INChO 2013 Chemistry Solution from India Indian National Chemistry Olympiad

11 Pages·2013·0.53 MB·English

by  HBCSE Chemistry Team

Indian National Chemistry Olympiad Theory 2013
Name of Student Roll No.

Problem 1 20 marks

Free radicals

1.1 c) X

1.2 b) X

1.3 C : C = 1: 3.85 or 7: 27 or 0.259: 1
1 2

1.4 d) X

1.5

1.6
OOH

(CH)COOH
(CH)COOH 27
27 and
CH (CH )
CH (CH ) 3 2 4
3 2 4

OOH
B

1.7 Me
.

Me

Me
Me

1.8 b) X X C

 HBCSE, 2nd February 2013 1
Indian National Chemistry Olympiad Theory 2013
1.9 i) EE

ii)
EE

1.10

or

G (C H O )

1.11
.

H
I

OH

HO

OH
OH

(C H O ) (C H O )

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Indian National Chemistry Olympiad Theory 2013
1.12

.
.

N (radical)

CH

(C H O)

P does not decolourise Br

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Problem 2 18 marks

Transition Metal Chemistry

2.1
.. .. .. ..
:

I follows octet rule
I I
2.2  *(CO)


b) I follows octet rule

X
I


I
2.3 I) Crx(CO)y x = 1 y = 6

II) [Mn (CO) ] x = 2 y = 1 0 or 11
x y

I II

2.4
1) b 2) c 3) a

2.5
[Ni(CN)] + CO → [Ni(CO) ] + [Ni(CN) ]
4 2

2.6 ῡ (CO) = 2143cm 1 or 6.424×1013 Hz
 = 1.1386×1026 kg

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Indian National Chemistry Olympiad Theory 2013
2.7
i) a) MCO b) M CO c) M CO
2 3
> >
a b c
ii) a) b) c)
> >
b c a
2.8

i)

ii)

2.9
Step No. Reaction type (Choose from Formal oxidation state of metal Number of d
the above list and write only in the product obtained at the electrons
the alphabet) end of the step
3 d +1 8
6 e +3 6
7 b +1 8

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Indian National Chemistry Olympiad Theory 2013
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Problem 3 14 marks

Synthesis of natural products

3.1 (b) ketonic carbonyl X

(d) no unsaturation
X

3.2
Number of acidic functional groups present in B = 02 and C =03

3.3
ii) mole CH I

3.5

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Indian National Chemistry Olympiad Theory 2013
3.6
ii) CH Br

COOH
or
or COOH CHBr
or

H

or

F (C H O )

3.7

or

B

3.8
1, 2, 2-Trimethylcyclopentane-1,3-dicarboxylic acid

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Problem 4 18 marks

Hydrogen as a fuel

A.

4.1 p = 6 4.58 kg m3

4.2
i) ΔH=  143 kJ g1 of hydrogen

ii) ΔH =  32.8 kJ g1 of carbon

4.3 (i) max work = 1.2 105 kJ or 1.18 105 kJ

(ii) Heat engine =  6.9 104 kJ

4.4 (i) 46.3 months or 45.7 months

(ii) I = 0.813 A

B.
CH (g) + H O (g) → CO(g) + 3H (g)
4.5 4 2 2

Conversion (methane) = 66%
4.6

4.7 Total pressure at 1100 K = 6.550 bar

Conversion (methane)  49 %

C.
1) T = 300K
1
4.8
2) T2 = 600K

3) T = 1200K
3

4) T = 600K
4

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Indian National Chemistry Olympiad Theory 2013
4.9 ( i) F or path 1→2:
ΔE = 3.74kJ
int,12

(ii) For path 2→3:
ΔE = 7.5kJ
int,23

(iii) For path 3→ 4:

ΔEint,34 = 7.48kJ
(iv) For path 4→1:
ΔEint,41 =  3.75kJ

4.10
The efficiency of the cycle is given by: ε 15%

4.11 (a) ΔS + ΔS + ΔS + ΔS = 0
1 2 3 system

(b) T = 267K
3

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Indian National Chemistry Olympiad Theory 2013
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Problem 5 15 marks
Acid-Base Equilibria
5.1 K' 7. 9107
a

[HCO ¯]/[CO (dissolved)] = 20/1
3 2
5.2 i)
[HCO ¯] = 20/21× 2.52×102 = 2.4×102 M
3

[CO (dissolved)] = 1.2×103 M
2

ii) pH = 6.58
iii)
pH = 7.29

5.3 [HCO ] = 25.8 ×10-3
3 CO2 rich blood
[CO ] = 1.39 ×10-3
2 CO2 rich blood

5.4 i) In presence of CO2, higher p oi2s needed for a given percent saturation X
iii) In absence of CO2, maximum saturation of haemoglobin occurs at lower po2 X

5.5
Normal Hb-Curve1: (0.98 – 0.17) mol × 4  3.2 mol

Abnormal Hb- Curve 2: (1.00 – 0.60) mol × 4  1.6 mol

Abnormal Hb- Curve 3: (0.73 – 0.01) mol × 4  2.9 mol

5.6 pH = 3.16

5.7 K = 3.11 × 102
1

5.8 Max. Concentration of "free" Ca2+ ions: [Ca2+] = 1.9×104 M
max

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