Question

The value of Kc for rxn below 2 H2S(g) <----> 2H2(g) + S2 (g) find value of [H2S] ?

The value of Kc for rxn below

2 H2S(g) <----> 2H2(g) + S2 (g)

is 2.2 x 10^-4 at 1400 K. A sample of [H2S] = 5.00 M is heated to 1400 K in a sealed vessel. After chemical equilibrium, find value of [H2S] No H2 or S2 was present original sample.

Answer 1

The equilibrium concentration of `H_2S` is `4.78M`

Explanation:

`K_c=2.2xx10^-4=([H_2]^2[S_2])/([H_2S]^2)`

Let

`n_(H_2S)^0` = the initial concentration of `H_2S=5M`

Assume that `x` molar concentration of `H_2S` decomposes at equilibrium. At equilibrium, the concentration of `H_2S` will be `n_(H_2S)^0-x`.

This will produce `x` molar concentration of `H_2` and `x/2` molar concentration of `S_2`. Thus at equilibrium we will have

`K_c=(x^2(x/2))/((n_(H_2S)^0-x)^2)`

Because the equilibrium constant is small, we will make the approximation that `n_(H_2S)^0-x~~n_(H_2S)^0`. This allows us to write

`K_c~~(x^3)/(2(n_(H_2S)^0)^2)`

Solving for `x` we have

`x=[2K_c(n_(H_2S)^0)^2]^(1/3)=[2*2.2xx10^-4(5)^2]^(1/3)~~0.22`

This would give the final concentration of `H_2S` as

`n_(H_2S)^0-x=5-0.22=4.78M`

As a check of our approximation we can calculate the equilibrium constant as

`(0.22^2(0.22/2))/((4.78)^2)~~2.3xx10^-4`

which is less than a 5% error from the given value.