RI 2024 Y5 Promotional Examination Suggested Solutions

2024 Y5 H2 Chemistry Promotion Examinations Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C D B C D C B B A C A D C A B Question 1 (C) Ar = � 60.9 100 � × 20 + � 1.2 100� × 21 + � 37.9 100 � × 22 = 20.8 Question 2 (D) Method 1: comparing mol ratio CxHy → xCO2 + 𝑦𝑦 2H2O 0. 06 mol 0.04 mol x : 𝑦𝑦 2 = 0.06 : 0.04 = 6 : 4 x : y = 6 : 8 = 3 : 4 = 9 : 12 M ethod 2: tri al and error CO2: H2O = 0.06 mol: 0.04 mol = 6 : 4 = 3 : 2 equations CO2: H2O A C3H6 → 3CO2 + 3H2O 3:3 B C6H10 → 6CO2 + 5H2O 6:5 C C6H12 → 6CO2 + 6H2O 6:6 D C9H12 → 9CO2 + 6H2O 9:6 = 3:2 Q uestion 3 (B) Atomic number of Sr = 38 Number of neutrons in Sr, W, X and Y = 84 – 38 = 46 Number of electrons in Sr2+, W, X–, Y2– = 38 – 2 = 36 W X Y proton number 36 35 34 nucleon number 82 81 80 N ote: W = Kr, X = Br, Y = Se Q uestion 4 (C) Q uestion 5 (D) Option A is incorrect : Ion is not planar as it is tetrahedral about C in –CH3. O ption B is incorrect : The π electrons are delocalised over the –COO– group. O ption C is incorrect: In the resonance hybrid, the two C−O bonds lengths are identical. O ption D is correct: The H−C−H bond angle (109.5o) is smaller than the C−C−O bond angle (120o). O2 O2 O2 O2 O2 This document is copyrighted, please do not reproduce it without permission

Question 6 (C) Definition of Bond Energy: The bond energy of a X–Y bond is the average energy required to break 1 mole of the X–Y bonds in the gaseous state. To us e the average bond energies in the table, we need to have the molecules be converted to the gaseous state. To convert 2 mol of H 2O(l) to H2O(g), that is 2 x enthalpy change of vaporisation of water. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) C(g) + 4H(g) + 4O(g) CO2(g) + 2H2O(g)4(410) + 2(496) 2(805) + 4(460) -890 By Hess’ Law, −890 + 2∆Hvap + 2(805) + 4(460) = 4(410) + 2(496) 2∆Hvap = +72 kJ mol−1 ∆Hvap = +36 kJ mol−1 Question 7 (B ) Graph A pV = nRT, constant n and T pV = constant Graph of V against pV should be vertical (not horizontal). Graph is incorrect. Graph B pV = nRT, constant n and V p = constant × T, hence p α T Graph of p against T is correct. Graph C pV = nRT, constant n and T p = constant/V, hence p α 1/V Gr aph of p against 1 V is incorrect. It should be a y = kx graph. G raph D pV = nRT, constant V and T p = constant × n, hence p α n Graph of p against n is incorrect. It should be a y = kx graph. Quest ion 8 (B) The lattice energy (LE) of an ionic compound is the energy released when 1 mole of the solid ionic compound is formed from its constituent gaseous ions under standard conditions. Question 9 (A ) ∆H > 0 and ∆S > 0 for this reaction. The entropy change is positive bec ause the reaction results in an increase in the number of moles of gaseous particles in the system (from 0 to 1 mol of gas). H ence, only at high temperatures can the negative −T∆S term outweighs the positive ∆H, causing the ∆G to be negative. Question 1 0 (C) Comparing Expt 1 to 3, rate = k[I−][S2O82−]. The total volume of Expt 4 is doubl e that of all the other experiments (e.g. Expt 2). In E xpt 4 (compared to Expt 2), when the volume of I− used is doubled, initial [I−] in the mixture of Expt 4 is the same as Expt 2. In Expt 4 (compared to Expt 2), when the volume of S2O82− used x 4, initial [S2O82−] in the mixture of Expt 4 is double that of Expt 2. When [S2O82−] is doubled, rate of reaction for Expt 4 is double that of Expt 2. rate ∝ 1/t, time taken for Expt 4 (compared to Expt 2) = 42.5 2 = 21.3 s Question 11 (A ) 2∆Hvap This document is copyrighted, please do not reproduce it without permission

