NJC 2019 H1 Promo Paper 2 solutions

1 NJC Promotional Examination 8873/02/19 [Turn over NJC 2019 H1 Chemistry Promo Paper 2 Solutions 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] Relative atomic mass is the weighted average isotopic mass of one atom of an element compared to 1/12 the mass of a 12C atom. (ii) Determine the value of a. Hence calculate the relative atomic mass of lead. Give your answer to two decimal places. [2] % abundance of 208Pb, a = 100 – 1.4 – 24.1 – 22.1 = 52.4% Relative atomic mass of lead = (204 × 1.4%) + (206 × 24.1%) + (207 × 22.1%) + (208 × 52.4%) = 207.24 (2 d.p.) Examiner’s comments: [1] must be 2 dp. ECF from value of a Some students used Pb’s Ar 207.2 from Data Booklet to find the value of a, which is incorrect. For those who gave gmol -1 for relative atomic mas s of lead, they are penalised under units. (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] Bi has a higher nuclear charge but same screening effect as Pb. Thus there is a stronger nuclear attraction for the most loosely held electrons in Bi, more energy is required to remove the most loosely held electron in Bi. Hence Bi has a higher first IE.

2 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]

3 NJC Promotional Examination 8873/02/19 [Turn over 2 One means of measuring toxicity is using LD, which stands for "Lethal Dose". LD 50 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 (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 the LD50 of the compound and state its toxicity rating. Toxicity rating:……………… [2] Mass of compound = 4.45 x 10 −4 x (48 + 5 + 14 + 16) = 0.03694 g Hence, LD50 = 36.9 mg/kg The toxicity rating is 2 (highly toxic)

4 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 Given that the LC50 for PH3 is 200 mg m ̶ 3 at room temperature and pressure, convert the LC50 to ppm and determine its toxicity rating. Toxicity rating:……………… [3] [Total:5] Volume of PH3 = 200 𝑥 10−3 34.0 x 24 dm3 = 0.141 dm3 LC50 = 0.141 𝑥 10−3 1 x 106 = 141 ppm The toxicity rating is 3 (moderately toxic). Examiner’s comments: Many students did not get this part correct. It’s either they didn’t calculate the volume of PH3 correctly or fail to do the units conversion from mg to g.

5 NJC Promotional Examination 8873/02/19 [Turn over 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– For Examiner’s Use (i) Write a half-equation for the reduction of NO2– in acidic conditions to give N2. ………………………………………………………………………………………………...[1] 2NO2− + 8H+ + 6e– → N2 + 4 H2O (ii) By combining the half -equations, or otherwise, balance the overall equation shown below. ......CH4 + ......NO2− + ......H+ → ......CO2 + ......N2 + ...... H2O [2] CH4 + 2H2O → CO2 + 8H+ + 8e– ----(1) x3  3CH4 + 6H2O → 3CO2 + 24H+ + 24e– 2NO2− + 8H+ + 6e– → N2 + 4 H2O ----(2) x4  8NO2− + 32H+ + 24e– → 4N2 + 16H2O ....3..CH4 + ...8...NO2− + ....8..H+ → ...3...CO2 + ...4...N2 + ...10... H2O (iii) Identify the oxidising agent in the reaction in (ii). Justify your answer using oxidation numbers. [2] NO2− is the oxidising agent, itself is reduced. N is reduced from +3 in NO2− to zero in N2. [Total:5]

6 NJC Promotional Examination 8873/02/19 [Turn over 4 Hematite is a common iron oxide with the formula Fe ₂O₃. 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) (a) (i) What is meant by the term heterogeneous catalyst? [1] The catalyst is of a different phase compared to the reactants. (ii) State the three stages involved in a typical reaction invo lving a heterogeneous catalyst. [1] Adsorption, reaction, desorption Examiner’s Comments: Adsorption is not absorption. Adsorption of reactant particles onto the surface of the catalyst. Weak interactions form between the reactant and catalyst. Reaction at the surface occurs at a faster rate as reactant molecules are brought closer together and existing interactions in the reactant molecules are weakened thus lowering activation energy Desorption of products from the catalyst surface. Catalyst is regenerated. (iii) With the aid of a Boltzman distribution curve, explain how Fe₂O₃ affect the rate of the Haber Process. [3] [3] In the presence of catalyst, the reaction proceeds with an a