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IB Chemistry HL Paper 2 Structured Practice Paper
IB Chemistry HL Paper 2 Structured
IB Chemistry HL Paper 2 structured practice paper: 10 original questions, 100 marks in 135 minutes, covering titration, enthalpy of combustion, Kc, weak acid pH.
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Paper details Level: IB DP2 | Subject: Chemistry HL | Time: 135 minutes | Marks: 100 | Questions: 10
Section A Stoichiometry, Energy and Electrochemistry (50 marks)
Questions 1-3
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 25.0 cm3 of 0.100 mol dm-3 sodium hydroxide is exactly neutralised by 20.0 cm3 of dilute sulfuric acid. (a) Calculate the amount, in mol, of sodium hydroxide used. (b) Given H2SO4 + 2NaOH → Na2SO4 + 2H2O, find the amount of sulfuric acid. (c) Find the concentration of the acid in mol dm-3. | (a) n(NaOH) = 0.0250 x 0.100 = 2.50 x 10-3 mol. (b) n(H2SO4) = 2.50 x 10-3 / 2 = 1.25 x 10-3 mol, because one mole of the acid neutralises two moles of the base. (c) c = 1.25 x 10-3 / 0.0200 = 0.0625 mol dm-3. Marking: the 1:2 ratio must come from the equation and units must be stated. |
| 2 | 2.30 g of ethanol (M = 46.0) burns completely in oxygen and raises the temperature of 200 g of water by 15.0 C (specific heat capacity of water = 4.18 J g-1 K-1). (a) Calculate the heat energy absorbed by the water in J. (b) Find the amount of ethanol and its enthalpy of combustion in kJ mol-1, with sign. (c) The literature value is -1367 kJ mol-1; calculate the percentage error and suggest one major source of error. | (a) q = 200 x 4.18 x 15.0 = 12540 J = 12.54 kJ. (b) n = 2.30 / 46.0 = 0.0500 mol; dH = -12.54 / 0.0500 = -250.8 kJ mol-1. (c) Error = (1367 – 250.8) / 1367 x 100 = 81.7%; the main cause is heat loss to the surroundings and apparatus, with some incomplete combustion and evaporation of water. Marking: the sign of dH must be negative. |
| 3 | In a 1.00 dm3 vessel, 1.00 mol of hydrogen is mixed with 1.00 mol of iodine vapour. At equilibrium the concentration of hydrogen iodide is 1.60 mol dm-3. H2(g) + I2(g) ⇌ 2HI(g). (a) Find the equilibrium concentrations of H2 and I2. (b) Calculate Kc. (c) If the vessel is compressed to 0.50 dm3 at constant temperature, do Kc and the amount of HI change? Explain. | (a) From the equation, forming 1.60 mol dm-3 of HI consumes 0.80 mol dm-3 each of H2 and I2, so [H2] = [I2] = 0.20 mol dm-3. (b) Kc = (1.60)2 / (0.20 x 0.20) = 2.56 / 0.040 = 64, no units. (c) Kc is unchanged because temperature is constant; since both sides have equal numbers of gas moles (dn = 0), pressure does not shift the position of equilibrium, so the amount of HI is also unchanged while its concentration doubles. |
Questions 4-5
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | A monobasic weak acid HA has Ka = 1.8 x 10-5. A 0.10 mol dm-3 solution is prepared. (a) Calculate the equilibrium concentration of hydrogen ions. (b) Find the pH to 2 decimal places. (c) Find the percentage dissociation of HA to 2 decimal places and explain why pH = -log c cannot be used directly. | (a) [H+] = sqrt(Ka x c) = sqrt(1.8 x 10-5 x 0.10) = 1.34 x 10-3 mol dm-3. (b) pH = -log(1.34 x 10-3) = 2.87. (c) Dissociation = 1.34 x 10-3 / 0.10 x 100 = 1.34%. Because HA is only partly dissociated, [H+] is far below c, so using -log c would overestimate [H+] and underestimate the pH. Marking: state the approximation [HA]eq = c and that it holds below 5% dissociation. |
| 5 | A current of 2.00 A is passed through copper(II) sulfate solution for 30.0 minutes. (F = 96500 C mol-1, A(Cu) = 63.5). (a) Calculate the charge passed. (b) Find the amount of electrons. (c) Find the mass of copper deposited at the cathode in g to 3 decimal places, and state what is observed at the anode. | (a) Q = I t = 2.00 x 1800 = 3600 C. (b) n(e-) = 3600 / 96500 = 3.73 x 10-2 mol. (c) Cu2+ + 2e- → Cu, so n(Cu) = 1.865 x 10-2 mol and m = 1.865 x 10-2 x 63.5 = 1.184 g. The anode copper dissolves and thins, the blue colour fades slightly, and the Cu2+ concentration stays roughly constant. |
