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 δhvap is 30.7 kj/mol?

Answer 1

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

Solution :

Formula used :

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

or,

`\Delta S=((w)/(M)* \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

w = mass of benzene = 17.5 g

M = molar mass of benzene = 78 g/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=((17.5g)/(78g/mole)* (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