Question

The vapour pressure of mercury is $0.002$mm at ${27}^{o}C$ .$Hg\left(I\right)\rightleftharpoons Hg\left(g\right)$. The value of equilibrium constant $K_C$ is:

• A. ${1.07\times 10}^{-7}$
• B. ${3.95\times 10}^{-5}$
• C. ${8.12\times 10}^{-5}$
• D. ${2.63\times 10}^{-6}$

Answer 1

Answer: (A)

${1.07\times 10}^{-7}$

Given $K_p=P_{Hg}=0.002$ $mmHg$ $=2.6\times {10}^{-6}$ $bar$

$K_p=\frac{K_p}{(RT{)}^{\Delta n}}$ $=\frac{2.6\times {10}^{-6}}{RT}=\frac{2.6\times {10}^{-6}}{0.08\times 314\times 299}$

$K_c=1.1\times {10}^{-7}$