VJC Prelim P1 P2 ANS

1 2009 H2 Chemistry 9746/2 Suggested Answers Answer all the questions in the spaces provided. 1 Hydrogen cyanide was first isolated from a blue dye (Prussian blue) which had been known since 1704 and is now known to be a coordination polymer, FeIII 4[FeII(CN)6]3. In the year 2000, 732,552 tonnes of HCN were produced in the US. The most important process for HCN production is the Andrusso v oxidation in which methane and ammonia react in the presence of oxygen at about 1200 °C: 2CH 4 + 2NH3 + 3O2 → 2HCN + 6H2O The energy needed for the reaction is provided by the partial oxidat ion of methane and ammonia. (1 tonne = 1 000 kg) HCN is also obtainable from fruits that have a pit, such as cherries, apricots, apples, and bitter almonds. Many of these pits contai n small amounts of cyanohydrins such as mandelonitrile and amygdalin, which slowly release hydrogen cyanide. An article mentioned that 100 g of crushed apple seeds can yield about 10 mg of HCN. (a) (i) Draw the structure of the anion of Prussian blue to sh ow its shape. Indicate on the structure the bond angle and give the name of the shape. Structure: Fe CN CN CN CN NC NC 4- 90 O Bond angle must be ON the structure. Correct structure drawn (pls draw dotted lines to show square plane). Shape: Octahedral (ii) Using specific examples fr om the information above, state two characteristic properties of transition elements. They are able to show variable oxidation states. (e.g. FeIII 4[FeII(CN)6]3 , FeIII is +3 O.S. but FeII is +2 O.S.) © VJC 2009 9746/02/PRELIM/09 [Turn over

2 They are able to form stable complex compounds or ions. (e.g. FeII(CN)6]3 4- complex ion) They are able to form coloured compounds and ions. (e.g. Prussian blue) [Only two properties required. For each property, specific example must be given.] (iii) Find the percentage by mass of cyanide in Prussian blue. Percentage by mass of cyanide in Prussian blue = ) x55.8 ( + ] . +3[(26.0x6) × x . 4 8 55 3 6 0 26 x 100% = 858.6 468.0 x 100% = 54.5 % [5] (b) (i) Assuming that the amount of HCN produced in the US annually is the same, calculate the number of moles of HCN produced in the US from 2000-2002. Total mass of HCN produced in 3 years = 732,552 x 3 = 2.20 x 106 tonnes Mr of HCN = 1.0 + 12.0 + 14.0 = 27.0 No. of moles of HCN = 0 . 27 x102.20x10 6 6 = 8.15 x 1010 mol (ii) What is volume in dm 3 of oxygen gas required via the Andrussov oxidation to form the amount of HCN mentioned in (i) at r.t.p.? 3 O2  2 HCN No. of moles of O2 = 3/2 x 8.15 x 1010 = 1.22 x 1011 mol Volume of O2 at r.t.p. = 24 x 1.22 x 1011 = 2.93 x 1012 dm3 [2] (c) An apple has an average of 4 seeds and eac h seed weighs an average of 0.20 g. Find out the mass of HCN that can be yielded from seeds from 10 apples. Mass of seeds from 10 apples = 10 x 4 x 0.20 = 8.00 g Mass of HCN that can be produced from 10 apples = 10 x 100 00 . 8 = 0.800 mg [1] © VJC 2009 9746/02/PRELIM/09 [Turn over

3 (d) (i) HCN gas is a highly poisonous gas t hat can cause permanent damage to the central nervous system. The toxic level of HCN gas in the air is about 0.001 mg dm-3. Express the toxic level of HCN gas in the air in terms of mol dm -3. Toxic level of HCN = 27.0 0.001x10 3 = 3.70 x 10-8 mol dm-3 (ii) How many molecules of HCN are pres ent at this toxic level in 2 dm3? No. of molecules of HCN at this toxic level = x2x6.02x1027.0 0.001x10 23 3 = 4.46 x 1016 molecules (iii) When it rains, HCN dissolves in water to give an acid solution. If 2 dm 3 of air containing the toxic level of HCN dissolves in 20 dm 3 of rainwater, find the concentration of HCN in the acid solution. No. of moles of HCN in 2 dm 3 of air = -83.70 x 10 x 2 = 7.40 x 10-8 mol [HCN] = 3 3 -8 10 x 10 x 20 10 x 7.40= 3.70 x 10-9 mol dm-3 [3] (e) Name and describe the mechanism for the formation of cyanohydr ins starting from propanal and HCN. State any additional reagent that is needed. Name: Nucleophilic addition Additional reagent: NaCN (catalyst) or NaOH (trace amount) CN– functions as a nucleophile and it attacks the electron deficient carbonyl carbon. Both curly arrows and partial charges on r.d.s. (slow step) must be shown. O CH3CH2C H+ – CN O CH3CH2C C H N + The tetrahedral intermediate is a strong base and quickly captures a proton from HCN. O CH3CH2C C H N CNH O CH3CH2C C H N H + + CN + – CN– catalyst is regenerated which then attacks another carbonyl molecule. [Total: 14] © VJC 2009 9746/02/PRELIM/09 [Turn over

4 © VJC 2009 9746/02/PRELIM/09 [Turn over 2 This question is about Group VII hydrides, HX. (a) The Group VII hydrides show clear trends in the H– X bond energy where X = C l, Br and I. They behave as weak acids in liquid ethanoic acid, dissociating to different extents. (i) Write an equation to show how HX dissociates in ethanoic acid. HX + CH3CO2H ⇌ CH3CO2H2 + + X- (ii) Rank the pKa of the three Group VII hydrides in ethanoic acid in decreasing order. Explain your answer with reference to the Data Booklet. pKa of acids: HCl > HBr > HI From the Data Booklet, Bond Bond energy / kJ mol -1 H—C l 431 H—Br 366 H— I 299 Since the strength of H-X bond decreases down the group due to the increasing atomic radii of the halogen atom (or poorer orbital overlap between the halogen and hydrogen atoms), it becomes easier to dissociate HX down the group. [2] (iii) Suggest why the Group VII hydrides are weak acids in ethanoic acid whereas they are strong acids in water. Ethanoic acid is a weaker base/stronger acid than water. Hence, ethanoic acid is less likely to accept a proton from HX. Or Since ethanoic acid is a st ronger acid than water, it will partially dissociate into H + ion in aqueous state, hence, causing the position of equilibrium of HX ⇌ H+ + X- to shift left by commo n ion effect/ suppress the dissociation of HX by common ion effect. Group VII hydrides are weak acids in ethanoic acid. [4] (b) Another Group VII hydride, HF, behaves as a weak acid. A solution is prepared by dissolving HF in a fixed volume of water. The addition of some barium hydroxide to the solution neutralizes one thir d of the HF present. The pH of the resulting solution is 2.88. (i) Calculate the concentration of hydrogen ions in the solution after barium hydroxide is added. [H+] = 10-2.88 = 1.32 x 10-3 mol dm-3

5 (ii) Using your answer to (i) and the given information, calculate the value of the acid dissociation constant, Ka, of HF. K a = [H+][F-] / [HF] = 1.32 x 10 -3 x (1/2) = 6.60 x 10 -4 mol dm-3 [3] (c) The figure below shows two cells connected in series. Cell 1 and Cell 2 involve the electrolysis of saturated solution of CuBr2 and concentrated HX respectively, where X is one of the Group VII elements. Pt Saturated CuBr2(aq) Pt Cell 1 Cell 2 Concentrated HX (i) Using E θ values, write ion-electron equations, for the reactions occurring at the cathode and ano