Bw32 Theorie l e

7/29/2019 Bw32 Theorie l e

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32nd Austrian Chemistry OlympiadNational Competition

Theoretical Part SolutionsJune 19th, 2006

1

28 bp 10 rp f = Task 1, 10 points

1.1. Element: silicon 1 bp

1.2. Name: silanes Formel: SinH2n+2 1 bp

1.3. formula and name: (CH3)2Si(OH)2 dimethylsilandiol1 bp

1.4. reaction equation:

n HO-Si(CH3)2-OH H-O(-Si(CH 3)2-O)n-H + (n-1) H2O

name of the product: silicones 2 bp

1.5. general formula for n: n = 4x-2y1 b

1.8. accompanying rock: lime2 bpargumentation: CaCO3 + 2 HCl CaCl 2 + H2O + CO2

CO2 + Ca(OH)2 CaCO 3 + H2O

1.9. reaction equation:

2 S3- + 2 H+ 5 S + H 2S 2 bp

1.10. LEWIS-formula: 1 bp

1.6. formula: Ag10[Si4O13]1 bp

1.7. formula of lapis lazuli: Na4[Al3Si3O12]S3 2 bp

S

S

S-


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2

1.11. Element: selenium2 bpjustification:

6219.096.126

96.78)Se(w?SeO:onverificati

seleniummol/g96,78)X(M2)X(n

)O(n

elementreasonablenomol/g48.39)X(M1)X(n

)O(n

)X(M.16.7116

84.28)X(m).O(M)X(M).O(m

)X(n)O(n

3 ==

==

==

==

1.12. formula of HaXOb: H2SeO3 formula of HcXOd: H2SeO4 1 bp

1.13. oxidant: MnO4-

redox equation:2 MnO4- + 5 SeO32- + 6 H+ 5 SeO 42- + 2 Mn2+ + 3 H2O 2 bp

1.14. argumentation for aromaticity: Frost-Musulin-scheme:24 valence electrons

8 -bond electrons 2 charge- 8 lone-pair electrons

6 -electrons

order of -bonding: 2/8= 2 bp1 bp

1.15. Element: xenon 1 bp

1.16. oxidation number: +2, +4, +6 1 bp


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3

1.17. geometry of the three compounds:

XeF2 linearXeF4 plane squareXeF6 pentagonal pyramidal 3 bp

1.18. reaction equation:XeF6 + 3 H2O XeO 3 + 6 HF 1 bp

1.19. speciality:

exceedance of the octet rule 1 bp


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4

Task 2, 15 points

2.7. molar initial concentration: 3 bp

TR

pc

V

nTRnVp

===

L/mol00488,0cm/mol876,4967314,8

39200)694(c 0

3 ==

=

L/mol00561,0cm/mol605,51030314,8

48000

)757(c 03

===

2.8. rate constants:

orderondsecc

1

0

0

A0AA c

1kk

t

1

c

1

c

1

==

s.mol/L842.0005605.0212

1)695(k

s.mol/L135.0004876.01520

1)695(k

=

=

=

= 3 bp

2.1. reaction equation: 2 C (s) + O2 (g) 2 CO (g) 1

2.2. BP:)O(p

)CO(pBP

2

2

= 1

2.3. Free standard enthalpie of the reaction: 2bp

kJ443J442950)101.5ln(1273314.8KlnTRG 18P1273 ====

2.4. reaction quotien + direction of the reaction: 0169.0113.0Q

2==

G1273 = -442950 + RTln(0.0169) = -486 kJG1273 < 0 left to right 2 bp

2.5. entropy of the reaction:

K/J2051273

486136225000

T

GHSSTHG =

+=

== 2 bp

2.6. reaction equation of the decomposition: 2 N2O 2 N 2 + O2 1 bp

33 bp 15 rp f =


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5

Task 3, 10 points 25 bp 10 rp f =

2.9. activation energy:

kJ241T

1

T

1

)T(k

)T(klnRE

T

1

T

1

R

E

)T(k

)T(kln

1

211

2A

21

A

1

2 =

=

=

2

2.10. resonance formulae: 2 bp

2.11. VSEPR: linear 1 bp 2.12.point group: C,v 1 bp

2.13. structure and name:CH3-CH2-OH, ethanol 1 bp

Aufgabe 2, 15 Name:

2.14. Newman-projection:

1bp

2.15. mean value of boiling point: HV,m = 86246 = 39652 J/mol

calculation (60C): K033.352T002840643.0T

12

2

==

calculation (70C): K562.351T002844445.0T

12

2

==

mean value: BP = 351.8 K = 78.7C 4bp

2.16. entropy of vaporisation: 2 bP

K/J1138.351

39652

T

HS

m,V

m,V ==

=

2.17. ebullioscopic constant:

mol/Kkg19.1396521000

46314.88.351

H1000

MRTK

2

V

2S

EB =

=

= 2

2.18. vapour pressure of the solution:n (ethanol) = 100/46 = 2.1740 mol,n (vanillic aldehyde) = 5/152 = 0.0329 mol,ng = 2.2069 mol x (ethanol) = 0.9851p (ethanol) = p0x = 46.0 kPa 2

bp

N N O N N O+

-

+

-

OH

H H

HH

H

)T(p

)T(p

lnH

R

T

1

T

1

T

1

T

1

R

H

)T(p

)T(p

ln1

2

V1221

V

1

2

=

=


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6

3.1. solubility of BaCrO4:

L(BaCrO4) = M102.9BaCrO(K 64L

= 1 bp

solubility of SrCrO4:

L(SrCrO4) = M100.6SrCrO(K3

4L= 1 bp

3.2. concentrationen of Cr2O72- and of CrO42-:

Cr2O72- + H2O 2 H + + 2 CrO42- K= 151050.1

1

= 6.6710-16

Cr2O72- H2O H

+ CrO4-

at the beginning 0.1 55.56 - -

equilibrium 0.1-x 55.56 10-3 2x

1626

2272

224

2

1067.656.55)x1.0(

x410

]OH[]OCr[

]CrO[]H[K

+

=

=

=

x1 = 3.0410-5 (x2 = -3.0410-5)[Cr2O72-] = 1.010-1 M[CrO42-] = 6.110-5 M 6 bp

Konzentrationen von Ba2+ und Sr2+:

[Ba2+] [CrO 42-] = 8.510-11 [Ba2+] =5

11

101.6

105.8

= 1.410-6 M

[Sr2+] [CrO 42-] = 3.610-5 [Sr2+] =5

5

101.6

106.3

= 0.59 M

2 bp

3.3. mass of sodium acetate:

B

SS

n

nlgpKpH =

3.00 = -lg(1.7810-5) -x1.0lg

x = 1.7810-3 molm = n M = 1 .7810-3 mol 82.03 g/mol = 0.146 g

2 bp


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7

3.4. standard potentials x and y:

3(-0.744) =-0.408 + 2y

Y = - 0.912 V

0.55+1.34+x-30.744 = 60.293x = 2,1 V

2 bp

3.5. Is there any tendency of disproportionation of Cr(IV) to Cr(III) andCr(VI)?:

Yes!

Explaination: E=2,1-0,5(1,34+0,55) = 1,155 > 0 G < 0 1 bp


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8

3.6. half equation for Cr2O72-/Cr3+:

Cr2O72-

+ 6 e

-

+ 14 H

+

2 Cr

3+

+ 7 H2O 1 bp

potential change per pH-step:

V138.0F6

143026.2RT

])OHlog[(

F6

143026.2RT

]Cr[

]OCr[ln

F6

RTEE

)]OHln[(F6

RT

]Cr[

]OCr[ln

F6

RTEE

]Cr[

]OH][OCr[ln

F6

RTEE

323

2

72

14323

2

72

23

143

2

72

=

+=

+=

+=

++

++

+

+

3 bp

3.7. coordination number: 6 1 bp

3.8. geometry: octahedral 1 bp

3.9. name of the ion: dichlorodioxalatochromat(III) 1 bp

3.10. stereoisomers:one trans-isomertwo enantiomeric cis-isomers 3 bp


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4.1.1. numbering 1 bp

N 1 2

3

4

5

6

7

8

4.1.2 3-tropanole 1 bp

4.2.4. configurationalformula of (S)-tropic acid

2bp

4.2.5. constitutional formula ofhyoscyamine

2 bp

4.3.1. reaction forming the Mannich base

4 bp

4.4.1. configurational formulae of A and B

4 bp

4.2.1. empiricalformula of

hydrotropic acid

C9H10O2 2bp

4.2.2. constitutionalformula of

hydrotropic acid

3 bp

4.2.3. constitutionalformula of tropic acid

3bp



9

Task 4, 15 points 40 bp 15 rp f =

N

OH

O

OH

HOHN

O

O

OH

C6H

5

OH

O

OH

O

OH

CHO

CHO

CH=NHCH3+

CHO

CH=NCH3

CHO

OH

CHO

N

O+ CH

3NH

2

H+

or

+

- H+


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4.4.2. isomeric relationenantiomeric 1

