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IB Chemistry SL Paper 2 Mock Paper (10 questions, 50 marks, 75 minutes)
IB Chemistry SL
IB Chemistry SL Paper 2 mock paper: 10 questions, 50 marks, 75 minutes, calculator allowed, covering stoichiometry, titration, the gas laws, energetics, equilibrium, acids and bases, electrolysis, structure and periodicity, organic chemistry and kinetics, with full worked answers and marking notes.
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Paper details Level: IBDP | Subject: Chemistry SL | Time: 75 minutes | Marks: 50 | Questions: 10 | Calculator allowed
Section A (Questions 1 to 5, 25 marks)
Answer ALL questions. A calculator is allowed. Unless told otherwise, give numerical answers to 3 significant figures and show your working, stating units in every final answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Calcium carbonate (Mr = 100.0) decomposes on heating: CaCO3(s) to CaO(s) + CO2(g). 25.0 g of calcium carbonate is completely decomposed. (a) Calculate the amount, in mol, of calcium carbonate. (b) Calculate the volume of carbon dioxide formed at room temperature and pressure (molar volume 24.0 dm3 mol-1). (c) Calculate the mass of calcium oxide formed (Mr = 56.08). | (a) n = 25.0 / 100.0 = 0.250 mol. (b) n(CO2) = 0.250 mol, so V = 0.250 x 24.0 = 6.00 dm3. (c) m = 0.250 x 56.08 = 14.0 g (3 s.f.). |
| 2 | 25.0 cm3 of 0.100 mol dm-3 sodium hydroxide is titrated with 0.125 mol dm-3 hydrochloric acid (HCl + NaOH gives NaCl + H2O). (a) Calculate the amount, in mol, of sodium hydroxide. (b) Calculate the volume of acid required, in cm3. (c) Calculate the mass of hydrogen chloride used (Mr = 36.46). | (a) n(NaOH) = 0.100 x 25.0 / 1000 = 2.50 x 10-3 mol. (b) The ratio is 1:1, so V = 2.50 x 10-3 / 0.125 x 1000 = 20.0 cm3. (c) m = 2.50 x 10-3 x 36.46 = 0.0912 g. |
| 3 | A 0.0250 mol sample of a gas occupies a certain volume at 298 K and 100 kPa and has a mass of 1.10 g. (R = 8.314 J K-1 mol-1) (a) Use pV = nRT to calculate the volume of the sample in cm3. (b) Calculate the molar mass and decide whether the gas is carbon dioxide (Mr = 44.0) or oxygen (Mr = 32.0). | (a) V = nRT / p = 0.0250 x 8.314 x 298 / 100000 = 6.19 x 10-4 m3, which is 619 cm3. (b) M = 1.10 / 0.0250 = 44.0 g mol-1, so the gas is carbon dioxide. |
| 4 | 0.0200 mol of a metal is dissolved in 50.0 g of water and the temperature rises from 21.0 to 27.5 degrees C. (Specific heat capacity of water = 4.18 J g-1 K-1; ignore the heat capacity of the container.) (a) Calculate the heat energy absorbed by the water in J and in kJ. (b) Calculate the enthalpy change of reaction in kJ mol-1 and state whether the sign shows an exothermic or endothermic change. | (a) q = mcDT = 50.0 x 4.18 x 6.5 = 1358.5 J = 1.36 kJ. (b) dH = -q / n = -1.36 / 0.0200 = -67.9 kJ mol-1; the negative sign means the reaction is exothermic, which agrees with the temperature rise. |
| 5 | Hydrogen reacts with iodine vapour in a 2.00 dm3 container: H2(g) + I2(g) is in equilibrium with 2HI(g). The initial concentration of both H2 and I2 is 0.250 mol dm-3; at equilibrium the concentration of HI is 0.350 mol dm-3. (a) Calculate the equilibrium concentration of H2. (b) Calculate Kc to 3 significant figures. (c) The container is compressed to 1.00 dm3. State and explain how the equilibrium concentration of HI changes. | (a) Forming 0.350 mol dm-3 of HI uses 0.175 mol dm-3 of H2, so at equilibrium [H2] = [I2] = 0.250 – 0.175 = 0.0750 mol dm-3. (b) Kc = [HI]2 / ([H2][I2]) = 0.3502 / 0.07502 = 21.8. (c) There are equal numbers of gas moles on both sides (2 and 2), so changing the pressure does not shift the position of equilibrium. The amounts in mol stay the same, so halving the volume doubles each concentration and [HI] becomes about 0.700 mol dm-3. |
Section B (Questions 6 to 10, 25 marks)
