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.

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

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

Formula used :

$\Delta S=\frac{n\times \Delta H_{vap}}{T_b}$

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.

$\Delta S=\frac{0.224mole\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K$

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