S1 CHEMISTRY

Name : ………………………………………………… Class : ………..…….…..

I.C Number : ……………………………........…………

CHEMISTRY 962
SEMESTE PAPER 1
NOVEMBER 2020
One And a Half Hours

KOLEJ TINGKATAN ENAM SHAH ALAM
JALAN TIMUN 24/1

SHAH ALAM SELANGOR DARUL EHSAN

PROGRAM PENINGKATAN PRESTASI SEMASA PKPB 2020
(Online)

CHEMISTRY 962/1 (MARKING SCHEME)

PAPER 1

DURATION: 1 HOUR 30 MINUTES

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE For examiner’s use
TOLD TO DO SO Section A

Write your name, I.C number and class in the spaces 1 - 15
provided. Section B

There are fifteen questions in Section A. For each 16
question, four choices of answers are given. Choose 17
one correct answer and indicate it on the Multiple-
choice Answer sheet provided. Answer all questions. Section C

Answer all the questions in Section B. Write your TOTAL
answers in the spaces provided.

Answer two out of the 3 questions in Section C. Write
your answer on your examination test pad. All
working should be shown. Numerical answers should
be given to an appropriate number of significant
figures; units should be quoted wherever appropriate.
Begin each answer on a fresh sheet of paper and
arrange your answers in numerical order.

This question paper consists of 8 printed pages

Prepared by: Checked by: Validated by:

………………………………...… …………………………………… ………………………….…….….
Pn Marin Bt Abd Ghani Pn Maziah Binti Muhamad Zain Puan Shirin Ahmad Sapiuddin
Head of Chemistry Unit Science and Mathematics Form 6 Senior Assistant
Head of Department

1

NAME:…………………………………………………
I.C NUMBER:…………………………………
INDEX NUMBER:………………………………………. SCORE:

962/1

KOLEJ TINGKATAN ENAM SHAH ALAM

PROGRAM PENINGKATAN PRESTASI SEMASA PKPB 2020
CHEMISTRY
PAPER 1

One hour 30 minutes

SECTION A

1. =A= =B= =C= =D=
2. =A= =B= =C= =D=
3. =A= =B= =C= =D=
4. =A= =B= =C= =D=
5. =A= =B= =C= =D=
6. =A= =B= =C= =D=
7. =A= =B= =C= =D=
8. =A= =B= =C= =D=
9. =A= =B= =C= =D=
10. =A= =B= =C= =D=
11. =A= =B= =C= =D=
12. =A= =B= =C= =D=
13. =A= =B= =C= =D=
14. =A= =B= =C= =D=
15. =A= =B= =C= =D=

2

Marking Scheme (PROGRAM PENINGKATAN PRESTASI SEMASA PKPB 2020 15
) B

Section A
Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Ans. C C D C D A B B B A D D B B

16. (a) .(a) i. PV = nRT @ MR = dRT / P [1]
Mr = 0.163 x 8.31 x 298 / 101x 103 x 1.0 x 10-3 [1]
Mr = 4.00 [1]
ii. gas M is helium [1]
it behave almost like an ideal gas under room temperature and pressure [1]

17.a) Pressure exerted by each individual gas (or any same meaning) [1],[1]
[1]
b) = ( 3)2
( 2)2 ( 2) [1]

c) = (0.364)2 [1]
(0.456)2(0.180)
[1] (ins. s.f. & unit)
Kp = 3.54 atm-1 [1]
[1]
d) i. - total mol of gas at left is more than right [1]
[1]
- P decrease, equilibrium shift to left

ii. -partial pressure of all gas/mixture is reduce @ volume in container increase

- Partial pressure decrease, equilibrium shift to left

19. [1]
(a) i) Dalton’s law states that for a mixture of non-reacting gases at a constant temperature,
the total pressure exerted is the sum of the partial pressures of the gases in the mixture.

3

ii) partial pressure of nitrogen [1]
=58.5-25.8 [1]
=32.7kPa
pV=nRT [1]
n= (32.7 x103)(2.50x10-3)
(8.31)(320) [1]……Total 4m
= 3.07 x 10-2mol
Mass of nitrogen [1]
=3.07x 10-2 x 28 [1]……Total 2m
= 0.86g

iii) using Boyle’s law:
P1V1= P2V2
(58.5)(2.50)=P2(1.50)
P2=97.5kPa

(b)

ii) Since solid helium and liquid helium have the same density, this means that there is [3m]
no change in volume during the phase change between solid and liquid helium. [2m]
Thus ,pressure will not affect melting point of helium.

(c )

20 (a) The rate law shows the mathematical relationship between the rate of a reaction [3m]
and its rate constants and concentration of the reactants. [1]
Rate = k[A]x [1]

Where k = Rate constant, x = Order of reaction. A is a reactant [1]
[1]
(b) (i) 2NO + O2 → 2NO2 [1]
(ii) Let the rate equation be as follows. Rate = k[NO]x [O2]y [1]
For experiment 1:
6.62 x 10-4 = k[0.800]x[0.160]y [1]

For experiment 2:
4.14 x 10-5 = k[0.200]x[0.160]y

For experiment 3:
1.33 x 10-3 = k[0.800]x[0.320]y

Comparing experiment 1 and experiment 2 ;

6.62 x 10-4 = k[0.800]x[0.160]y
4.14 x 10-5 k[0.200]x[0.160]y

16 = 4x x= 2

Comparing experiment 1 and experiment 3 ;
6.62 x 10-4 = k[0.800]x[0.160]y
1.33 x 10-3 k[0.800]x[0.320]y

4

0.5 = 0.5y y=1 [1]
[1]
The rate equation is as follows. Rate = k[NO]2 [O2]1
[1]
Using experiment 1:
6.62 x 10-4 = k(0.800)2(0.160) [1]
k = 6.47 x 10-3 mol-2 dm6 [1]

(iii) A change in concentration does not affect the rate constant as long as there is [1]
no change in the temperature (or catalyst). [1]
[1]
Let the original rate be w mol dm-3 s-1 ; w = k[NO]2 [O2]1

Under the new conditions:
New rate = k[3NO]2 [1/8 O2]1
= 9/8k[NO]2 [O2]1
= 1.125 w

Therefore, the rate increases 1.125 times

5