2018 Paper 4 Ans

2018 A level H2 Chem Paper 4 Answer

Expt VFA1 / cm3 Vwater / cm3 t / s 1/t / s−1 lg(1/t) lg(VFA1) 1 25.00 0.0 27.9 0.0358 −1.45 1.40 2 22.50 2.5 30.2 0.0331 −1.48 1.35 3 20.00 5.0 34.0 0.0294 −1.53 1.30 4 17.50 7.5 38.9 0.0257 −1.59 1.24 5 15.00 10.0 45.7 0.0219 −1.66 1.18 Note: Measuring cylinder is used for distilled water hence precision is up to 0.5cm3.

Gradient = (−1.400)−(−1.730) 1.450−1.100 = 0.943 Rate = k[S2O32−]x[HCl]y, Rate ∝ 1 𝑡 and [S2O32−] ∝ VFA1 1 𝑡 = k (VFA1)x[HCl]y lg(1 𝑡) = x lg(VFA1) + lgk + ylg[HCl] gradient of lg (1 𝑡) vs lg(VFA1) graph is the order w.r.t S2O32− 0.943 First order t will increases The water left in the flask would lead to dilution of the reactants. With lower concentration of reactants, rate of reaction decreases and a longer time is required to produce the fixed amount of sulfur solid. We would need to know the amount of thiosulfate left in the reaction mixture when the fixed amount of sulfur solid is produced to make the solution opaque. This would allow us to find the change in [S2O32−].

8 S2O32− → S8 + 8 SO32− H+ acts as a catalyst in the reaction. It is reacted in step 1 and regenerated in step 9

Average vol of FA1 used = 24.80+24.80 2 = 24.80 cm3 24.80 cm3

6.30 g Percentage by mass of Cu in alloy = 6.299 11.54 × 100% = 54.6% 54.6% 2.02 : 1 1) Reduction of Cu2+ to Cu+ 2) Precipitation of CuI from Cu+ and I− 3) Lewis acid-base reaction of I2 + I− to give I3− Any two of the above

The acids present in FA2 would react with Na2CO3(aq) to give effervescence of CO2(g). When Na2CO3 is in slight excess, a white ppt of metal carbonate is produced. When CH3COOH (aq) is added dropwise, the metal carbonate dissolves due to acid- carbonate reaction that also gives effervescence of CO2(g) The acids present in FA2 would react with S2O32− to give sulfur solid (Reaction in experiment 1, kinetics). If the acids are not removed, more S2O32− would be required to completely react with the I2 and H+ present and this would lead to larger titre values.

White ppt, soluble in excess NaOH(aq) to give a colourless solution White ppt, soluble in excess NH3 (aq) to give a colourless solution White ppt CuI dissolves in NH3(aq) to give a colourless filtrate that turns dark blue rapidly. When H2SO4(aq) is added dropwise, pale blue ppt is formed. When more H2SO4(aq) is added, the pale blue ppt turns into a brown suspension. Upon leaving to stand, we can observe white ppt in the brown solution.