RI H2 Chemistry 2016 Paper 2 TYS Solutions

© Raffles Institution 2016 A level Paper 2 Suggested Solutions Q1 Planning (a) Effervescence would be seen; colourless gas which forms a white ppt with Ca(OH) 2(aq) would be evolved. The deep blue azurite would dissolve completely to give a blue solution. (b) Calculation of a suitable mass of powdered rock to react with about 75% of the acid in the conical flask Cu3(CO3)2(OH)2  3H2SO4 Amt of H2SO4 in 50.00 cm3 = 1.00 x 50.00 x 103 = 0.0500 mol Amt of azurite that reacts with 75% of H2SO4 = 1 3 x 0.75 x 0.0500 = 0.0125 mol Mass of azurite in 0.0125 mol = 0.0125 x 344.5 = 4.306 g Mass of powered rock to be used = 4.306 0.90 = 4.78 g Dilution and Volume of unreacted sulfuric acid required for titration Amt of excess H2SO4 in reaction mixture = 0.25 x 0.0500 = 0.0125 mol Assume that average volume of NaOH required for titration is 25.00 cm3 Amt of excess H2SO4 used for titration = ½ x 0.100 x 25 x 103 = 0.00125 mol (ie 10% of 0.0125 mol) Hence, 10% of excess H2SO4 is needed for titration with NaOH(aq). So, the final reaction mixture can be diluted to 250 cm3 and 25.0 cm3 (ie 10% of 250 cm3) will be pipetted for titration with NaOH. Procedure 1. Using an electronic balance, weigh out accurately about 4.78 g of the powdered rock in a clean and dry weighing bottle. Record the mass of the weighing bottle and powdered rock. 2. Transfer the powdered rock sample into the 250 cm3 conical flask containing 50.00 cm 3 of sulfuric acid and swirl the contents. Place a glass filter funnel on the mouth of the conical flask to prevent acid spray. 3. Reweigh the emptied weighing bottle and record its mass. 4. When effervescence has ended and all the powdered rock has dissolved, t ransfer the final reaction mixture quantitatively into a 250 cm 3 graduated flask with the aid of a funnel and a glass rod. Rinse the conical flask and the glass filter funnel a few times with small volumes of deionised water each time and transfer all the washings into the graduated flask. 4. Fill the graduated flask to the 250 cm 3 mark with more deionised water. Use a teat pipette (or dropper) to add the deionised water drop by drop when nearing the mark. 5. Stopper the graduated flask and shake the solution thoroughly to ensure that it is homogeneous. Label the solution FA 3. 6. Pipette 25.0 cm3 of FA 3 into a 250 cm3 conical flask. Add 2 drops of phenolphthalein indicator. 7. Fill the burette with the 0 .100 mol dm –3 NaOH(aq) provided. Titrate the solution in the conical flask with the standard dilute NaOH(aq) placed in the burette.

© Raffles Institution 8. Stop the titration when one drop of the NaOH(aq) added changes the colour of the solution in the conical flask from blue to light purple. 9. Repeat the titration until at least two consistent results are obtained, i.e. the two titre volumes do not differ by more than 0.10 cm3. Calculations Let that the average values of two consistent titres be 1000b cm3 or b dm3. and mass of powdered rock sample and weighing bottle / g = d mass of emptied weighing bottle / g = e mass of powdered rock sample /g = d  e = c Amt of excess sulfuric acid in 25.0 cm3 of FA3 = ½ x b x 0.100 = 0.0500b mol Amt of excess sulfuric in reaction mixture = 0.0500b x 250 25.0 = 0.500b mol Amt of sulfuric acid reacted with azurite = 0.0500  0.500b mol Amt of pure azurite present in the powdered rock sample = 1 3 x (0.0500  0.500b) mol Mass of pure azurite present in the powdered rock sample = 1 3 x (0.0500  0.500b) x 344.5 = 114.8 x (0.0500  0.500b) g Percentage by mass of pure azurite present in the powdered rock sample = 114.8 x (0.0500  0.500b) 𝑐 x 100 %

