SRJC 2008 JC2 Prelim H2 P3 Solutions

SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 CHEMISTRY 9746/03 Preliminary Examination Paper 3 Free Response 19th August 2008 2 hours Candidates answer on the separate paper. Additional Materials: Answer Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your name and class on all the 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 work. 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 the 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 11 printed pages and 1 blank page SRJC 2008 9746/03/Prelim/2008 [Turn over

2 SRJC 2008 9746/03/Prelim/2008 [Turn over Answer any four questions. 1. The Haber Process is the nitrogen fixation reac tion of nitrogen and hydrogen over magnetite, iron oxide, to produce ammonia. This process is important because ammonia is difficult to be produced on an industrial scale, and ammonia is the precursor for many nitrogen-containing products such as nitrate fertilizers and munitions. N2(g) + 3H2(g) 2 NH3(g) ΔHo = – 92.4 kJ mol-1 (a) Write an expression for Kp for the above reaction, stating its units. [2] ( ) () () 3 2 22 3 H N NH p P P PK = atm–2 (or Pa-2 or torr-2 or kg m-1 s-2 or N-2m4) (b) When a 1:3 mixture by volume of N 2 and H2 at an initial pressure of 200 atm is allowed to reach equilibrium at 500 oC, it was found that the percentage yield of NH3 is 30 %. ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ×= % 100% amount l theoretica amountactualyield (i) Calculate the equilibrium partial pressures of N2 and H2. (ii) Hence calculate a value of Kp at 500 oC. (iii) Suggest, with a reason, whether or not a higher temperature would favour the formation of ammonia. (iv) When some H2 is added to the existing equilibrium mixture, the partial pressure of ammonia is increased to 50 atm. Calculate the new equilibrium partial pressure of H2. (v) The value of Δ G is 0 kJ mol -1 at 196 °C. Use this information, and any other information to calculate the value of ΔS. (vi) Calculate ΔG for the forward reaction at 500 oC. (vii) Based on the answer in (vi), comment on the feasibility and relate it with the reaction conditions of 200 atm and 500 oC. [8] i. Using mole fraction, atm PN 50 2004 1 2 = × = atm PH 150 2004 3 2 = × =

3 SRJC 2008 9746/03/Prelim/2008 [Turn over ii. N2(g) + 3H2 (g) 2 NH3 (g) Initial amount/atm 50 150 0 Change /atm -15 -45 +30 Final amount/atm 35 105 30 100100 30 = × ( ) () () 3 2 22 3 H N NH p P P P K = () () () 3 2 105 35 30=pK = 2.22 x 10-5 atm-2 iii. Low temperature. Forward reaction is exothermic. iv. N 2(g) + 3H2 (g) 2 NH3 (g) Initial amount/atm 35 105 + x 30 Change /atm -10 -30 +20 Final amount/atm 25 75 + x 50 ( ) () () 3 2 22 3 H N NH p P P P K = ( ) () ( ) 3 2 5 - 75 25 50 10 x 2.22 x+ = x = 90.1 atm v. ΔG = ΔH – TΔS 0 = (–92.4) – (196+273) Δ S ΔS = -197 J mol-1 K-1 vi. ΔG = ΔH – TΔS ΔG = (–92.4) – (500+273)(-0.197) ΔG = +59.9 kJ mol-1 vii. Positive ΔG value shows that the reaction is thermodynamically unfeasible / non-spontaneous. Thus, condition of high pressure and moderate temperature is needed to push the reaction forward. (c) Ammonia can be used as a reagent to make nitrogen-containing products. (i) State the reagents and conditions that is needed for the conversion of ammonia

