Questions on measuring an enthalpy change: calorimetry

Question 1

Describe how to determine the enthalpy change for the reaction of zinc with copper(II) sulfate solution, including how the results are processed.

Accepted Answer

Indicative content

a measured volume of copper(II) sulfate solution, for example 25.0 cm³ by pipette, in a polystyrene cup with a lid, supported in a beaker
zinc powder weighed out in excess, so that copper(II) sulfate is the limiting reagent
the temperature recorded at regular intervals, such as every 30 s, for a few minutes before adding the zinc, to establish the starting temperature
the zinc added at a recorded time, with stirring, and the temperature recorded at regular intervals until it has been falling steadily for several minutes
a graph of temperature against time, with the cooling line extrapolated back to the time of mixing to find ΔT
q = mcΔT with m the mass of the solution, divided by the amount of copper(II) sulfate in moles, gives ΔH in kJ mol⁻¹ with a negative sign

Levels

Level 3 (5–6 marks) the measured volume of solution in an insulated cup, zinc in excess, temperature readings before and after mixing at regular intervals, the extrapolated cooling curve to find ΔT, and the processing through q = mcΔT and the moles of the limiting reagent to a negative ΔH are all present
Level 2 (3–4 marks) the solution, the excess zinc and the temperature rise are measured and q = mcΔT used with moles of copper(II) sulfate; there is no cooling-curve extrapolation, or no readings before mixing
Level 1 (1–2 marks) zinc is added and the temperature rise measured; the mass in q = mcΔT is often taken as the mass of zinc, and ΔH is left positive

Question 2

Calculate the enthalpy of neutralisation from the readings in the table. A student put 25.0 cm³ of 2.00 mol dm⁻³ hydrochloric acid in a polystyrene cup and, at 3.0 minutes, added 25.0 cm³ of 2.00 mol dm⁻³ sodium hydroxide solution at the same starting temperature. Use the cooling readings to find ΔT at the time of mixing. c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

M1 the cooling line runs through the readings from 4.0 minutes, falling (32.5 − 30.9) ÷ (8.0 − 4.0) = 0.40 °C per minute A1 extrapolated to 3.0 minutes it reads 32.5 + 0.40 = 32.9 °C, so ΔT = 32.9 − 19.6 = 13.3 °C; accept 13.2 to 13.4 °C read from a graph M1 q = 50.0 × 4.18 × 13.3 = 2780 J, using the total mass of 50.0 g M1 amount of water formed = 0.0250 dm³ × 2.00 mol dm⁻³ = 0.0500 mol A1 ΔH = −2.780 ÷ 0.0500 = −55.6 kJ mol⁻¹ to three significant figures; a ΔT of 12.9 °C, from the highest reading, loses the second mark but allow error carried forward to −53.9 kJ mol⁻¹

Question 3

State the equation used to calculate the heat energy transferred in a calorimetry experiment, giving the unit of each quantity, and use it to calculate the heat needed to raise the temperature of 75.0 g of water by 8.4 °C. c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

B1 q = mcΔT: heat energy = mass × specific heat capacity × temperature change B1 q in J, m in g, c in J g⁻¹ K⁻¹ and ΔT in K; accept °C for ΔT, since a change is the same size in both M1 q = 75.0 × 4.18 × 8.4 A1 q = 2.6 × 10³ J, or 2.6 kJ, to two significant figures, matching the 8.4 °C

Question 4

Calculate the enthalpy of neutralisation from the results in the table, for H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l). Take the density of the solution as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

M1 q = 75.0 × 4.18 × 9.1 = 2853 J, using the total mass of 75.0 g M1 amount of water formed = amount of NaOH = 0.0500 dm³ × 1.00 mol dm⁻³ = 0.0500 mol; accept 2 × n(H₂SO₄) M1 ΔH = −2.853 ÷ 0.0500 kJ mol⁻¹ A1 ΔH = −57 kJ mol⁻¹ to two significant figures; −114 kJ mol⁻¹, per mole of acid, scores the first mark only

