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 $1.09\times 10^2J/mol$

Explanation :

Formula used :

$\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}$

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:

$\Delta S^o=\frac{\Delta H^o_{vap}}{T_b}$

$\Delta S^o=\frac{40.7kJ/mol}{373K}$

$\Delta S^o=1.09\times 10^2J/mol$

Therefore, the value of change in entropy for vaporization of water is $1.09\times 10^2J/mol$