Statement 2 (incorrect) An increase in the partial pressure (hence concentration) of Cl2 will increase the reaction rate since the order of reaction w.r.t. Cl2 is ½. Statement 3 (incorrect) The units of rate constant ((mol dm−3)−½ time−1) are different from the units for equilibrium constant of step 1 (mol dm−3). Question 1 2 (D) Given that the mole fraction of HI is 2x, mol fraction of H2 + mol fraction of I2 = 1 – 2x, Given H2 and I2 are equimolar, mole fraction of H2 = mole fraction of I2 = ½(1–2x) Let the total pressure of the system at equilibrium be P T. partial pressure of HI = 2x(PT) partial pressure of H2 = partial pressure of I2 = 1 − 2x 2 (PT) Kp = PHI 2 PH2PI2 = (2x(PT))2 ÷ ( 1 − 2x 2 (PT)) 2 = 16x2 (1 − 2x)2 Question 1 3 (C) Option A is incorrect. Adding a catalyst will increase the rate of both forward and backward reaction to the same extent. Hence equilibrium position is not shifted and percentage composition of NH 3 is not increased. Option B is incorrect. Decreasing the temperature will cause the equilibrium position to shift right to favour the forward exothermic reaction and increase the percentage composition of NH 3. However, decreasing the temperature will reduce the rate of formation of NH 3 as the rate of reaction will be reduced. Option C is correct. Reducing the total volume at constant temperature leads to an increase in total pressure, which will shift the equilibrium position to the right, favouring the forward reaction as it produces less gaseous molecules . This results in the increase in the percentage composition of NH 3. Also, the partial pressures of the individual reactants increase, resulting in the increase in the rate of formation of NH 3. Option D is incorrect. Since argon is added at constant volume, the partial pressure of the reacting gases remains the same . Hence, the rate of formation of NH3 and percentage composition of NH3 will remain the same. Question 1 4 (A) Only A has a chiral centre as shown below. * Note: C has 2 chiral centres while B and D have none. Question 1 5 (B) Statement 1 (correct) Both E & F are polar. Note: In E, the region around O is not l inear. Statement 2 (c orrect) They have the same molecular formula (C4H10O) but different structural formula. Statement 3 (inco rrect) Only F has a chiral carbon. OH * This document is copyrighted, please do not reproduce it without permission

Section B B1 (a) (i) [B] remains approximately constant so that the order of reaction with respect to A can thus be found. Examiners’ Comments • A handful of students gave an incomplete answer. Note that there are two parts to this answer. • Examiners accepted answers that described a pseudo first-order reaction wrt A. However, in actual fact, when B is added in excess, we do not yet know if the reaction is a pseudo first - order or a pseudo second- order reaction. It would be more accurate to simply describe the reaction as a pseudo-order reaction. • While it is true that using excess B allows all the A to react, it is not a valid answer here as the kinetics of the reaction can be studied without monitoring till the completion of the reaction. (ii) If the reaction were to go to completion, final [C] = initial [A] = 0.120 mol dm−3 1st t½ is time taken for [C] to increase from 0.00 to 0.06 mol dm−3 while 2nd t½ is time taken for [C] to increase from 0.06 to 0.09 mol dm−3. From Experiment 1 graph, 1st t½ = 3.8 min and 2nd t½ = 3.8 min. Since 1st t½ = 2nd t½, t½ is constant, the reaction is first-order with respect to A. (Note: Results from Expt 2 can also be used.) Examiners’ Comments • Students must learn how to determine order wrt a reactant given a [product] -time graph. Revisit Section 5 of Kinetics lecture notes. • Clear working includes showing all construction lines & labelling at least two t½ on the graph. This document is