2022 RVHS JC2 H2 CM Prelim P1 (Worked Solutions)

1 2022 JC 2 H2 Chemistry Prelim Paper 1 Worked Solutions Qn Ans Solutions 1 D Number of proton > electron => cation Number of proton > neutron => likely contain D (highlighted by question) No. of proton No. of neutron No. of electron A. NO2+ 7 + 2(8) 7 + 2(8) 7 +2(8) -1* B. ND2H 7 + 2(1)+ 1 7 + 2(1) + 0 7 + 2(1) + 1 C. NDH− 7 + 1 + 1 7 + 1 + 0 7 + 1 + 1 +1* D. ND3H+ 7 + 3(1) + 1 7 +3(1) + 0 7 + 3(1) + 1 -1* 2 A A. V− → V2− [Ar] 4s23d5 B. N → N− 1s2 2s2 2p4 C. Se+ → Se [Ar] 4s23d104p4 D. Ti → Ti− [Ar] 4s23d3 3 B Significant increment from 5th to 6th I.E. Element M from period 3 and has 5 valence electrons. => Phosphorous Equation 2 is formation of PCl3 ,Equation 3 is formation of PCl5 4 D Option A: The dipole moment on the HF molecule should be larger (HF dipole moment is larger and factoring in partial cancellation of 2 smaller OH dipole moments). Option B: As fluorine is more electronegative, the hydrogen bond formed between HF molecules will be stronger than the hydrogen bond formed between H 2O molecules. Option C: The change in id -id interactions is negligible. id-id interactions are not the predominant IMF for HF and H2O molecules. Option D: H2O forms more extensive hydrogen bonds than HF. 5 C Option A: In BF3, B shared its 3 valence electrons to form 3 covalent bonds with 3 F atoms. B is electron deficient. Option B: In CO, C shares its 2 valence electrons with 2 valence electrons on O (to maintain an octet electronic configuration for O). To achieve an octet electronic configuration for C, O will donate a lone pair of electrons to C and form a dative bond. Option C: In NO, N shares its 2 valence electrons with the 2 valence electrons on O (to maintain an octet electronic configuration for O). N is electron deficient. Option D: In SiO2, which has a giant molecular structure, Si shares its 4 valence electrons with 4 oxygen atoms (which will further extends to obtain a macromolecule).

2 6 B Deviation depends on IMF and size of particles. All 4 compounds has id-id, SO3 has the highest Mr. 7 C When an ideal gas is compressed from 20 atm to 80 atm with no further reaction, the volume is expected to change from 67.0 cm3 to 16.75 cm3 (instead of 15.5 cm3 as stated in the question). Option C is not a valid explanation since the dissociation of a gas would result in an observed gas volume of larger than 16.75 cm 3 (instead of 15.5 cm3 as stated in the question). 8 A From the information given, 20.0 cm3 of 0.20 mol dm −3 acid reacts exactly with 40 cm3 0.10 mol dm−3 of base, indicating that reacting mole ratio of acid and base is 1:1. This would be inconsistent with option A since CH3COOH and Ba(OH) 2 would react in a 2:1 ratio. Hence, option A is false, making A the answer. 9 A H must be negative for the overall process must be exothermic since heat must be generated by the reaction to enable the heating process. S must be negative as indicated by the reaction equation as the reaction goes from reactants of (1 mol solid and 1 mol liquid) to products of 1 mol solid. G must be negative since the process must be spontaneous as described by the question context. 10 C For the decomposition of H2O2: H = ½[–572 – (–188)] = –192 kJ mol–1 S = ½[–0.325 – (–0.225)] = –0.050 kJ mol–1 K–1 G = H – TS = (–192) – (25 + 273)(–0.05) = –117.1 kJ mol–1 11 B a is twice the enthalpy change of neutralisation of calcium hydroxide (option 1) b is enthalpy change of reaction of calcium with acid (option 2) 12 A Let the formula of hydrocarbon be CxHy. Molar ratio of carbon dioxide : hydrocarbon is 2 : 1 ⇒ x = 2 C2Hy (g) + (2 + 𝑦 4 ) O2 (g) 2CO2(g) + 𝑦 2 H2O(l) Since the remaining O2 can burn up exactly 30 cm3 of the same hydrocarbon, this means that 100 cm3 of O2 can burn 40 cm3 of the hydrocarbon.