Section B Gases, Kinetics, Yield and Buffers (50 marks)
Questions 6-8
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | 0.500 g of a gas occupies 400 cm3 at 300 K and 100 kPa (R = 8.31 J K-1 mol-1). (a) Calculate the amount of gas in mol to 5 decimal places. (b) Find its molar mass to 1 decimal place. (c) Deduce the gas and explain why the calculation becomes inaccurate at high pressure. | (a) n = pV / RT = (100000 x 4.00 x 10-4) / (8.31 x 300) = 1.604 x 10-2 mol. (b) M = m / n = 0.500 / 1.604 x 10-2 = 31.2 g mol-1. (c) Most likely oxygen, O2 (M = 32.0). At high pressure the molecular volume is not negligible and intermolecular forces matter, so the ideal gas law overestimates pressure and underestimates molar mass. |
| 7 | A reaction is first order with respect to A. Experiment shows the concentration of A falls from 0.20 mol dm-3 to 0.10 mol dm-3 in 58 s. (a) Find the rate constant k in s-1 to 4 decimal places. (b) State the half-life and explain the characteristic of half-life for a first order reaction. (c) Calculate the time for the concentration to fall from 0.40 mol dm-3 to 0.050 mol dm-3. | (a) For a first order reaction k = ln2 / t1/2 = 0.693 / 58 = 1.20 x 10-2 s-1. (b) The half-life is 58 s and is independent of the starting concentration. (c) 0.40 → 0.20 → 0.10 → 0.050 is three half-lives, so t = 3 x 58 = 174 s. Marking: use ln2/t1/2, not (c0 – c)/t, and count the half-lives. |
| 8 | Limestone decomposes on heating: CaCO3 → CaO + CO2 (M: CaCO3 = 100, CaO = 56, CO2 = 44). (a) Calculate the atom economy of this reaction in %. (b) Find the theoretical yield of calcium oxide from 25.0 g of calcium carbonate. (c) Only 12.6 g of calcium oxide is obtained; find the percentage yield and suggest one reason why it is below 100%. | (a) Atom economy = Mr of desired product / total Mr of reactants x 100 = 56 / 100 x 100 = 56%. (b) n(CaCO3) = 25.0 / 100 = 0.250 mol, so theoretical yield = 0.250 x 56 = 14.0 g. (c) Percentage yield = 12.6 / 14.0 x 100 = 90%; reasons include incomplete decomposition, the product absorbing moisture to form calcium hydroxide, or loss on transfer. |
Questions 9-10
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | 25.0 cm3 of iron(II) sulfate solution is fully oxidised in acid by 24.0 cm3 of 0.0200 mol dm-3 potassium manganate(VII). MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. (a) Calculate the amount of MnO4- used. (b) Find the amount of Fe2+. (c) Find the concentration of the Fe2+ solution and state the colour change at the end point. | (a) n(MnO4-) = 0.0240 x 0.0200 = 4.80 x 10-4 mol. (b) The equation gives MnO4- : Fe2+ = 1 : 5, so n(Fe2+) = 2.40 x 10-3 mol. (c) c(Fe2+) = 2.40 x 10-3 / 0.0250 = 0.0960 mol dm-3; the solution turns from colourless to a persistent pale purple at the end point. |
| 10 | Equal volumes of 0.10 mol dm-3 ethanoic acid and 0.10 mol dm-3 sodium ethanoate are mixed (Ka = 1.8 x 10-5). (a) Calculate the pH of the mixture to 2 decimal places. (b) Explain why the pH is almost unchanged after adding a small amount of hydrochloric acid; write the equations. (c) Find the mole ratio of acid to salt needed for a buffer of pH 4.44. | (a) After mixing equal volumes each concentration is 0.050 mol dm-3; pH = pKa + log([salt]/[acid]) = 4.74 + 0 = 4.74. (b) Added H+ reacts with ethanoate ions: CH3COO- + H+ → CH3COOH, so the acid is removed rather than accumulating and the pH hardly changes. (c) At pH 4.44, log([salt]/[acid]) = -0.30, so [salt]/[acid] = 0.50 and acid : salt = 2 : 1. |
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