RI 2021 Year 6 June TP Suggested Solutions

2021 H2 Chemistry Y6 Term 3 Common Test Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C B C D B A C D D A A A D C B Question 1 (C) nestradiol = 76/1000 (18×12.0+24×1.0+2×16.0) = 2.794 × 10−4 mol ROH + Na → RO–Na+ + ½H2 1 mol of alcohol reacts with Na to produce 0.5 mol of H2. Since there are 2 –OH groups in estradiol, 1 mol of estradiol produces 1 mol of H2. amt of H2 formed = 2.794 x 10–4 mol volume of H2 formed at s.t.p. = 2.794 x 10–4 x 22.7 = 0.00634 dm3+ = 6.34 cm3 Question 2 (B) no. of protons no. of electrons no. of neutrons 2 1D 1 1 1 11 5B 5 5 6 12 6C 6 6 6 14 7N 7 7 7 16 8O 8 8 8 species no. of electrons no. of neutrons A 11BD4– 10 10 B CD3– 10 9 C ND4+ 10 11 D D3O+ 10 11 Question 3 (C) butan-1-ol (C4H10O) 2-methylpropan-2-ol (C4H10O) Butan-1-ol is a straight-chain isomer of C4H10O and has a greater surface area for instantaneous dipole–induced dipole (id -id) interactions to occur compared to 2 -methylpropan-2-ol which is a branched-chain isomer of C4H10O. Thus, id -id interactions are stronger in butan-1-ol than in 2-methylpropan-2-ol. (Option 1 is correct) The strength of the hydrogen bonds in both molecules are the same as the hydrogen atom involved in hydrogen bonding is bonded to the same element, oxygen. The extent of hydrogen bonding is also the same as they have same number of lone pairs and number of H atoms attached to O. (Options 2 and 3 are incorrect) Question 4 (D) The enthalpy change of the neutralisation reaction between KOH and HNO3 is −57.3 kJ mol−1. Since KOH and HNO 3 have the same concentrations and KOH has a lower volume, KOH is the limiting reagent. Hence, 57300 2( )1000 114.6 J qqH yn qy  =− − =− = Also, q = mcT = (25.0 + y)(4.18)(10.0) Therefore, (25.0 + y)(4.18)(10.0) = 114.6y y = 14.35 cm3 Question 5 (B) By combining the 2 equations, you can obtain the overall equation for the decomposition of 1,2-dioxetanedione (C2O4). C2O4 → 2CO2 + light ∆H : (–). Due to formation of the more stable CO2, the reaction is exothermic. ∆S : (+). Due to the formation CO 2 gas from liquid 1,2-dixetanedione, there is an increase in entropy. ∆G: ( –). Since the reaction occurs when the chemicals are mixed, it is spontaneous. Alternatively, you can derive the sign of ∆G using ∆G = ∆H – T∆S and the signs for ∆H and ∆S deduced previously. This document is copyrighted, please do not reproduce it without permission

