JPJC Prelim 2025 Paper 3 Answers H2 Chem

© Jurong Pioneer Junior College [Turn Over Answer Scheme Paper 3 Section A 1 Nitrogen, a Period 2 element, forms a wide variety of compounds with diverse structures, oxidation states, and properties. These include ammonia, ammonium salts, amines, and nitrates, all of which play important roles in industry, agriculture, and the environment. (a) The enthalpy changes involved in the formation of ionic and covalent compounds vary across the period. (i) Explain why the first ionisation energy of nitrogen is more endothermic than that of carbon and oxygen. [2] C: 1s2 2s2 2p2 N: 1s2 2s2 2p3 O: 1s2 2s2 2p4 Comparing C and N: N has a higher nuclear charge, while the shielding effect caused by inner shell electrons remains approximately constant. Therefore, it has a smaller atomic radius, and valence electrons experience a stronger nuclear attraction. Consequently, N has a higher first ionisation energy compared to C. Comparing N and O: Less energy is required to remove one of the paired 3p electrons in oxygen due to the larger interelectronic repulsion. Thus, O has a lower first ionisation energy than N. (ii) Some Period 3 elements can react with chlorine to produce chlorides, for example, NaCl and SiCl4. Describe and explain what happens when NaC l and SiCl4 are separately added to excess water. State the pH of the resultant mixtures and give relevant equations. [3] NaCl is an ionic chloride and will dissolve/hydrate in water to form a solution of pH 7. NaCl → Na+(aq) + Cl– SiCl4 is a covalent chloride and has energetically accessible vacant 3d orbitals that allow the water molecule to form dative bonds. Thus SiCl4 hydrolyse in water to form an acidic solution of pH 1-2. SiCl4 + 4H2O → Si(OH)4 + 4HCl

2 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 1 (b) Ammonia (NH3), ammonium nitrate (NH4NO3), and ethanamide (CH3CONH2) are important compounds that can be used as fertilisers due to their high nitrogen content. Arrange these three compounds in order of increasing pH value of their aqueous solution. Explain your answer. [3] NH4NO3 < NH2CONH2 < NH3 NH4NO3 will dissociate into NH4+ and NO3- ions in solution. NH4+ will undergo hydrolysis to produce H + ions, forming an acidic solution. Thus, the pH value of NH4NO3 solution is less than 7. NH3 will be basic due to the availability of lone pair electrons on N for protonation. Thus, the pH value of the solutions will be higher than 7. CH3CONH2 will be neutral due to the delocalisation of the lone pair of electrons on N into the neighbouring  electron cloud of the electron - withdrawing C=O group, the lone pair of electrons in urea is not available for protonation than NH 3. Thus, it is a weaker base compared to NH 3, hence, lower pH value. (c) Aqueous iron(III) nitrate, Fe(NO3)3(aq), is yellow in colour. (i) Explain why the solution of iron(III) nitrate is yellow in colour. [3] The presence of ligands cause the d orbitals to split into 2 different energy levels with a small energy gap, E, that corresponds to the visible light region of the electromagnetic spectrum. Visible light is absorbed when an electron transits from a lower energy d orbital to a higher energy d orbital which is unfilled or partially filled. Hence, the colour observed is the complement of the colours absorbed. (ii) Using data from the Data Booklet , predict what will be observed when aqueous iron(III) nitrate is added to potassium iodide, KI, giving a balanced equation for the reaction that occurs. [3] From Data Booklet: Fe3+ + e- ⇌ Fe2+ E = +0.77 V I2 + 2e- ⇌ 2I- E = +0.54 V ( ) ( ) 0.77 - 0.54 0.23 0 ( ) cellE V reaction is energetically feasible  = + + = +  Obs: Yellow solution of Fe 3+ turn pale green and brown solution of I 2 formed Eqn: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(aq)

3 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 [Turn Over (iii) A sample of Fe(NO3)3(aq) was added to aqueous sodium hydroxide. A red- brown precipitate formed immediately and remained unchanged upon standing. When a solution of SCN −(aq) is added to the resulting precipitate, the precipitate dissolves and blood red solution of [Fe(SCN)(H2O)5]2+ complex is formed. The formation of the blood red complex is non-reversible. Explain the above observations and include relevant equations. [3] Fe3+ reacts with OH- to form red-brown precipitate of Fe(OH)3 Fe3+(aq) + 3OH−(aq) → Fe(OH)3(s) When SCN− is added, a ligand exchange reaction occurs. The stronger SCN− ligand replaces the weaker H2O ligand to form a stronger dative bond and a more stable complex. Fe(H2O)63+ + SCN−(aq) → Fe(H2O)5(SCN)2+ + H2O The precipitate in the solution is sparingly soluble. Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH−(aq) --- (1) This decreases concentration of Fe 3+ in the solution, causing position of equilibrium (1) to shift left and thus precipitate dissolves. [Total: 17]

4 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 2 Lucas Test, named after an American chemist, Howard Luca, is a simple qualitative test, using concentrated HCl in the presence of ZnCl2, to classify alcohols by observing the rate of turbidity, indicating the formation of insoluble alkyl halides. (a) The rate of the Lucas Test is investigated, using tert-butanol, (CH3)3COH. When 1 dm 3 of 0.0100 mol of tert -butanol is reacted with 5.00 mol of HC l in the presence of ZnCl2, the concentration of tert-butyl chloride, (CH3)3CCl, formed over time is shown in the graph below. (a) (i) Define the term order of reaction. [1] order of reaction with respect to a reactant is the power to which the concentration of the reactant is raised to in the rate equation OR In an experimentally determined rate equation: rate = k[A]m[B]n where A and B are the reactants, the orders of reaction with respect to [A] and [B] are m and n respectively 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0 10 20 30 40 [tert-butyl chloride] / moldm-3 time / min

5 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 [Turn Over (ii) The same graph was obtained when the test was repeated with 7.5 mol of HCl. Using the information provided and the graph, deduce the orders with respect to (CH 3)3COH and HC l. Show clearly your working and any construction lines on the graph. [3] Since 1 st t½ = 2 nd t½ = 10 mins, the half life of the reaction is (approximately) constant. The reaction is 1st order w.r.t (CH3)3COH and When concentration of HCl is increased by 1.5 times/changed, the rate of reaction remains the same, thus, the reaction is 0 order w.r.t HCl. (iii) Hence, write the rate equation for the Lucas Test, and calculate a value for the rate constant. Include units in your answer. [3] Rate = k[(CH3)3COH] Since it is a overall first order reaction k = ln 2 / t½ = ln 2 / 10 = 0.0693 min-1 OR Initial rate method: y-intercept at 16 mins Initial rate = 0.011 / 16 = 0.000688 mol dm-3 min-1 K = 0.000688 / 0.01 = 0.0688 min-1 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 0.011 0 5 10 15 20 25 30 35 40 t½ = 10 mins t½ = 10 mins Max. pdt formed ½ total pdt formed ¾ total pdt formed 0.75

6 © Jurong Pioneer Junior College 9729/03/J2 PRELIMINARY EXAM/2023 2 (b) The Lucas Test, as shown above, follows a nucleophilic substitution reaction mechanism as follows: • Tert-butanol reacts with HC l to generate the following reaction intermediate in the first step. • This intermediate subsequently loses water to form a carbocation, which reacts further to form the product, tert-butyl chloride. • HCl is acting as a Lewis acid in the first step and as a source for nucleophile in the subsequent steps. (i) Define the term, nucleophile. [1] Nucleophile is defined as electron-rich sp