Question

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm? Please report your answer two points past the decimal with the unit J/molK. ∆H˚fus = 3.17 kJ/mol.

Answer 1

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K