Question

What is the % dissociation of $H_{2}S$ if one mole of $H_{2}S$ is introduced in 1 litre vessel at $1000K$ if $K_c$ for the reaction, $2H_{2}S\left(g\right)\rightleftharpoons 2H_2\left(g\right)+S_2\left(g\right)$ is $4\times {10}^{-6}$ ?

Answer 1

Since, total volume is 1 litre, the number of moles corresponds to molar concentration.

$2H_{2}S⟶2H_2+S_2$

1 0 0 -------- initial

1-2x 2x x ------ eqbm

The expression for the equilibrium constant is $K_c=\frac{\left[H_2{]}^2\right[S_2]}{H_{2}S}$

Substitute values in the above expression.

$4\times {10}^{-6}=\frac{(2x{)}^{2}x}{(1-2x{)}^2}$

Since, the value of the equilibrium constant is very small, $(1-2x)\simeq 1$

Hence, the equilibrium constant expression becomes

$4x^3=4\times {10}^{-6}$

$x=0.01$.

The percentage dissociation for $H_{2}S$ is $\frac{2x}{1}\times 100=\frac{2\times 0.01}{1}\times 100=2$.