NJC 2019 H1 Promo Paper 2 (QP)

1 NJC Promotional Examination 8873/02/19 [Turn over NATIONAL JUNIOR COLLEGE SH 1 PROMOTIONAL EXAMINATIONS Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Additional Materials: Data Booklet 8873/02 Tues 1 Oct 2019 1 hour 30 min READ THE INSTRUCTIONS FIRST Write your name, subject class and registration number on all the work you hand in. Write in dark blue or black ink on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Structured Questions Answer ALL questions on the question paper. The use of an approved calculator is expected where appropriate. A Data Booklet is provided. The number of ma rks is given in brackets [ ] at the end of each question or part question. Appropriate significant figures and units are expected for final numerical answers. For Examiner’s Use 1 /7 2 /4 3 /5 4 /14 5 /7 6 /5 7 /10 8 /8 Penalty s.f. units Total /60 This document consists of 15 printed pages and 1 blank page.

2 NJC Promotional Examination 8873/02/19 [Turn over Answer ALL questions in the space provided. 1 (a) A sample of lead contains four stable isotopes with the following percentage abundances. Isotope Percentage abundance / % 204Pb 1.4 206Pb 24.1 207Pb 22.1 208Pb a For Examiner’s Use (i) Define the term relative atomic mass. ………………………………………………………………………………………………..… …………………………………………………………………………..……………………… [1] (ii) Determine the value of a. Hence calculate the relative atomic mass of lead. Give your answer to two decimal places. [2] (iii) Bismuth is on the right side of lead in the Periodic Table. Predict and explain whether bismuth has a higher or lower first ionisation energy compared to lead. ………………………………………………………………………………………………..… ………………………………………………………………………………………………..… …………………………………………………………………………..……………………… [2]

3 NJC Promotional Examination 8873/02/19 [Turn over (b) When the atomic orbitals from two atoms overlap a chemical bond may result. The p orbitals can overlap to form sigma (σ) or pi (π) bonds. When two atoms overlap the z-axis is used to define the internuclear axis. For Examiner’s Use (i) On the diagram below draw two p orbitals (one orbital on each atom) that could overlap to produce a sigma (σ) bond. atom 1 atom 2 [1] (ii) On the diagram below draw two p orbitals (one orbital on each atom) that could overlap to produce a single pi (π) bond. atom 1 atom 2 [1] [Total: 7]

4 NJC Promotional Examination 8873/02/19 [Turn over 2 One means of measuring toxicity is using LD, which stands for "Lethal Dose". LD50 is the amount of a material which causes the death of 50% of a group of test animals. LD50 value is expressed as the mass of a chemical administered per kg body mass of a test animal. Another means of measuring toxicity is using LC, which stands for "Lethal Concentration". The concentration of the chemical in air that kills 50% of the test animals during the observation period is the LC50 value. The table below shows the values for the LD50 and LC50 along with the toxicity ratings. (1 g = 1000 mg) Toxicity Rating Commonly used term LD50: Oral (mg kg−1) LC50: Inhalation (ppm) 1 Extremely Toxic 1 or less 10 or less 2 Highly Toxic >1 – 50 11– 100 3 Moderately Toxic 51 – 500 101 – 1000 4 Slightly Toxic 501 – 5000 1001 – 10,000 5 Practically Non-toxic 5001 – 15,000 10,001 – 100,000 For Examiner’s Use (i) 4.45 × 10–4 mol of a toxic compound, C4H5NO, was found to cause death in 50 % of test animals weighing 1 kg. Calculate LD50 of the compound and state its toxicity rating. Toxicity rating:……………… [2]

5 NJC Promotional Examination 8873/02/19 [Turn over (ii) Phosphine gas, PH3, is widely used in the semi-conductor industry as a dopant. The concentration of a small quantity of gas is usually expressed in parts per million (ppm) as shown below: Concentration in ppm = volume of gas volume of air × 106 For Examiner’s Use Given that LC50 for PH3 is 200 mg m ̶ 3 at room temperature and pressure, convert LC50 to ppm and determine its toxicity rating. Toxicity rating:……………… [2] [Total:4] 3 (a) Some bacteria can oxidise methane to carbon dioxide in the absence of oxygen. It has recently been reported that the mechanism involves a reaction between methane and nitrite ions in acidic conditions (reported in Nature, 2010). The half-equation for the oxidation of methane is: CH4 + 2H2O → CO2 + 8H+ + 8e– (i) Write a half-equation for the reduction of NO2– in acidic conditions to give N2. ………………………………………………………………………………………………...[1] (ii) By combining the half -equations, or otherwise, balance the overall equation shown below. ......CH4 + ......NO2− + ......H+ → ......CO2 + ......N2 + ...... H2O [2] (iii) Identify the oxidising agent in the reaction in (ii). Justify your answer using oxidation numbers. ………………………………………………………………………………………………….. ………………………………………………………………………………………………….. [2] [Total:5]

6 NJC Promotional Examination 8873/02/19 [Turn over 4 Hematite is a common iron oxide with the formula Fe₂O₃. It is a very important naturally occurring compound that finds widespread use as a heterogeneous catalyst. Fe ₂O₃ is used in the Haber Process which combines nitrogen with hydrogen into ammonia. N2 (g) + 3 H2 (g) Fe2O3 2 NH3 (g) For Examiner’s Use (a) (i) What is meant by the term heterogeneous catalyst? …………………………………………………………………………………………………. …………………………………………………………………………………………….…… [1] (ii) State the three stages involved in a typical reaction involving a heterogeneous catalyst. 1. .............................................................. 2. .............................................................. 3. .............................................................. [2] (iii) With the aid of a Boltzman distribution curve, explain how Fe₂O₃ affect the rate of the Haber Process. …………………………………………………………………………………………………. ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… [3]

7 NJC Promotional Examination 8873/02/19 [Turn over (b) Fe₂O₃ can also react with CO to produce Fe. Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ∆Hr = −27 kJ mol−1 For Examiner’s Use (i) Draw a labelled reaction pat hway diagram for this reaction, given that the activation energy of the reaction is 37 kJ mol−1. [2] (ii) Hence, or otherwise, determine the activation energy for the following reaction. 2Fe(s) + 3CO2(g) →