RI 2021 Y5 Promo Solutions

1 © Raffles Institution 2021 9729/S/21 2021 Y5 H2 Chemistry Promotion Examination – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C D D D B B D A A A C B C B A 1 Amount of E = 1 24000 mol (Note: 1 mole of gas occupies 24 dm3 at room temperature and pressure.) No. of molecules in 1 cm3 of E = = 1 24000  L = L 24000 No. of atoms in 1 cm3 of E = L 24000  3 = 3L 24000 (Note: E is a triatomic molecule.) 2 Reduction half-equation: SO2 + 4H+ + 4e– ⎯→ S + 2H2O Oxidation half-equation: H2S ⎯→ S + 2H+ + 2e– Overall balanced equation: SO2 + 2H2S ⎯→ 3S + 2H2O After reaction is completed, the mixture contains unreacted H2S. Hence H2S is present in excess and SO2 is the limiting reagent  Options B and C are incorrect. When 5 mol of H2S reacts with 1 mol of SO2, SO2 is the limiting reagent. 3 mol of H2S will remain unreacted and 3 mol of S will be formed. 3 Since the biggest jump in ionisation energy is from the 5th to 6th ionisation energy, element M has five valence electrons. Since element M is in Group 15, it reacts with chlorine to form the covalent compound MCl3. 4 Rb+, Br− and Kr are isoelectronic species (i.e. with same total number of electrons)  their outermost e− experience the same shielding effect. However, nuclear charge of Br− < Kr < Rb+. Hence, effective nuclear charge of Br− < Kr < Rb+. Electrostatic attraction between the nucleus and the outermost e− of Br− < Kr < Rb+. Energy required to remove the outermost e− increases in the order: H2 < H3 < H1

2 © Raffles Institution 2021 9729/S/21 5 molecule structure shape bond dipoles cancel out? polar / non-polar SF4 see-saw No polar XeF2 linear Yes non-polar CH2F2 tetrahedral No polar SO2 bent No polar 6 Statement 1 Incorrect. In the aldehyde group ( ) of butanal, the hydrogen atom is bonded to carbon (not oxygen). Hence there is no intermolecular hydrogen bonding in butanal. Only intermolecular instantaneous dipole- induced dipole and permanent dipole -permanent dipole interactions exist in butanal. Recall the criteria for hydrogen bonding: • A hydrogen atom covalently bonded to F, O and N. • A lone pair of electrons on a F, O and N atom in a neighboring molecule bearing a − which can attract the + charge on the H atom. Statement 2 Incorrect. All three compounds are simple molecules. Although C=C bond is stronger and requires more energy to break compared to C−C bond, boiling of simple molecules involves overcoming intermolecular forces of attraction, NOT covalent bonds. Statement 3 Correct. Butanal and 2-methylpropanal differs only in their hydrocarbon chain. Both compounds have the same aldehyde functional group and the same number of electrons (as they have the same molecular formula). Butanal is a straight-chain aldehyde, which has greater surface area for intermolecular interactions as compared to its branched isomer, 2-methylpropanal. Hence, the instantaneous dipole –induced dipole interactions in butanal are stronger than that in 2-methylpropanal.

3 © Raffles Institution 2021 9729/S/21 7 The carbon atoms in graphite are sp 2 hybridised while those in diamond are sp 3 hybridised. Since sp 2 hybrid orbitals have greater s -character, there is more effective overlap between the sp2 hybrid orbitals to form stronger carbon-carbon bond in graphite.  Statement 1 is correct. With reference to the energy level diagram on the left, t he enthalpy change of atomisation of diamond is less endothermic than that of graphite.  Statement 2 is correct. (Note: ∆Hatomisation > 0) With reference to the energy level diagram on the left , the enthalpy change of combustion of diamond is more exothermic than that of graphite.  Statement 3 is correct. (Note: ∆Hcombustion < 0) 8 Since the amount of gas and temperature are kept constant, pV = nRT = constant as time progresses.  The graph of pV against time is a horizontal line. As equal masses of C l2 and Ne are used and C l2 has a higher molar mass than Ne, a smaller amount of Cl2 is present in the syringe. Hence, the value of pV (= nRT) of Cl2 is smaller than that of Ne.  Option A is correct. As the volumes of the syringes are increased at the same constant rate, V  time and p = nRT V  nRT time  graph of p against time is a downward sloping curve. Since there is a smaller amount of Cl2 present in the syringe, the partial pressure of Cl2 is also smaller. Hence the correct graph of p against time is:

