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The enthalpy of vaporisation of benzene ...

The enthalpy of vaporisation of benzene `(C_(6)H_(6))` is `30.8 kJ mol^(-1)` at its boiling point `(80.1^(@)C)`. Calculate the entropy change in going from:
a. liquid to vapour and
b. vapour to liquid at `80.1^(@)C`.

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Verified by Experts

The correct Answer is:
(i)`87.2 J K^(-1) "mol"^(-1)` (ii) `-87.2 J K^(-1) "mol"^(-1)`
`Delta_(vap)S = (Delta_(vap)H)/(T) (T = 273 + 80.1 = 353.1 K)`
or `= (30.8 xx 10^(3))/(353.1) = 87.2 JK^(-1) "mol"^(-1)`
(ii) Since `Delta_(cond)H = -Delta_(vap) H`
`Delta_(cond)S = -87.2 JK^(-1) "mol"^(-1)`.
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