The enthalpy of vaporization of water at 100∘C is 40.63 kJ mol−1. Its entropy of vaporization would be

The enthalpy of vaporization of water at $100^{\circ} C$ is $40.63 kJ {mol}^{- 1}$ . Its entropy of vaporization would be

A
$\begin{aligned} 406.3 {J K}^{- 1} {mol}^{- 1} \end{aligned}$
B
$108.9 J K^{- 1} {mol}^{- 1}$
C
$606.3 J K^{- 1} {mol}^{- 1}$
D
$808.3 J K^{- 1} {mol}^{- 1}$

Solution:

We know that entropy change in phase transformation is given as:

$Δ S = \frac{Δ H}{T} = \frac{40.63 \times 10^{3} J {mol}^{- 1}}{373 K}$

So,

$Δ S$ $= 108.9 J K^{- 1} {mol}^{- 1}$

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