Search any question & find its solution
Question:
Answered & Verified by Expert
If the enthalpy of vaporization of water is 186.5 $\mathrm{kJmol}^{-1}$, the entropy if its vaporization will be:
Options:
Solution:
1194 Upvotes
Verified Answer
The correct answer is:
$0.5 \mathrm{kJK}^{-1} \mathrm{~mol}^{-1}$
Given enthalpy of vaporization,
$$
\Delta \mathrm{H}=186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Boiling point of water
$$
=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}
$$
Entropy change,
$$
\begin{array}{r}
\Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}=\frac{186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}}{373 \mathrm{~K}} \\
=0.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{array}
$$
$$
\Delta \mathrm{H}=186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Boiling point of water
$$
=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K}
$$
Entropy change,
$$
\begin{array}{r}
\Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}=\frac{186.5 \mathrm{~kJ} \mathrm{~mol}^{-1}}{373 \mathrm{~K}} \\
=0.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.