8.1.4 Entropy of Vaporisation

Objective:

To determine the entropy change of vaporisation ($∆S°$) of water by measuring the enthalpy change during boiling.

Background:

At the boiling point, water and water vapour exist in equilibrium, meaning the free energy change ($ΔG°$) is zero. The relationship between the Gibbs free energy change, enthalpy change ($ΔH°$), and entropy change ($ΔS°$) can be expressed as:

$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 0$$

This rearranges to:

$$\Delta S^\circ = \frac{\Delta H^\circ}{T}$$

Where:

• $\Delta H^\circ$ is the enthalpy change of vaporisation,
• $T$ is the temperature in Kelvin.

Apparatus:

• Electric kettle (with a power rating of approximately 3 kW)
• Top-pan balance (accuracy ±0.1 g, capacity of at least 2.5 kg)
• 1 dm³ of water
• Stopwatch

Procedure:

1. Weigh the kettle: Weigh the empty kettle using the balance and record the mass.
2. Add water: Pour exactly 1 dm³ (1000 cm³) of water into the kettle.
3. Reweigh the kettle: Measure and record the combined mass of the kettle and water.
4. Boil the water: Switch the kettle on and allow it to boil for exactly 100 seconds, ensuring the kettle's cut-off mechanism is deactivated to allow continuous boiling.
5. Reweigh after boiling: After 100 seconds, switch off the kettle and reweigh it to determine the mass of water remaining.

Data and Calculations:

1. Power supplied: Use the power rating of the kettle (3 kW) to calculate the total energy supplied during the 100 seconds of boiling.

$$\text{Energy supplied (q)} = \text{Power} \times \text{Time} = 3 \, \text{kW} \times 100 \, \text{seconds} = 300 \, \text{kJ}$$

2. Mass of water evaporated: Subtract the mass of the kettle and water after boiling from the mass before boiling to find the mass of water evaporated.

$$\text{Mass evaporated} = \text{Initial mass} - \text{Final mass}$$

Example: If 130 g of water is lost, convert this to moles (using

$$M = 18 \, \text{g mol}^{-1}$$ $$\text{Moles of water} = \frac{130 \, \text{g}}{18 \, \text{g mol}^{-1}} = 7.22 \, \text{mol}$$

3. Calculate $\Delta H^\circ$ (enthalpy change): The enthalpy change per mole of water is calculated by dividing the energy supplied by the number of moles evaporated:

$$\Delta H^\circ = \frac{300 \, \text{kJ}}{7.22 \, \text{mol}} = 41.55 \, \text{kJ mol}^{-1}$$

4. Calculate entropy change $\Delta S^\circ$: Using the formula

$$\Delta S^\circ = \frac{\Delta H^\circ}{T}$$

where $T$ is the boiling point of water (373 K):

$$\Delta S^\circ = \frac{41.55 \, \text{kJ mol}^{-1}}{373 \, \text{K}} = 111.4 \, \text{J mol}^{-1} \, \text{K}^{-1}$$

Conclusion:

The calculated entropy change for the vaporisation of water is approximately

$$111.4 \, \text{J mol}^{-1} \, \text{K}^{-1}$$

Practical Considerations:

• Ensure the kettle has a reliable power rating of around 3 kW for accurate energy calculations.
• Use a balance with sufficient precision to accurately measure small differences in mass after evaporation.
• The volume of water added is not critical, but ensure that the amount added covers the heating element of the kettle.