Question

${\Delta }_f{G}^{\circ }$ at $500K$ for substance ' $S$ ' in liquid state and gaseous state are $+100.7kcalmo{l}^{-1}$ and $+103kcalmo{l}^{-1}$ respectively. Vapour pressure of liquid ' $S$ ' at $500K$ is approximately equal to_____ atm. $\left(R=2cal{K}^{-1}mo{l}^{-1}\right)$

Answer 1

Answer: 0.1

$S_{(l)}⟶S_{(g)}$
$\Delta G^{\circ }=\Delta f{G}^{\circ }\left(S_{(g)}\right)-{\Delta }_f{G}^{\circ }\left(S_{(l)}\right)$
$=+103-100.7=2.3kcalmo{l}^{-1}$
$\Delta G^{\circ }=-2.303RT{\log }_{10}K$
$2.3=-2.303\times 2\times 10^{-3}\times 500\times {\log }_{10}K$
${\log }_{10}K=\frac{-2.3}{2.303\times 2\times 10^{-3}\times 500}=-1$
$\therefore K=10^{-1}=0.1$
$K=p_{S(g)}$
( $\because$ Partial pressure of liquid is not considered in calculating $K$ )
$\therefore$ Vapour pressure of liquid ' $S$ ' at $500K=0.1atm$