EJC 2021 GCE A-Level H1 Chemistry Paper 1 Suggested Solution

Eunoia Junior College 8873 H1 Chemistry 2021 Paper 1 Suggested Solution 1 Atom Q has 70 n + p and 10 more n than p. Thus there are 40 n and 30 p  Q is 30Zn. Atom R has 70 n + p and 2 fewer n than Q. Thus there are 38 n and 32 p  R is 32Ge.  A 2 Small jump between 4 th and 5 th I.E. and large jump between 6th and 7th I.E.  ns2 np4 configuration  Group 16  C 3 A : B : C : D ✓:  D 4 Since N is in Group 15 with 5 valence electrons, N gains 3 electrons to give the monatomic nitride anion, N3–, with an octet. For electrical neutrality, the formula of the solid is (Zy+)3(N3–)y. If the solid is ZN 3, y = 9 which is not possible since Z9+ does not exist. If the solid is Z 3N, y = 1 which is possible for Z+ cations.  B 5 Cyanogen is 1 ✓: There are two  bonds about each C, hence a linear molecule. 2 ✓: There are 4  bonds and 3  bonds. 3 : There are only two lone pairs of electrons, one on each N atom.  B 6 Boiling point of Br 2 is higher than that of CH3CH2Br  stronger IMF in Br2 The Br –C bond in Br –CH2CH3 is polar, while the Br–Br bond in Br2 is non-polar  id-id and pd-pd in CH3CH2Br, and id-id in Br2 [statement 1] Br2 (Mr = 159.8) has a larger and more polarisable electron cloud [statement 4] compared to CH3CH2Br ( Mr = 108.9)  stronger id-id in Br2  B 7 1 ✓: When pressure is increased, the gas particles are forced closer together and the strength of IMF increases. If the IMF becomes sufficiently strong such that the potential energy overcomes the kinetic energy at room temperature, the gas will liquefy. 2 : At high pressure, the gas particles are forced closer together and the strength of IMF becomes significant. When the potential energy overcomes the kinetic energy, the gas will liquify. 3 ✓: At low temperature, the kinetic energy of the gas particles is lower. If the kinetic energy becomes sufficiently low, such that the potential energy due to the IMF becomes more significant, the gas will liquify.  B 8 The weaker bond is more reactive as it would takes less energy to break the bond, leading to a decrease in the activation energy. A more polar bond is more reactive as the partial charges will be sites of reactivity, attracting species of the opposite charge, again lowering the activation energy.  D 9 A : HCl(aq) is an Arrhenius acid as it produces H+(aq) in solution. But CuO is not an Arrhenius base as it is insoluble and does not produce OH–(aq). B : The reaction does not take place in aqueous medium, hence does not involve Arrhenius acid and base. C : HCl(aq) is an Arrhenius acid as it produces H +(aq) in solution. But CaCO3 is not an Arrhenius base as it is insoluble and does not produce OH–(aq). D ✓: HCl(aq) is an Arrhenius acid as it produces H +(aq) in solution, and KOH(aq) is an Arrhenius base as it produces OH–(aq) in solution.  D 10 ( ) ( ) ( )2H O H aq OH aq+− +l   w H OHK +−= ( ) ( ) ( ) ( )3 2 4NH aq H O NH aq OH aq+−++ l     4 b 3 OH NH NHK −+=  B 11 pH of blood decreases  [H+] increases 1 ✓: 2 3 3H CO HCO H −++ shifts to the left, causing [H 2CO3] to , which in turn causes 2 2 2 3CO H O H CO+ to shift left 2 : 2 3 3H CO HCO H −++ shifts left, where HCO 3– acts as a Brønsted - Lowry base, reacting with H+ 3 : 2 3 3H CO HCO H −++ shifts left, resulting in the [HCO3–] to   C 12 A : I– is a stronger reducing agent than Cl–, hence reducing C l2 to Cl–, itself oxidised to I2. B : Solubility of Cl2 and I2 does not affect the redox reaction. C ✓: Cl2 is a stronger oxidising agent than I2, hence oxidising I– to I2, itself reduced to Cl–. D : 2I– → I2 + 2e–; I– loses an electron.  C 13 2H–X → H2 + X2 Thermal decomposition involves cleavage of H–X bond.  