2020 Paper 4 Ans

40.0 17.0 45.0 25.8 20.0 12.9 2020 A level H2 Chem Paper 4 Answer

= 25.8 17.0 = 1.5 (1d.p.) = 25.8 12.9 = 2.0 (1d.p.) 1 rate of reaction ( 1 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛), rate 1 = 0.0588, rate 2 = 0.0388 Comparing experiments 1 & 2, [H2O2] and [H+] is kept constant. when [I−] × 1.5, the rate of reaction also × 1.5. Hence, we can deduce it is first order w.r.t. I−

1 Let rate = k’[I−] [H2O2]b and rate = (1/time taken) comparing expt 2 & 3, 𝑟𝑎𝑡𝑒 3 𝑟𝑎𝑡𝑒 2 = 1/12.9 1/25.8 = 𝑘′ × 5 ×(40)𝑏 𝑘′ ×10 ×(10)𝑏 4 = (40)𝑏 (10)𝑏 b = 1 [I−] [H2O2] The reaction mixture has a buffer solution containing CH 3COOH and its conjugate base CH3COO− (from FA4). When [H+] decreases, CH3COOH dissociates to produce some H+ to maintain the [H+] in the mixture. Note: The question is asking about why [H+] remains constant in ALL the experiments, implying that it is referring to the entire duration of each experiment. The question is NOT about why it is important to keep initial [H+] constant across the experiments.

Experiment 1 2 3 4 FA6 Final Burette Reading /cm3 5.50 26.50 36.50 15.00 FA6 Initial Burette Reading /cm3 0.00 5.50 26.50 0.00 Volume of FA6 used /cm3 5.50 21.00 10.00 15.00 FA7 Final Burette Reading /cm3 26.20 40.70 20.20 34.30 FA7 Initial Burette Reading /cm3 0.00 35.00 0.00 21.00 Volume of FA7 used /cm3 26.20 5.70 20.20 13.30 Note: Labelling of headers with units Readings recorded to 2 d.p. for burette readings

Gradient = 33.50 - 0.00 0.00 - 25.50 = −1.31 Note: Can use any two points from the straight line to calculate gradient. Must include -ve sign for the gradient • Gradient is Vol FA7 Vol FA6, magnitude is 1.31 Total amount of H+ from 33.50 cm3 of HCl in FA7 = Total amount of H+ from 25.50 cm3 of H2SO4 in FA6 Note: 1 mol of H2SO4 produces 2 mol of H+ [FA7] × Vol FA7 = [FA6] × Vol FA6 × 2 [FA7] [FA6] = Vol FA6 Vol FA7 × 2 = 1 1.31  2 = 1.53 25.50 cm3 33.50 cm3

25.0 cm3 of FA5 requires 33.50 cm3 of FA7 for complete neutralisation Ba(OH)2 + 2 HCl ⎯→ BaCl2 + 2H2O Total amount of HCl used = 0.150 × 33.50 1000 = 0.005025 mol Amount of Ba(OH)2 = ½ × 0.005025 = 0.002513 mol Concentration of Ba(OH)2 = 0.002513 25.01000⁄ = 0.101 mol dm−3 [FA7] [FA6] = 1.53 [FA6] = 0.150 1.53 = 0.0980 mol dm−3 Alternatively, Ba(OH)2 + H2SO4 ⎯→ BaSO4 + 2H2O 25.0 cm3 of 0.101 mol dm−3 Ba(OH)2 reacts with 25.50 cm3 of H2SO4 Amount of Ba(OH)2 used = 0.101 × 25.0 1000 = 0.002525 mol = Amount H2SO4 Concentration of H2SO4 = 0.002525 25.501000⁄ = 0.0990 mol dm−3 Total amount of H+ required to neutralise Ba(OH)2 = H+ from FA6 + H+ from FA7 As volume of FA6 used increases, greater amount of H+ is contributed by FA6. Lesser amount of FA7 is required to complete the neutralisation and hence vol of FA7 required decreases. White precipitate is formed in the colourless solution. Ba(OH)2 (aq) + CO2 (g) ⎯→ BaCO3 (s) + H2O (l) similar to reaction of Ca(OH)2 with CO2(g)

Reaction with FA7: BaCO3 (s) + 2HCl (aq) ⎯→ BaCl2