Question

Calculate the percent dissociation of $H_{2}S\left(g\right)$ if $0.1\text{mole}$ of $H_{2}S$ is kept in $0.4L$ vessel at $1000K$. For the reaction, the value of $K_c$ is $1.0\times {10}^{-6}$.

• A. $2.0$
• B. $4.0$
• C. $6.0$
• D. None of the above

Answer 1

The correct option is A $2.0$

At equilibrium
$2H_{2}S\left(s\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)Atequi.(0.1-x)x\frac{x}{2}$
$conc.\frac{(0.1-x)}{0.4}\frac{x}{0.4}\frac{x}{0.8}$

Equilibrium constant,
$K_c=\frac{\left[H_2{]}^2\right[S_2]}{[H_{2}S{]}^2}$

putting the values,
$K_c=\frac{\left[\frac{x}{0.4}{]}^2\right[\frac{x}{0.8}]}{[\frac{(0.1-x)}{0.4}{]}^2}$
or $\frac{x^3}{0.8(0.1-x{)}^2}=1.0\times {10}^{-6}$

Now, as x is very small
$\Rightarrow 0.1-x\to 0.1$
$\Rightarrow x=2\times {10}^{-3}$
So, percent dissociation $=\frac{2\times {10}^{-3}}{0.1}\times 100=2.0$