Question:
18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?
Solution:
Enthalpy change of vapourisation for 1 mole = 40.79 kJ mol–1 enthalpy change of vapourisation for 2 moles of water = (40.79 × 2) = 81.58kJ mol–1
for water ∆Hvapourisation, ⁰ will be equal to =40.79 kJ mol–1
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