2019 TMJC H2 Chem Prelim P3 ANS

1 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Section A Answer all the questions from this section. 1 Hydrocarbons, CxHy, are used extensively as sources of fuel in our modern day civilisation. (a) 10 cm3 of a gaseous hydrocarbon C5Hy was allowed to burn in an excess of oxygen at 300 ºC and 1 atm. In the reaction, there was an expansion of volume by 20 cm3. (i) Write a balanced equation, with state symbols, for the reaction at 300 ºC and 1 atm. [2] C5Hy (g) + 5 4   y O2 (g)   5 CO2 (g) + 2 y H2O (g) (ii) Determine the value of y. [2] Using Avogadro’s Law, mole ratio  volume ratio Equation: C5Hy (g) + 5 4   y O2 (g)   5 CO2 (g) + 2 y H2O (g) Vol of gases: 10 cm3 ( 5 4   y x 10) cm3 50 cm3 ( 2 y x 10) cm3 Volume of gases used up in the reaction = 10 + ( 1050 4 y ) = 1060 4   y cm3 Volume of gases produced in the reaction = 50 + 2   y x 10 =  50 5 y cm3 Volume of gas produced – Volume of gas used up = 10 cm3  50 5 y – 1060 4   y = 20 y = 12 TAMPINES MERIDIAN JUNIOR COLLEGE 2019 JC2 H2 Chemistry Prelim Exam Paper 3 (Suggested Answers)

2 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Methane, CH4, is sometimes used in the production of hydrogen via a process known as steam reforming. CH4(g) + H2O(g) CO(g) + 3H2(g) (b) The steam reforming process is an endothermic reaction. (i) Using data from the Data Booklet , calculate a value for the enthalpy change of the forward reaction. [2] Hr = BE (reactants) – BE (products) = 410 x 4 + 460 x 2 – (1077 + 436 x 3) = +175 kJ mol –1 (ii) The actual value of the enthalpy change of reaction is found to be +206 kJ mol–1. Suggest a reason why your calculated answer in (i) differs from this value. [1] The bond energy calculation is an approximation method as the bond energy values given in the Data Booklet are average values . (c) At 600 K, t he value of the equilibrium constant, Kp, for the steam reforming reaction is 7.20 x 10–4. (i) Write the Kp expression for this reaction, giving its units. [2] Kp = )P)(P( )P)(P( OHCH 3 HCO 24 2 atm2 (accept Pa2) (ii) Gaseous CH4, H2O and CO are introduced into a closed container at 600 K and their initial partial pressures are 1.20 atm, 2.10 atm and 1.80 atm respectively. Determine the partial pressure of H2 when equilibrium is reached. (You may assume that the extent of the forward reaction is small.) [3] CH4(g) + H2O(g) CO(g) + 3H2(g) Initial pp/atm 1.20 2.10 1.80 0 Change pp/atm –x –x +x +3x Eqm pp/atm 1.20 – x 2.10 – x 1.80 + x 3x Kp = )P)(P( )P)(P( OHCH 3 HCO 24 2 7.20 x 10–4 = 3(1.80 )(3 ) (1.20 )(2.10 ) xx xx   Assuming x is small, 7.20 x 10–4 = 3(1.80)(3 ) (1.20)(2.10) x

3 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry Solving for x, x = 0.0334 Partial pressure of H2 at equilibrium = 3 x 0.0334 = 0.100 atm (iii) Using information from (b), suggest how the temperature of the reaction can be changed so as to increase the yield of H2. Explain your answer. [2] The temperature should be raised to increase the yield of H2. When the temperature is raised, by Le Chartelier’s Principle, the equilibrium position shifts to the right towards the endothermic reaction to absorb heat, increasing the yield of H2. (d) Additional hydrogen can be recovered using the carbon monoxide produced in another reaction known as the water–gas shift reaction. CO(g) + H2O(g) CO2(g) + H2(g) (i) Name the type of hybridisation in the carbon atom in CO. Draw the hybrid orbitals around the carbon atom. [2] sp hybridisation (ii) Given that the above reaction was conducted at 300 oC and 1 atm, calculate the volume of H2 that can be recovered from 5 kg of CO. [3] Amount of H2 that can be recovered = 5000/28 = 178.6 mol pV = nRT V = p nRT V = 178.6 8.31 (300 273) 101325    = 8.39 m3 (iii) The volume of CO2 collected in the water-shift reaction should be the same as that of H2. However, the actual volume of CO2 collected was smaller. Suggest a reason why this is so. [1] CO2 deviates more from ideal gas behaviour as it has more significant intermolecular instantaneous dipole–induced dipole interaction . Hence it will occupy a smaller volume than H2. [Total: 20]