Down Group 17, atomic radius es, hence H–X bond length es  H–X bond strength es  Thermal stability es  B 14 NaCl is neutral in water (pH 7), while Na2O dissolves in water to give NaOH (pH 14). Element X is sodium metal. A : Sodium is a metal consisting of cations immersed in a sea of delocalised 3s electrons, hence is a good conductor of electricity. B : Being a met al, sodium exists in a giant metallic lattice. C ✓: The sodium atom readily loses its single 3s electron to form Na+, hence a strong reducing agent. D : Being a metal, Na2O is an ionic oxide, consisting of Na+ and O2– ions.  C 15 A : There are 24 atoms in each glucose molecule. 1 mole of glucose contains 24 × 6.02 × 1023 atoms. B : There are 6 carbon atoms in each glucose molecule. 1 mole of glucose contains 6 × 6.02 × 10 23 carbon atoms. C : There are 12 hydrogen atoms in each glucose molecule. 1 mole of glucose contains 12 × 6.02 × 10 23 = 7.224 × 1024 hydrogen atoms. D ✓: One mole of glucose contains 6.02 × 1023 molecules.  D 16 Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O Balancing the number of e–s transferred. 1 : 6Br– → 3Br2 + 6e– 2 ✓: 6Fe2+ → 6Fe3+ + 6e– 3 ✓: 3SO2 + 6H2O → 3SO42– + 12H+ + 6e–  D 17 2Na + 2H2O → 2NaOH + H2 Na 4.60 0.200 mol23.0n == 2H Na 1 0.100 mol2nn == 22 3 H H m 0.100 22.7 2.27 dmV n V=  =  =  A

18 ( )( ) ( )200 1.00 4.18 26.6 21.5 4263.6 J Q mc T= −  = −    − =− 2Pb 100 1.0 0.100 mol1000nn +−= =  =I Pb2+ + 2I– → PbI2 KI is the limiting reagent. 2Pb 1 0.0500 mol2nn −==I I 2 reaction Pb 1 1 4263.6 0.0500 85272 J mol 85.3 kJ mol QH n − − − = = =− − I  A 19 XY(g) → X+(g) + Y–(g) L.E. L.E. qqH rr +− +−  =− =  + As q+ and q– are the same for NaCl, NaBr, KCl and KBr, H is most endothermic for the smallest ( )rr+−+ . Down the group, and rr+− es.  Na K C Br and r r r r+ + − − l  NaCl has the smallest ( )rr+−+  A 20       2 2 2 2 3 rate H NO rate H NO mol dmunit of k k k − = = = 1 3 s mol dm − −( )( ) 23 2 6 1 mol dm mol dm s − −−=  C 21 Using expt 1 and 2, when [O2]  by 2.5×, initial rate  by 2.5× hence reaction is first order w.r.t. O2 Using expt 2 and 3, when [NO ]  by 2× and [O2]  by 3×, initial rate  by 12× so when [NO ]  by 2×, initial rate  by 4× hence reaction is second order w.r.t. NO rate = k[NO]2[O2] using expt 2 and 4, when [NO] is halved and [O 2] is halved, initial rate will be 2 1 1 1 2 2 8  = times that in expt 2.  B 22 111 2220.400 0.200 0.100 0.050 ttt ⎯⎯→ ⎯⎯→ ⎯⎯→ since 11 22 2 600 s, 300 stt == total time = 1 2 3 900 st =  C 23 Since pepsin catalyses the hydrolysis of the amide bond in X, i.e. the rate of hydrolysis increases with pepsin, the rate constant is larger when pepsin is present. Since pepsin only catalyses the hydrolysis of the amide bond in X but not in Y, pepsin has specific activity.  A 24 ( ) ( )2 4 2N O g 2NO g initial amt/mol 2.0 0 change in amt/mol 1.0 2.0 eqm amt/mol 1.0 2.0 −+     2 2 2 3 c 24 2.0 NO 1.0 4.0 mol dm1.0NO 1.0 K −  = = =  D 25 For a system in dynamic equilibrium , the rates of the forward and reverse reactions are the same, but non-zero. Hence there is no net change in the concentrations of the reactants and products.  A 26  A 27 1 ✓: planar 2 : tetrahedral 3 ✓: planar  B 28 Substitution of H by Cl occurs: Mr = 30 Mr = 64 Mr = 98  B 29 A : B : C ✓: D :  C 30 Compound Y reacts with compound Z to give ester, CH3CH2CO2CH2CH2CH3. Since Y is the reduced form of X, while Z is the oxidised form of X, Y must the alcohol, CH3CH2CH2OH, while Z must be the acid, CH3CH2CO2H Hence, X is the aldehyde, CH3CH2CHO  D Answer Key Qn Ans Qn Ans Qn Ans 1 A 11 C 21 B 2 C 12 C 22 C 3 D 13 B 23 A 4 B 14 C 24 D 5 B 15 D 25 A 6 B 16 D 26 A 7 A 17 A 27 B 8 D 18 A 28 B 9 D 19 A 29 C 10 B 20 C 3