Answer ALL questions. A calculator is allowed. Unless told otherwise, give numerical answers to 3 significant figures and show your working, stating units in every final answer.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | (a) Calculate the pH of 0.0100 mol dm-3 hydrochloric acid. (b) A weak acid HA has concentration 0.0100 mol dm-3 and pH 3.20. Calculate Ka to 2 significant figures. (c) Calculate the pH of 0.0500 mol dm-3 sodium hydroxide at 25 degrees C. | (a) [H+] = 0.0100 mol dm-3, so pH = 2.00. (b) [H+] = 10-3.20 = 6.31 x 10-4 mol dm-3, so Ka = (6.31 x 10-4)2 / 0.0100 = 4.0 x 10-5. (c) [OH-] = 0.0500 mol dm-3, so pOH = 1.30 and pH = 14.00 – 1.30 = 12.70. |
| 7 | A current of 2.00 A is passed through copper(II) sulfate solution for 30.0 minutes. (Faraday constant 96 500 C mol-1; cathode: Cu2+ + 2e– to Cu) (a) Calculate the charge passed in C and the amount of electrons in mol. (b) Calculate the mass of copper deposited at the cathode (Mr = 63.55). (c) Calculate the volume of oxygen formed at the anode at room temperature and pressure (molar volume 24.0 dm3 mol-1). | (a) Q = It = 2.00 x 1800 = 3600 C, so n(e-) = 3600 / 96500 = 0.0373 mol. (b) n(Cu) = 0.0373 / 2 = 0.0187 mol, so m = 0.0187 x 63.55 = 1.19 g. (c) At the anode 2H2O gives O2 + 4H+ + 4e-, so n(O2) = 0.0373 / 4 = 9.33 x 10-3 mol and V = 9.33 x 10-3 x 24.0 = 0.224 dm3, that is 224 cm3. |
| 8 | (a) Explain why the melting point of magnesium oxide (2 852 degrees C) is much higher than that of sodium chloride (801 degrees C). Refer to structure and bonding. (b) Explain why ammonia is very soluble in water whereas methane is almost insoluble. Refer to intermolecular forces. (c) Write the equation for the reaction of magnesium with cold water and state how its vigour compares with that of sodium. | (a) Magnesium oxide is a giant ionic lattice in which both Mg2+ and O2- carry two charges and are small, so the electrostatic attraction (ionic bonding) is very strong. In sodium chloride each ion carries only one charge and the ions are larger, so the attraction is weaker and the melting point is lower. (b) Ammonia molecules form hydrogen bonds with water molecules and can react with water to give NH4+ and OH-. Methane is non-polar and interacts with water only through weak van der Waals forces, so it is almost insoluble. (c) Mg(s) + 2H2O(l) gives Mg(OH)2(aq) + H2(g). Magnesium reacts only very slowly with cold water, whereas sodium reacts violently and floats on the surface, because sodium is more metallic and has a lower first ionisation energy. |
| 9 | An organic compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass and has a molar mass of 180 g mol-1. (a) Determine its empirical formula. (b) Determine its molecular formula. (c) The compound gives a brick-red precipitate with Benedict solution. Suggest one possible structure by name and explain your reasoning. | (a) Mole ratio: C 40.0/12.01 = 3.33, H 6.7/1.008 = 6.65 and O 53.3/16.00 = 3.33, which simplifies to 1:2:1, so the empirical formula is CH2O (formula mass 30.03). (b) n = 180/30.03 = 6, so the molecular formula is C6H12O6. (c) Glucose (or fructose). It reduces Benedict solution, which shows an aldehyde group or an alpha-hydroxy ketone group, so it is a reducing sugar. |
| 10 | At a certain temperature the rate of decomposition of compound A was measured: when [A] = 0.10 mol dm-3 the rate is 2.0 x 10-3 mol dm-3 s-1, and when [A] = 0.20 mol dm-3 the rate is 4.0 x 10-3 mol dm-3 s-1. (a) Deduce the order with respect to A and explain. (b) Calculate the rate constant k, including its unit. (c) Calculate the initial rate when [A] = 0.50 mol dm-3 and hence the half-life under these conditions. | (a) Doubling [A] doubles the rate, so the reaction is first order with respect to A and rate = k[A]. (b) k = 2.0 x 10-3 / 0.10 = 2.0 x 10-2 s-1. (c) rate = 2.0 x 10-2 x 0.50 = 1.0 x 10-2 mol dm-3 s-1; for a first order reaction t(half) = ln2 / k = 0.693 / 0.020 = 34.7 s, independent of the starting concentration. |
A TutorZone practice paper; original questions and passages, not publisher material.