Question 5

Calculate the enthalpy of combustion of propan-1-ol from the spirit burner results in the table. c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

M1 q = 150.0 × 4.18 × 31.0 = 19 437 J = 19.44 kJ, using the mass of water; reject the mass of fuel in q M1 amount of propan-1-ol = 0.720 ÷ 60.0 = 0.0120 mol M1 ΔcH = −19.44 ÷ 0.0120 kJ mol⁻¹ A1 ΔcH = −1.62 × 10³ kJ mol⁻¹, that is −1620 kJ mol⁻¹, to three significant figures

Question 6

Explain why the enthalpy of combustion of ethanol measured with a spirit burner and a copper can is less exothermic than the data-book value.

Accepted Answer

B1 heat is lost to the surroundings, the air, the can and the clamp, so not all of it warms the water and ΔT is too small B1 combustion is incomplete, giving carbon or carbon monoxide, which releases less energy than complete combustion to CO₂ B1 ethanol evaporates from the wick without burning, so the mass loss overstates the amount burned B1 the water formed leaves as steam rather than liquid, so the heat of condensation included in ΔcH° is not collected

Question 7

Calculate ΔH for Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s) from the results in the table, in which magnesium powder is in excess. c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

M1 q = 50.0 × 4.18 × 22.8 = 4765 J M1 amount of CuSO₄, the limiting reagent = 0.0500 × 0.200 = 0.0100 mol A1 ΔH = −4.765 ÷ 0.0100 = −477 kJ mol⁻¹ to three significant figures

Question 8

Calculate the percentage uncertainty in a temperature rise of 6.8 °C measured with a thermometer that has an uncertainty of ±0.1 °C in each reading.

Accepted Answer

M1 ΔT is the difference between two readings, so its uncertainty is 2 × 0.1 = ±0.2 °C M1 percentage uncertainty = 0.2 ÷ 6.8 × 100 A1 percentage uncertainty = ±2.9% to two significant figures; 1.5%, from a single reading, scores zero

Question 9

Explain why a reaction in solution is carried out in a polystyrene cup with a lid rather than in a glass beaker.

Accepted Answer

B1 polystyrene is a poor conductor of heat, so less heat is lost through the walls to the surroundings B1 the lid reduces heat lost from the surface of the solution, by convection and evaporation B1 polystyrene has a very small heat capacity, so it absorbs little of the heat released, which q = mcΔT ignores

Question 10

Explain why the temperature rise in required practical 2 is found by extrapolating the cooling line back to the time of mixing rather than taken from the highest temperature recorded.

Accepted Answer

B1 the reaction takes time to finish, and heat is lost to the surroundings throughout B1 so by the time the highest temperature is recorded some heat has already been lost, and that reading is too low B1 extending the cooling line back to the time of mixing gives the temperature that would have been reached with no heat lost, so ΔT is larger and more accurate

Question 11

Suggest three changes to the spirit burner method that would give an enthalpy of combustion closer to the data-book value, and explain how each one helps.

Accepted Answer

B1 a draught screen around the burner and can, which keeps the flame under the can and reduces heat lost to moving air B1 a shorter gap between the flame and the base of the can, so less heat warms the air between them B1 keeping the cap on the burner except while it burns, so ethanol does not evaporate and the mass loss is only fuel that burned; accept insulating the can, with a lid

Question 12

Compare the polystyrene cup method with the spirit burner method in terms of heat loss and how close their results come to data-book values.

Accepted Answer

B1 in the cup the heat is released directly into the solution, whereas from a burner it must cross an air gap and the can B1 so the burner loses a much larger fraction of the heat, and also suffers incomplete combustion and evaporation B1 cup results for neutralisation and displacement are usually within about 10% of the data book, whereas burner results are often much further from it

Question 13

Explain why the enthalpy of neutralisation of hydrocyanic acid, HCN, by sodium hydroxide is much less exothermic than that of hydrochloric acid.

