Answer: The change in entropy for the system is 140.3 J/K
Explanation:
The chemical equation follows:
[tex]CH_3OH(l)\rightleftharpoons CH_3OH(g)[/tex]
To calculate the entropy change for vaporization of methanol, we use the equation:
[tex]\Delta S=\frac{n\Delta H_{vap}}{T}[/tex]
where,
[tex]\Delta S[/tex] = entropy change of the reaction
n = number of moles = 1.00 mole
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 35.21 kJ/mol = 35210 J/mol (Conversion factor: 1 kJ = 1000 J)
T = temperature of the system = [tex]64.6^oC=[64.6+273]=337.6K[/tex]
Putting values in above equation, we get:
[tex]\Delta S=\frac{1.00mol\times 35210J/mol}{337.6K}\\\\\Delta S=140.3J/K[/tex]
Hence, the change in entropy for the system is 140.3 J/K
