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 )  HBCSE, 2nd February 2013 2 Indian National Chemistry Olympiad Theory 2013 1.12 . . N (radical) CH (C H O) P does not decolourise Br  HBCSE, 2nd February 2013 3 Indian National Chemistry Olympiad Theory 2013 Name of Student Roll No. 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  HBCSE, 2nd February 2013 4 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  HBCSE, 2nd February 2013 5 Indian National Chemistry Olympiad Theory 2013 Name of Student Roll No. 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  HBCSE, 2nd February 2013 6 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  HBCSE, 2nd February 2013 7 Indian National Chemistry Olympiad Theory 2013 Name of Student Roll No. 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  HBCSE , 2nd February 2013 8 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  HBCSE, 2nd February 2013 9 Indian National Chemistry Olympiad Theory 2013 Name of Student Roll No 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  HBCSE, 2nd February 2013 10