Question

With this problem I understand the ICE table used to solve the problem as well as setting up the Kc. I just don't understand how the small-x approximation applies to this problem since it doesn't seem to follow the rules for x is small approximation. I tried comparing the value of 0.097 M/9.30*10^-8 which was 1.0*10^6 which is smaller than 400 so the x is small doesn't apply, however I can't seem to find another method that helps me simplify the problem.

Hydrogen sulfide decomposes according to the following reaction, for which K. = 9.30 ~ 10-8 at 700°C: 2 H2S(g) = 2 H2(g) + S2(g) If 0.29 mol of H2S is placed in a 3.0-L container, what is the equilibrium concentration of H2(g) at 700° C?

Answer 1

Given reaction :

2 H₂S <====> 2 H₂ + S₂

Given that, K_c = 9.30 x 10^-8

Initial concentration of H₂S = [H₂S] = (0.29 mol) / (3 L) = 0.0967 M

Apply ICE table:

Let X be the change.

[H₂S] = 0.0967 - 2X

[H₂] = 2X

[S₂] = X

So the equilibrium constant expression is given by:

Kc = [Hz]2.[S2] [H2S]2

Kc (2x)2.(X) (0.0967 -2X)2

4X3 Kc = 0.0967 (0.0967_2X12 = 9.3 x 10-8

Since X <<< 0.0967:

4X3 (0.0967)2 = 9.3 x 10-8

4x3 = 9.3 x 10-8 x (0.0967)

X = 9.3 x 10-8 % (0.0967)2

X = 0.00059

Therefore, Equilibrium concentration of H₂:

(H2) = 2X = 2 x 0.00059 M

[H2= 0.00118 M