RI 2019 Y5 Promo Solutions

1 2019 Y5 H2 Chemistry Promotion Exam – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D C A B B C C B A D D A A C B MCQ worked solutions 1 (Ans: D) A : Incorrect. (A mole of substance is the amount of that substance which contains 6.02  1023 elementary entities of that substance.) Since 1 mole of compound (e.g. CO2) contains more than 1 mole of atoms (e.g. 1 mole of C and 2 moles of O atoms), it will not contain the same number of atoms as there are atoms in 12 g of carbon-12 (i.e. 1 mole of 12C). B : Incorrect. Relative molecular mass = (weighted) average mass of 1 molecule of the substance 1 12  mass of 1 atom of carbon-12 C : Incorrect. Relative isotopic mass = mass of 1 atom of the isotope 1 12  mass of 1 atom of carbon-12 D : Correct. Relative atomic mass = (weighted) average mass of 1 atom of the element 1 12  mass of 1 atom of carbon-12 2 (Ans: C) 2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g) Mass of carbon dioxide and water vapour lost = 0.85 g Let the amount of NaHCO3 be x mol. Amount of CO2(g) = amount of H2O(g) = (x/2) mol (x/2)(44) + (x/2)(18) = 0.85  x = 0.02742 % purity by mass = [(0.02742 x 84.0) / 3.50] x 100 = 65.8% Note: The sample of NaHCO3 is impure. After decomposition, the mass of Na 2CO3 in the remaining residue is NOT = (3.50 – 0.85) g due to the impurities present. 3 (Ans: A) No of protons and neutrons in X = 289 + 3(1) – 244 = 48 No of protons in X = 114 + 3(0)  94 = 20 No of neutrons in X = 48  20 = 28

2 4 (Ans: B) CO32− ion exhibits resonance. ⇒ All C–O bonds present in the ion are of the same bond length. CO32− ion has 3 regions of electron density around the central C atom (sp2 hybridisation). Since there are 3 bond pairs of electrons present, the CO32− ion has a trigonal planar shape and a bond angle of 1200. SO32− ion has 4 regions of electron density around the central S atom. Since there are 3 bond pairs and 1 lone pair of electrons present, the SO 32− ion has a trigonal pyramidal shape and a bond angle of approximately 1070. Hence, only options 1 and 2 are correct. 5 (Ans: B) A: Correct There are 2 𝜋 bonds in the C≡C triple bond in ethyne (H–C≡C–H). There are 2 𝜋 bonds in the C≡N triple bond in butanenitrile (CH3CH2CH2C≡N). ⇒ Both molecules have the same number of 𝜋 bonds B: Incorrect. Since butanenitrile is polar (with a net overall dipole moment ), the molecules are held together by permanent dipole-permanent dipole (pd-pd) interactions. On the other hand, ethyne is non-polar (with no net dipole moment) and the molecules are held together by the weaker instantaneous dipole–induced dipole (id-id) interactions. In addition, since butan enitrile has a larger and more polarisable electron cloud than ethyne, its id-id interactions are stronger than that of ethyne. ⇒ Butanenitrile forms stronger intermolecular forces of attraction than ethyne. C: Correct. Since the C atoms in ethyne are sp hybridised while the C atom bonded to H a in butanenitrile is sp3 hybridised, there is higher s -character in the hybrid orbitals of the C atoms in ethyne. ⇒ In ethyne, the less diffuse sp hybrid orbital of C overlap more effectively with the 1s orbital of H. Hence, C–H bond energy in ethyne is greater than that of C–Ha in butanenitrile.

