PJC H1 CHEM P2 ANS

1 Section A Answer all the questions in this section in the spaces provided. 1 This question is about aluminium, an elem ent in the third peri od of the Periodic Table. (a) (i) On the grid below, sketch the trend of the first ionisation energies of elements of Period 3 from sodium to argon. [1] (ii) Write an equation for the first ionisation energy of aluminium. Al(g) → Al+(g) + e- (iii) Explain why the first ionisation ener gy of aluminium is lower than magnesium. first electron of A l is removed from a higher sub-energy level as compared to that of Mg [3] (b) A beam of particles of 27Al+ and 9Be+ is passed through an electric field. If the angle of deflection of the 27Al+ particle is 3o, state the angle of deflection for the 9Be+ particle. 9 o [1] For Examiner’s Use first ionisation energy / kJ mol -1 Na Mg Al Si P S Cl Ar element

2 A lCl3 is a covalent compound that exists as a white solid at room temperature and pressure. When A lCl3 is heated to a sufficiently high temperature, it sublimes to gives the Al2Cl6 dimer. (c) Draw a dot-and-cross diagram for the A l2Cl6 dimer. Name the type of bond responsible for the formation of the dimer. Type of bond: _dative bond _ [2] (d) 2.0 mol of gaseous A lCl3 is heated at 800 K in a 750 cm 3 vessel. When dynamic equilibrium is achieved, 0.971 mol of gaseous Al 2Cl6 is formed in the vessel. (i) Define the term dynamic equilibrium. rate of forward reaction equals rate of reverse reaction (ii) Write a balanced equati on for the equilibrium. 2AlCl 3 ⇌ Al2Cl6 (iii) Calculate the value and state the units of the equilibrium constant, K c, for this reaction at 800 K. equation 2AlCl3 ⇌ Al2Cl6 initial amount / mol 2 0 change in amount / mol -2(0.971) +0.971 equilibrium amount / mol 0.058 0.971 equilibrium [ ] / mol dm-3 0.058 / 0.75 = 0.07733 0.971 / 0.75 = 1.2947 [1 (amount) +1 (concentration)] Kc = 2 3 6 2 ] [ ] [ AlCl Cl Al = 2) 07733 . 0 ( ) 2947 . 1 ( = 216 mol-1 dm3 [1] [5] For Examiner’s Use

3 (e) AlCl3 dissolves in water to give an acidic solution, as shown by the following equation. [Al(H2O)6]3+(aq) + H2O(l) ⇌ [Al(H2O)5(OH)]2+(aq) + H3O+(aq) (i) Define the term Bronsted base . Hence, identify the two Bronsted bases and state one conjugate acid-base pair in the equation above. Definition of Bronsted base: _proton acceptor _. The two Bronsted bases are _H2O_ and _[Al(H2O)5(OH)]2+ One conjugate acid-base pair: [Al(H 2O)6]3+ and [Al(H2O)5(OH)]2+_. OR H3O+ and H2O (ii) Write an expression for the acid dissociation constant, Ka, of [Al(H2O)6]3+. Ka = ] ) ( [ ] ][ ) ( ) ( [ 3 6 2 3 2 5 2 + ++ O H Al O H OH O H Al (iii) When 0.1 mol of A lCl3 is dissolved in 1 dm 3 of solution, the resulting solution is found to have a pH of 3. Using your equation in (ii), calculate the Ka of [Al(H2O)6]3+. [H+] = [Al(H2O)5(OH)]2+ = 10-3 = 1.0 x 10-3 mol dm-3 Ka = 1 . 0 ) 10 0 . 1 (2 3−x = 1.0 x 10-5 mol dm-3 O R Ka = ) 10 0 . 1 1 . 0 ( ) 10 0 . 1 ( 3 2 3 − − − x x = 1.01 x 10-5 mol dm-3 [6] [Total: 17] For Examiner’s Use

4 2 Cyanohydrins can be made by reacting ketones with an acidified solution of sodium cyanide. When propanone is reacted as described, compound T is formed. (CH3)2CO + H+ + CN- → T (a) Draw the displayed formula of compound T. On your diagram, state the type of hybridisation of all the carbon atoms in T. CC C O C H H H H H H H N [2] For Examiner’s Use sp3 sp An experiment is performed in which 2.00 mol dm -3 of propanone is reacted with an excess of an acid and sodium cyanide. The following results are obtained. time / s [(CH3)2CO] / mol dm-3 0 2.00 15 1.30 30 0.84 45 0.55 60 0.35 75 0.23 (b) (i) On the grid provided below, plot a graph of these values on suitable axes.

