# Physical Sciences Grade 12 P2 KZN June 2023 Paper & Memo

> Grade 12 Physical Sciences Paper 2 question paper and memo from the KZN June Exam 2023, 27 pages. Read online or get the free PDF download.

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- Updated: 2026-09-28
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 27

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Physical Sciences · Grade 12 · KZN June Exam · 2023. Question paper and memorandum, 27 pages. Read online or download the PDF.

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### Page 1

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SR SEERERERHEHKZKRREREE

@

a PHYSICAL SCIE MISTRY)

.

rrrrrrrririitt ty Seessesesesssee%
MARKS : 150
TIME : 3Hours

This question paper consists of 14 pages and 2 data sheets.

Seeseseasaaes

Peeseesseeee”

### Page 2

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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.

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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)

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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)

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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)

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June Common Test

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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.

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(2)
[20]

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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.

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(3)
(3)

(1)

(1)

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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]

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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)

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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.

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(3)

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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]

|=)

|

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June Common Test

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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)

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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]

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QUESTION 7 (Start on a new page.)

7.1

7.2

7.3

74

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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]

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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

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MARKS: 150

This. memorandum consists of 11 page

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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)

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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]

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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]

### Page 21

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

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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)

### Page 22

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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)

### Page 23

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]

### Page 24

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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

### Page 25

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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]

### Page 26

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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)

### Page 27

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

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