OCR 2018 HCI H2 Chemistry Promo Answers

HWA CHONG INSTITUT~ON 2018 C1 H'2 ~ CHI EMISTRY PIROMOTIONAL EXAM~NAT~ ~ ONS .SUGGESTED SOILUTIO INS Paper 1 1 2 3 4 5 6 7 8 9 10 D B D ,c B ,c B .A A c 11 12 13 14 15 16 17 18 19 20 D ,c A B D B A D ,c D Comments 1 ID The Ia rge diP' between D and E shows ttl at ttl'e electron to be remo .ved from E is. in the next quantum shell . Hence D has the e~e ctron ic oonfig uration of a noble g.as ns2 n p6 and belongs to' Group 18. 2 IB To do this question quickly ~ you shou lid clhedk tlhe group number of the given e~ e1me nts a.nd recall tile gro up,'s val ,ence e~ ectJronie conflg uration. Ga - Group 13 (ns 2 np1) Se - Group 16 (ns2 np4 ) Cs - Group 1 1(ns 1) H!enae on~ Ga .and s ,e ·fulfil bo~h criteria stated . Cs does not fulfil the 1st crit,erion . 3 ID For this que stl on 1· you wi II ha1ve to wolik out the sllru ctures for all tlhe given species and ~h e1i r number of bond pairs (bp) vs number of lone pairs (I p) to deduce their shapes and bond angles . ] ]I! A N03- 3 bp NO- 2: bp, 1 lp (120°) .2 (<1120°) B GlE.- 4 bp·, 2 lp SIFG 6 bp (QOO) (90°) c XeF4 4 bp, 2 lp BrFs 5 bp, 1 lp (QOa) (<QOa) D Bro~- 3 bp', 1 lp ClDl- 3 bp, 1 lp (104°) (1 OSO.) For option D ~ although both have the same shape and number of b p and lp· ~ Br is less ~el ~ectronegative than Cl. H.enoo its bonding electrons are furttler from the central atom . Hence the bond angle is small ,er as mpu~sion between the bond pa.irs is sm.aller . 2018 HC I C 1 H2 Chemistry Pro motional Exam 4 Cl CJ H Option A: H HI has a dipo1 e moment and he nee is polar . There are permanent dipal ,e-permanent dipo~e interactions betwe .en the molecul ,es. in addition to C! IH H d is.pers.ion foroes.. lrhe r'e is on ~y dispersion forces between molecul ,es of H CJ as the mol,ecule has no, dipo~e moment and is non-polar. Hence tile tis isomer has higher bailing point (bp) than the tr:ans isomer. Option B: Hydrogen bonding (CHJCH120H) is stronger than permanent dipol ,e­ '0 II CH .... C v \ permanent dipole ( H ) interacUons. Hence C HJCH 20 H has higher boiling paint . Option tC: I Bo~h m ol,ecul ,es have hydrogen bonds pres ·ent. One r~ eason for why ethano ic acid has higher boiling point is because mo~erules of ethanoic acid form dim em and this also ' leads to' stronger dis pen~~ on forces lbetwee n its m·ol,ecul ,es .. 1(refer to Chapter 21 pa.ge 95 of Foundation Topic boo ld,et) OHI OH Option 1!) : has intr.amo ~ecu lar hyd rag en bonding between the two OH g mu ps,t hen ae it will have ~ess extens ive intermol eru lar hydrogen bonds and ~ower lbp. 5 B For C02~ since bolll C=O bonds have dipo lles o,f ~ equa. l m.agnitude and in opposit~e d irectio llS 1 they ca noel each oilier off and r~ esulted in overall 0 po~a rity . For COS~ the overa ll! polarity is lower than that ofCOSe . That means thatC=S bond is more pol.ar (larg er dipole moment) than C=Se as it is able to cance~ out mo:re of C=O dipo 'e moment This leads t~o a lower o,verallpolarity o,f 0. 71 . This also follows the trend o,f ~ el ~ectronegativity decr,easing do·wn the group (0", S and Se are a II Group ~ 16 e ~e ments) I hence I eadi ng to decreasing di iffer~en ce in elie otroneg .ativity values between C .and ttl'e G1ro up, 16 ellements (C =X bonds are l ~ess polar down the group~ ).