4 SRJC 2008 9746/03/Prelim/2008 [Turn over into I propanamine II propanamide (ii) It is noted that in the formation of propanamine, a by-product is formed which has a melting point of 270 °C and gives an immediate precipitate with aqueous silver nitrate. Suggest a structure for this by-product. [3] (i) I 1-chloropropane or CH 3CH2CH2Cl Heat II Propanoyl chloride, room temperature (ii) N CH2CH2CH3 CH2CH2CH3 CH2CH2CH3 H3CH2CH2C + X- (d) The compound propanamide, which is a product of ammonia, can be used to make other organic compounds. (i) Propanamide can undergo hydrolysis to form propanoic acid. How would you expect the rate of this reaction to compare to that of the hydrolysis of propanoyl chloride? Explain your answer using their respective reaction conditions. (ii) State the intermediates and suggest the reagents and conditions needed for the conversion of propanoic acid into butanoic acid. [5] (i) Rate of hydrolysis of propanamide is slower than that of propanoyl chloride. Propanamide only undergoes hydrolysis in the presence of aqueous acid or alkali with heating/reflux Propanoyl chloride readily undergo hydrolysis with water at room temperature. (ii) Propanoic acid Æpropan–1–ol LiAlH4 in dry ether, room temperature Propan-1-ol Æ 1–chloropropane PCl5... rtp

5 SRJC 2008 9746/03/Prelim/2008 [Turn over Chloropropane Æ butanitrile Alcoholic KCN, reflux Butanitrile Æbutanoic acid Dilute H2SO4, heat (e) When propanamide, propanoyl chloride and propanoic acid are added into separate portions of water, solutions of pH 1.0, 2.8 and 7.0 are formed. (i) Suggest which pH value is associated with each compound. (ii) Give a reason for your choice of pH of propanamide. [2] (i) For propanamide, pH = 7.0 For propanoic acid, pH = 2.8 For propanoyl chloride, pH = 1.0 (ii) Amide is neutral. Lone pair of electron on the nitrogen in amide group is not available to accept a proton due to presence of electron withdrawing (oxygen on the) carbonyl group. OR Lone pair on nitrogen interacts with π electrons on the carbonyl group. [Total: 20]

6 SRJC 2008 9746/03/Prelim/2008 [Turn over 2 Trichloroethanoic acid (TCA), CC l3COOH, widely used in biochemistry for the precipitation of macromolecules such as proteins. Solutions co ntaining trichloroethanoic acid are also used in tattoo removal. (a) Trichloroethanoic acid is prepared by the reaction of chlorine with ethanoic acid. CH3COOH + 3Cl2 → CCl3COOH + 3HCl (i) State the type of reaction for the preparation of trichloroethanoic acid. (ii) Suggest a suitable condition for this reaction. (iii) A by-product of this synthesis has a molecular formula of C 4H4O4Cl2. Draw a suitable structure for this compound. [3] i Free Radical Substitution ii Ultraviolet (uv) light / radiation iii HO CCHCHC O Cl Cl OH O OR CHCH2C Cl OH OCl CHO O (b) In a laboratory experiment, the decomposition of CCl3COOH was conducted at 70 oC and the volume of carbon dioxide produced was monitored using 0.100 mol dm -3 and 0.200 mol dm-3 of trichloroethanoic acid separately. CCl3COOH → CO2 + CHCl3

7 SRJC 2008 9746/03/Prelim/2008 [Turn over 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 0 . 00 . 51 . 01 . 52 . 02 . 53 . 03 . 5 Volume of CO2 / dm3 Time / hour Graph of volume of CO2 / dm3 against time / hour [TCA] = 0.100 mol dm-3 [TCA] = 0.200 mol dm-3 (i) From the graph, determine the order of reaction and the rate constant, k. (ii) Determine the half life of reaction using rate constant that you have calculated in (i). (iii) Explain, using Maxwell Boltzmann distribution curve, the effect of lowering temperature to 60 0C. [7] i From graph A: initial rate = 1.5/1.2 = 1.25 dm3 h-1 From graph B: initial rate = 4.0/1.5 = 2.66 dm3 h-1 As [CCl3COOH] increase two times, initial rate of reaction increases approximately two times. Hence order of reaction w.r.t. CCl3COOH is one. Using graph A,(or us