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. 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?
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. 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?
This is a Short Answer Type Questions as classified in NCERT Exemplar
Enthalpy of a reaction is the energy change per mole for the process.
18 g of H2O = 1 mole ΔHvap = 40.79 kJ/ mol
Enthalpy change for vapourising 2 moles of H2O = 2 x 40.79 = 81.58 kJ ΔH°vap = 40.79 kJ mol-1.
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