Question

1 Calculate the change in entropy that occurs in the system when 1.00 mol of methanol (CH3OH) vaporizes from a liquid to a gas at its boiling point (64.6 °C). For methanol, ΔHvap=35.21 kJ/mol.

Answer 1

Answer: The change in entropy for the system is 140.3 J/K

Step-by-step explanation:

The chemical equation follows:

`CH_3OH(l)\rightleftharpoons CH_3OH(g)`

To calculate the entropy change for vaporization of methanol, we use the equation:

`\Delta S=(n\Delta H_(vap))/(T)`

where,

`\Delta S` = entropy change of the reaction

n = number of moles = 1.00 mole

`\Delta H_(vap)` = heat of vaporization = 35.21 kJ/mol = 35210 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system =
`64.6^oC=[64.6+273]=337.6K`

Putting values in above equation, we get:

`\Delta S=(1.00mol* 35210J/mol)/(337.6K)\\\\\Delta S=140.3J/K`

Hence, the change in entropy for the system is 140.3 J/K