Question 6 (A) Since the decomposition of X follows first -order kinetics, t½ = ln 2 / k. t½ = ln 2 / k = ln 2 / (7  10–2) = 9.90 years Let n be the number of half-lives in 3 years. n  t½ = 3 years n = 3 / 9.90 = 0.303 [X] = (½)n [X]initial [X] / [X]initial = (½)0.303 = 0.811 Hence 81.1% of X remains after 3 years. Question 7 (C) A Incorrect. The chemical equation does not provide any information about the order of a reaction. No other information is provided as well. B Incorrect. If the order of reaction w.r.t. A is zero (or negative), increasing [A] does not increase the reaction rate. C Correct. A catalyst provides an alternative pathway of lower activation energy, increasing the rate of the reaction. D Incorrect. Removing B, a product, does not impact the reaction rate. Question 8 (D) A Incorrect. There are 0.015 mol of CH3COO– and 0.030 mol of H + from HCl i.e. H + is in excess. Since CH 3COO– + H + → CH3COOH, the resultant mixture contains 0.015 mol of H + and 0.015 mol of CH3COOH. This is not a buffer solution. B There are 0.020 mol of CH3COO– and 0.010 mol of H + from HC l i.e. CH 3COO– is in excess. Since CH 3COO– + H + → CH3COOH, the resultant mixture contains 0.010 mol of CH3COO– and 0.010 mol of CH3COOH. This is a buffer solution. C Incorrect. There are 0.010 mol of CH3COOH and 0.020 mol of OH – from NaOH i.e. OH– is in excess. Since CH3COOH + OH– → CH3COO– + H2O, the resultant mixture contains 0.010 mol of OH– and 0.010 mol of CH3COO–. This is not a buffer solution. D There are 0.030 mol of CH 3COOH and 0.015 mol of OH– from NaOH i.e. CH3COOH is in excess. Since CH3COOH + OH– → CH3COO– + H2O, the resultant mixture contains 0.015 mol of CH3COOH and 0.015 mol of CH3COO–. This is a buffer solution. Comparing the buffer solutions in B and D, the solution in D contain greater amounts of the buffering species (i.e. CH3COOH and CH3COO–) to react with any small amounts of acid or base added. Hence, for the same amount of acid or base added, the pH change in D will be smaller, making it more effective in resisting pH change. Question 9 (D) Silver chloride is a sparingly soluble salt. AgCl(s) ⇌ Ag+(aq) + Cl–(aq) – eqm (1) A Incorrect. AgNO3(aq) provides a high [Ag+(aq)], causing the position of eqm (1) to shift to the left, reducing the amount of AgCl that can dissolve. B Incorrect. NH4Cl(aq) provides a high [Cl–(aq)], causing the position of eqm (1) to shift to the left, reducing the amount of AgCl that can dissolve. C Incorrect. HNO3(aq) does not affect [Ag+(aq)] and [Cl–(aq)]. Thus, it does not affect the solubility of AgCl. D Correct. The presence of NH 3 causes another equilibrium to be established which forms a soluble complex, [Ag(NH3)2]+. Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+(aq) – eqm(2) The position of eqm (2) lies very much to the right, significantly lowering [Ag +(aq)], causing the position of eqm (1) to shift to the right. This results in more AgCl dissolving. This document is copyrighted, please do not reproduce it without permission

Question 10 (A) Since E (NiOOH/Ni(OH)2) is more positive than E(Cd(OH)2/Cd), NiOOH undergoes reduction and Cd undergoes oxidation. Cathode: NiOOH + 2H2O + 2e− → Ni(OH)2 + 2OH− Anode: Cd + 2OH− → Cd(OH)2 + 2e− Since electrons are released at the Cd(OH)2/Cd half-cell, Cd is the negative electrode. Cd(OH) 2 is formed at the negative electrode. Question 11 (A) 1 Correct. • M (Al) has the highest melting point of the 3 elements. • Oxide of K (P4O10) is an acidic oxide that dissolves in dilute KOH. • pH of the chlorides in water decrease in the following order. L > M > K MgCl2 > AlCl3 > PCl5 6.5 3 2 2 Incorrect. • K (Si) has the highest melting point of the 3 elements. • Oxide of K (SiO2) is only dissolves in concentrated OH–. • pH of the chlorides in water decrease in the following order. L > M > K NaCl > MgCl2 > SiCl4 7 6.5 2 3 Incorrect. • M (Si) has the highest melting point of the 3 elements. • Oxide of K (Al2O3) is an amphoteric oxide that dissolves in dilute KOH. • pH of the chlorides in water decrease in the following order. L > K > M MgCl2 > AlCl3 > SiCl4 6.5 3 2 Question 12 (A) 1 Correct. Both the phenol and alkene reacts with Br2(aq) to form the following major product. 2 Incorrect. The –COOH group, which reacts with Na2CO3 to form CO2 gas, is absent. Phenol does not react with Na 2CO3 to form CO2 gas. 3 Incorrect. LiAlH4 reduces only the ketone functional group in the compound to form the following product, which contains only 1 chiral centre. Question 13 (D) The energy profile diagram has one peak and represents a reaction with only one step in the mechanism. A Incorrect. Chlorobenzene does not undergo nucleophilic substitution with cyanide ion due to the partial double bond character of the C–Cl bond, which strengthens the bond. B Incorrect. The nucleophilic addition of HCN to propanone occurs via a two -step mechanism. C Incorrect. The electrophilic addition of HBr to propene occurs via a two -step mecha