bp

4.4.3 type of reaction:SN2 1

bp

4.5.1. constitutional formula of tropinic acid

3 bp

4.6.1. D

2 bp

4.6.1. E

2bp

4.6.1. F

2bp

N

O

O

OOH

OMe

4.6.2. reagent ?

2 bp

4.7.1. chiral centres

2 bp

O Cl4.7.3. configuration(s)C-1: R; C-2: RC-3: S C-5: S 2 bp

4.7.2. stereoisomers 16 1



10

N

C6H

5

O

N

O

C6H

5

Br- Br-

+ +

N

O

O

OMeN

O

OMe

OH

N

O

O

C6H

5

COOMe

*

*

**

NCOOH

COOH

N

COOHHOOC

or


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11

Task 5, 5 Punkte 12 bp 5 rp f =

5.1. Leu-Val-Ser:

3 bp

5.2. flight time:

3 bp

6.3. Lebensdauer der Sonne:

g10461,17346,010988,17346,0mm 303SonneSonne,H

===+

a10697,7s10429,2s/g10014,6

g10461,1t 1018

14

33

===

1 bp

6.4. Massendefekt der Sonne:

( )kg10279,4

s/m109979,2

J10846,3

c

Em 9

28

26

2=

==

1 bp

s1025.376963

50.2vst 5===

2

mvE

2

= s/m76963J1032

m

E2v

123 mol10022.6

mol/kg610.0

15

=

==

5.3.masses of amino acids possible: 71,73,131,147,188in the case of an AS within the chain the molar mass is reduced by 18 u.

71 73 131 147 188

within the chain (+18) 89 (Ala) 91 149 (Met) 165 (Phe) 206

Therefore: Ala, Met and Phe. 3 bp

5.4.masses of amino acids possible: 73 and 188At the C-terminal the mass of the AS is reduced by 2 u, at the N-terminal by16 u.

73 188

C-terminal (+2) 75 (Gly) 190

N-terminal (+16) 89 (Ala) 204 (Trp)

AS at teh N-terminal: TrpAS at the C-terminal: Gly

3 bp

H3N

NH

NH

O

O

O

O

OH

+


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12

Task 6, 5 Punkte 11 bp 5 rp f =

6.1. mass of the sun:

m1096.6km6960002

dr 8Sonne ===

3273

Sonne m10412.13

r4V ==

Sonne = 1.408 g/cm3=1408 kg/m3

mSonne = VSonne Sonne = 1.9881030 kg = 1.9881033 g2 bp

6.2. mass of hydrogen per second: 3 bp

E = 26.72 MeV/4 H+ = 6.68 MeV/H+ = 1.07010-12 J/H+ == 6.4451011 J/mol H+

s/mol10967.5mol/J10445.6

s/J10846.3n 14

11

26

H=

=+

s/g10014.6mol/g0078.1s/mol10967.5Mnm 1414HHH

=== +++

6.3. Lebensdauer der Sonne:

g10461,17346,010988,17346,0mm 303SonneSonne,H

===+

a10697,7s10429,2s/g10014,6

g10461,1t 1018

14

33

===

1 bp

6.4. Massendefekt der Sonne:

( )kg10279,4

s/m109979,2

J10846,3

c

Em 9

28

26

2=

==

1 bp

6.3. lifetime:

g1046117346010988.17346.0mm 3333SonneSonne,H ===+

a106977s104292s/g100146

g10461.1t 1018

14

33

==

=

6.4. mass defect:

( )kg10279.4

s/m109979.2

J10846.3

c

Em 9

28

26

2

=

==

1 bp

6.5. solar constant:

Oorbit = 4 (1.4961011 m)2 = 2.8121023 m2

2

223

26

ms/J1368m10812.2

s/J10846.3E =

= 3 bp

Bw32 Theorie l e

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