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 :

Options

A. `250 K`
B. `400 K`
C. `450 K`
D. `600 K`

Answer

B

Explanation

`DeltaH=30 kJ mol^(-1)=30,000 J mol^(-1)`
`DeltaS=75 J mol^(-1)K^(-1)`
At boiling point, the reversible process
`"liquid" hArr "vapour"`
is in equilibrium at one atmospheric pressure
`:. DeltaG=0`
`DeltaG=DeltaH-TDeltaS`
`:. T=(DeltaH)/(DeltaS)=(30,000 Jmol^(-1))/(75 J mol^(-1)K^(-1))=400 K`