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.

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asked Mar 7, 2018 in Class XI Chemistry by nikita74 (-1,032 points)

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 ? 

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answered Mar 7, 2018 by vijay Premium (530 points)

Enthalpy of a reaction is the energy change per mole for the process.
18 g of H2O = 1 mole (∆Hvap = 40.79 kJ moE1)
Enthalpy change for vapourising 2 moles of H2O = 2 x 40.79 = 81.58 kJ ∆H°vap = 40.79 kJ mol -1

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