Question

Hydrogen sulfide decomposes according to the following reaction, for which Kc = 9,30 x 10^-8 at 700 degree C.2 H2S(g)leftrightharpoons 2H2(g) + S2(g)It 0.42 mol H2S is placed in a 2.8 L container, what is the equilibrium concentration of H2(g) at 700 degree C? M

Answer 1

$K_c = \frac{[H_2]^2[S_2]}{[H_2S]^2} = 9.3*10^{-8}$

By Law of mass action:

$\ \ \ \ \ \ \ \ \ 2H_2S_{(g)} \rightleftharpoons 2H_2_{(g)} + S_2_{(g)} \\ \\ intial: \ \ \ \ \ \ 0.42\ \ \ \ \ \ \ \ \ \ \ \ - \ \ \ \ \ \ - \\ \\ rxn: \ \ \ \ \ \ \ \ \ -x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\ \ \ x/2 \\ \\ eq: \ \ \ \ \ \ \ \ \ \ \ \ 0.42-x\ \ \ \ \ \ x\ \ \ \ \ \ x/2 \\ \\ conc: \ \ \ \ \ \ \frac{0.42-x}{2.8}\ \ \ \ \ \ \ \ \ \frac{x}{2.8}\ \ \ \frac{x/2}{2.8} \\ \\ 9.3*10^{-8}= \frac{[x/2.8]^2[x/5.6]}{[(0.42-x)/2.8]^2} \\ \\ x = 4.5*10^{-3}$

so :

$[H_2] = \frac{x}{2.8} = \frac{4.5*10^-3}{2.8}=1.6*10^{-3}$