Question

The heat of vaporization of 1–pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K•mol. What is the approximate boilingpoint of 1–pentanol?

Answer 1

375 K

Step-by-step explanation:

Using the experssion shown below as:

`\Delta G^0=\Delta H^0_(vap)-T\Delta S^0_(vap)`

At vaporization point, the liquid and the gaseous phase is in the equilibrium.

Thus,

`\Delta G^0=0`

So,

`Delta H^0_(vap)=T\Delta S^0_(vap)`

Given that:

`Delta H^0_(vap)=55.5\ kJ/mol`

Also, 1 kJ = 10³ J

So,

`Delta H^0_(vap)=55500\ J/mol`

`\Delta S^0_(vap)=148\ J/K.mol`

So, temperature is :

`T=(Delta H^0_(vap))/(\Delta S^0_(vap))`

`T=(55500)/(148)`

T= 375 K