Question

In the balanced equilibrium reaction: 2 H2S <----> S2 + 2 H2, with K = 3.7 x 10-6, if the initial concentration of H2S is 0.080 M, what is the final concentration of H2?

Answer 1

In the balanced equilibrium reaction: 2 H2S <----> S2 + 2 H2, with K = 3.7 x 10-6, if the initial concentration of H2S is 0.080 M, what is the final concentration of H2?

Given : K = 3.7 E -6

Reaction :

2 H2S <----> S2 + 2 H2

Solution:

To calculate final concentration of H2, we need to set up an ICE chart

               2 H2S <----> S2 + 2 H2

I              0.080 M          0       0

C           - 2x                    +x     +2x

E           (0.080-2x )         x         2x

Lets set up equilibrium constant expression.

K = [S2][H2]²/ [H2S]²

We plug value of K and equilibrium concentration from the ICE table

3.7 E -6 = x * (2x)²/ (0.080-2x)²

Lets solve for x

Since value of equilibrium constant is too small we can use 5% approximation. In that we neglect 2x at the denominator.

So, equation becomes.

3.7 E -6 = x * (2x)²/ (0.080)²

3.7 E -6 = x * (2x)²/ 0.0064

2.368 E -8 = x * (2x)²

4x³ = 2.368 E -8

x³ = 5.92 E -9

x = 0.00181

We check 5 % approximation

Percent dissociation = 2x / 0.080 * 100

= 2 * 0.00181 / 0.080 * 100 %

= 4.52 %

Percent dissociation is less than 5 so approximation is valid.

Now we know equilibrium concentration of H2 is 2x

[H2] = 2 * 0.00181 = 0.00362 M

So final concentration of H2 will be 0.00362 M