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 :

$\Delta q=\Delta H\times n$

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:

$\Delta q=29.82kJ/mole\times 0.35mole=10.437kJ$

Now we have to calculate the entropy change.

Using third law of thermodynamic :

$\Delta S=\frac{\Delta q}{T}$

where,

$\Delta S$ = entropy change

$\Delta q$ = heat change

T = boiling temperature

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

$\Delta S=\frac{10.437kJ}{349.7K}=0.0298455kJ/K=29.8455J/K$

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