RI 2019 Prelim P1 Ans

1 2019 Year 6 H2 Chemistry Preliminary Examinations Paper 1 Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D A B B C A D A C D C D C B B Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer D C C A D B D D A C A B C A B 1(D) A Incorrect. The number of protons determine the identity of the atom/ion. Since both 65Cu2+ and 65Cu+ are ions of copper, they both have 29 protons. B Incorrect. 63Cu has 29 electrons and has the electronic configuration [Ar]3d 104s1. Hence, 63Cu+ has the electronic configuration [Ar]3d10. C Incorrect. No. of neutrons in 63Cu+ = 63 – 29 = 34 No. of neutrtons in 65Cu3+ = 65 – 29 = 36 D Correct. 63Cu+ : [Ar]3d10. With a fully filled d-subshell, it has no unpaired electrons. 65Cu3+ : [Ar]3d8 i.e. has 2 unpaired electrons. Q2(A) chemical formula structure H2O2 H2S2 N2H4 1 Correct. O, S and N in the respective molecules have 4 regions of electron density and are sp 3 hybridised. Hence, the O –H, S–H and N –H bonds are formed from the overlap between the sp3 hybrid orbitals and the s orbital of H. 2 Correct. Since O is more el ectronegative than S, it draws the O–H bond−pair in H–O–O closer to itself, compared to S drawing the S –H bond−pair in H–S–S to itself. The O–H bond–pair in in H –O–O are thus nearer to the nucleus of the central O, and exert more repulsion than those in H–S–S. 3 Incorrect. S ince H 2S2 has the largest number of electrons, it has the largest and most polarisable electron cloud out of the 3 compounds. Hence, H2S2 has the strongest id-id interactions. 4 Incorrect. N2H4 has 2 lone-pairs of electrons and 4 hydrogens, it forms an average of 2 hydrogen bonds per molecule. H 2O2 has 4 lone -pairs of electrons and 2 hydrogens, it also forms an average of 2 hydrogen bonds per molecule. Q3(B) CH3CH2CO2Na dissociates in water to form CH3CH2CO2– and Na + ions which form strong ion-dipoles interactions with water, allowing it to have high solubility in water. The dissolution is also accompanied by an increase in entropy, favouring the dissolution While CH 3CH2CH2NH2 forms hydrogen bonds with water, CH 3CH2CH2Cl forms pd -pd interactions with water and CH3CH2CH2CH3 forms id-id interactions with water, these are not as strong as the ion -dipole interactions CH3CH2CO2Na form with water. Q4(B) The number of moles of air, n air, in the dented and undented plastic ball are the same. Using data from the dented ball, 6(100000)(27.5 10 ) (8.31)(273 22) 0.001122 mol air pVn RT −== + = At 57 °C when the ball is restored, 6( )(30.0 10 )0.001122 = (8.31)(273 57) 102542 Pa 103 kPa air pn p −= + == Q5(C) 0.010 mol of A lCl3, SiC l4 and PC l5 were added to 1.0 dm3 (i.e. excess) water. AlCl3 + 6H2O → [Al(H2O)6]3+ + 3Cl– • a weakly acidic solution due to [Al(H2O)6]3+. Highest pH of the three resultant solutions. SiCl4 + 2H2O → SiO2 + 4HCl • a strongly acidic solution due to formation of HC l which dissolves in water. PCl5 + 4H2O → H3PO4 + 5HCl • a strongly acidic solution due to formation of HC l which dissolves in water. Since more HC l is produced compared to SiCl4, the resultant solutuion is more acidic and pH is the lowest of the three resultant solutions.

