PJC H2 Chem 2013 Prelim P3 Soln

© PJC 2013 9647/03/Prelim/13 [Turn over PIONEER JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION HIGHER 2 CHEMISTRY 9647/03 Paper 3 Free-Response 23 September 2013 2 hours Candidates answer on separate paper. Additional Materials: Writing Papers Data Booklet Cover Page READ THESE INSTRUCTIONS FIRST Write your Centre number, index number and name on all work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough workings. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer any four questions. A Data Booklet is provided. You are reminded of the need for good English and clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. This document consists of 13 printed pages. Name: Index No.: CT Group: 12 1012

2 © PJC 2013 9647/03/Prelim/13 [Turn over Answer any four questions. 1 (a) With the aid of a diagram, describe how a polypeptide chain is held in the shape of an alpha helix. [3] The helix has a regular coiled spiral polypeptide chain held in place by hydrogen bonds between peptide C=O group of one amino acid residue and peptide N–H group of another amino acid residue in the covalently bonded sequence. There are 3.6 amino acids per helical turn. The –R groups (side chains) point outside of the helix or label R group in diagram. (b) Haemoglobin is the oxygen-carrying protein in red blood cells. (i) With reference to the haemoglobin molecule, describe and explain what is meant by the term quaternary structure of proteins. In your answer, you should state the type of bonding or interaction involved. Quaternary structure refe rs to the specific or ientation or spatial arrangement of polypeptide chains wit h respect to one another and the nature of interactions that stabilise s this orientation. Haemoglobin consists of four separate polypeptide chains of two types: two α chains and two β chains. The haemoglobin molecule is nearly spherical with the four polypeptide chains packed closely together. Each polypeptide carries a haem group to form one of the four subunits in a molecule of haemoglobin. Each subunit folds and coils into a compact globular shape and joined together by van der Waals’ forces, hydrogen bonds and ionic linkages. 1013

3 © PJC 2013 9647/03/Prelim/13 [Turn over (ii) A typical polypeptide chain of haemoglobin consists of 141 amino acids. The R groups of 3 of the amino acids are given below. amino acid formula of side chain (R in RCH(NH2)CO2H) glutamic acid CH2CH2CO2H valine CH(CH3)2 lysine CH2CH2CH2 CH2NH2 Use all three amino acids above to construct the displayed formula of a possible section of the polypeptide. NC H H CCC H H H H H H C O N C H CHH C HH C O O H C O NC H C O C C C C H H HH H H HH N H H H H H Displayed formula should be an open ended structure with N terminus of the left and C terminus on the right. (iii) Uncoiling of the protei n structure is favoured w hen temperature is higher than physiological conditions. With the aid of the equation, ∆G = ∆H – T∆S suggest why this process is favoured at higher temperatures. ΔH should be positive due to the overcome/disruption of various interactions (e.g. hydrogen bonding, disulfide bridges, van der Waals’ forces, ionic linkages) between the R-groups of the constituent amino acids, which is an endothermic process. ΔS should be positive when protein structure uncoils and increases in disorderliness. In order for reaction to be spontaneous, ΔG has to be negative and it is favoured by high temperature. [ 8 ] 1014

4 © PJC 2013 9647/03/Prelim/13 [Turn over (c) One molecule of haemoglobin can bi nd up to four molecules of oxygen, according to the following equation. Hb(aq) + 4O2(aq) Hb(O 2)4(aq) Kc = 3 x 1020 mol-4 dm12 (i) Write an expression for Kc for this reaction. 24 c 4 2 [Hb(O ) ]= [Hb][O ]K (ii) Calculate the percentage of Hb(O 2)4 in an Hb-Hb(O 2)4 mixture when [O2] = 7.6 x 106 mol dm3. [3] 24 c 4 2 20 24 -6 4 24 [Hb(O ) ]= [Hb][O ] [Hb(O ) ]3? 0 = [Hb][7.6? 0 ] [Hb(O ) ] = 1.001 [1/2][Hb] K Fraction of Hb(O2)4 in mixture = ] Hb [ ] ) O ( Hb [ ] ) O ( Hb [ 4 2 4 2  % o f H b ( O 2)4 = % 100] Hb [ ] Hb [ 001 . 1 ] Hb [ 001 . 1 = 50% (d) (i) Carbon monoxide, CO, mainly causes adverse effects in humans by combining with haemoglobin to form carboxyhaemoglobin in the blood. This prevents haemoglobin from rel easing oxygen in tissues, effectively reducing the oxygen-carrying capacity of the blood. The equations below show the reac tions of haemoglobin with oxygen and carbon monoxide respectively. Fe N N N N O N H2 Fe N N N N O2 N + O2 + H2O Equilibrium 1 haemogloblobi n oxyhaemoglobin 1015

5 © PJC 2013 9647/03/Prelim/13 [Turn over Fe N N N N O N protein H2 Fe N N N N CO N protein + CO + H2O Equilibrium 2 haemoglobin carboxyhaemoglobin CO is found to have 245 times mo re affinity for haemoglobin than O 2. Explain how excessive inhalation of CO will give rise to poisoning in the blood. CO is a stronger ligand than H 2O (or O2). . In the presence of CO, haemoglob in undergoes ligand exchange. CO displaces H 2O (or O 2) from haem to form ca rboxyhaemoglobin. Unlike H2O and O 2, CO binds strongly and irreversibly to Fe as carboxyhaemoglobin is a more stable complex than oxyhaemoglobin. (ii) Suggest a method for the treatment of carbon monoxide poisoning. Treatment of carbon monoxide poisoning largely consists of administering pure / high concentration of oxygen so as to shift the position of equilibrium 1 to the right. [3] (e) Oxygen can be liquefied by pressure al one if its temperat ure is below the 'critical temperature' of -118°C. Above this critical temper ature no amount of pressure will liquefy oxygen. Suggest a suitable value for the critical temperature of ammonia. Give your reasoning. The critical temperature of a substance is determi ned by the strength of its intermolecular forces. Oxygen molecules are held together by weak intermolecular van der Waals' forces and so, it has a very low critical temperature. Below the critical temperature , this force is sufficiently strong to hold the molecules together (under some appropriate pressure) in a liquid. Above the critical temperature , the higher kinetic ener gy of the oxygen is sufficient to overcome the intermolecular forces of attraction. Ammonia molecules have hydrogen bonding between molecules that need to be overcome, which are stronger than the van der Waals’ forces attraction. Therefore, critical temperature of ammonia > -118°C. [3] [Total: 20] 1016

6 © PJC 2013 9647/03/Prelim/13 [Turn over 2 (a) The diagram below shows the reactions of a salt, A, in aqueous medium. orange solution B brown solution C colourless solution colourless solution cream precipitate yellow precipitate colourless A(aq) AgNO3(aq)AgNO3(aq) A has the following composition by mass: K, 41.1%; S, 33.7%; O, 25.2%. The relative formula mass, M r, of A is 190.4. One formula unit of A contains only one type of anion. (i) Determine the formula of the salt A . Let the mass be 100g. K S O Mass / g 41.1 33.7 25.2 Amount / mol 1.0512 1.0498 1.575 Ratio 1.001