2023 NJC H2 Chem Prelim P3 (Ans)

1 NJC SH2 Preliminary Examination 9729/03/23 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 3 Free Response Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/03 14 September 2023 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. A Data Booklet is provided. The use of an approved scientific calculator is expected, where appropriate. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 /20 2 /16 3 /24 Section B 4 /20 5 /20 Paper 3 Total /80 This document consists of 28 printed pages.

2 NJC SH2 Preliminary Examination 9729/03/23 Section A Answer all the questions in this section. 1 (a) Table 1.1 list physical properties of some Period 4 elements. Table 1.1 property K Ca Fe As Se relative atomic mass 39.1 40.1 55.8 74.9 79.0 atomic radius (metallic) / nm 0.197 0.126 melting point / K 1112 1808 density / g cm−3 1.54 7.86 1st I.E. / kJ mol−1 944 941 2nd I.E. / kJ mol−1 3070 1150 (i) Explain why the atomic radius of Fe is less than that of Ca. [2] (ii) Use relevant data from the table to explain why the density of Fe is significantly greater than that of Ca. (no calculations are required) [2] (iii) Suggest why the melting point of Fe is significantly higher than the melting point of Ca. [2] (iv) With reference to the electronic configuration of K and Ca, explain why the 2nd I.E. of Ca is lower than that of K. [2] (v) With reference to the electronic configurations of As and Se, explain why the 1st I.E. of Se is lower than that of As. [2] (i) Fe has higher nuclear charge than Ca as it has more protons. Although Fe has more inner-shell electrons than Ca, the additional 3d electrons are poor shielding electrons. Thus, the increase in nuclear charge outweighs the increase in shielding effect. Overall, the valence electrons in Fe experience a stronger nuclear attraction, resulting in smaller atomic radius than Ca. (ii) Density = mass/volume Fe has a smaller atomic radius than Ca and thus more atoms per unit volume. Furthermore, Fe has a larger relative atomic mass than Ca. Hence, Fe has greater mass per unit volume than Ca. (iii) The small energy difference between 3d and 4s subshells allow Fe to contribute both the 3d and 4s electrons to the sea of delocalised electrons for metallic bonding, whereas Ca contributes only 2 valence electrons (from 4s subshell) to the sea of delocalized electrons. More energy is required to overcome the stronger metallic bonds in Fe and hence the melting point of Fe is significantly higher than that of Ca.

3 NJC SH2 Preliminary Examination 9729/03/23 [Turn over (iv) K: 1s2 2s2 2p6 3s2 3p6 4s1 Ca: 1s2 2s2 2p6 3s2 3p6 4s2 The second most loosely held electron to be removed from K is from the 3 p subshell whereas the second most loosely held electron to be removed from Ca is from the 4s subshell, which experiences greater shielding effect since it has one more filled principal quantum s hell and is further away from the nucleus . These factors outweigh the higher nuclear change in Ca. Thus, the nuclear attraction for the second most loosely held electron in Ca is weaker and less energy is required to remove it, resulting in lower 2nd I.E for Ca as compared to K. (v) As: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 Se: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4 The most loosely held electron to be removed from Se is a paired 4p electron whereas the most loosely held electron to be removed from As is an unpaired 4p electron. Less energy is required to remove the paired electron since it experiences inter-electronic repulsion, hence, Se has a lower 1st I.E than As. (b) The Fischer-Tropsch process involves converting a mixture of carbon monoxide and hydrogen, known as syngas, into hydrocarbons, in the presence of catalysts. One such reaction is shown below: CO(g) + 3H2(g) CH4(g) + H2O(g) ∆H < 0 (i) Write an expression for the equilibrium constant, Kp, for this reaction, and state its units. [2] (ii) A mixture of CO and H 2 was introduced into a sealed vessel and heated to 1200 K. At equilibrium, it was found that the total pressure was 32 atm, and the mole fractions of CO and CH4 were 0.5 and 0.12 respectively. Calculate the equilibrium partial pressures of all gases, and hence calculate the value of Kp. [3] (iii) Higher temperatures and higher pressures can lead to faster reactions. However, in commercial facilities that use the Fischer-Tropsch process, this was avoided. Explain why. [2] Transition m etals such as iron or cobalt, are commonly used as catalysts in the Fischer-Tropsch process. (iv) State the type of catalysis in the Fischer-Tropsch process. [1] (v) Outline the mode of action of the catalyst in the Fischer-Tropsch process. [2] (i) Kp = 𝑃𝐶𝐻4×𝑃𝐻2𝑂 𝑃𝐶𝑂×(𝑃𝐻2)3 Units = atm−2

4 NJC SH2 Preliminary Examination 9729/03/23 (ii) When position of equilibrium shifts to the right, equal amount of CH4 and H2O are formed Mole fraction of CH4 = H2O = 0.12 PCH4 = PH2O = 0.12 × 32 = 3.84 atm Mole fraction of CO = 0.5 PCO = 0.5 × 32 = 16 atm Mole fraction of H2 = 1 – 0.5 – 0.12 – 0.12 = 0.26 PH2 = 0.26 × 32 = 8.32 atm Kp = 3.84  3.84 16  (8.32)3 = 1.60  10−3 atm−2 (iii) The forward reaction is exothermic, using a high temperature would favour the backward endothermic reaction instead, resulting in lower yield, hence avoided. Using a higher pressure will incur higher costs of maintaining the equipment, hence avoided. (iv) Heterogeneous catalysis (v) Step 1: Adsorption of reactant particles CO(g) and H 2(g) onto the active sites of Fe(s) catalyst surface through weak interactions. Step 2: Reaction at the surface occurs at a faster rate as reactant molecules are brought closer together in the correct orientation for reaction and existing bonds within the reactant molecules are weakened, thereby reducing E a. Step 3: Desorption of products CH4(g) from the Fe(s) catalyst surface. Catalyst is regenerated and there are vacant active sites available for adsorbing other reactant molecules. [Total: 20]

5 NJC SH2 Preliminary Examination 9729/03/23 [Turn over 2 Iodine is found naturally in compounds in many different oxidation states. (a) Iodide ions, I−, react with acidified H2O2(aq) to form iodine, I2, and water. The resultant mixture is then shaken with cyclohexane, C6H12