5 (ii) Using your graph, deduce the order of reaction with respect to propanone. workings of two t½ on graph t ½ is constant at 24.5 s order of reaction w.r.t. propanone is 1 (iii) Why are the concentration of the acid and the concentration of sodium cyanide used in excess? to ensure their concentration rema in relatively constant during the reaction and not affect the change in rate of reaction [5] (c) When the experiment is repeated using 1.00 mol dm -3 of the acid and an excess of propanone and sodi um cyanide, the following Graph 1 of concentration of the acid against time is obtained. Graph 1 Graph 2 O n Graph 2 above, sketch a graph for the rate of reaction of the acid against the concentration of the ac id. Hence, state the order of reaction with respect to the acid. Order of reaction with respect to the acid: _ zero order [2] (d) The order of reaction with respect to sodium cyanide is one. Together with your answers in (b)(ii) and (c), write the rate law for this reaction. rate = k [(CH 3)2CO] [CN-] [1] rate [acid] / mol dm-3 For Examiner’s Use [acid] / mol dm-3 time / s

6 (e) (i) Propanal is a functional group isomer of propanone. Suggest a simple chemical test to distinguish be tween propanal and propanone. State any observations seen. Test Propanal Propanone I2(aq), NaOH(aq), heat no yellow ppt seen yellow ppt of CHI3 seen K2Cr2O7(aq) / KMnO4(aq), H2SO4(aq), heat purple MnO4 - decolourised / orange Cr2O7 2- turned green Tollens’ reagent, warm / heat silver mirror seen Fehling’s solution, warm / heat brick red ppt of Cu2O seen no observable change (ii) Propanal can be converted to pro panone in the laboratory as follows. CH 3CH2CHO X C H 3CH=CH2 C H 3COCH3 CH 3CH(OH)CH3 Identify compound X. State the reagent and conditions for Step I and III. Compound X: CH3CH2CH2OH Step I: LiAlH4 in dry ether OR H2(g), Ni, 200 oC Step III: (i) concentrated H 2SO4 (ii) H2O(l), heat OR concentrated H3PO4, H2O(g), 300 oC, 60 atm (iii) Draw the structural formula of t he compound formed when propanal reacts with 2,4-dinitrophenylhydrazine. CH 3CH2CH=N-NH2C6H3(NO2)2 [6] [Total: 16] For Examiner’s Use I II III IV

7 3 Olive oil is a popular flavouring us ed in Mediterranean cuisine. The main constituent of olive oil is oleic acid, CH3(CH2)7CH=CH(CH2)7CO2H. (a) Name the functional groups that are present in oleic acid. alkene and carboxylic acid [1] (b) State the reagent and condition to synthesise oleic acid from CH3(CH2)7CH=CH(CH2)7CH2OH. K 2Cr2O7(aq), H2SO4(aq), heat [1] (c) State the reagents and conditions needed to convert oleic acid into each of the following. compounds reagents and conditions CH 3(CH2)7CH(OH)CHBr(CH2)7CO2H Br2(aq) CH 3(CH2)7CH(OH)CH(OH)(CH2)7CO2H KMnO 4(aq), H2SO4(aq), cold (if NaOH(aq), no full credit] CH 3(CH2)7CHCl(CH2)8CO2H HCl(g) [3] (d) Another constituent of o live oil is linoleic acid. Linoleic acid has the same functional groups as oleic acid. When linoleic acid is oxidised, the following compounds are obtained in equimolar quantities. CH3(CH2)4CO2H, HO2CCH2CO2H, HO2C(CH2)7CO2H Write the two possible structural formula of linoleic acid. CH 3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H a n d C H 3(CH2)4CH=CH(CH2)7CH=CHCH2CO2H [2] [Total: 7] For Examiner’s Use

8 Section B Answer two questions f