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 :

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

or,

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

where,

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

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

n = number of moles of benzene

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.

$\Delta S=\frac{\frac{17.5g}{78g/mole}\times (30.7KJ/mole)}{353.1K}=0.0195kJ/K=0.0195\times 1000=19.5J/K$

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

Answer 2

1 mole of benzene (78g) requires 30.8 kJ/ of heat, so 11.5g will need ..... (it's a proportion calculation.) Temperature does not change at BPt and is not relevant if the temp of the liquid is already at the BPt ne definition of entropy is qrev/T, where qrev is the heat added in reversible operation (for complicated reasons pertaining to heat as a path function) and T is the temperature at which this is done. Phase changes are particularly good examples for calculations of changes in entropy, since temperature will not change will the bonds of a state are being broken. The calculations required boils down to: 1) finding the moles of benzene given from molar mass. 2) multiplying that moles by the heat of vaporization. 3)diving the heat energy required by the temperature of boiling point.