ΔG??= ΔH??– TΔS??= 0
ΔH??= TΔS?
ΔS??=?ΔH??/?T

Determining?the entropy of vaporisation by measuring the enthalpy change of boiling water
Steps in the procedure
Practical tips
Specimen Results
Specimen Results Table

Analysis
moles of water evaporated = 130 / 18 = 7.22 mol
enthalpy of vaporisation of water = 300 / 7.22 = 41.55 kJ mol-1
enthalpy of vaporisation of water = 41.55 x 1000 = 41 550 J mol-1
temperature in Kelvin =? 100 + 273 =? 373 K
ΔS?=?ΔH??/?T
ΔS?= 41 550?/ 373 = 111.4 J mol-1
轉載自savemyexams
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