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Calculate the enthalpy of vaporisation per mole for ethanol. Given `DeltaS = 109.8 J K^(-1) mol^(-1)` and boiling point of ethanol is `78.5^(@)`. |
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Answer» The entropy change for vaporisation, `Delta_(vap)S^(Theta) = (Delta_(vap)H)/(T_(b))` or `Delta_(vap)H^(Theta) = Delta_(vap)S^(Theta)T_(b)` `DeltaS^(Theta) = 109.8 J K^(-1)mol^(-1), T_(b) = 273 +78.5 = 351.5K` `:. DeltaH^(Theta) = (109.8 xx 351.5) = 38.595 kJ mol^(-1)` |
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