3 6 (Ans: C) pV = nRT = m MRT pM = m VRT = ρRT p = ρRT M pressure of gas A pressure of gas B = ( density of A density of B)( temp of A in K temp of B in K)( relative molecular mass of B relative molecular mass of A) = ( 3 1)( 273+30 273+20)( 4 1) = 12.4 7 (Ans: C) A: Correct. The enthalpy change for the conversion of E to F, ∆Hr, is –30 kJ mol–1. B: Correct. The 1st Ea is 100 kJ mol–1. C: Incorrect. There are two transition states in the reaction. D: Correct. Since 1st Ea is greater than 2nd Ea, the first step is the rate-determining step. 8 (Ans: B) Statement 1 is correct. The degree of covalent character in an ionic compound depends on the polarising power of the cation and polarisability of the anion. Both JY and LY have the same anion, hence we compare the polarising power of J+ and L+. J+ has a higher polarising power as J+ has a higher charge density (same charge but smaller radius) than L+. Hence, JY has higher covalent character than LY. Statement 2 is correct. + + qqLE rr       2 x 2LE of 0.18 0.14 MZ >   1 x 1LE of 0.08 0.12 JX Hence, the LE of MZ is more exothermic than that of JX. (Recall: The lattice energy is always exothermic as it is the energy released when one mole of the solid ionic compound is formed from its constituent gaseous ions at 298 K and 1 bar.) Statement 3 is incorrect. Since J+ has the same charge but smaller ionic radius than L+, J+ has a larger q/r value, and hence a more exothermic hydration energy. (Recall: The hydration of gaseous ions by water molecules is an exothermic process due to the formation of ion-dipole interactions between the ions and water molecules.) transition states intermediate ∆Hr 1st Ea 2nd Ea |Hhyd|  q r

4 9 (Ans: A) For a reaction which measures the time taken for a coloured reactant to disappear, rate  VI2 t . reaction mixture volume of aqueous CH3COCH3 / cm3 volume of aqueous I2 / cm3 volume of aqueous HCl / cm3 volume of starch solution / cm3 volume of water / cm3 time taken for disappearance of deep blue colour / s VI2 t / cm3 s−1 1 20 10 10 1 5 54 10 54 2 15 10 10 1 10 72 10 72 3 20 5 10 1 10 y 5 y 4 15 10 15 1 5 48 10 48 Statement 1 is correct. Comparing reaction mixtures 1 and 3: Since order of reaction with respect to I2 is zero, rate of reaction mixture 1 = rate of reaction mixture 3  10 54 = 5 y . Solving, y = 27. Statement 2 is correct. Comparing reaction mixtures 1 and 2: When VCH3COCH3 is increased from 15 to 20 cm3, i.e. VCH3COCH3 4 3, rate  4 3  order of reaction with respect to CH3COCH3 is 1. Statement 3 is incorrect. Comparing reaction mixtures 2 and 4: When VHCl is increased from 10 to 15 cm3, i.e. VHCl  1.5, rate  1.5  order of reaction with respect to H+ is 1. When HCl (monobasic acid) is replaced by H2SO4 (dibasic acid) of the same concentration,  [H+]  2 and rate  2. Hence time taken for the disappearance of deep blue colour should be halved. 10 (Ans: D) Based on the mechanisms, the rate equations and overall equations are shown below. rate equation overall equation for reaction A rate = k[P]2[Q2] P + Q2 + R  PQR + Q B rate = k[P]2 2P + Q2 + R  P2Q + QR C rate = k[P][Q2] 2P + Q2 + R  P2Q + QR D rate = k[P]2[Q2] 2P + Q2 + R  P2Q + QR Only D has the correct rate equation and overall equation for the reaction.

5 11 (Ans: D) A: Incorrect. Water is a solvent and not a catalyst in this reaction. Also, a catalyst increases the rates of both the forward and backward reactions to the same extent. B and C: Incorrect. Even though H2O is added and the number of moles/amount of H2O increases, the [H2O] stays almost the same/does not change appreciably as H2O is the solvent and is already present in large excess. Hence, the shift in position of equilibrium is not due to change in [H2O]. D: Correct. Addition of water reduces the total ion concentration in the system. Hence the system tries to counteract this change by favouring the reaction that produces more ions (which is the forward reaction in this case), resulting in the solution turning pink. 12 (Ans: A) Decreased proportion of products imply that the position of equilibrium shifted left after the following changes were introduced. P is increased at constant T (reaction that produces fewer gas particles is favoured) T is decreased at constant P (exothermic reaction is