4 © Raffles Institution 2021 9729/S/21 9 Option A H2(g) + 1 2 O2(g) ⎯→ H2O(l) ∆Hformation(H2O(l)) H2(g) + 1 2 O2(g) ⎯→ H2O(l) ∆Hcombustion(H2(g)) The equation that represents both processes is the same. Hence, the two enthalpy changes have the same value. Option B 1 2 I2(s) ⎯→ I(g) ∆Hatomisation(I2(s)) 1 2 I2(g) ⎯→ I(g) 1 2  BE(I–I) Since the two processes are different, as each involves different states of I2, the enthalpy changes are of different values. Option C C(s) + O2(g) ⎯→ CO2(g) ∆Hcombustion(C(s)) CO2(g) ⎯→ C(g) + 2O(g) 2  BE(C=O) Since the two processes are different, the enthalpy changes are of different values. Option D O(g) ⎯→ O+(g) + e– 1st IE of O O(g) + e– ⎯→ O–(g) 1st EA of O Since the two processes are different, the enthalpy changes are of different values. Note: The energy required to remove 1 mol of electrons from 1 mol of O(g) to form 1 mol of O+(g) is equal in magnitude to the energy released when 1 mol of electrons is gained by 1 mol of O+(g) to form 1 mol of O(g). 10 Statement 1 Correct. The equation that represents ∆Hformation(KI(s)) is: K(s) + 1 2 I2(s) ⎯→ KI(s) The standard enthalpy change of formation of a substance is the energy change when 1 mole of the pure substance in a specified state is formed from its constituent elements in their standard states under standard conditions. (Note: Standard state of I2 is solid). Statement 2 Correct. The equation that represents the lattice energy of potassium iodide is: K+(g) + I–(g) ⎯→ KI(s) The lattice energy 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. Statement 3 Incorrect. The equation that represents the hydration energy of potassium iodide is: K+(g) + I–(g) ⎯→ K+(aq) + I–(aq) Statement 4 Incorrect. ∆H2 = 1st IE of potassium + 1st EA of iodine

5 © Raffles Institution 2021 9729/S/21 11 G = H − TS For a reaction to be less spontaneous at higher temperatures, G has to be less negative (or more positive) at higher temperatures. Hence, S must be negative so that the −TS term is more positive at higher temperatures. The reactions in options A, B and D produce more gaseous products compared to the reactants, hence S is positive. Only the reaction in option C produces less gaseous products compared to the reactants, hence S is negative. 12 At constant room temperature and pressure, Vgas  ngas. X2(g) ⎯→ 2X(g) Initial volume / cm3 50 0 Change in volume /cm3 – a +2a Final volume / cm3 50 – a 2a Total volume at time, t = 50 – a + 2a = 50 + a = 87.5 a = 37.5 Volume of X2(g) = 50 – 37.5 = 12.5 cm3 t½ t½ Since 50 cm3 ⎯⎯⎯→ 25 cm3 ⎯⎯⎯→ 12.5 cm3 Time taken = 2(t½) = 2(20) = 40 min 13 initial rate  VA t Since total volume is kept constant, VA  [A] (same for B and C). Expt VA / cm3 VB / cm3 VC / cm3 VH2O / cm3 t / s VA t 1 10 10 5 25 10 1 2 10 10 10 20 10 1 3 10 20 10 10 5 2 4 20 10 5 15 10 2 5 30 5 5 10 y 30 y Comparing expt 1 and 2, when volume of C  2, initial rate is unchanged. Hence the reaction is zero-order with respect to