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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?

User Ecs
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1 Answer

3 votes

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

User WeakPointer
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