YJC H1 CHEM P1 soln

Yishun Junior College H1 Chemistry Preliminary Examinations 2016 Paper 1 Solutions 1 Volume of O2 reacted = 50 cm3; volume of CO2 = 30 cm3 CxHy + (x+y/4)O2  xCO2 + y/2H2O 10 x = 30 x = 3 3 + y/4 = 5 y = 8 C 2 2Br−  Br2 + 2e; Pb4+ + 2e−  Pb2+ 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑃𝑏𝐶𝑙4 = 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵𝑟2 = 6.98 349 = 0.02 𝑚𝑜𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵𝑟2 = 0.02 × 159.8 = 3.196 𝑔 C 3 A Cl2 + 2OH−  OCl− +Cl− +H2O Oxidation state of Cl decrease from 0 in Cl2 to -1 in OCl- and Cl- Change in oxidation state = -1 B 3Cl2 + 6OH−  ClO3− + 5Cl− + 3H2O Oxidation state of Cl decrease from 0 in Cl2 to -1 in Cl- and increase to +5 in ClO3- Change in oxidation state = +5 C 3MnO42− + 4H+  MnO2 + 2MnO4− + 2H2O Oxidation state of Mn decrease from +6 in MnO42- to +4 in MnO2 and increase to +7 in MnO4- Change in oxidation state = -2 D Cr2O72− + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ + 7H2O Oxidation state of Cr decrease from +6 in Cr2O72- to +3 in Cr3+ while oxidation state of Fe increase from +2 in Fe2+ to +3 in Fe3+. Change in oxidation state = -3 B 4 compound no. of unpaired electrons A CH3Cl 3 (around Cl atom) B O=C=O 4 (2 around each O atom) C H2N-NH2 2 (1 around each N atom) D NH4CN 2 (1 each around C and N atoms) B 5  bonds are formed from the head-on overlap of 2 s orbitals / 1 s orbital with 1 p orbital / 2 p orbitals.  bonds are formed from the sideways overlap of 2 p orbitals. D 6 Atom no. of electron pairs bond angle C1 4 (bond pairs) 109o C2 3 (bond pairs) 120o O3 4 (2 lone pairs & 2 bond pairs) 105o C

7 Standard enthalpy change of formation of CO: C(s) + ½ O2(g)  CO (g) A ½ C(s) + ½ O2(g)  ½ CO2(g) B ½ C(s) + ½ O2(g)  ½ CO2(g) C C(s) + O2(g)  CO2(g) CO2(g)  CO(g) + ½ O2(g) (inverse of CO(g) + ½ O2(g)  CO2(g)) C(s) + ½ O2(g)  CO (g) D C(s) + O2(g)  CO2(g) ½ CO2(g)  ½ C(s) + ½ O2(g) (inverse of ½ C(s) + ½ O2(g)  ½ CO2(g)) ½ C(s) + ½ O2(g)  ½ CO2(g) C 8 ⇒ Reaction is exothermic Enthalpy change of formation can be endothermic or exothermic (i.e. not always exothermic). Enthalpy change of combustion, enthalpy change of neutralization and lattice energy are always exothermic. B 9 ∆𝐻𝑓 = Σ∆𝐻𝑐(𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) − Σ∆𝐻𝑐(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) = 4(−394) + 5(−286) − (−2877) = −129 𝑘𝐽 𝑚𝑜𝑙−1 B 10 A catalyst provides analternative route with lower activation energy. D 11 Comparing expts 1 & 2, 6 ℎ = (2 1) 2 ℎ = 1.5 Comparing expts 1 & 3, (0.5 1 ) (2 1) 2 = 6 𝑖 𝑖 = 3 Comparing expts 1 & 4, (0.5 1 ) (2 𝑗) 2 = 6 0.75 𝑗 = 0.5 B ∆H < 0

