Question

Calculate the percent dissociation of $H_{2}S\left(g\right)$ if 0.1 mol of $H_{2}S$ is kept in a 0.4 litre vessel at 1000 K. For the reaction,
$2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$ the value of $K_c$ is $1\times {10}^{-6}$.

• A. $2\%$
• B. $20\%$
• C. $200\%$
• D. $50\%$

Answer 1

The correct option is A $2\%$

$\begin{array}{cccccc}& 2H_{2}S\left(g\right)& \rightleftharpoons & 2H_2\left(g\right)& +& S_2\left(g\right)\\ \text{(Initially)}& 0.1& & 0& & 0\\ \text{(At equilibrium)}& 0.1-x& & x& & \frac{x}{2}\end{array}$
$K_c=\frac{\left[H_2{]}^2\right[S_2]}{[H_{2}S{]}^2}=\frac{{\left(\frac{x}{0.4}\right)}^2\left(\frac{x}{2\times 0.4}\right)}{{\left(\frac{0.1-x}{0.4}\right)}^2}$
Assuming
$0.1-x\approx 0.1$
Since, ${}^{\prime }{x}^{\prime }$ is small because $K_c={10}^{-6}$
$\therefore \frac{x^3}{2\times 0.4\times (0.1{)}^2}={10}^{-6}$ or,
$x=2\times {10}^{-3}$
This is degree of dissociation for 0.1 mol
$\therefore$ Degree of dissociation for 1 mol $=10\times (2\times {10}^{-3})=2\times {10}^{-2}$
Hence, $\%$ dissociation $=2\times {10}^{-2}\times 100=2$