Question

At what temperature liquid water will be in equilibrium with water vapour?
$\Delta H_{vap}=40.73kJmo{l}^{-1}$, $\Delta S_{vap}=0.109kJ{K}^{-1}mo{l}^{-1}$,

• A. 282.4 K
• B. 373.6 K
• C. 100 K
• D. 400 K

Answer 1

Answer: (B)

373.6 K

The given reaction is :-

$H_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(g\right)$

Now, at equilibrium, $△G=0$

Now, from fibb's free energy change equation

$△G=△H-T△S$

$\Rightarrow △H=T△S$

$\Rightarrow T=\frac{△H}{△S}=\frac{40.73}{0.1079}=373.6K$

$(\because △Hrap=40.73KJ/mol,△Srap=0.109KJ{K}^{-1}{mol}^{-1})$