanklike molle(n(Y¥2) ewewig
+ n(¥2)verander/gereageer).”
« Formulen (Y¥2)= mY

« Finale antwoord. ¥

* Correct use of mole ratio between Y2: XY3 - 3:2 to find reacted

e Initial moles (n(Y2) equilibrium + n(Y2)change/reacted).v

¢ Correct substitution into n(Y2)initial= av

* Korrekte gebruik van molverhouding tussenY2:XY3- 3:2

*  Korrekte vervanging in n(Y2)initial= am”

Xe Y2 XY3
Initial/Aanvanklik (mol) 1,05 2,53V 0
ee
Change/Verandering (mol) 0,25 0,75 0,5”
Equilibrium/Ewewig (mol) 0.8 1,784 0,5
Equilibrium concentration/ 0,4 0,89 0,25
Ewewigskonsentrasie
c == ( mol-drv’)
Vv
NY )* iv
2 sie
we 24

V = 56,67 dm’ v
Range/ Reeks: (56,5 — 56,8)

7.5 Increase in temperature.”

(6)

e Increase in temperature favours the endothermic reactionY

Reverse reaction is favoured.”

e The numbers of mole XY3 decreases and the number of moles of X2and

Y2 increases. ¥
Toename in temperatuur.

(4)

e Toename in temperatuur bevoordeel die endotermiese reaksie.

e Omgekeerde reaksie word bevoordeel.

e Die aantal mol XY3 neem af en die aantal mol X2 en Y2 neem toe.

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[20]

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QUESTION 8/ VRAAG 8

8.1.1 Activation energy v/Aktiveringenergie (1)
8.1.2 Shifts to the leftv
It provides an alternative pathway of lower activation energyY (2)
Skuif na links
Dit bied 'n alternatiewe pad van laer aktiveringsenergie
8.1.3 (a) (curve/kromme) 3 ¥ (1)
(b) = (curve/kromme) 1 ~ (1)

8.2.1 Exothermic v, AH is less than zero/energy of the reactants is more than
the energy of the products “ (2)
Eksotermies, 4H is minder as nul/energie van die reaktante is meer as
die energie van die produkte

8.2.2 Remains the same “/ Bly dieselfde (1)

8.2.3 AH =Hp—Hr
—150 ¥ = Hp — 100 ¥

Hp = -50 kJ.mol!¥ (3)
8.2.4 Eac=Eat+ Hr
=90 + 100 ¥
= 190 kJ.mot'” (2)
8.2.5 f
2 aa
BP iy %
Po Bg \
2 = |Reactants//’ i;
UW @ 2 |Reaktante \
Za2 roducts/ Produkte
on =
22o~
oo
ag .
Course of reaction/
Verloop van reaksie
CRITERIA /KRITERIA
Energy of products and reactants stays the same/ vi
Energie van produkte en reaktante bly dieselfde
Lower activation energy for the catalysed reaction/ v (2)
_Laer aktiveringsenergie vir die gekataliseerde reaksie

[15]
GRAND TOTAL GROOTTOTAAL: 150

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