Question

For the process $H_{2}O\left(l\right)(1bar,373K)\to H_{2}O\left(g\right)$, the correct set of thermodynamic parameters are:

• A. $△G=0,△S=+ve$
• B. $△G=0,△S=-ve$
• C. $△G=+ve,△S=0$
• D. $△G=-ve,△S=+ve$

Answer 1

Answer: (D)

$△G=-ve,△S=+ve$

$H_2{O}_{\left(l\right)}\to H_2{O}_{\left(g\right)}$ is a spontaneous reaction and the entropy of water molecules should be high in order to get transformed to gaseous state.

More is the entropy$\to$ more the molecules are free to move.

Hence, $△G=-ve$ which means to carry out the reaction no extra energy is required and $△S=+ve$ means the intermolecular attraction should be least.