4 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry 2 Propanoic acid, CH 3CH2COOH was initially known as propionic acid based on the Greek words, protos, meaning ‘first’ and pion, meaning ‘fat’. The pKa values of CH3CH2COOH and CH3CH(Cl)COOH are listed below. (a) (i) Explain why acid 2 has a lower pKa than acid 1. [2] The electron-withdrawing Cl group in acid 2 (OR CH3CH(Cl)COOH) further reduces the intensity of the negative charge on CH3CH(Cl)COO– to a greater extent . CH3CH(Cl)COO– is more stable and CH3CH(Cl)COOH has a greater tendency to dissociate and therefore a stronger acid. . (ii) Suggest a pKa value for acid 3. [1] Accept value between 2.8 to 4.9 (b) A 25.0 cm 3 solution of 0.10 mol dm –3 CH3CH2COOH was titrated against 0.20 mol dm–3 sodium hydroxide, NaOH. (i) Calculate the pH of the 0.10 mol dm–3 CH3CH2COOH. [1] [H+] = √0.10 x 10–4.9 = 0.00112 mol dm–3 pH = –log10(0.00112) = 2.95 (ii) Calculate the volume of NaOH required for complete neutralisation. [1] Amount of CH3CH2COOH = (25 ÷ 1000) x 0.1 = 2.50 x 10–3 mol CH3CH2COOH ≡ NaOH Amount of NaOH required = 2.50 x 10–3 mol Volume of NaOH required = 2.50 x 10–3 ÷ 0.20 = 0.0125 dm3 OR 12.5 cm3 acid structural formula pKa 1 CH3CH2COOH 4.9 2 CH3CH(Cl)COOH 2.8 3 CH2(Cl)CH2COOH

5 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry (iii) Write a suitable equation to explain why the pH at equivalence point is greater than 7. [1] CH3CH2COO– + H2O ⇌ CH3CH2COOH + OH– Concept: CH3CH2COO– undergoes hydrolysis to form OH –. At equilibrium, [OH –] > [H +] pH at equivalence point is basic (> 7). (iv) Sketch the expected titration curve for this titration given that a total volume of 25.0 cm 3 of NaOH was added. On the titration curve, indicate the initial pH value and the equivalence volume. [2] (v) A buffer involving CH 3CH2COOH and its salt was formed during the progress of the titration. Circle the buffer region on the sketched curve in (iv) and indicate the corresponding pH value and volume at the maximum buffering capacity. [2] Volume of NaOH / cm3 pH 0 2.95 12.5 25.0 buffer 4.9 6.25

6 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry There are many organic compounds such as amino acids and drug molecules that are derivatives of propanoic acid. (c) The starting material to synthesise 2-aminopropanoic acid, also known as alanine can be either pyruvic acid or ethanal. (i) The proposed synthesis for the deprotonated form of alanine from pyruvic acid is shown below. Suggest the reagents and conditions for steps 1, 2 and 3. [3] Step 1: NaBH4 in ethanol Step 2: HCl (g), heat Step 3: excess conc. NH3 in ethanol, heat in sealed tube (ii) State the two types of reaction that had occurred in step 3 of the above proposed synthesis. [2] Nucleophilic substitution Acid-base reaction (OR neutralisation)

7 Tampines Meridian Junior College 2019 JC2 Prelim H2 Chemistry (iii) Write