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

M H2S2 initially = 0.0125/0.5 = 0.025M

............2H2S ==> 2H2 + S2
I...........0.025.....0.....0
C............-2x......2x.....x
E.........0.025-2x....2x.....x

Substitute the E line into the Kc expression and solve for x.

I got this equation but now I am lost

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

To find the equilibrium concentrations of H2 and S2, we can use the equilibrium constant expression and set up an ICE table.

The balanced equation for the reaction is:
2 H2S <---> 2 H2 + S2

Let's assume that at equilibrium, the concentration of H2S is x mol/L, and the concentrations of H2 and S2 are y mol/L.

For the reaction: 2 H2S <--> 2 H2 + S2, the initial concentration of H2S is given as 0.0125 mol/L.

Using the ICE table, we have:

Initial: 0.0125 0 0
Change: -2x +2x +x
Equilibrium: 0.0125-2x 2x x

The equilibrium constant expression (Kc) is given as 1.67 × 10^-7. For this reaction, Kc is defined as:

Kc = [H2]^2[S2] / [H2S]^2

Substituting the equilibrium concentrations into the equation, we have:

Kc = (2x)^2 * x / (0.0125 - 2x)^2

Simplifying the equation, we get:

1.67 × 10^-7 = 4x^3 / (0.0125 - 2x)^2

To solve this equation, you can use a numerical method such as the Newton-Raphson method or substitute different values of x and find the value that satisfies the equation. However, for the purpose of this step-by-step explanation, I will assume that the value of x is negligible compared to 0.0125. This approximation is justified by the fact that the equilibrium constant (Kc) is very small, indicating that there will be very little dissociation of H2S.

By making this approximation, we can simplify the equation to:

1.67 × 10^-7 ≈ 4x^3 / 0.0125^2

Simplifying further:

1.67 × 10^-7 ≈ 4x^3 / 0.00015625

Rearranging the equation:

x^3 = (1.67 × 10^-7) * (0.00015625) / 4

x^3 ≈ 6.5104 × 10^-12

Taking the cube root of both sides:

x ≈ 1.790 × 10^-4

Therefore, the equilibrium concentration of S2 (y) is approximately 1.790 × 10^-4 mol/L, and the equilibrium concentration of H2 (x) is also approximately 1.790 × 10^-4 mol/L.

To find the equilibrium concentrations of H2 and S2, we need to set up an expression using the equilibrium constant (Kc) and the stoichiometry of the reaction.

Given the reaction:
2 H2S <--> 2 H2 + S2

The equilibrium constant expression is:
Kc = ([H2]^2 * [S2]) / ([H2S]^2)

Here, [H2] represents the equilibrium concentration of H2, [S2] represents the equilibrium concentration of S2, and [H2S] represents the equilibrium concentration of H2S.

We are given the initial amount (in moles) of H2S (0.0125 mol) and the volume of the vessel (0.500 L). From this, we can calculate the initial concentration of H2S:

Initial concentration of H2S = initial amount of H2S / volume of vessel
= 0.0125 mol / 0.500 L
= 0.025 mol/L

Let's denote the equilibrium concentrations of H2 and S2 as [H2]eq and [S2]eq, respectively. We can set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentrations during the reaction:

2 H2S <--> 2 H2 + S2
Initial 0.025 0 0
Change -2x +2x +x
Equilibrium 0.025 - 2x 2x x

According to the stoichiometry, 2 moles of H2S produce 2 moles of H2 and 1 mole of S2. Therefore, the change in the concentration of H2S is -2x (since 2 moles of H2S decreases by 2x), the change in the concentration of H2 is +2x (since 2 moles of H2 is formed by 2x), and the change in the concentration of S2 is +x (since 1 mole of S2 is formed by x).

At equilibrium, we can express the concentrations in terms of the initial concentration, [H2S]eq = 0.025 - 2x, [H2]eq = 2x, and [S2]eq = x.

Substituting these values into the equilibrium constant expression, we have:
Kc = ([H2]^2 * [S2]) / ([H2S]^2)
1.67 x 10^-7 = (2x)^2 * (x) / (0.025 - 2x)^2

Now, we can solve this equation to find the value of x, which corresponds to the equilibrium concentration of S2. Once we have the value of x, we can calculate the equilibrium concentration of H2 using [H2]eq = 2x.

Please note that solving the equation involves some algebraic manipulation and possibly using numerical methods if the equation cannot be solved analytically.