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2H2S(g)⇌2H2(g)+S2(g),Kc=1.67×10−7 at 800∘C is carried out at the same temperature with the following initial concentrations: [H2S]=0.100M, [H2]=0.100M, and [S2]=0.00 M. Find the equilibrium concentration of S2. Express the molarity to three significant figures.

Respuesta :

Answer : The concentration of [tex]S_2[/tex] at equilibrium will be, [tex]1.67\times 10^{-7}M[/tex]

Explanation :  Given,

Equilibrium constant = [tex]1.67\times 10^{-7}[/tex]

Initial concentration of [tex]H_2S[/tex] = 0.100 M

Initial concentration of [tex]H_2[/tex] = 0.100 M

Initial concentration of [tex]S_2[/tex] = 0.00 M

The balanced equilibrium reaction is,

                      [tex]2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)[/tex]

Initial conc.    0.1                0.1          0

At eqm.         (0.1-2x)       (0.1+2x)      x

The expression of equilibrium constant for the reaction will be:

[tex]K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}[/tex]

Now put all the values in this expression, we get :

[tex]1.67\times 10^{-7}=\frac{(0.1+2x)^2\times (x)}{(0.1-2x)^2}[/tex]

By solving the term 'x' by quadratic equation, we get two value of 'x'.

[tex]x=1.67\times 10^{-7}M[/tex]

Concentration of [tex]S_2[/tex] at equilibrium = [tex]x=1.67\times 10^{-7}M[/tex]

Therefore, the concentration of [tex]S_2[/tex] at equilibrium will be, [tex]1.67\times 10^{-7}M[/tex]

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