Question

For water ∆H°vap = 40.7 kJ/mol at 100.°C, its boiling point. Calculate ∆S° for the vaporization of 1.00 mol water at 100.°C and 1.00 atm pressure.

Answer 1

Answer : The value of change in entropy for vaporization of water is
`1.09* 10^2J/mol`

Explanation :

Formula used :

`\Delta S^o=(\Delta H^o_(vap))/(T_b)`

where,

`\Delta S^o` = change in entropy of vaporization = ?

`\Delta H^o_(vap)` = change in enthalpy of vaporization = 40.7 kJ/mol

`T_b` = boiling point temperature of water =
`100^oC=273+100=373K`

Now put all the given values in the above formula, we get:

`\Delta S^o=(\Delta H^o_(vap))/(T_b)`

`\Delta S^o=(40.7kJ/mol)/(373K)`

`\Delta S^o=1.09* 10^2J/mol`

Therefore, the value of change in entropy for vaporization of water is
`1.09* 10^2J/mol`