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