ASRJC H2 Chem 2022 Prelim P1 Soln

ASRJC JC2 PRELIMS 2022 9729/01/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2022 JC 2 PRELIMINARY EXAMINATION SOLUTIONS NAME:______________________________ ( ) CLASS: 22 /____ CHEMISTRY 9729/01 Paper 1 Multiple Choice 21 September 2022 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Please look through the marks scheme and understand the concepts tested in each question. Do pay attention as the tutor go through selected questions in lecture. If you still have further questions, you may check with your tutor at your own time. Be proactive in asking questions. If you do not ask now, then when? Remember the power of tiny gains and be a little better every day!

2 ASRJC JC2 PRELIMS 2022 9729/01/H2 1 Ans: D Statement A is incorrect. Electrons has a larger e/m ratio and hence will be deflected to a larger extent than protons. Statement C is incorrect. Electron beam should be deflected towards the positively charged plate. Statement B is incorrect and D is correct. Proton beam travels in a parabolic path towards the negatively charged plate. 2 Ans: C There is a large increase in the 7th to 8th ionisation energy. This means that the 8 th electron is removed from the inner principal quantum shell and element D is from Group 17. Its valence configuration will be ns 2np5. Hence the 5 th electron is from the p subshell and 6 th electron is from the s subshell. While it is possible that D is fluorine (i.e. in Period 2), it cannot be conclusively inferred from the data given as element D may have more than 9 electrons. However, if the 10th I.E. is given, we can conclude that D is not fluorine in Period 2. 3 Ans: D 1: C-Cl bond is polar and there are 2 C l bonds in molecule I, thus it is more polar than molecule II. 2: The dipole moments associated with the polar C -Cl bond is canceled out in the trans molecule. Thus, the cis isomer is more polar. 3: The dipole moments associated with the polar C -Cl bond is canceled out in the trans molecule. Thus, molecule I is more polar. 4: C-F bond is more polar than C-Cl bond. Thus, molecule I is more polar. 4 Ans: A E has simple molecular structure with low m.p because little amount of energy is needed to overcome the weak intermolecular forces of attraction. It is unable to conduct electricity in any state due to absence of mobile charge carriers. F has giant ionic lattice structure with high m.p because large amount of energy is needed to overcome the strong electrostatic forces of attraction between the oppositely charged ions. It conducts electricity in aqueous and molten state due to the presence of mobile ions. G has giant metallic lattice structure with high m.p because large amount of energy is needed to overcome the strong metallic bonds. It conducts electricity in solid and molten state due to the presence of delocalised electrons. H is SiO 2. It has giant molecular structure with high m.p because large amount of energy is needed to overcome the strong and extensive covalent bonds between the atoms. It cannot conducts electricity in solid state because there is no mobile charge carriers.

3 ASRJC JC2 PRELIMS 2022 9729/01/H2 [Turn over 5 Ans: D Gas J deviates more than gas K from ideality and hence have stronger intermolecular forces of attraction. A: O2 has a greater electron cloud size and has greater ease of distortion of electron cloud. Hence it has stronger instantaneous dipole-induced dipole interactions between O2 molecules and should deviate more. B: HI has a greater electron cloud size and has greater ease of distortion of electron cloud. Hence it has stronger instantaneous dipole-induced dipole interactions between HI molecules and should deviate more. HI should be gas J. C: H2O has stronger intermolecular hydrogen bonds and should deviate more. D: HBr is polar and F2 is non-polar. Hence, HBr has stronger permanent-dipole induced dipole interactions and deviate more. 6 Ans: C pV = nRT V = 𝑛𝑅𝑇 𝑝 V = 𝑚 𝑀𝑟𝑅𝑇 𝑝 Since R, mass, pressure and temperature are constant, volume is proportional to 1 𝑀𝑟 NH3 CH3Cl CH2Cl2 HCOOH Mr 17 50.5 85 46 7 Ans: B Lattice energy  │ q+ × q– r++ r– │ Cationic charge: Mg2+ > Na+ Anionic charge: O2– > Cl– Cationic radius: Mg2+ < Na+ Magnitude of lattice energy: MgO > MgCl2 > NaCl 8 Answer: B During vapourisation, molecules need to overcome intermolecular forces of attraction to move further apart to convert to gaseous states. Atomisation is to convert molecules into atoms and that would need to break strong covalent bonds within molecules instead.

4 ASRJC JC2 PRELIMS 2022 9729/01/H2 9 Answer: B comparing expt I and II, when [sucrose] increased 1.5times, the initial rate increased by 1.5times. Since rate of reaction is directly proportional to [sucrose], order of reaction wrt sucrose is 1. comparing expt I and III, when [HCl] is tripled, the initial rate tripled. Since the rate of reaction is directly proportional to [HCl], order of reaction wrt HCl is 1. rate = k [HCl] [sucrose] rate = k’ [sucrose] where k’ = k [HCl]  HCl is a catalyst and its concentration remains constant Since the half–life of expt I is 3.0 s and t½ = 'k 2ln = I]HCl[k 2ln = 3.0s for 1st order reaction wrt sucrose, expt II Since [HCl] is constant, half–life for expt II will remain as 3.0 s. expt III As [HCl] is tripled, half–life for expt III will be 1.0 s (t½ for expt III = III]HCl[k 2ln = ln 2 (3[ ] ) Ik HCl = 1 ln 2()3 [ ] Ik HCl = 1 (3.0)3 = 1.0 s) 10 Answer: C The rate equation / order of reaction of a reaction can be derived from the stoichiometric coefficients of the species involved in the slow step, including preceding fast step(s). 1 is consistent. 1 mol of H2O2 reacts with 1 mol of I– in the slow step (r.d.s.), which is also the first step. Hence the rate equation is as shown. 2 is inconsistent. Correct rate equation should be rate = k2 [H2] since only 1 mol of H2 is involved in the slow step. 3 is consistent. As the slow step is not the first step, you need to consider the preceding fast step. Based on the elementary step, rate = k3 [HBrO] [HBr] However, HBrO is an intermediate and its concentration is dependent on the reaction between HBr and O2 so you need to include this in the rate equation. Shortcut for consideration: If the intermediate is the ONLY product formed in the preceding fast step, you can just change its stoichiometric coefficient to be the same as that in the slow step. Similarly, multiply by the same factor for the reactants in the fast step. In this case, the mechanism can be rewritten as HBr + 1/2O2 HBrOfast HBrO + HBr slow H2O + Br2 Hence the rate equation is rate = k3 [HBr] [HBrO] = k3’ [HBr] [HBr] [O2]½ = k3’ [HBr]2 [O2]½

5 ASRJC JC2 PRELIMS 2022 9729/01/H2 [Turn over 11 Answer: D Since this is a first order reaction based on the rate constant units, only the half-life remains constant. Option A: Rate should decrease as the pressure of reactant decreases. Option B: The total pressure should increase as the volume remains constant but more gas products are formed. Option C: The rate constant is different without the use of gold as reaction rate will decrease. 12 Answer: D Options 2 and 3 are incorrect. Option 2: Decreasing the partial pressure of nitrogen will decrease the rate of reaction even though the POE will eventually shift to the left. Option 3: Adding a catalyst should affect the rate constant since it increases the rate of reaction without concentration change. 13 Answer: D H+(aq) + F– (aq) HF(aq) ------------(1) MgF2(s) Mg2+(aq) + 2F–(aq)-----------(2) Towards lower pH as [H+] increases, POE for equation (1) shifts right. This results in a decrease in [F–] in equation (2).