Question

The molar enthalpy of vaporization of carbon tetrachloride (CCl) 29.82 kJ/mol and the boiling point of carbon tetrachloride is 76.7 °C. Calculate AS for the vaporization of 0.35 moles of carbon tetrachloride.

Answer 1

Answer : The value of entropy change will be, 29.8455 J/K

Explanation : Given,

Moles of
`CCl_4` = 0.35 mole

Boiling temperature =
`76.7^oC=273+76.7=349.7K`

conversion used :
`(0^C=273K)`

Molar enthalpy of vaporization of
`CCl_4` = 29.82 kJ/mole

First we have to calculate the heat change.

Formula used :

`\Delta q=\Delta H* n`

where,

`\Delta q` = heat change

`\Delta H` = molar enthalpy of vaporization

n = number of moles

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

`\Delta q=29.82kJ/mole* 0.35mole=10.437kJ`

Now we have to calculate the entropy change.

Using third law of thermodynamic :

`\Delta S=(\Delta q)/(T)`

where,

`\Delta S` = entropy change

`\Delta q` = heat change

T = boiling temperature

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

`\Delta S=(10.437kJ)/(349.7K)=0.0298455kJ/K=29.8455J/K`

Therefore, the value of entropy change will be, 29.8455 J/K