3 ⇒ Molar ratio of hydrocarbon : O2 = 40 : 100 = 1 : 5 2 ⇒ 2 + 𝑦 4 = 2.5 ⇒ y = 2 Formula of hydrocarbon is C2H2. 13 B Amount of SO32– = 25.0 1000 × 0.10= 0.0025 mol Amount of electrons lost by 0.0025 mol of SO32– = 2 x 0.0025 = 0.005 mol Amount of metallic salt = 50.0 1000 × 0.10 = 0.005 mol Amount of electrons gained by 0.005 mol of metal ion = 0.005 mol Amount of electrons gained by 1 mol of metal ion = 1 mol Oxidation state of metal in product = +3 – 1 = +2 14 D Since the concentration of acid in the solution remains constant, rate = k’[CH3CO2CH2CH3] where k’ = k[H+] t½ = ln 2 𝑘′ = ln 2 𝑘[𝐻+] Expt 1 When [HCl] = [H+] = 0.2 mol dm-3, t½ = 31 min (given in question) Expt 2 When [HCl] = [H+] = 0.1 mol dm-3 (halved), t½ = 62 min For [CH3CO2CH2CH3] to fall from 0.2 mol dm-3 to 0.05 mol dm-3, it would take two t½. Time taken = 2 × 62 = 124 min 15 B Option A: N 2O2 is an intermediate since it is produced in the first step and consumed in the second step. Option B: Rate equation is rate = k[NO]2[H2], hence the overall order of reaction is 3. Option C: Refer to the rate equation above. The order of reaction with respect to H2 is 1. Option D: Refer to the rate equation above. The order of reaction with respect to NO is 2. 16 C 3Fe + 4H2O ⇌ Fe3O4 + H2 initial amt 3.0 2.0 0 0 change − 3𝑥 4 − x + 𝑥 4 + 𝑥 4

4 Equilibrium amt 3.0 − 3𝑥 4 2.0 − x + 𝑥 4 + 𝑥 4 17 B salt Ksp = IP Ag2S 6.8 × 10−50= [Ag+(aq)]2(0.1) [Ag+(aq)] = 8.25 × 10−25 mol dm−3 CuS 6.3 × 10−36= [Cu2+(aq)](0.1) [Cu2+(aq)] = 6.3 × 10−35 mol dm−3 SnS2 1.0 × 10−70= [Sn4+(aq)](0.1)2 [Sn4+(aq)] = 1.0 × 10−68 mol dm−3 18 D The lower the pKa value, the stronger the acid. Strength of acid: CH3COOH < CH2BrCOOH < CH2FCOOH The three compounds given are carboxylic acids. The more stable the conjugate base, the stronger the carboxylic acid. CH3COOH ⇌ CH3COO− + H+ CH2BrCOOH ⇌ CH2BrCOO− + H+ CH2FCOOH ⇌ CH2FCOO− + H+ Stability of conjugate base: CH3COO−  CH2BrCOO−  CH2FCOO− Statement 1: Correct, electronegative Br and F will draw electrons away from the O−H bond in COOH, resulting in the deprotonation of H from O −H bond to be easier for the halogenated carboxylic acids. Thus, the two halogenated carboxylic acids are more acidic than CH 3COOH. Also, F is more electronegative than Br thus CH2FCOOH is more acidic than CH2BrCOOH. Statement 2: Correct, electronegative Br and F will help to stabilise the negative charge on the COO − hence resulting in these two halogenated carboxylic acids to be more acidic than CH3COOH. Also, the more electronegative F will stabilise the conjugate base to a greater extent than Br thus CH 2FCOOH is more acidic than CH2BrCOOH. Statement 3: Correct, the methyl group in CH 3COOH is electron donating and intensify the negative charge on CH 3COO−, destabilising CH 3COO− and hence making CH3COOH the weakest acid amongst the three. 19 A Standard hydrogen electrode (S.H.E) consists of H 2(g) at 1 bar bubbling over platinum electrode coated with finely divided platinum which is dipped into 1 mol dm−3 H+(aq) at 298 K.

5 By convention, the standard electrode potential for this reference hydrogen half - cell is taken to be 0.00 V. Statement 1: Correct because the hydrogen gas needs to be at 1 bar. Statement 2: Correct because the standard electrode potential of S.H.E is taken to be 0.00 V. Statement 3: Incorrect because at pH 1.0, [H +] = 10 −1.0 = 0.100 mol dm −3. This condition does not fulfil the requirement of [H+] to be 1.00 mol dm−3. 20 C -carotene undergoes oxidative cleavage with hot, acidified KMnO4. Statement 1 is correct. • All sections of -carotene labelled A formed an organic product. • All sections labelled B formed another organic product. • Final oxidised products from section labelled C are inorganic. Statement 2 is correct. • All organic products have a COCH3 group that gives a positive iodoform test with warm aqueous alkaline iodine. Statement 3 is incorrect. • No stereoisomers or constitutional isomers present. • All products from the same se