Answered

Hydrogen sulfide decomposes according to the following reaction, for which Kc=9.30E-8 at 700 degrees Celsius. 2 H2S(g) --> 2 H2(g) + S2(g) If 0.29 moles of H2S is placed in a 3.0-L container, What is the equilibrium concentration of H2(g) at 700 degrees Celsius?

Answer :

Answer: The equilibrium concentration of [tex]H_2(g)[/tex] at 700 degrees Celsius is 0.0012 M

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_c [/tex]

Moles of  [tex]H_2S[/tex] = 0.29 mole

Volume of solution = 3.0 L

Initial concentration of [tex]H_2S[/tex] = [tex]\frac{0.29mol}{3.0L}=0.097M[/tex]

The given balanced equilibrium reaction is,

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

         Initial conc.         0.097 M           0M          0M

At eqm. conc.    (0.097-2x) M            (2x) M   (x) M

The expression for [tex]K_c[/tex] is written as:

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

[tex]K_c=\frac{(2x)^2\times x}{(0.097-2x)^2}[/tex]

[tex]9.30\times 10^{-8}=\frac{(2x)^2\times x}{(0.097-2x)^2}[/tex]

[tex]x=0.00060[/tex]

Equilibrium concentration of [tex][H_2][/tex]= 2x= [tex]2\times 0.00060=0.0012M[/tex]