2025 ASRJC Prelim H2Chem P1 Solution

9729/01/H2 [Turn over ANDERSON SERANGOON JUNIOR COLLEGE 2025 JC2 PRELIMINARY EXAMINATION Paper 1 Solutions 1 Sodium azide, NaN3 is an explosive used to inflate airbags in cars when they crash. It consists of positive sodium ions and negative azide ions. What are the number of electrons in the sodium ion and the azide ion? sodium ion azide ion A 10 20 B 10 22 C 12 20 D 12 22 Ans: B Na+ (11 – 1 = 10) and N3- (7 x 3 + 1 = 22) 2 The graph shows the variation of the first ioni sation energy with proton number for some elements. The letters used are not the actual symbols for the elements. Which statement about the elements is correct? A P and X are in the same period in the Periodic Table. B The general increase from Q to X is due to increasing atomic radius. C The small decrease from R to S is due to decreased shielding. D The small decrease from U to V is due to repulsion between paired electrons. Ans: D A – same Group B – increasing nuclear charge (and relatively constant shielding effect) C – higher energy level (ns2 vs np1)

2 9729/01/H2 3 The table identifies the shape and polarity of four molecules. Which row is correct? molecule molecular shape polarity A boron trichloride trigonal pyramidal polar B nitrogen trichloride trigonal planar non-polar C sulfur dioxide bent polar D trichloromethane tetrahedral non-polar Ans: C BCl3 NCl3 SO2 CHCl3 No. of bp, lp on central atom 3 bp, 0 lp 3 bp , 1 lp 2 bp, 1 lp 4 bp, 0 lp Molecular shape trigonal planar trigonal pyramidal bent tetrahedral B Cl Cl Cl N Cl Cl Cl S O O C Cl Cl Cl H polarity non–polar polar polar polar 4 The element tin exists in two forms, grey tin and white tin. Some properties of grey tin and white tin are shown. grey tin white tin boiling point 2543 oC 2533 oC electricial conductivity none in solid or liquid good in solid and liquid malleability brittle malleable Which structural change might take place when grey tin changes to white tin? A giant covalent to giant ionic B giant covalent to giant metallic C giant ionic to giant covalent D giant ionic to giant metallic Ans: B White tin must have giant metallic structure as it conducts electricity in solid and liquid state and its malleable. Grey tin cannot have giant ionic structure as it does not conduct electricity when in liquid state, thus, it has to be giant covalent structure.

3 9729/01/H2 [Turn over 5 0.01 mol of K IOn reacts with 0.05 mol of K I stoichiometrically to produce I2 under acidic conditions. In this reaction, all the iodine containing reactants were converted to I2. What is the value of n? A 1 B 2 C 3 D 4 Ans: C Let the oxidation state of iodine in KIOn be In+ (this approach allow you to focus on the species that is undergoing Redox.) Reduction: (2n) e + 2 In+ → I2 x 2 Oxidation: 2I− → I2 + 2e x 2n Overall equation: 4In+ + (4n)I− → 2I2 + (2n)I2 Comparing mole ratio n+amt of I 4 0.01 amt of I 4 0.05− == n , solving n = 5 Since IOn−, considering the oxidation number and overall charges of the ion, +5 + n(-2) = −1 n = 3 OR IOn−: I− 1 : 5 oxidation: 2I− → I2 + 2e− 5I− → 5 2 I2 + 5e− [R]: 2IOn− + 4nH+ + (4n−2)e− → I2 + 2nH2O 2n −1 = 5 n = 3

4 9729/01/H2 6 X and Y are two different samples of the same ideal gas. Given that X contains a higher mass than Y, which graph shows the correct ideal gas relationship for the two samples of gas? (T is measured in K.) A B C D Ans: D Manipulation of ideal gas equation would give • V = mR pM(T)  graph of V against T gives a straight line passing thru the origin. [option D] Since X has a larger mass, the line for X has a steeper gradient. 7 A student mixes 25.0 cm 3 of 0.350 mol dm –3 sodium hydroxide solution with 25.0 cm 3 of 0.350 mol dm –3 hydrochloric acid. The temperature increases by 2.5°C. No heat is lost to th e surroundings. The final mixture has a specific heat capacity of 4.2 J g–1 K–1. What is the molar enthalpy change for the reaction? A –150 kJ mol–1 B –60 kJ mol–1 C –30 kJ mol–1 D –0.15 kJ mol–1 Ans: B Limiting reagent is hydrochloric acid = (0.35 x 0.025) = 0.00875 mol q = (25.0 + 25.0)(4.2)(2.5) = 525 J ΔH = –(525)/(0.00875) = –60 kJ mol–1

