Home
Class 11
CHEMISTRY
Calculate the enthalpy of vaporisation p...

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^(@)`.

Text Solution

Verified by Experts

The entropy change for vaporisation,
`Delta_(vap)S^(@) = (Delta_(vap)H^(@))/(T_(b)^(@))`
or `Delta_(vap)H^(@) = Delta_(vap)S^(@) XT_(b)^(@)`
`DeltaS^(@) = 109.8 J K^(-1) mol^(-1), T_(b)^(@) = 273 + 78.5 = 351.5 K`
`therefore Delta_(vap)H^(@) = (109.8 J K^(-1) mol^(-1)) xx (351.5 K)`
`= 38595 J mol^(-1) = 38.595 kJ mol^(-1)`.
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS

    MODERN PUBLICATION|Exercise Practice Problems|65 Videos

  • THERMODYNAMICS

    MODERN PUBLICATION|Exercise Advanced Level Problems|15 Videos

  • STRUCTURE OF ATOM

    MODERN PUBLICATION|Exercise Unit Practice Test|13 Videos

Similar Questions

Explore conceptually related problems

Ethanol boils at 78.4^@C and the enthalpy of vaporisation of ethanol is 42.4kJ mol^(-1) . Calculate the entropy of vaporisation of ethanol.

It the enthalpy of vaporization of benzene is 308/ kJ mol^(-1) at boiling point (80^(@)C) , calculate the entroy (j mol^(-1) K^(-1)) in changing it from liquid to vapor.

The value for DeltaH_(vap) and Delat S_(vap) for ethanol are respectively 38.594 kJ mol^(-1) and 109.8 JK^(-1) . The boiling point of ethanol will be

The enthalpy of vaporisation of a liquid is 30 kJ mol^(-1) and entropy of vaporisation is 75 J mol^(-1) K^(-1) . The boiling point of the liquid at 1atm is :