Question

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

• 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

Answer 1

Answer: (A)

$Hg\left(\right.l\left.\right)\rightleftharpoons Hg\left(\right.g\left.\right)$

$\Delta n_{g}=1$

$K_{p}=\frac{0.002}{760}=2.63\times 10^{- 6}$ atm

$K _{ p }= K _{ c }( RT )^{\Delta n }$

$K _{ c }=\frac{ K _{ p }}{ RT }=\frac{2.63 \times 10^{-6}}{0.0821 \times 300}$

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