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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 enthalphy of vapourisation for water?
Solution
Enthalpy of a reaction is defined as the energy change per mole for the process. Therefore,
18 g of \[\ce{H2O}\] = 1 mole (ΔHVap = 40.79 kJ Mol–1)
Enthalpy change for vaporising 2 moles of \[\ce{H2O}\] = 2 × 40.79 = 81.58 kJ
ΔHVap = 40.79 kJ Mol–1.
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