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?IF THE CORRECT OPTION ISNT THERE PLEASE GIVE CORRECT ANSWER AS THERE MAY BE AN ERROR IN THE HOMEWORK PROBLEM. THANKS!

A.3.70 x 10^-9

B.2.10 x 10^-3

C.9.25 x 10^-3

D.6.10 x 10^-4

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?

Solution:

We set up Ice chart to get final concentration of H2

I= Initial concentration

C = change in concentration

E = Equilibrium concentration

The reaction is

               2 H2S <----> S2 + 2 H2

I              0.080 M          0       0

C           - 2x                    +x     +2x

E           (0.080-2x )         x         2x

Equilibrium constant is given and we use equilibrium equation.

K = [S2][H2]²/ [H2S]²

3.7 X 10 ^-6 = x (2x)²/(0.080-2x)²

We have to solve for x

5 % approximation :

(0.080-2x) = 0.080 is assumed

3.7 X 10 ^-6 = x * (2x)²/ (0.080)²

4x³ / 0.0064=3.7 X 10 ^-6

4x³ = 3.7 X 10 ^-6 *0.0064

     = 2.37 X 10^-8

x³ = 5.9 X 10

x = 0.001809

Approximation check.

= 2*x / concentration of H2S

= 2*0.001809 / 0.080 * 100 = 45.2 %

This percent dissociation <5

So approximation is correct.

[ H2] =2x = 0.00362 M

= 3.62 X 10^-3 M

So 3.62 X 10^-3 M is the correct option .