2019 MI Prelim H2 Chem P1 ANS

Q1 Option 1: Amount of H2SO4 = 0.5 x 1 = 0.5 mol H2SO4 → 2H+ + SO42- Amount of H+ = 0.5 x 2 = 1 mol Option 2: Amount of SO3 molecules = 6 / 24 = 0.25 mol 1 SO3 molecule has 4 atoms (1 S and 3 O atoms) Amount of atoms = 0.25 x 4 = 1 mol Option 3: Amount of SO2 molecule = 64.1 / (32.1 + 16.0 x 2) = 1 mol 1 SO2 molecule has 16 + 8 x 2 = 32 electrons Amount of electrons = 1 x 32 = 32 mol Option 4: Amount of S8 molecule = 1.6 /(32.1 x 8) = 0.00623 mol 1 S8 molecule has 16 x 8 = 128 neutrons Amount of neutrons = 0.798 mol Ans: 1 and 2 are correct => A Q2 CH3CH2CH2OH + 4O2 → 2CO2 + CO + 4H2O Since mole ratio = volume ratio of gases, Vol of O2 consumed = 4y Ans: B Q3 [R] MnO4- + 4H+ + 3e- → MnO2 + 2 H2O [O] Mn2+ + 2H2O → MnO2 + 4H+ + 2e- 2MnO4- + 3Mn2+ + 2H2O → 2MnO2 + 3MnO2 + 4H+ 2 mol of MnO4- reacts with 3 mol of Mn2+ 1 mol of MnO4- reacts with 1.5 mol of Mn2+ Ans: C

Q4 Z2+ has 12 electrons and 15 neutrons Z has 14 electrons, 14 protons and 15 neutrons=> Ar = 14 + 25 = 39 Option A: angle of deflection ∝ charge mass charge mass for Z2+ = +𝟐 𝟏𝟒+𝟏𝟓 = +𝟐 𝟐𝟗 charge mass for Mg2+ = +𝟐 𝟐𝟒.𝟑 charge mass for Mg2+ > charge mass for Z2+ Angle of deflection for Mg2+ > Angle of deflection for Z2+ Option A is correct Option B: Based on the proton number, Z is referring to Silicon. Si has a higher IE than Al (Group 13 element in the same period as Z) => You can check from Data Booklet or recall from the trend across the period. Option C: SiCl4 has a lower melting point than MgCl2 as SiCl4 has a simple molecular structure while MgCl2 has a giant ionic structure. Larger amount of energy is needed to overcome the stronger electrostatic forces of attraction between Mg2+ and Cl- than the weaker instantaneous dipole-induced dipole forces of attraction between SiCl4 molecules. Option D: Element Z has 14 electrons while neon has 10 electrons. Hence, they are not isoelectronic.

Q5 All options have simple molecular structures. Option 1 and 4 have instantaneous dipole-induced dipole forces of attraction between molecules Option 2 has hydrogen bonding between molecule Option 3 has permanent dipole-permanent dipole forces of attraction between molecules. Hence option 2 will be the highest followed by option 3 For option 1 and 4, the two molecules are isomers of each other. Thus, the surface area determines the boiling point. Since option 1 is a straight-chained molecule, it has a larger surface area between molecules which results in stronger instantaneous dipole-induced dipole forces of attraction than that of option 4 since option 4 is a branched-chain molecule. Hence option 4 as the lowest boiling point. Boiling point: 4 < 1 < 3 < 2 => Option D is correct Q6 The N highlighted in yellow has 6 electrons from N hence this N has a charge of -1. The N highlighted in turquoise has 5 electrons from N hence thus N has no charge. Since the complex has no overall charge, M has a charge of +2. Bond between M2+ and N (in turquoise) is due to dative bond. Bond between M2+ and N (in yellow) is due to ionic bond. Ans: 1 and 2 => B is correct

Q7 Converting from graphite to diamond has ΔH > 0 and k is small refers to large Ea. Thus C is the answer. Q8 Reaction 1: Evaporation of ethanol: C2H5OH(l) → C2H5OH(g) ΔH > 0 => energy is taken in to overcome the hydrogen bonding between the molecules to convert liquid to gas. Recall: Liquid is more closely packed than gas ΔS > 0 => increase in number of gas molecules from zero to one. More ways of arranging the particels. Entropy increases. ΔG = 0 as it is an equilibrium reaction Reaction 2: Atomisation of magnesium: Mg(s) → Mg(g) ΔH > 0 => energy is taken in to overcome the forces of attraction to convert solid to gas. Recall: solid is more closely packed than gas ΔS > 0 => increase in number of gas molecules from zero to one. More ways of arranging the particels. Entropy increases. Reaction 3: Initiation step for the free radical substitution between chlorine and ethane Cl2(g) → 2Cl•(g) uv ΔH > 0 => energy is taken in to break the covalent bond between the Cl atoms ΔS > 0 => increase in number of gas molecules from zero to one. More ways of arranging the particels. Entropy increases. ΔG > 0 as it is not a spontaneous reaction since uv light is supplied for the reaction to occur. Option 1 and 2 are correct=> C graphite diamond

