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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?

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Solution

No of mole = Given mass in gMolar mass=1818 = 1 moleFor 1 mole of water the enthalpy change for the vapourisation = 40.79 kJ/molFor two moles of water the enthalpy change for the vapourisation = 2 × 40.79 =81.58 kJStandard enthalpy of vapourisation Hf0 = 40.79 kJ/molIts same as the given value. Because the given condition is the standard condition - 1 bar

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