7 C To p ns~d ict ~he graph ical m ~atio ns hip between two' variab les1 you can rearrange the ideal gas equation p V=n RT in tlh e ·form J' = mT + c ~ according to what is given in the .x and y· axis,. A p= nR.lr _!_ Hence th is • y- kx graph where k=nRT. At higher IS ,a ...... .... .. . v te1mperatu r'e T2~ tJhe grad i,ent oflle line should be :steeper . (wrong graph) B pV - nRT. Since pV • ,always constant and not affected by value o,f P ~ - IS a this , y=k graph where k=nRT. At T:z, the line should be at higher IS 8 a value . (wrong graph ) c pV nRT. Them • • relationsh ip, between and v For • - IS an Inverse p ' a grven - . volume ., the pressur 'e should be higher at T.2. Hence the graph should :shift outwards . (correct graph ) D niRT Hence this , y=lkx gr,aph where Rl At 1"2, the gradient of the v- I:S a k=-- - . . p p line should be steeper . (wrong graph) IB n(INH2CONH :z) = 0.150 -:- 60 = 0.00250 mol TJh,e volu m~ es of gases was measur ,ed at room te1mp eratu r'e and pressure ~ thus H:zO exists as a I iq uid. C02 is an acidic gas ~ ain d wil ~ be r~ em o:ved by the a.lkal ine INaOHL TJh , erefor,e ~ the remaining volume cons ists only of N2 gas .. n( N2) = 0 .00250 x 2 = 0.00500 mol Vo ~ of g,as after p~as sing tlh mug h NaOH ::::::: 0.00500 X 24 = 0.12 dm3 .,.2o··· 3 = 1 .. em · 8 A Tlh e mola.r mass of the sample of pallad ium is higher than the mol,ar mass of isotopically pure 100 Pd. This outcome can only occur when a heavier isotope of Pd is present in the sample . Since 108 Pd is the on~y option which fits th is criteria, , 108 Pd must be the source of the d is·crepancy . Note that no callcul,artion is r~ eq ui red I nor should it be attempted , in tlh is qu es ~ion .. 2018 HC I C 1 H2 Chemistry Pro motional Exam 9 A 1 Given that &W r· , negative ,and AS"'r . negative 1(becaus ·e o,f the decrease I:S IS . number of mo~es of gas), this reaction . feasib1e at ~ow T. The - lrA .. SS' 1n IS term . m 1s, pos vel and at high T mary become la.rger ~han lle negative AW r_ 2 Students may use this formula to' calcu llate tlh e e ntlhal py change of reaction : .d. ~r = LAH£tc (reactants )- LAH£tc{products ).. 3 Nickel and the r~ eaotants exist in different phases . INi is a so lid whi ~e etlhanal and hydrogen ,are both gases in this reaction . 4 Both the alkene and a. ~dehyde funct ional groups win be reduced by hydrogen under the . conditions . lrhe correct product should be g1ven OH 11 C Option A: Tlherma .l decomposition of solid CaC0 :3 willl form CO:z(g) and incre ,ase ~he number of gas ~eous p~, artJides .. Hence AS i:s po~ sitive as them are more ways, that partic~es ,and the energy can be distributed in gases a.s com pared to sol ids . 11 Jj) Option B~ : Sublimat ion causes so lid C02 to become gaseous C0:2. He nee AS is positive . Opt ion 'C: lin fu e process of rustn ng I 0.2 (g) reacts with sol id iron to form solid iron oXJide. Hence there · s a decrease in the number of gaseous particles and AS is n ~ egaUve , . Opt ion li): The react ion between NaO H and am moni urn ch ~or ide produces N H 3 1(g ). Hence AS is positive . 11 H+ does 1n10t fulfil II the criteria of a1 ~cata ~yst as it is consumed without being regenerated . 2 Both are formed from reactants and consumed to fo1nm products iin subsequent steps . Yes ~ tlh ey a.re intermed iates of the reacllioin. 3 Step 1 is tlh'e slow step1, ther,efore it must have the highest Ea. Step, 2 therefore must have a lower Ea than step 1. 4 B~ ased on the mechan " sm ~ the s)ow step (step 1 ) indicates tlhat the rea,ctio n is first order with r~espect to' H202 and first order with respect to 1-. 1.2 'C You shou ~d work out the in ilia I p r~ essu re of 02 for e,ach of the ~experiments by subtracting initial pressu r'e of NO from total in mal pressu m . Experiment Tota l initial pressure I atm Initial pressure o'f NO I atm llnilti!al pressu fie of 0:2 I atrn In itia 1 mte of r&action I atm s-1 II 11.00 0.40 0.60 11.08 Ill 1.6'0 0.40 1m20 2.16 IIIII IIV 2 .00 X 0.80 0.2{) 1.20 x - 0.20 8.64 1J)8 2

13 14 A IB From experiment I and n ~ the pns~ssum of NO rema ined uncha nged wh ile pmss.ure of 0 2 is. doubled . Rate is d ou bl,ed and this indicates 1h at read:i on is. first order with m spect to 0 2 .. From experiment 1111 and Ill , the pr,essum of 0 2 remai rui~d undhanged whHe pmssur~e of NO is doub l,ed. Rate is quadrupled .and this indicates tlhat reaction is second order with respect to' NO. A IB c ID 1 R.eactJi,on is second o:rde r with r~ es peel to N 0 . The rate equation shou lid be rate = lcf''No2 Po2. To fin