Question

The enthalpy change, for the transition of liquid water to steam, $\Delta {H_{vapour}}$ is 40.8 KJ $mo{l^{ - 1}}$ at 373 K. calculate the entropy change for the process.

Answer 1

Since the enthalpy change, for the transition of liquid water to steam is given in the question
( $\Delta {H_{vapour}}$ is 40.8 KJ $mo{l^{ - 1}}$ ), and we know that the entropy change for the vaporization of water is given by the equation, $\Delta {S_{vapour}} = \dfrac{{\Delta {H_{vapour}}}}{{{T_{vapour}}}}$. So, by putting the value of enthalpy change and temperature (373 K), we can calculate the value of entropy change for the process

Complete step by step answer:
We know that the entropy of vaporization is the increase in the entropy when liquid is vaporised.
The transition that we are considering here is,
${H_2}O(l) \to {H_2}O(g)$
Given in the question,
Enthalpy change, $\Delta {H_{vapour}} = 40.8KJmo{l^{ - 1}}$
Temperature, $T = 373K$
We know that,
The entropy change for the vaporization of water is given by,
$\Delta {S_{vapour}} = \dfrac{{\Delta {H_{vapour}}}}{{{T_{vapour}}}}$
Putting the values in the above equation, we get
$\Delta {S_{vapour}}40.8 \times 1000\dfrac{{Jmo{l^{ - 1}}}}{{373K}}$

$\Rightarrow \Delta {S_{vapour}} = 109.38\,J{K^{ - 1}}mo{l^{ - 1}}$

So, the entropy change $(\Delta {S_{vapour}})$ for the process is 109. 38 $J{K^{ - 1}}mo{l^{ - 1}}$

Note: Whenever we are vaporising water at 373 K, we add heat, while temperature remains constant. But both enthalpy and entropy will increase. During vaporization the randomness of the system increases, so the change in entropy will always have a positive value and cannot be zero since the degree of disorder increases in the transition from a liquid in a relatively small volume to a vapour or gas occupying a much larger space. Change in temperature, change in volume and change in phase will lead to change in entropy.