2 Q6(A) 1 Incorrect. H I is less volatile than HBr and the difference in their volatility (i.e. boiling point) is due to the different strengths of their intermolecular forces and not the H–I and H–Br bond strengths. 2 Correct. The stronger H –F bond means that more energy is required to break the bond during thermal decomposition. Hence, HF is more thermally stable. 3 Incorrect. Since E(Cl2/Cl–) = +1.36V is more positive than E(Br2/Br–) = +1.07V, C l2 has a greater tendency to be reduced than Br 2 i.e. Cl2 is a stronger oxidising agent than Br 2. The stronger bond energy of Cl–Cl (+244 kJ mol–1) compared to Br–Br (+193 kJ mol –1) see ms to contradict the conclusion from the E values. This is because there are other factors which contribute more significantly to the oxidising power of C l2 than simply the bond strength of the Cl–Cl bond. Q7(D) heat released during combustion (1000)(4.18)(85 25) 250800 J Since process is 60% efficient, theoretical amt of heat released 250800= 418000 J0.6 of ethyl mercaptan = 2(12.0)+6(1.0)+32.1 r mc T M = =− = = 1 1 = 62.1 per mole of ethyl mercaptan 418000= 1527000 J mol17.062.1 1527 kJ mol Since reaction 1 involved the combustion of 2 mol of ethyl mercaptan, of reaction 1 H H − −  −= =−  1= 2 x 1527 = 3054 kJ mol −−− Q8(A) Since the total volume of every experiment is the same (100 cm3), Vreactant  [reactant]. Using expression for rate provided, experiment rate of reaction 1 20.0 / 1 = 20.0 2 20.0 / 2 = 10.0 3 20.0 / 2 = 10.0 4 10.0 / 0.5 = 20.0 Comparing experiment 1 and 2, • when volume of H+ x 2, rate x 2 • order of reaction w.r.t. H+ = 1 Comparing experiment 1 and 4, • when volume of I2 x 2, rate remains the same • order of reaction w.r.t. I2 = 0 Comparing experiment 1 and 3, • when volume of CH3COCH3 x 2, rate x 2 • order of reaction w.r.t. CH3COCH3 = 1 Therefore, rate = k[CH3COCH3][H+]. 1 Correct. Calculating units for rate constant: rate = k[CH3COCH3][H+] mol dm–3 min–1 = k (mol dm–3)2 k = mol–1 dm3 min–1 2 Correct. When volume of H+ x 4 from experiment 5 to 4 , rate should x 4 i.e. rate of experiment 5 = ¼(20.0) = 5.0. x = time taken in expt 5 = 10.0 / 5 = 2.0 min. 3 Incorrect. Since the order of reaction w.r.t iodine is 0, changing the concentration of iodine does not change the reaction rate. However, since double the concentration of iodine is consumed at the same rate, the time taken for each experiment will be doubled. Q9(C) Since steps 1 and 2 are part of a chain reaction, steps 1 and 2 continue in succession during the reaction. 1 Correct. Since Cl• is consumed in step 1 and regenerated in step 2, it is a catalyst of the chain reaction. 2 Correct. Steps 1 and 2 are part of a chain reaction. step 1: Cl• + O3  ClO• + O2 step 2: ClO• + O3 Cl• + 2O2 step 1: Cl• + O3  ClO• + O2 step 2: ClO• + O3 Cl• + 2O2 Since ClO• is consumed in step 2 and regenerated in step 1, it is a catalyst of the chain reaction. 3 Incorrect. Since step 2 is the slow step, it has a higher activation energy than step 1. 4 Correct. Since the reaction is catalysed, the activation energy of every step of the mechanism is lower than the activation energy of the uncatalysed reaction. Q10(D)

3 From the slow step, rate = k2[ClO•][O3]. The overall rate equation cannot contain any intermediate s, such as ClO•. Find a mathematical equation which will allow [ClO•] to be substituted. From the fast step, =  = 132 1 32 [Cl][O ][ClO][O ] [ClO][Cl][O ] [O ] KK Therefore, rate = k2[ClO•][O3] becomes = 2 1 3 3 2 3 2 1 22 [ ][ ] [ ][ ]rate = k [ ] k[ ] [ ] K Cl O Cl OOKOO Q11(C) When the temperature is increased from 480 K to 550 K, the rate s of both the forward and backward reactions increase because both the rate constants of the forward and backward reactions increase i.e. kf and kb increase. Since the forward reaction is exothermic, increasing the temperature favours the backward endothermic reaction, decreasing the yield and the value of Kp. Q12 (D) =  33 1 p NH BF K PP We use pV = nRT to convert the partial pressure of NH3 to [NH3] in mol dm–3. Let V be volume in dm3.  = ==   33 3 3 3 1000 1000 1000 [ ] NH NH NH NH VP n RT n P RT RT NH V Substituting the above into the Kp expression. =  = 3 3 3 3 1 1000 [ ] 1[] (1000)(8.31)(298) p BF BF p K P RT NH NH PK Q13(C) Even though HF and HCl have the same pH (i.e. the [H+] at equilibrium is the same), t he solution of HF required a greater amount of NaOH f