In this experiment you will determine the enthalpy change of solution, ΔHsol, for hydrated sodium thiosulfate, Na₂S₂O₃•5H₂O. To do this you will measure the temperature change when a known mass of hydrated sodium thiosulfate is dissolved in a known volume of water. FB 5 is hydrated sodium thiosulfate, Na₂S₂O₃•5H₂O.
- (a)Method • Support the cup in the 250 cm³ beaker. • Use the 25 cm³ measuring cylinder to transfer 20.0 cm³ of distilled water into the cup. • Weigh the stoppered container of FB 5 and record the mass. • Measure and record the initial temperature of the water in the cup. • Add all the FB 5 to the water in the cup. • Stir the mixture and record the minimum temperature that is reached. • Reweigh the stoppered container. Record the mass. • Calculate and record the mass of FB 5 added to the water and the change in temperature. I II III IV[4]
- (b)(i)Calculations Calculate the energy change of the reaction. (Assume that 4.2 J of heat energy changes the temperature of 1.0 cm³ of solution by 1.0 °C.) Show your working. energy change = ................................. J[1]
- (b)(ii)Calculate the enthalpy change of solution, ∆Hsol, for hydrated sodium thiosulfate. ΔHsol for Na₂S₂O₃•5H₂O = ................................. kJ mol⁻¹ sign value[2]
- (b)(iii)Assume that under the same conditions, the enthalpy change of solution, ΔHsol, for anhydrous sodium thiosulfate, Na₂S₂O₃, is −7.7 kJ mol⁻¹. Construct a Hess's cycle and determine the enthalpy change for the following reaction. (If you were unable to calculate an answer to (b)(ii), assume a value of +32.2 kJ mol⁻¹. Note this is not the correct value.) Na₂S₂O₃(s) + 5H₂O(l) → Na₂S₂O₃•5H₂O(s) ΔH = ................................. kJ mol⁻¹ sign value[2]
- (c)How would your temperature change in (a) be affected if your sample of FB 5 contained a small amount of anhydrous sodium thiosulfate? Explain your answer.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [1] I Table to include initial and final mass of stoppered container with FB 5, initial and final temperature, mass added and temperature change. All with correct units.
- [1] II All masses recorded to the same precision and all temperatures recorded to .0 or .5 °C.
- [1] III Award if the difference in the ratio between candidate and Supervisor is within 0.20 °C g⁻¹
- [1] IV Award if the difference in the ratio between candidate and Supervisor is within 0.40 °C g⁻¹
- [1] correctly calculates 4.2 × temp change × 20
- [1] correct expression showing 248.2
- [1] moles of thiosulfate = candidate mass / 248.2. answer to 2-4 sf
- [1] ∆H = + ans (i) / moles of thiosulfate
- [1] answer to min 2 sf penalise
- [1] 2 arrows linking to Na2S2O3 (aq) with correct values indicated for each step
- [1] ΔH = −7.7 – (ans (b)(ii))
- [1] default value +39.9
- [1] anhydrous is exothermic, therefore the temperature drop would be less / final temp would be higher / temperature change would be less
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