5 9729/01/H2 [Turn over 8 Silane, SiH4, exists as a gas at standard temperature and pressure. Hess’ Law can be used to calculate the average Si–H bond energy in gaseous SiH4. Which information is needed to perform the calculation? A ∆Hoformation(SiH4), ∆Hoatomisation(Si), ∆Hocombustion(H2) B ∆Hoatomisation(Si), ∆Hoatomisation(H), ∆Hoformation(SiH4) C ∆Hoatomisation(H), ∆Hocombustion(Si), ∆Hocombustion(SiH4) D ∆Hocombustion(Si), ∆Hocombustion(H2), ∆Hoformation(SiH4) Ans: B 9 Group 2 hydroxides undergo thermal decomposition in a similar fashion to Group 2 carbonates. Barium hydroxide undergoes decomposition as shown in the equation below: Ba(OH)2(s) → BaO(s) + H2O(g) H > 0 Which statements about this reaction are correct? 1 The Gibbs free energy change can be positive or negative depending on the temperature. 2 The decomposition is spontaneous only at high temperature. 3 The entropy change is negative. A 1 and 2 only B 1 and 3 only C 2 and 3 only D 1, 2 and 3 Ans: A S is expected to be positive since there is formation of gaseous H2O, thus statement 3 is wrong. G can be positive or negative depending on the temperature. For reactions to be spontaneous, G needs to be negative. As G = H − TS, T has to be high so that |TS| > H. Thus statement 2 is correct.

6 9729/01/H2 10 The reaction between NO and Br2 is proposed to proceed via the following mechanism: Step 1: NO + Br2 NOBr2 (fast) Step 2: NOBr2 + NO 2NOBr (slow) Which statements are correct? 1 NOBr2 is a radical. 2 The rate equation for this reaction is rate = k[Br2][NO]2. 3 NOBr2 is formed at the transition state. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans: B Statement 1: Correct. There is one unpaired electron on the N atom in the molecule. Statement 2: Correct. From the slow step, Rate = k[NOBr2][NO] --- (1) From step 1, rate of forward reaction = rate of reverse reaction kf[NO][Br2] = kr[NOBr2] 2 2 [NOBr ] [NO][Br ] f r k k = Kc = 2 2 [NOBr ] [NO][Br ] [NOBr] = Kc[NO][Br2] --- (2) Sub. (2) into (1), Rate = k(Kc[NO][Br2])[NO] = k’[NO]2[Br2] Statement 3: Incorrect. NOBr2 is an intermediate as it appears in the reaction mechanism but does not appear in the overall equation.

7 9729/01/H2 [Turn over 11 The kinetics of the following reaction is investigated, and the experimental data is given in the table below. 2R + 2S ⎯→ T + U [R] / mol dm−3 [S] / mol dm−3 initial rate / mol dm−3 s−1 0.015 0.010 5.10  10−4 0.030 0.020 4.08  10−3 0.045 0.010 1.53  10−3 What is the numerical value of the rate constant for this reaction? A 0.00294 B 3.40 C 227 D 340 Ans: D Comparing the first and third experiments, When [R]  3 while keeping [S] constant, initial rate  3  rate  [R]. Hence, order of reaction with respect to R is 1. Comparing the first and second experiments, When [R]  2 and [S]  2, initial rate  8. Since the order of reaction with respect to R is 1, rate  [S]2. Hence, order of reaction with respect to S is 2. The rate equation is: rate = k[R][S]2 Using data from the first experiment, k = 5.10  10−4 ÷ (0.015  0.0102) = 340 mol−2 dm6 s−1

8 9729/01/H2 12 Ethanol is produced industrially by reacting ethene and steam. C2H4(g) + H2O(g) ⇌ C2H5OH(g) H = –50.3 kJ mol–1 Kp has a value of 1.8 × 10–5 and the partial pressures