Question

The vapour pressure of mercury is $0.002mmHg$ at ${27}^{o}C.Hg\left(l\right)\leftrightharpoons Hg\left(g\right),$ the value of equilibrium constant $K_c$ is :

• A. $1.068\times {10}^{-7}M$
• B. $0.002M$
• C. $8.12\times {10}^{-5}$
• D. $3.9\times {10}^{-5}M$

Answer 1

The correct option is A $1.068\times {10}^{-7}M$

Given,

Vapour pressure of Hg = 0.002mm

$R=0.0821$

$T={27}^{\circ C}$

$\Rightarrow 27+273=300K$

We know that $K_p=K_c(RT{)}^{\Delta n}$ $...\left(i\right)$

Where $\Delta n=$ number of gaseous moles of products – number of gaseous moles of reactants

For the given reaction $Hg\left(l\right)\iff Hg\left(g\right)$, $\Delta n=1$, since the number of gaseous moles of products is $1$ and there are no gaseous reactants present.

For $1mm$ of $Hg$ the equivalent pressure in bar will be $1mmHg=0.0013bar$

Therefore, the value of $K_p$ will be

$K_p=P_{Hg}=0.002mmHg$

$\Rightarrow K_p=2.6\times {10}^{-6}bar$

Substituting the given values and above calculated value of $K_p$ in equation $\left(i\right)$ we get,

$K_p=K_c(RT{)}^{\Delta n}$

$2.6\times {10}^{-6}bar=K_c(0.0821\times 300{)}^1$

Solving the above equation for $K_c$ we get,

$K_c=\frac{2.6\times {10}^{-6}bar}{0.0821\times 300}$

$K_c=1.06\times {10}^{-7}M$

Hence, the correct option is option A.