Practice Examination for G.C.E A/L Students- 2020 ...

AL/2020/02/T

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Chemistry

Practice Examination for G.C.E A/L Students- 2020

Conducted by Tamil students of Faculty of Engineering

University of Moratuwa

Serial- 03

2020

Marking scheme

CHO

N2O4(g)

jhf;fp

NO2(g)

tpisT

G G

B

A

C .

.

.

CH3−CH=CH−CH−C≡C−CHO

│ CH3

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Answer for MCQ

Q Ans Q Ans Q Ans Q Ans Q Ans

01. 5 02. 4 03. 4 04. 4 05. 4

06. 5 07. 3 08. 3 09. 2 10. 1

Marks Distribution

Part-I=10×2=20 Marks

Part-II=250 Marks

Part-II (Structured essay question) =100 Marks

Part-II (Essay question) =150 Marks

Total; = Part -I + Part –II

= 20 + 𝟐𝟓𝟎 /𝟓𝟎

= 70 Marks = 𝟕𝟎/ 𝟕𝟎 × 𝟏𝟎𝟎

= 100 Marks

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Part-II

Structured essay

(a)

(b)

1.

(05)

2. 4-methyl-2-hexene (04)

3. 4-methyl-3-hexanal \ 4-methylhexan-3-ol (04)

4.

(05)

5. Nucleophilic substitution reaction. (03)

6. Positional isomers ;. (04)

CH3 – CH2 – C – CH3

CH3

OH

C→ H– C – C – CH3

CH3 H

CH3 OH

A→

CH3

CH3 – CH2 – CH – CH2OH B→ (i)

CH3 – CH2 – CH–O –CH3

CH3

E→

CH3

CH3 – CH – CH2CH2OH D→

(7×5=35 Marks)

CH3 – CH – CH2CH2OH

CH3

CH3 – CH – CH=CH2

CH3

OH

CH3 – CH – CHCH3

CH3

CH3 – CH = CH–CH3

CH3

OH

CH3 – C – CH2CH3

CH3

(3)

(4)

(3)

(4)

(3)

(3)

(4)

(ii)

(24 Marks)

CH3

CH3 – CH2 – CH – CH=CHCH3

CH3

CH3 – CH2 – CH – CH2CHBrCH3

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7.

(05)

(05)

8. G, the one which gives immediate turbidity,

C, the one which gives turbidity after some time (06)

Part -II

CH3

CH3 – CH2 – CH = CHCH2CH3 F→

CH3

Br

CH3 – CH2 – CH – CH2CH2CH3 G→

(Total=100 Marks)

PCl5 (g) PCl3(g) + Cl2(g)

initial 0.5𝑚𝑜𝑙 Equilibrium mol; 0.5(1 − 𝛼)𝑚𝑜𝑙 0.5𝛼 0.5𝛼 𝛼-degree of dissociation

𝑃𝑉 = 𝑛𝑅𝑇 (system show ideal behavior)

𝑃 𝛼 𝑛𝑇 (𝑉, 𝑅 constants )

𝑃1 𝛼 0.5 × 𝑇1

𝑃2 𝛼 0.5(1 + 𝛼) × 𝑇2

𝑃1

𝑃2=

𝑇1

(1 + 𝛼)𝑇2

𝑃1𝑇2𝛼 = 𝑃2𝑇1 − 𝑃1𝑇2

𝛼 =𝑇1𝑃2 − 𝑇2𝑃1

𝑇2𝑃1

……….(10)

………(5)

……….(5)

graph I→wrong →concentrations must be[PCl3]=0.6M, [Cl2]=0.6M

Graph II→ wrong →when [Cl2]increases [PCl3] decreases

Graph III→ wrong →[PCl3]has changed by 0.5M but

[PCl5]is not changed by0.5M.

Graph IV→right

……….(5)

……….(5)

……….(5)

……….(5)

5) (a) (i)

(ii)

(40Gs;spfs;)

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KC =[ Y(g)][Z(g))]

[X(g)]=

0.4moldm−3 × 0.4moldm−3

0.2moldm−3

= 0.8moldm−3

PV = nRT ,y; P ∝ 1

V (n, R, T constants)

P → P

2 Mf P → 2V MFk;

C = n

V ⟹ concentration becomes

1

2 times

[X(g)] = 0.2moldm−3, [Y(g)] = [Z(g)] = 0.4moldm−3

QC =[ Y(g)][Z(g))]

[X(g)]=

0.2moldm−3 × 0.2moldm−3

0.1moldm−3

= 0.4moldm−3

KC=0.8moldm-3, QC=0.4moldm-3

here KC > QC

/ reaction takes place in forward direction

……….(3)

……….(2)

……….(5)

……….(5)

……….(3)

……….(2)

b) (i)

(ii)

(iii)

(iv) X (g) Y(g) + Z(g)

Initial concentration 0.1 0.2 0.2 moldm-3

final concentration 𝑥 - - moldm-3

equilibrium concentration 0.1−𝑥 0.2+𝑥 0.2+𝑥 moldm-3

T constant so, KC = 0.8moldm-3

(0.2 + 𝑥)moldm−3 × (0.2 + 𝑥)moldm−3

(0.1 − 𝑥)moldm−3= 0.8moldm−3

(0.2 + 𝑥)2 = (0.1 − 𝑥)0.8

𝑥2 + 1.2𝑥 − 0.04 = 0

𝑥 =−1.2 ± ξ1.44 − 0.16

2

𝑥 = −0.6 ± ξ0.1 × 2

𝑥 = 0.632 − 0.6 = 0.032

/at equlibrium

[X(g)]=0.68moldm-3

[Y(g)]= [Z(g)]=0.232moldm-3

……….(5)

……….(5)

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KP=KC(RT) Δn

Δn=2-1=1

Kp=0.8moldm-3 x 8.314Jmol-1x400K

=2.66X 103 Pa

P→ 2P ⟹ concentration doubles

𝑄𝑐 =0.8moldm−3 × 0.8moldm−3

0.4moldm−3= 1.6moldm−3

QC>KC /reaction takes place in backward direaction

……….(3) ……….(2)

(v)

(vi) X (g) Y(g) + Z(g)

Initial concentration 0.4 0.8 0.8 moldm-3

final concentration 𝑦 𝑦 moldm-3

Equilibrium concentration 0.4+𝑦 0.8−𝑦 0.8−𝑦 moldm-3

𝐾𝑐 =(0.8 − 𝑦)moldm−3 × (0.8 − y)moldm−3

(0.4 + y)moldm−3= 0.8moldm−3

(0.8 − y)2 = (0.4 + y)0.8

𝑦2 − 2.4𝑦 + 0.32 = 0

𝑦 =2.4 ± ξ5.76 − 1.28

2

𝑦 = +1.2 ±1

2ξ4.48

𝑦 = 0.142

/at equilibrium

[X(g)]=0.542moldm-3

[Y(g)]= [Z(g)]=0.658moldm-3

……….(5)

……….(5)

……….(5)

……….(3)

……….(2) (55

Gs;spfs;)

(vii)

……….(5)

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……….(2)

……….(2)

……….(3)

……….(2)

……….(2)

……….(2)

……….(2)

……….(2)

……….(2)

(c)

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……….(2)

……….(2)

……….(2)

……….(2)

……….(5)

……….(5)

(35

Gs;spfs;)

……….(5)

(d)

(i)

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therefore point c

……….(5)

……….(2)

……….(3)

……….(5) (20

Gs;spfs;)

nkhj;jk;-150 Gs;spfs;

(ii)

(iii)

(iv)

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Practice Examination for G.C.E A/L Students- 2020 ...

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