Accepted Answer

B1 hydrochloric acid is a strong acid, fully dissociated, so its neutralisation is only H⁺(aq) + OH⁻(aq) → H₂O(l) B1 hydrocyanic acid is a weak acid, mostly undissociated, so more HCN molecules dissociate as OH⁻ removes the H⁺ ions B1 that dissociation takes in energy, so less heat is released overall per mole of water formed; 'weak acids release less energy' alone is not enough

Question 14

Calculate the enthalpy of neutralisation of nitric acid by potassium hydroxide from the results in the table. Take the density of the solution as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹.

Accepted Answer

M1 q = 60.0 × 4.18 × 6.8 = 1705 J, using the total mass of 60.0 g M1 n(HNO₃) = 0.0300 mol and n(KOH) = 0.0400 mol, so the acid is limiting and 0.0300 mol of water forms A1 ΔH = −1.705 ÷ 0.0300 = −57 kJ mol⁻¹ to two significant figures; dividing by 0.0400 mol gives −43 kJ mol⁻¹ and scores the first mark only

Question 15

Suggest how the temperature change should be found when ammonium nitrate is dissolved in water in a polystyrene cup, an endothermic change, with readings taken every 30 s.

Accepted Answer

B1 after mixing the temperature falls to a minimum and then rises slowly as heat flows in from the warmer surroundings B1 the warming line after the minimum is extrapolated back to the time of mixing B1 ΔT is the fall from the starting temperature to that extrapolated value, and ΔH is positive

Question 16

Give two assumptions made when q = mcΔT is applied to a solution in a polystyrene cup.

Accepted Answer

B1 the density of the solution is 1.00 g cm⁻³, so the mass in grams equals the volume in cm³ B1 the specific heat capacity of the solution equals that of water, 4.18 J g⁻¹ K⁻¹; allow 'the cup absorbs no heat' as the second assumption

Question 17

Identify the limiting reagent when 3.00 g of zinc is added to 25.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate, and justify the answer. Ar of Zn = 65.4.

Accepted Answer

B1 amount of zinc = 3.00 ÷ 65.4 = 0.0459 mol and amount of CuSO₄ = 0.0250 × 0.500 = 0.0125 mol B1 the reacting ratio is 1:1, so copper(II) sulfate is the limiting reagent and zinc is in excess

Question 18

Suggest why, in a displacement reaction, the mass of the zinc is not added to the mass of the solution when q = mcΔT is used.

Accepted Answer

B1 the heat is absorbed almost entirely by the water of the solution, which is the substance whose temperature is measured B1 the zinc is a small mass of solid with a much smaller specific heat capacity than water, so the heat it absorbs is negligible

Question 19

Suggest why the temperature continues to rise for about a minute after zinc powder is added to copper(II) sulfate solution.

Accepted Answer

B1 the reaction is not instantaneous; the zinc takes time to mix through the solution and react B1 heat is still being released during that time faster than it is lost, so the temperature keeps rising until the reaction is nearly complete

Measurement	Value
Volume of 1.00 mol dm⁻³ sulfuric acid	25.0 cm³
Volume of 1.00 mol dm⁻³ sodium hydroxide	50.0 cm³
Temperature rise	9.1 °C

Measurement	Value
Mass of propan-1-ol burned	0.720 g
Mass of water in the can	150.0 g
Temperature rise of the water	31.0 °C
Mr of C₃H₇OH	60.0

Measurement	Value
Volume of copper(II) sulfate solution	50.0 cm³
Concentration of copper(II) sulfate	0.200 mol dm⁻³
Extrapolated temperature rise	22.8 °C

Measurement	Value
Volume of 1.50 mol dm⁻³ nitric acid	20.0 cm³
Volume of 1.00 mol dm⁻³ potassium hydroxide	40.0 cm³
Temperature rise	6.8 °C

Time / min	Temperature / °C
0.0	19.6
1.0	19.6
2.0	19.6
3.0	alkali added; no reading
3.5	31.8
4.0	32.5
5.0	32.1
6.0	31.7
7.0	31.3
8.0	30.9