MI Prelim 9647 P3 Answers

Class Adm No Candidate Name: This question paper consists of 11 printed pages and 1 blank page. 2017 Preliminary Examination II Pre-University 3 H2 CHEMISTRY 9647/03 Paper 3 Free Response 15th Sept 2017 2 hours Candidates answer on separate paper. Additional materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Do not turn over this question paper until you are told to do so Write your name, class and admission number on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer any four questions. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need for good English and clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. Question 1 2 3 4 5 Total Marks 20 20 20 20 20 80

2 1 (a) Graphite is a naturally-occurring form of crystalline carbon and is found in metamorphic and igneous rocks. Graphite is extremely soft and cleaves with very light pressure yet it is extremely resistant to heat and nearly inert in contact with almost any other material. These extreme properties give it a wide range of uses in metallurgy and manufacturing. (i) With reference to the structure, account for the following properties of graphite.  Soft  Heat resistant and inert [3] Graphite has a giant covalent structure (;). The bonding between the layers is weak temporary dipole – induced dipole forces(;) which can be easily overcomed, allowing the layers to slide over each other easily and thus graphite is soft. Large amount of thermal energy is required to break the strong covalent bonds(;) between the carbon atoms within the layers which account for its resistance to heat. (ii) The following table shows some thermochemistry data. Reaction H / kJ mol-1 Standard enthalpy change of atomisation of carbon +715 Enthalpy change of 4H(g) + O(g) + C(g) → CH3OH(l) -2069 With the use of relevant data from the Data Booklet and the above information, calculate the enthalpy change of formation of methanol. [3] ∆Hf = +715 +½(+496) + 2(+436)+(-2069) = -234 kJ mol-1 (b) A butane burner is used to heat the air in a hot air balloon. The hot air balloon has a volume of 2.1 m3 and its volume does not change when the enclosed air is heated. (i) Using the ideal gas equation, calculate the amount of gas molecules the balloon contains at temperature 800 K and a pressure of 1.0 x106 Pa. [1] pV=nRT 1.0 x 106 x 2.1= n x 8.31 x 800 n= 316 mol C(s) + ½O2(g) + 2H2(g) CH3OH(l) +715 +½(+496) + 2(+436) C(g) + O(g) + 4H(g) -2069 ∆Hf

3 [Turn over (ii) Hence calculate the mass of air it contains, assuming an average relative molecular mass of 29. [1] Mass = 315.9x29 = 9160g (iii) The standard enthalpy change of combustion of butane is -2877.5 kJ mol-1. It requires 1.0 J of energy to raise the temperature of 1.0 g of air by 1.0 K. Using your answer in (b)(ii), calculate the mass of butane that needs to be burnt to raise the temperature of the air in the balloon by 2 0 K. Assume that the hot air balloon is a closed system. [2] Q=mc∆T=9160x1.0x20=183213 J ∆H=− 𝑄 𝐴𝑚𝑡 𝑜𝑓 𝑏𝑢𝑡𝑎𝑛𝑒 × 𝑐𝑜𝑒𝑓𝑓 -2877.5= − 183212÷1000 𝑎𝑚𝑡 𝑜𝑓 𝑏𝑢𝑡𝑎𝑛𝑒 × 1 Amount of butane =0.06367 mol Mass of butane = 0.06367x(4x12.0+1.0x10) = 3.69g (iv) The actual mass of butane that needs to be burnt to raise the temperature of the air in the balloon by 20 K was found to be 3.81 g. Suggest why this differs from your answer in (b) (iii). [1] More butane needs to be burnt due to heat loss. (c) Paracetamol and aspirin are effective at pain and fever relief due to their ability to dissolve quickly in the blood stream and are soluble in fatty compounds found in cell membrane. Paracetamol Aspirin (i) Account for these properties based on the structure and bonding of aspirin. [2] Aspirin has a simple molecular structure. The p resence of COOH forms ion-dipole interactions(when hydrolysed to form COO-) / hydrogen bonding with water. The p resence of benzene ring forms favourable van der Waals’ forces with the fatty compounds which results in the solubility.

4 (ii) One of the pain relievers cause more stomach irritation than the other. With reference to the functional groups present, s uggest and explain the pain reliever that you will recommend to someone who suffers from gastric bleeding. [2] Paracetamol causes less stomach irritation and is recommended(;) as the acidic –COOH group of aspirin will attack the lining of the stomach walls, causing irritation.(;) (iii) Write the structural formula of the organic product(s) formed when paracetamol tablet is refluxed with sodium hydroxide. [2] O - Na + NH2 and CH3COO-Na+ (iv) Extensive research has been made to improve the effectiveness of the pain -relievers. Two proposals were made to modify aspirin. Drug A Drug B Given that the melting point s of Drugs A and B are 179 °C an d 154 °C respectively, account for the melting point in terms of structure and bonding. [2] Both drugs have simple molecular structure. Drug A has a higher melting point as the van der Waals’ forces of attraction is more extensive and stronger than the hydrogen bonding in Drug B . More energy is needed to overcome the stronger van der Waals’ forces of attraction between Drug A. (v) State the relative solubility of Drugs A and B in water. [1] Drug A is less soluble in water than drug B. [Total: 20]

5 [Turn over 2 Halogens and their compounds can be toxic but some are essential for the human body's functioning and are used in daily products. The oxidising power of chlorine allows it to act as a good disinfectant. (a) With the use of Data Booklet, explain why FeCl3 exists but FeI3 does not. [4] Fe3+ + e- ⇌ Fe2+ E= +0.77V Cl2 + 2e- ⇌ 2Cl- E= +1.36V For FeCl3, initial species present: Fe3+ and Cl- [O] 2Cl- → Cl2 + 2e- [R] Fe3+ + e- → Fe2+ Ecell= +0.77 – (+1.36) = -0.59V <0 Hence, the species cannot undergo further redox. Thus FeCl3 exists. Fe3+ + e- ⇌ Fe2+ E= +0.77V I2 + 2e- ⇌ 2I- E= +0.54V For FeI3, initial species present: Fe3+ and I- [O] 2I- → I2 + 2e- [R] Fe3+ + e- → Fe2+ Ecell= +0.77 – (+0.54) = +0.23V >0 Hence, the species can undergo further redox to form Fe2+ and I2. Thus FeI3 does not exist. (b) Grignard reaction is an important reaction which helps in lengthening the carbon chain. A Grignard reagent has a general formula of R -MgX where R is an alkyl or aryl group and is formed via the reaction of an alkyl or alkyl halide with magnesium powder. R−X + Mg → R−MgX Grignard reagent Carbonyl compounds react with Grignard reagent to increase the carbon chain. (i) In the formation of Grignard reagent, it is important to carry out the reaction in a dry environment. Suggest a reason. [1]