Potassium hydrogencarbonate, KHCO3, decomposes when strongly heated to form potassium carbonate, K2CO3. 2KHCO3(s) → K2CO3(s) + CO2(g) + H₂O(I) A student plans to determine the value for the enthalpy change for this reaction, ΔΗ₁, which cannot be determined directly. The student carries out two separate experiments using the following apparatus. [Figure showing polystyrene cup, thermometer, solid, 2.0 mol dm-3 hydrochloric acid] Experiment 1 uses solid KHCO3. Experiment 2 uses solid K2CO3. The following method is used for both experiments: • Transfer 50.00 cm³, an excess, of 2 mol dm⁻³ hydrochloric acid into a cup. • After 2 minutes, record the temperature of the acid. • Weigh approximately 0.0250 moles of solid. • Add the solid to the acid, stir the mixture using a thermometer and record the temperature throughout the reaction. Hazard information: 2 mol dm⁻³ hydrochloric acid is irritant, solid potassium hydrogencarbonate and solid potassium carbonate may cause irritation to the skin and eyes. The equations for the two reactions are: reaction 1 KHCO3(s) + HCl(aq) → KCl(aq) + CO2(g) + H₂O(I) reaction 2 K2CO3(s) + 2HCl(aq) → 2KCl(aq) + CO2(g) + H₂O(I)
- (a)Suggest why it is not possible to measure ∆H, for the decomposition reaction directly.[1]
- (b)(i)Calculate the mass of 0.0250 moles of each solid. Give your answers to three decimal places. [Α,: Κ, 39.1; Η, 1.0; C, 12.0; Ο, 16.0] mass of KHCO3 = ..................................... g mass of K2CO3 = ..................................... g[1]
- (b)(ii)The masses of solid are measured using a three decimal place balance. Calculate the percentage error in the measurement of the mass of KHCO3. Show your working. percentage error = .....................................[1]
- (c)(i)The student obtained the following results. | solid | initial temperature/°C | maximum/minimum temperature/°C | temperature change, ∆T/°C | |---------|------------------------|----------------------------------|---------------------------| | KHCO3 | 17.5 | 14.0 | | | K2CO3 | 19.0 | 20.5 | | Complete the table by calculating temperature change. Use the formula q = mc∆T to determine the energy change, q, that took place during experiment 1. Use q to calculate the enthalpy change of reaction 1, ∆H₁, in kJ mol⁻¹. Include a sign in your answer. Assume 1.00 cm³ of solution has a mass of 1.00g. c = 4.18 J g⁻¹ K⁻¹ ΔΗ₁ = ............................. kJ mol⁻¹[2]
- (c)(ii)Use the formula q = mc∆T to determine the energy change, q, that took place during experiment 2. Use q to calculate the enthalpy change of reaction 2, ∆H₂, in kJ mol⁻¹. Include a sign in your answer. Assume 1.00 cm³ of solution has a mass of 1.00 g. c = 4.18 J g⁻¹ K⁻¹ ΔΗ₂ = ............................. kJ mol⁻¹[2]
- (d)Use the following cycle to calculate ΔΗ₁. [Figure showing enthalpy cycle: 2KHCO3(s) --(ΔH₁)--> K2CO3(s) + CO2(g) + H2O(l); both reactants and products react with +2HCl to give 2KCl(aq) + 2CO2(g) + 2H2O(l)] ΔΗ₁ = ............................. kJ mol⁻¹[2]
- (e)A textbook states the value of the enthalpy change for the decomposition of potassium hydrogencarbonate as +76.0 kJ mol⁻¹. Suggest two reasons why the experimental value is different to the actual value. 1 2[2]
- (f)Suggest one improvement to the apparatus which would reduce the difference between the experimental value and the actual value.[1]
- (g)Name a suitable piece of apparatus which should be used to measure the volume of acid used in experiment 1.[1]
- (h)Apart from wearing safety glasses and a lab coat, state one safety precaution which must be taken during experiment 1. Explain your answer.[1]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [1] It is not possible to measure a temperature when heating.
- [1] Mass of KHCO3 = 2.503 g AND Mass of K2CO3 = 3.455 g
- [1] 2×0.0005×100 / 2.503 = 0.04(00)% Correct working must be shown
- [2] M1 3.5 °C for KHCO3 and 1.5 °C for K2CO3 in results table AND q = 50 × 4.18 × 3.5 = 731.5 (J) Μ2 ΔΗ₁ = Μ1 / (0.0250 × 1000) = + 29.26 kJ mol⁻¹
- [2] M1 q = K2CO3 = 50 × 4.18 × 1.5 = 313.5 (J) Μ2 ΔΗ2 = M1 / 0.025×1000 = -12.54 kJ mol⁻¹
- [2] M1 2 × 1(c)(i) = 2 × 29.26 = 58.52 kJ mol⁻¹ M2 M1 – 1(c)(ii) = +58.52 – (-12.54) = +71.06 kJ mol⁻¹
- [2] Any two from: • Heat loss OR heat gain (to / from surroundings) • Amount of solid not exactly 0.0250 moles • Experiment not carried out under standard conditions.
- [1] Add a lid to the polystyrene cup.
- [1] Burette.
- [1] Wear chemically resistant gloves AND Reagent(s) are irritant(s).
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