RI 2023 Y6 CT Suggested Solutions

1 2023 Year 6 H2 Chemistry Common Test Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A C D A D B B A A D C B B D B Question 1 (A) Angle of deflection ⍺ charge mass Since the electron has the smallest mass ( 1 1840 that of proton), it has the largest angle of deflection. Question 2 (C) There is a large increase from the 7th to 8th ionisation energy. Significantly more energy is required to remove the 8th electron as it is located in an inner shell. Hence, the element has 7 valence electrons i.e. chlorine. Question 3 (D) Since the dipole moment of the highly polar C=O is not completely cancelled out in X and Y, X and Y have greater polarity than W and Z, which have no highly polar bonds i.e. polarity of W, Z < X, Y. Since the dipole moment of the C=O in Y is partially cancelled out by the dipole moments of the 2 C–Cl bonds, the overall dipole moment in Y is lower than in X i.e. polarity of Y < X. The dipole moments in Z are exactly cancelled out, while the dipole moments in Y are not. Hence, Z is less polar than W i.e. polarity of Z < W. Question 4 (A) Comparing expts 2 and 3, when [H+] x 2, rate x 4 i.e. order of reaction wrt H+ = 2. Comparing expts 3 and 4, when [BrO3–] x 2, rate x 2 i.e. order of reaction wrt BrO3– = 1. Comparing expts 1 and 3, when [Br–] x 2 , rate x 2 i.e. order of reaction wrt Br– = 1.

2 Question 5 (D) 1 Correct. As HBr is formed, [HBr] in the denominator of the rate equation increases, causing the rate to decrease i.e. rate slows down. 2. Incorrect. In a propagation step, a radical is consumed as a reactant and another radical is produced as a product. Hence, step 2 is also a propagation step, in addition to steps 3 and 4. 3. Incorrect. [Br2] appears in both the numerator and denominator . Do ubling its concentration does not double the rate due to the powers which each [Br2] is raised to. Question 6 (B) From the initial buffer provided, 4.46 = pKa + lg [ethanoate] [ethanoic acid] 4.46 = pKa + lg( 1 2) --- (1) Let x be the no. of mols of NaOH added to the buffer to achieve pH 4.60. Amt / mol CH3COOH + NaOH CH3COO– Na+ + H2O Initial 2(2) x 2(1) -- Change –x –x +x -- final 4–x 0 2+x After adding x mol of NaOH, 4.60 = pKa + lg( 2 + x 4 - x ) --- (2) Equation (2) – (1) gives: 4.60 – 4.46 = lg( 2 + x 4 - x ) – lg( 1 2) 0.14 = lg( 2 + 𝑥𝑥 4 − 𝑥𝑥 ÷ 1 2 ) 100.14 = 2(2 + x) 4 - x x = 0.4503 mol mass of NaOH = 0.4503 x 40.0 = 18.0 g. Question 7 (B) A Incorrect. If HA is a weak acid, then Ka(HA) x Kb(conjugate base of HA) = Kw. This is because Ka = [H+][A−] [HA] and Kb(conjugate base of HA) = Kb(A–) = [HA][OH−] [A−] . Ka(HA) x Kb(conjugate base of HA) = [H+][A−] [HA] x [HA][OH−] [A−] = [H+][OH–] = Kw. For this statement involving water to be correct, it should read either • K a(H2O) x Kb(OH–) = Kw, or • Ka(H3O+) x Kb(H2O) = Kw. B Correct. When