Enthalpy of vapourization of benzene is `+35.3 kJ mol^(-1)` at its boiling point of `80^(@)C`. The entropy change in the transition of the vapour to l

Question

Enthalpy of vapourization of benzene is `+35.3 kJ mol^(-1)` at its boiling point of `80^(@)C`. The entropy change in the transition of the vapour to l

Enthalpy of vapourization of benzene is `+35.3 kJ mol^(-1)` at its boiling point of `80^(@)C`. The entropy change in the transition of the vapour to liquid at its boiling points [in `K^(-1) mol^(-1)`] is
A. `-441`
B. `-100`
C. `+441`
D. `+100`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`C_(6)H_(6)(l)toC_(6)H_(6)(g)` , `DeltaH=+35.3 kJ mol^(-1)`
Reverse reaction
`C_(6)H_(6)(g)toC_(6)H_(6)(l)`, `DeltaH=-35.3 kJ mol^(-1)`
`T=80+273=353 K`
`DeltaS=(DeltaH)/(T)=(-35.5xx1000)/(350)=-100 JK^(-1) mol^(-1)`

4 views Answered 2023-02-05 15:29