The enthalpy of vaporization of benzene is +3 | vedclass

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Question

(A) The enthalpy of vaporization $(\Delta H_{vap})$ is $+35.3 \ kJ/mol = +35300 \ J/mol$.
The boiling point $(T)$ is $80^{\circ} C = 80 + 273 = 353 \

Vedclass · Accepted Answer

$-100 \ J/mol \cdot K$

Vedclass · Answer

(A) The enthalpy of vaporization $(\Delta H_{vap})$ is $+35.3 \ kJ/mol = +35300 \ J/mol$.
The boiling point $(T)$ is $80^{\circ} C = 80 + 273 = 353 \ K$.
For the transition of liquid to vapour,the entropy change is $\Delta S_{vap} = \frac{\Delta H_{vap}}{T} = \frac{35300 \ J/mol}{353 \ K} = +100 \ J/mol \cdot K$.
The question asks for the entropy change in the transition of vapour to liquid (condensation),which is the reverse process.
Therefore,$\Delta S_{cond} = -\Delta S_{vap} = -100 \ J/mol \cdot K$.

The enthalpy of vaporization of benzene is $+35.3 \ kJ/mol$ at its boiling point,$80^{\circ} C$. The entropy change in the transition of vapour to liquid at its boiling point is

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