ASR 2020 J2Prelim H2Chem P2 MS

1 ASRJC JC2 PRELIM 2020 9729/02/H2 Anderson Serangoon Junior College 2020 JC2 Preliminary Examinations H2 Chemistry (9729) Suggested Mark Scheme for Paper 2 1 One early nineteenth century Periodic Table had copper placed in the same group as potassium because they both formed +1 ions. (a) State the full electronic configurations of potassium and copper atoms. [1] K: 1s2 2s2 2p6 3s2 3p6 4s1 Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (b) (i) The mass of an atom is the sum of the masses of all subatomic particles it contains. Calculate the mass, in kg, of one atom of 64 29Cu . Quote relevant values from the Data Booklet and give your answer to three significant figures. [2] No of protons in 64 29Cu = 29 No of electrons in 64 29Cu = 29 No of neutrons in 64 29Cu = 35 Mass of one atom 64 29Cu = (29x1.67x10–27) + (29x9.11x10–31) + (35x1.67x10–27) = 1.07 x 10–25 kg (ii) During the process of ionisation, a Cu atom loses an electron. Cu(g)  Cu+(g) + e− State the electronic configuration of Cu+. Explain your answer. [2] 1s2 2s2 2p6 3s2 3p6 3d10 explanation: It is the highest energy orbital / the electron in this orbital is less attracted to the nucleus.

(iii) On the Cartesian axes given in Fig . 1.1, sketch the shape of the following three 3d orbitals found in copper atom. Fig. 1.1 [2] x y z x y z x y z 3dxz 3dz2 3dx2–y2

ASRJC JC2 PRELIM 2020 9729/02/H2 (c) (i) Define the term standard enthalpy change of formation of a substance. [1] The amount of heat evolved or absorbed when one mole of a substance is formed from its constituent elements, all in their standard states at 298 K and 1 bar. (ii) Table 1.1 enthalpy change value / kJ mol–1 standard enthalpy change for P(s) + 2O2(g) + 3e–  PO43–(aq) –1284 standard enthalpy change for K(s)  K+(aq) + e– –251 standard enthalpy change for K3PO4(s)  3K+(aq) + PO43–(aq) –2 Using a labelled energy cycle, and the enthalpy values given in Table 1.1, determine the standard enthalpy change of formation of solid potassium phosphate, K3PO4. [3] –2 K3PO4 (s) 3K+ (aq) + PO43– (aq) ? 3(–251) –1284 3K (s) + P (s) + 2O2 (g) –2 + Hf (K3PO4) = [3 x (–251) + (–1284)] = –2035 kJ mol–1 (iii) The value of GO at 298 K for K(s)  K+(aq) + e– is –284 kJ mol–1. Calculate SO for the reaction, and explain its sign. [2] GO = HO - TSO –284 = –251 – (298) SO SO = +0.111 kJ mol-1 K-1 = +111 J mol-1 K-1 SO is positive, indicating an increase in ways of arranging the particles and distributing energy as the reaction results in a change from regular arrangement in the solid state to ions in the aqueous state. [Total: 13]

2 2–methylpropane, (CH 3)2CHCH3, is an important precursor for petrochemical industry. Butane, CH3(CH2)2CH3, can be converted to 2–methylpropane in the presence of a suitable heterogeneous catalyst. CH3(CH2)2CH3(g) (CH3)2CHCH3(g) equilibrium 1 (a) Explain briefly how a heterogeneous catalyst increases the rate of equilibrium 1. [1] A heterogeneous catalyst provides active sites where the butane molecules could be adsorbed via temporary bonds. This weakens the covalent bonds in butane and lowers the activation energy which increases the rate of reaction. OR The adsorption increases the concentration of the reactant molecules at the catalyst surface and allows more reactant molecules to come into close contact with proper orientation for reaction to take place. (b) Butane gas was added to an enclosed vessel, at 373 K and at constant pressure. The concentration of butane and 2 –methylpropane was measured at regular time intervals and a graph was plotted as shown in Fig. 2.1. Fig. 2.1 (i) Using information from Fig. 2.1, calculate the partial pressures of butane and 2–methylpropane in the mixture at equilibrium. You may assume that both gases behave ideally. [2] pV = nRT p = ( n V )RT (IMPT: V is in m3!) pbutane = 3 (0.30)(8.31)(373) 1 10  0.30 × (10−1)−3 × 8.31 × 373 = 930000 Pa (3 sig fig) p2–methylpropane = 3 (0.70)(8.31)(373) 1 10  = 2170000 Pa (3 sig fig) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00concentration (mol dm–3) time (CH3)2CHCH3 CH3CH2CH2CH3

ASRJC JC2 PRELIM 2020 9729/02/H2 (ii) Write the Kp expression of equilibrium 1 and calculate its value. [2] 2-methylpropane p butane = pK p 2170000= 930000 = 2.33 (c) State how the partial pressure of butane will change when argon is added to the enclosed vessel at constant volume. [1] partial pressure of butane remains unchanged / constant. (d) Bromine reacts with alkanes in the presence of light. (i) Outline the mechanism of the reaction between 2–methylpropane and bromine to form 2–bromo–2–methylpropane. [3] (Free radical substitution) Initiation: Propagation: Termination: (ii) Suggest why it is not possible to make iodoalkanes by this method. [1]  Iodoalkanes cannot be made by free –radical substitution as the first propagation step is endothermic (H = 410 – 299 = +111 kJ mol–1)  OR Less energy is evolved from the formation of the weak H–I bond (299 kJ mol–1) compared to that of H–Br (366 kJ mol–1).  OR Enthalpy change for the reaction is endothermic (+22 kJ mol–1). uv

(e) Iodoalkanes can be made by warming a chloroalkane with a solution of sodium iodide in propanone, in which sodium chloride is almost insoluble. CH3CH2Cl + NaI CH3CH2I + NaCl equilibrium 2 (i) By considering the equation for equilibrium 2, suggest why the reaction goes almost to completion, despite the C–I bond being weaker than the C–Cl bond. [1] Sodium chloride being insoluble can be easily removed, causing the [NaCl] to be low. This will cause the equilibrium position to shift to the right to form more NaCl, allowing the reaction to go almost to completion. A student performed two experiments to investigate the effect of changes in concentration on the rate of this reaction. The initial concentration and rate data obtained for each experiment is given in Table 2.1. Table 2.1 experiment initial [CH3CH2Cl] / mol dm–3 initial [NaI] / mol dm–3 initial rate / mol dm–3 s–1 1 1.0 x 10–3 2.0 x 10–3 3.0 x 10–11 2 3.0 x 10–3 4.0 x 10–3 1.8 x 10–10 (ii) Determine the order of reaction with respect to each reactant and hence deduce the rate equation for this reaction. Use data from Table 2.1 to explain your answers. [3] 3 3 10 2 3 3 11 1 (3.0 10 ) (4.0 10 ) 1.8 10 (1.0 10 ) (2.0 10 ) 3.0 10             xy xy rate k rate k 3 2 18 1 1 3            xy (3x)(2y) = 6 x = 1, y = 1 Therefore order of reaction with respect to CH3CH2Cl is 1 and order of reaction wi