© Raffles Institution Question 2 (a)(i) Individual fatty acids have intermolecular hydrogen bonds which are stronger than the permanent dipole -permanent dipole intermolecular forces between triester. Fatty acids have higher boiling points than triester as more energy is needed to overcome the strong hydrogen bonding between fatty acid molecules, resulting in its higher boiling point. (ii) stearic acid has longer hydrocarbon chain (2 more carbon atoms) than palmitic acid. As the chain length of alkyl g roup, R, increases, the instantaneous dipole-induced dipole forces between RCO2H molecules become stronger. More energy is needed to overcome these forces. (iii) The presence of C=C bonds in fatty acids with the same number of carbon atoms lowers the melting point of the fatty acids. [m.p. of linolenic acid < linoleic acid < oleic acid < stearic acid] (b)(i) Electrophilic addition (b)(ii) Mass of iodine that would react with 100g of olive oil = (100/0.256) x 0.237 = 92.6 g (b)(iii) Amt of iodine in 92.6 g = 92.6 / 254 = 0.3646 mol Amt of C=C bonds in olive oil = 0.3646 mol Amt of olive oil in 100 g = 100/782 = 0.1279 mol Average number of C=C bonds = 0.3646/0.1279 = 2.85 (b)(iv) Triesters containing 3 oleic acids conta in 3 C=C bonds per molecule. As the average number of C=C bond in each molecule of olive oil is 2.85 which is less than 3 and triesters of olive oil are mainly formed from oleic acid, olive oil should contain triesters of palmitic acid and/or stearic acid as the other significant component. Since the average Mr of olive oil is smaller than that of triester of oleic acid ( Mr = 282 x 3 + 12 x 3 + 2 = 884), the other major component of olive oil shoul d be triester of palmitic acid which has a lower Mr than that of stearic acid. (c)(i) H2, Ni catalyst, heat (c)(ii) (d) ester undergoes hydrolysis alkene(C=C) undergoes oxidative cleavage R1CH=CHR2CO2H + 4[O] R1COOH + HO2CR2CO2H or CH3(CH2)7CH=CH(CH2)7CO2H + 4[O] CH3(CH2)7COOH + HO2C(CH2)7CO2H Question 3 (a) Silicon carbide has a giant molecular structure. Each silicon atom is covalently bonded to four other carbon atoms arranged tetrahedrally around it. Each carbon atom is also covalently bonded to four other silicon atoms. This tetrahedral arrangement is repeated throughout the whole molecule. Melting requires a lot of energy to break the strong covalent bonds between all atoms, hence silicon carbride has very high melting point. (b)(i) CxHy + (x + 𝑦 4 ) O2  x CO2 + 𝑦 2 H2O (b) (ii) Heterogeneous catalyst is used, whereby the catalyst and the reactants are in different phases. For heterogeneous catalysis to occur, the reactant molecules need to be readily adsorbed onto the catalyst surface. The adsorption of the reactant molecules at the catalyst surface increases the reaction rate because it

© Raffles Institution 1. weakens the covalent bonds within the reactant molecules, thereby reducing the activation energy for the reaction. 2. increases the concentration of reactant molecules at the catalyst surface a nd allows the reactant molecules to come into close contact with proper orientation for reaction. (c) (i) N(-3) in NH3 to N(0) in N2 N(+4) in NO2 to N(0) in N2 (ii)8NH3 + 6NO2 7N2 + 12H2O (d) (i)NO2 + SO2 SO3 + NO (I) NO + ½ O2 NO2 (II) Homogeneous catalyst; oxidizes SO2 to SO3, and is regenerated (II) (ii) Cause breathing difficulties; form acid rain, corrodes buildings (iii) Kp =  2 4 2 4 2 N O N O 322 NO = (1.5 10 ) pp xp = 6.25 x 10-5 Pa1 24NOp = 1.41  10-10 Pa Question 4 (a) (i) K: 1s22s22p63s23p64s1 Cu: 1s22s22p63s23p63d10 4s1 (ii) Cu2+ + 2eˉ Cu E = +0.34 V K+ + eˉ K E = -2.92 V First ionisation energy of K is +418 kJmolˉ 1 while that for Cu is +745 kJmolˉ1. Much more energy is required to remove an electron from Cu than K. (Not in syllabus: Cu has a higher nuclear charge than K and t here is a minimal increase in shielding effect from K to Cu. This causes the valence electron in Cu to be more strongly attracted to the nucleus and les