Question

The x % dissociation of $H_{2}S$ if 1 mole of $H_{2}S$ is introduced into a 1.10 litre vessel at $100K$? $K_c$ for the reaction : $2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$ is ${10}^{-6}$. Then value of $10x$ is:

Answer 1

$2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$

Initial mole $100$ 0 0

Mole at equilibrium $1-\alpha$ $\alpha$ $\alpha /2$

where $\alpha$ is the degree of dissociation of $H_{2}S$

Also, volume of container $=1.10litre$

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

$=\frac{{\left(\alpha /1.1\right)}^2(\alpha /2\times 1.1)}{{\left[(1-\alpha )/1.1\right]}^2}$

$=1\times {10}^{-6}$

$\because 1-\alpha \approx 1$ Since, $\alpha$ is small because $K_c={10}^{-6}$

$\therefore \frac{{\alpha }^3}{2\left(1.1\right)}={10}^{-6}$

or $\alpha =1.3\times {10}^{-2}$

$=1.3$%

$\therefore \%x=1.3$

$\therefore$ value of $10x=13$