12 When bromocresoll-green is blue, the pH of the soltuion is greater than 5.5. When phenol- red is yellow, the pH of the solution is less than 6.8. Hence, the pH range is 5.5 < pH < 6.8. A pH < 7 B pH = 7 C pH > 7 D pH > 7 A 13 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻+ = 2 × 0.002 = 0.004 𝑚𝑜𝑙 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂𝐻− = 0.003 𝑚𝑜𝑙 Hence, OH- is limiting and there is an excess of 0.001 mol of H+. [𝐻+] = 0.001 ÷ 2 = 0.0005 𝑚𝑜𝑙 𝑑𝑚−3 𝑝𝐻 = − lg 0.0005 = 3.3 B 14 A Aluminum is readily oxidized by oxygen to form aluminum oxide which prevents further attack by oxygen. Aluminum oxide is amphoteric. B Magnesium oxide burns with a bright white flame to form white solid, MgO which is basic. C When heated strongly, silicon forms white solid, SiO2 which is acidic. D Sulfur burns with a blue flame to form a colourless gas, SO2 which is acidic. D 15 A Electronic configuration of Cl- (18 e-): 1s22s22p63s23p6 Electronic configuration of Na+ (10 e-): 1s22s22p6 Hence, Cl- has one more occupied shell than Na+ and so is larger. B True statement but does not explain why Cl- is larger than Na+ C Incorrect statement. Ionic radius decreases across the period for the cations and anions. Sharp increase in ionic radius when there is a change from cations to anions. D True statement but does not explain the trend in ionic radius. A 16 Trend in melting point: Si > Al > Mg > P Trend in IE: Al < Mg < Si < P C 17 A Products formed should be non -toxic but toxic CO is formed in this reaction. Also, C8H16 is not octane. B C8H16 is not octane C Products formed should be non-toxic but toxic CO is formed in this reaction. D Correct equation. D 18 A From the equation given, the C=C bond is used and the product’s side -chain is saturated (i.e. all single bonds) and cyclic. B The side chain is still unsaturated. C The side chain is still unsaturated. D The side chain is not cyclic. A 19 B COOH dil. H2SO4 heat CN alcoholic KCN reflux Br 2-bromopropane

20 Alkene groups in limonene undergo addition reaction with Br2. CH2BrBr Br Br D 21 A CH3(CH2)3CH2OH  CH3CH2CH2CH=CH2 (1 alkene formed) B CH3C(CH3)2CH2OH  cannot undergo elimination C CH3(CH2)2CH(OH)CH3  CH3CH2CH2CH=CH2 + CH3CH2CH=CHCH3 (exists as a pair of cis-trans isomers) (Hence, 3 alkenes formed) D CH3CH2CH(OH)CH2CH3  CH3CH2CH=CHCH3 (exists as a pair of cis-trans isomers) C 22 Reduce Tollens’ reagent to form a silver mirror  aldehyde group present CH3CH2CH2CHO + CH3CH(CH3)CHO Hence, there are 2 structural isomers. B 23 CH3CH2CH2CH=CHCHO contains alkene and aldehyde groups. CH3CH2CH(CH3)COCH2CH3 contains a ketone group. A Aldehyde group is the product from the oxidation of a primary alcohol while ketone group is the product from the oxidation of a secondary alcohol. B Only the alkene group can decolourise bromine. C The ketone group cannot decolourise KMnO4. D Only the aldehyde group gives a positive test with Fehling’s reagent. A 24 A The ester groups undergo hydrolysis with hot dilute sulfuric acid. The structure of the main product formed is OH O OH O OH Hence, the product has only 1 carboxylic acid group. (The other product formed is ethanoic acid.) D Br2 in CCl4 alkene 1 ester 1 ester 2 ketone alkene 2

B The alkene groups undergo oxidative cleavage. Alkene 1 forms 1 carboxylic acid group while alkene 2 does not form any carboxylic acid groups. The ester groups undergo hydrolysis (due to the hot acidified conditions). (Refer to option A for the products of hydrolysis of the esters). Ester 1 produces a carboxylic acid group and an alcohol group. Ester 2 produces an alcohol group and ethanoic acid. Both alcohols formed from the hydrolysis of the esters are secondary alcohols so they will be oxidized to ketones. Hence, the product has 2 carboxylic acid groups (from the oxidative cleavage of alkene 1 and hydrolysis of ester 1). C HBr is added to the C=C double bonds, i.e. C = C + HBr  - C – C - Hence, only 1 Br atom is added to each C=C double bond. Since there are 2 C=C double bonds, 2 Br atoms are added. D Only the ketone group will undergo addition with HCN and NaCN to form a hydroxynitrile, i.e. C = O + HCN  - C - OH Since there is only 1 ketone group, the product will contain only 1 N atom (from the –CN group). 25 CH3CH2OCOCH2CHCH2CH3 CH3 C Hence, the alcohol is CH3CH2OH and the carboxylic acid is CH3CH2CH(CH3)CH2COOH. 26 Uuq Uuh proton number 114 116 nucleon number 289 292 no. of neutrons 175 176 1 Refer to the above table. (No. of neutrons = nucleon no – proton no) 2 No. of electrons in 1 Uuq2− ion = 114 +2 = 116 (same as the no. of electrons in 1 Uuh atom) 3 No. of electrons in 1 Uuh+ ion = 116 – 1 =115 No. of electrons in 1 Uuq− ion = 114 + 1 = 115 A H Br CN from alcohol from acid

27 1 Position of equilibrium will shift to the right, favouring the forward reaction that produces lesser amount of gases so as to reduce the increased pressure. Hence , the yield of SO3 increases. 2 Increasing the temperature will increase the rate of the forward and backward reactions. 3 No. of moles of SO3 = no. of moles