Q. The vapour pressure of mercury is 0.002 mm Hg at $27^\circ C$
$Hg\left(\right.l\left.\right)\rightleftharpoons Hg\left(\right.g\left.\right)$
The value of equilibrium constant $K_{C}$ is

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A

$1.068\times 10^{- 7}$ M

B

$0.002$ M

C

$8.12\times 10^{- 5}$ M

D

$3.9\times 10^{- 5}$ M

Solution:

$Hg\left(\right.l\left.\right)\rightleftharpoons Hg\left(\right.g\left.\right)$
$\Delta n_{g}=1$
$K_p=K_c(R T$
$K_{p}=K_{c}\left(\right.RT\left(\left.\right)^{\Delta n}$
$K_{c}=\frac{K_{p}}{R T}=\frac{2.63 \times 1 0^{- 6}}{0.0821 \times 300}$
$=1.068\times 10^{- 7}$ M

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