I posted this last night, A 0.500L vessel initially contains 0.0125 mol of H2S2 . Find the equilibrium concentrations of H2 and S2. Kc = 1.67 x 10^-7

2 H2S <--> 2 H2 + S2
I have come up with this equation so far, but I am lost now because I don’t even know if this is right.

1.67x10^-7 = ((4x^2)(x)) / (0.000625-4x^2)

It's close but not right.

1.67E-7 = (2x)^2(x)/(0.025-2x)^2 is what you started with and that's correct. Your error was in expanding the denominator.
1.67E-7 = 4x^3/(6.25E-4 -0.1x +4x^2)
The next step is to multiply 1.67E-7 x the denominator, then collect the terms in descending order and you end up with a cubic equation. I haven't checked this but I ended up with (and you should go through the math to make sure it's right)
4x^3 -6.68E-7x^2 +1.67E-8x -1/04E-10 = 0
and solve that for x.

I made a typo in that cubic equation.

4x^3-6.68E-7x^2+1.67E-8x-1.04E-10 = 0

To find the equilibrium concentrations of H2 and S2, you need to use the given equilibrium constant (Kc) and the initial amount of H2S2 in the vessel.

First, let's define the initial concentration of H2S2 as [H2S2]0 = 0.0125 mol / 0.500 L = 0.025 M

Next, let's assume that at equilibrium, the concentrations of H2 and S2 are x M.

Using the stoichiometry of the balanced equation, we can express the concentrations of H2S2, H2, and S2 in terms of x:

[H2S2] = [H2S2]0 - 2x
[H2] = 2x
[S2] = x

Now, we can substitute these concentrations into the expression for the equilibrium constant:

Kc = ([H2]^2[S2]) / [H2S2]
= ((2x)^2)(x) / ([H2S2]0 - 2x)

Substituting the values:

1.67 x 10^-7 = (4x^2)(x) / (0.025 - 2x)

Now, you can rearrange this equation to solve for x and find the equilibrium concentrations of H2 and S2. To do this, multiply both sides by (0.025 - 2x) and then move all terms to one side so you have a quadratic equation, which can be solved using standard methods like factoring or the quadratic formula.

After solving for x, you can find the equilibrium concentrations of H2 and S2 by substituting the value of x back into the expressions [H2] = 2x and [S2] = x.