Question

What is the entropy change of the system when 17.5 g of liquid benzene (C6H6) evaporates at the normal boiling point? The normal boiling point of benzene is 80.1°C and ΔH vap is 30.7 kJ/mol.

Answer 1

Answer : The entropy change of the system is, 19.5 J/K

Solution :

First we have to calculate the moles of benzene.

`\text{Moles of }C_6H_6=\frac{\text{Mass of }C_6H_6}{\text{Molar mass of }C_6H_6}=(17.5g)/(78.11g/mole)=0.224moles`

Now we have to calculate the entropy change of the system.

Formula used :

`\Delta S=(n* \Delta H_(vap))/(T_b)`

where,

`\Delta S` = entropy change of the system = ?

`\Delta H` = enthalpy of vaporization = 30.7 kJ/mole

n = number of moles of benzene = 0.224 mole

`T_b` = normal boiling point of benzene =
`80.1^oC=273+80.1=353.1K`

Now put all the given values in the above formula, we get the entropy change of the system.

`\Delta S=(0.224mole* (30.7KJ/mole))/(353.1K)=0.0195kJ/K=0.0195* 1000=19.5J/K`

Therefore, the entropy change of the system is, 19.5 J/K