KCET · Chemistry · Thermodynamics (C)
Enthalpy of vaporization of benzene is \(+35.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at its boiling point, \(80^{\circ} \mathrm{C}\). The entropy change in the transition of the vapour to liquid at its boiling point in \(\left[\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right]\) is
- A \(-441\)
- B \(-100\)
- C \(+441\)
- D \(+100\)
Answer & Solution
Correct Answer
(B) \(-100\)
Step-by-step Solution
Detailed explanation
Entropy change in the transition of the vapour to liquid,
\[
\begin{aligned}
\Delta_{\text {cond }} \mathrm{S} &=\frac{\Delta \mathrm{H}_{\text {cond }}}{\mathrm{T}_{\mathrm{b}}} \\
&=\frac{-35.3 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{(80+273) \mathrm{K}} \\
&=-100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
\]
\[
\begin{aligned}
\Delta_{\text {cond }} \mathrm{S} &=\frac{\Delta \mathrm{H}_{\text {cond }}}{\mathrm{T}_{\mathrm{b}}} \\
&=\frac{-35.3 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{(80+273) \mathrm{K}} \\
&=-100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
\]
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