RI 2018 Y5 Promo MCQ Suggested Solutions

17 2018 Year 5 Promotion Examinations Section A Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B D B D B D A C C A C D B A C Q1(B) Oxidation half-equation 6OH– + S ⟶ SO32– + 3H2O + 4e– Reduction half-equation 4e– + 2S ⟶ 2S2– Overall equation 6OH– + 3S ⟶ 2S2– + 1SO32– + 3H2O Q2(D) Balanced equation: CxHySz(g) + (x + z + y/4)O2(g) ⟶ xCO2(g)+ zSO2(g) + y/2H2O(l) After cooling to room temperature, the gases which remain are CO2, SO2 and unreacted O2 i.e. 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 + 𝑉𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑂2 = 90𝑐𝑚3 The acidic gases, CO2 and SO2, react with the aqueous alkali. Therefore, 𝑉𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑂2 = 70𝑐𝑚3. 𝑉𝑂2 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 100 − 70 = 30𝑐𝑚3 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 + 70 = 90 ⇒ 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 = 20𝑐𝑚3 CxHySz(g) + (x+z+y/4)O2(g) ⟶ xCO2(g) + zSO2(g) + y/2H2O(l) V / cm3 10 30 20 -- mole ratio 1 3 2 -- x+z+y/4 = 3 and x +z = 2 ⇒ y = 4 x = z = 1. Therefore, compound is CH4S. Q3(B) A Incorrect. Using data from Data Booklet, radius of Ni2+ = 0.069nm and radius of Fe2+ = 0.061nm. B Correct. Using data from Data Booklet, 1st IE of Ni = 736 kJ mol–1 and 1st IE of Fe = 762 kJ mol–1. C Incorrect. Number of electrons in Fe 2+ = 24. Number of electrons in Ni atom = 28. They do not have the same no. of electrons i.e. they are not isoelectronic. D Incorrect. 2+ 2+ Angle of deflection 2(Ni ) 0.034158.7 2(Fe ) 0.035855.8 q m q m q m    Fe2+ will be deflected more than Ni2+. Q4(D) From question, presence of unpaired e – ⇒ paramagnetic. A Electronic configuration of Cl+ : 1s2 2s2 2p6 3s2 3p4. 2 unpaired electrons in 3p subshell ⇒ Cl+ is paramagnetic. B Electronic configuration of V 3+ : [Ar] 3d 2. 2 unpaired electrons in 3d subshell ⇒ V3+ is paramagnetic. C Electronic configuration of O – : 1s2 2s2 2p5. 1 unpaired electron in 2p subshell ⇒ O– is paramagnetic. D Electronic configuration of Cu + : [Ar] 3d 10. No unpaired electrons in Cu + ⇒ Cu+ is not paramagnetic. Q5(B) 1 Incorrect. Hence, the molecule is not planar. 2 Correct. From question: “…form the planar cyclopentadienyl anion…”. In order for the anion to be planar, every carbon atom in the anion is sp 2 hybridised. 3 Incorrect. Due to the resonance stabilisation in the anion, the pi electron density is evenly distributed over the entire anion. Hence, the carbon -carbon bond lengths in the anion are equal. Q6(D) The line is represented by an equation of the form y = mx where m is the gradient. If pV is the y -axis and T is the x -axis, the ideal gas equation can be manipulated to fit the form y = mx. This document is copyrighted, please do not reproduce it without permission

18 Q7(A) Refer to the defi nitions in Energetics I lecture notes to help you remember whether a process is endothermic or exothermic. It is also helpful to consider from a Chemical Bonding perspective whether some processes are endothermic or exothermic. A Enthalpy change of atomisation is always positive. B Enthalpy change of combustion is always negative. C Enthalpy change of formation can be positive or negative, depending on the reaction. D Enthalpy change of solution can be positive or negative, depending on the reaction. Q8(C) G = H – TS For a reaction to be spontaneous at all temperatures, H < 0 (narrow down options to B and C) and S > 0. The reaction in C involves the production of gaseous products which are significantly more disordered than the liquid reactants i.e. S > 0 for reaction in C. Q9(C) D Incorrect. At a given time, the gradient of the graph gives the rate of reaction. The initial rate for the reaction at T1 is greater than that for the reaction at T2. Hence, T1 is greater than T2. A Incorrect. At the higher temperature (T1), more H2 is formed i.e. more products are formed. When the temperature is increased from T 2 to T1, the system favours the endothermic reaction which, in this case, is forward reaction which produces more H 2. Hence the forward reaction is endothermic. B Incorrect. The addition of solid carbon does not alter the concentrations of any species and does not affect the position of equilibrium. C Correct. From B, increasing the temperature from T2 to T 1 causes the position of equilibrium to shift right, consuming more reactants to form more products. Hence Kc increases with increasing temperature. Q10(A) The two peaks in the reaction profile diagram indicate that the given reaction, A + 2B ⟶ 2C, proceeds via a two-step mechanism. The reactions involved in the two steps are: Step 1 : A + 2B ⟶ AB + B Step 2 : AB + B ⟶ 2C 1 Correct. Since the activation energy for step 2 is greater than that of step 1 (deduced from the reaction profile diagram), step 2 is the slow step. From step 2 (the slow step), rate = k[ AB][B] = k[A][B]2. The reaction is 2nd order with respect to B. 2 Correct. Since step 2 is the rate determining step and involves the reaction between 1 molecule of AB and 1 molecule of B, the rate determining step involves a bimolecular reaction i.e. involving two molecules. 3 Incorrect. Since A and B are both consumed but not regenerated, A and B are reactants and not catalysts. There are no catalysts in this reaction. Q11(C) 238U ⟶ Pb Molar ratio Initial amt / mol 1 0 - Amt after 1 half-life / mol ½ ½ 1 : 1 Amt after 2 half-lives / mol ¼ ¾ 1 : 3 Time taken = 2(4.5 x 109) = 9.0 x 109 years. Q12(D) When [X]eqm < [Y]eqm, the position of equilibrium lies t o the right ⇒ G< 0 i.e. at points 3 and 4. Q13(B) Equal amounts of reactants with total initial pressure p were reacted. Therefore, initial partial pressure of PCl3 = initial partial pressure of Cl2 = ½ p. Eqm partial pressure of PCl5 is given to be 1/5 p. PCl3(g) + Cl2(g) ⇌ PCl5(g) Initial P ½ p ½ p 0 Change in P –1/5 p –1/5 p +1/5 p Eqm p 3/10 p 3/10 p 1/5 p 1 205 33 9 10 10 c p K ppp         This document is copyrighted, please do not reproduce it without permission

19 Q14(A) Q15(C) When bromine is reacted with chlorocyclopropane in a free radical substitution reaction, two different radicals of chlorocyclopropane can be formed: A Possible. This product can be formed from B Possible. This product can be formed from C Not possible. This product can only be formed from the following reaction: The bromocyclopropyl radical comes from bromocycloproane, which is formed from the following reaction. Hr = BE(C–Cl) – BE(C–Br) = 340 – 280 = +60 kJ mol–1 Sr ≈ 0 kJ mol–1 K–1 Therefore, Gr = +60 kJ mol–1 > 0 i.e. the formation of bromocyclopropane from chlorocyclopropane is not spontaneous. D Possible. This product can be formed from This document is copyrighted, please do not reproduce it without permission