Equilibrium Constant

1. Nov 1, 2006

Fusilli_Jerry89

The value of Keq for the reaction: 2H2S(g) <-> 2H2(g) + S2(g) is 4.20x10^-6 at 830 degrees celsius.
a) What is favoured in this reaction, reactants or products?
b) What concentration of S2 can be expected at equilibrium after 0.200 mol of H2S is injected into an empty 1.00 L flask?

2. Nov 1, 2006

therealkellys

a) Do you know how to write out Keq? (Hint: Use the law of mass action).
b) Make an ice box
2H2S(g) <--> 2H2(g) + S2 (g)
i .200 0 0
c -2x +2x +x
e .200 - 2x 2x x

Keq = [((2x)^2)(x)] / [(.200 - 2x)^2]

Last edited: Nov 1, 2006
3. Nov 2, 2006

Fusilli_Jerry89

i solved x to equal 1.431 but when u plug it back into the original question, H2S comes out to a negative number

4. Nov 2, 2006

Fusilli_Jerry89

oh nm i made a mistake

5. Nov 2, 2006

Fusilli_Jerry89

k got down to 4x^3-1.68x10^-5x2+3.36x10^-6-1.68x10^-7=0 now what

6. Nov 2, 2006

geoffjb

What does that mean?

Look to your rules of equilibrium to determine which side of the reaction is favoured.

As for x, you can solve for it.

7. Nov 2, 2006

Fusilli_Jerry89

i put the products are favoured, but i do not know how to solve this equation cuz it has an 4^3 x^2 and an x

8. Nov 2, 2006

geoffjb

Why do you have $4x^{3}$ in your equation? Solve for x.