PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 1 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 1 (a) [ glucose] o – [ glucose] = kt @ (100) – (90) = k (22) k = 0.4545 Ms-1 [ glucose]o – [ glucose] = kt (100) – (0) = 0.4545t t = 220s 1 1 1 1 4 1 (b) i ii m = -147 ( -140 to -150) -m = - Ea R Ea = 1222 Jmol- @ 1.222 kJmol- ( 1164 Jmolto 1247 Jmol- ) Catalyst. Addition catalyst will lower the activation energy and increase rate of reaction Axis – 1 Linear graph – 1 1 1 1 1 6 TOTAL 10 -1.6 -1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0 0 0.001 0.002 0.003 0.004 0.005 0.006 ln k 1/T
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 2 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 2 (a) M(s) + Cl2 (g) MCl2 (s) M(g) M + (g) M 2+(g) + 2Cl- (g) 2Cl (g) 1-2 √ -- 1m 3-4 √ … 2 m All √ --- 3m ∆H f =∆Ha +IE1 + IE2 + Lattice energy @ -799 kJ = 178.2 kJ + 590 kJ + 1145 kJ + 121.5 (2) kJ + (-348.6 x 2) kJ + LE Lattice energy of MCl2 = -2258 kJmol- lattice energy of PCl2 > lattice energy of MCl2 Thus M has bigger ionic radii / bigger in size 3 1 1 1 1 7 2(b) Mass ethanol = density x volume = 0.79 gcm-3 x 1000 cm3 = 790 g Mole ethanol = 17.174 mol ∆H c of ethanol = -1430 kJmol- = q n q = 1430 kJ X 17.174 mol q = 24558 kJ @ 1 mol ethanol ≡ released 1430 kJ heat …….. 1m So, 17.174 mol ≡ 24558 kJ heat ……… 1m 1 1 1 1 4 TOTAL 11 MAX 10 Ha = +178.2 kJ IE1 = +590 kJ IE2 = +1145 kJ Ha = +121.5 kJ (2) EA = - 348.6 kJ x 2 Lattice energy Hf = -799 kJ
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 3 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 3 (a) Anode : 2H2O (l) → O2 (g) + 4H+ (aq) + 4e Cathode : Au3+(aq) + 3e → Au(s) 4 1 1 6 3 (b) Q= It = 7200 C 3F ≡ 3x 96500 C ≡ 1 mol Au deposited ≡ 197 g 7200 C ≡ 4.9 g 1 1 2 TOTAL MAX 8 7 Anode Cathode Anode & Cathode --- 1m Battery + ammeter --- 1m Electrodes ( Pt & silver spoon) ---1m AuCl3 (aq) ---------- 1m
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 4 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 4 (a) (b) (c) CH2 A O CH3 B C CH2 A O ii. Zn, H2O + CH2O CH2 CH3 A H2 , Pd CH3 Br 1+1+1 1 1 1 6 TOTAL 6 B C
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 5 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 5 (a) CH CH3 CH3 P CH CH3 CH3 Br Q CH CH3 CH3 Br R 1+1+1 3 (b) Bromination test Benzene CH CH3 CH3 P Observation Reddish brown of bromine remains Reddish brown of bromine decolourised Equation Br2 , CH2Cl2 uv No observable change CH CH3 CH3 Br2 , CH2Cl2 uv C CH3 CH3 Br 1 2 Equation 2 5 TOTAL MAX 8 7
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 6 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 6 (a) G CH3CH2C=CH2 H CH3CH2C=CH2 H HBr, H2O2 CH3CH2CH2CH2Br 1 2 (b) CH3CH2CH2CH2OH H CH3CH=CHCH3 I 1 1 (c) C Br H H H3CH2CH2C HOC H H CH2CH2CH3 HO Br HO C H H CH2CH2CH3 Br- + Type of reaction: Nucleophilic substitution 2 1 TOTAL 8 MAXIMUM 7
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 7 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 7 (a) AA= @ BB= Cl CC= O CCH3 O X = KMnO4 , H+ , heat O - Na+ DD= 1 1 1 1 1 TOTAL 5
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 8 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 8 (a) O O - Na+ + CHI3 M N 2 (b) I = LiAlH4 II= H3O + 2 (c) Test Brady’s test Equation J: H2N NH NO2 NO2 + CH3 O CH3 C=N N NO2 NO2 H H Observation Orange precipitate formed Equation L : + CH(OH)CH3 H2N NH NO2 NO2 no observable change Observation No orange precipitate formed. Or Lucas test Test-1 Both observatio n-1 Equation1 TOTAL 7
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 9 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 9 (a) i. ii. iii. CH3CH2C OH CH3 H R C O CH(CH3 )CH2CH3 O Q = C OH O + CH3CH2C OH CH3 H H + C O CH(CH3 )CH2CH3 O CH3CH2C OH CH3 H I2 ,NaOH CH3CH2C OO + CHI3 Observation: yellow precipitate formed 2 1 Reagent1 Product-2 1 (b) 1-butanol< butanoic acid< 2-fluorobutanoic acid<2,3-difluorobutanoic acid • 2,3-difluorobutanoic acid has more halogen (2 Cl atom) which is EWG. More EWG weaker the O-H bond, easier to produce H + . Acidity higher. • 1-butanol has alkyl group which is EDG. EDG destabilized alkoxide ion. Acidity lower. 1 1 1 (c) H2 N (CH2 ) 5 COOH 1 TOTAL 11 MAXIMUM 10
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 10 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 10 (a) NO2 NH2 conc HNO3 , conc H2SO4 SnCl2 , H + Reagent-2 Product nitrobenze ne-1 (b) NaNO2 , HCl 0 - 50 C 1 (c) Test Bromine water test Equation NH2 Br (aq) NH2 Br Br Br Observation white precipitate formed Equation H3C N H3C CH3 Br (aq) no observable change Observation No white precipitate formed. @Hinsberg’ test @ reaction with nitrous acid Test-1 Both observatio n-1 Equation1 TOTAL 7
PRA PSPM 2 KOLEJ MATRIKULASI KEDAH 11 CHEMISTRY SK025 2022/2023 SET 2 NO ANSWER SCHEME MARKS 10 (a) CH2CH(NH3 )COO Phenylalanine + - 1 (b) C H CH2 H2N C = O N = C H CH2 C OH H O dipeptide 1 (c) Peptide bond 1 (d) CH2CH(NH2 )COONa + 1 TOTAL 7