Question

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 1

Answer: The equilibrium concentration of
`H_2(g)` at 700 degrees Celsius is 0.0012 M

Step-by-step 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
`K_c`

Moles of
`H_2S` = 0.29 mole

Volume of solution = 3.0 L

Initial concentration of
`H_2S` =
`(0.29mol)/(3.0L)=0.097M`

The given balanced equilibrium reaction is,

`2H_2S(g)\rightleftharpoons 2H_2(g)+S_2(g)`

Initial conc. 0.097 M 0M 0M

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

The expression for
`K_c` is written as:

`K_c=([H_2]^2* [S_2])/([H_2S]^2)`

`K_c=((2x)^2* x)/((0.097-2x)^2)`

`9.30* 10^(-8)=((2x)^2* x)/((0.097-2x)^2)`

`x=0.00060`

Equilibrium concentration of
`[H_2]`= 2x=
`2* 0.00060=0.0012M`