Q9 pV = nRT Since nR and V are constant, p=kT p T = k 𝑝1 𝑇1 = 𝑝2 𝑇2 𝑝1 𝑇1 = 1.2𝑝1 𝑇1+273 𝑇1+273 𝑇1 = 1.2 0.2T1 = 273 T1 = 1365 K Ans: D Q10 H2O + ClO– ⇌ HOCl + OH– Acid and its conjugate base is a difference of H+ Hence H2O is the acid and OH- is its conjugate base ClO- is the base and HOCl is the conjugate acid. Hence, option A is the correct answer Q11 rate equation depends on the slow step and it cannot include intermediate. To determine if the species is an intermediate, write the overall equation by summing up the steps in the mechanism. If the species appear on either side of the overall equation, it is not an intermediate. Overall equation: A + 2B → AB2 Based on slow step, rate= k[AB][B] Since AB is an intermediate, it has to be re-expressed by using KC. Kc = [𝐴𝐵] [𝐴][𝐵] => [AB] = Kc[A][B] rate= k[AB][B] = k {Kc[A][B]} [B] = k’ [A] [B]2 Ans: D

Q 12 [Rxt] 100% → 50% → 25% … [Pdt] 0% → 50% → 25% … 2A(g) → B(g) + C(s) I x 0 0 C -x +½x - F 0 ½x - This is based on completion of reaction, hence A will be completely used up. Hence ½x = 200 Pa x = 400 Pa Note: pressure only applies to gas, hence carbon solid is ignored. At the 1st half-life, 50% of A will be used up. 2A(g) → B(g) + C(s) I x 0 0 C -½x +¼x - F ½x ¼x - =200 =100 At the 1st half-life, the total pressure = 200 + 100 = 300Pa t1/2 = 10 min => B is correct 13 ΔHrxn = [-685.3 + 2(-92.3)] – [2(-485.3)] = +127.5 kJ mol-1 As temperature increases, the system is disturbed. By LCP, the system will oppose the change by decreasing the temperature and favours the endothermic reaction to absorb the heat. Position of equilibrium shifts right and produces more products. Hence, [CHClF2] decreases with increasing temperature. Ans: C

14 Consider if there is a reaction from the mixtures and determine the resultant species left in the solution to decide if it is a buffer. Recall that a buffer must contain either: a weak acid and its conjugate base OR a weak base and its conjugate acid. Option A is made up of a salt and acid which does not react. resultant species: NaCl and HCl => not a buffer as it contains a strong acid Option B: a reaction will take place CH3COOH + NaOH → CH3COONa + H2O I 0.75 0.25 0 0 C -0.25 -0.25 +0.25 +0.25 F 0.50 0 0.25 0.25 resultant species: CH3COOH and CH3COONa => a buffer since it contains a weak acid (CH3COOH) and its conjugate base (CH3COO-) Option B is correct Option C: a reaction will take place 2NH3 + H2SO4 → (NH4)2SO4 I 0.6 0.4 0 C -0.6 -0.3 +0.3 F 0 0.1 0.3 resultant species: H2SO4 and (NH4)2SO4 => not a buffer as it contains a strong acid Option D: a reaction will take place. 2NaOH + H2SO4 → Na2SO4 + 2H2O I 0.4 0.6 0 0 C - 0.4 -0.2 +0.2 +0.4 F 0 0.4 0.2 0.4 resultant species: H2SO4 and Na2SO4 => not a buffer as it contains a strong acid

Q15 [H+] from pH 2 = 10-2 [H+] from pH 3 = 10-3 When equal volumes are mixed, each concentration is halved. [H+]total after mixing= ½ [10-2 + 10-3] = 0.00550 mol dm-3 pH = -log 0.00550 = 2.26 Ans: B Q16 MgO has the highest melting point across Period 3 oxides. => R is Mg Al has the highest electrical conductivity => Q is Al Ar has the lowest melting point => S is Ar Ans: Na is not represented. Option A is the answer. Q17 HF is a we