At some temperature, Keq = 33 for the reaction H2 + I2 „³ 2HI. If initially, [H2] = .0600 M and [I2] = .0300 M, what are all three equilibrium concentrations?

............H2 + I2 ==> 2HI

I......0.0600...0.0300...0
C.........-x.....-x......+2x
E.......0.06-x...0.03-x...2x

Substitute the E line into the Keq expression and solve for x. Then
H2 = 0.06-x
I2 = 0.03-x
HI = 2x

To determine the equilibrium concentrations of H2, I2, and HI, we can use the equation for the equilibrium constant (Keq) and the stoichiometry of the reaction.

The given information is as follows:
- Keq = 33
- [H2] = 0.0600 M (initial concentration of H2)
- [I2] = 0.0300 M (initial concentration of I2)

The balanced equation for the reaction is:
H2 + I2 ⇌ 2HI

To solve for the equilibrium concentrations, we can assign the change in concentrations for each species as follows:
- Change in [H2] = -2x (to account for the stoichiometric ratio)
- Change in [I2] = -x (to account for the stoichiometric ratio)
- Change in [HI] = +2x (to account for the stoichiometric ratio)

Let's set up an expression for Keq using the equilibrium concentrations:
Keq = ([HI]eq^2) / ([H2]eq * [I2]eq)

Substituting the given values, we get:
33 = ([HI]eq^2) / ((0.0600 - 2x) * (0.0300 - x))

Now, we can solve for x using the quadratic equation:
33 = (2x)^2 / ((0.0600 - 2x) * (0.0300 - x))

Rearranging and simplifying the equation, we have:
33(0.0600 - 2x)(0.0300 - x) = 4x^2

Expanding and rearranging further, we get:
0.198 - 0.099x - 1.98x + 0.996x^2 = 4x^2

Combining like terms, we have:
0.996x^2 + 1.881x - 4x^2 - 0.099x + 0.198 = 0

Simplifying the equation, we get:
-3.004x^2 + 1.782x + 0.198 = 0

We can now solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

For this equation, a = -3.004, b = 1.782, and c = 0.198. Plugging in these values, we can find the values for x.

Using the quadratic formula, we find two possible values for x: x = 0.113 or x = -0.024.

Since we cannot have a negative concentration, we discard the negative value.
Thus, x = 0.113.

Substituting this value back into our expressions for change in concentrations, we get:
Change in [H2] = -2x = -2(0.113) = -0.226 M
Change in [I2] = -x = -(0.113) = -0.113 M
Change in [HI] = +2x = 2(0.113) = 0.226 M

To find the equilibrium concentrations, we add the changes to the initial concentrations:
[H2]eq = [H2] + Change in [H2] = 0.0600 + (-0.226) = 0.0600 - 0.226 = 0.0338 M
[I2]eq = [I2] + Change in [I2] = 0.0300 + (-0.113) = 0.0300 - 0.113 = 0.0169 M
[HI]eq = [HI] + Change in [HI] = 0 + 0.226 = 0.226 M

Therefore, the equilibrium concentrations are:
[H2]eq = 0.0338 M
[I2]eq = 0.0169 M
[HI]eq = 0.226 M

To find the equilibrium concentrations of all three species, you can use the concept of the equilibrium constant (Keq) and the stoichiometry of the reaction. Here's how you can approach this problem step by step:

Step 1: Write down the balanced chemical equation for the reaction: H2 + I2 → 2HI.

Step 2: Express the equilibrium constant Keq in terms of the concentrations of the reactants and products. In this case, Keq = [HI]^2 / ([H2] * [I2]), based on the stoichiometry of the reaction.

Step 3: Substitute the given values into the equation for Keq. In this case, Keq = 33.

33 = [HI]^2 / ([H2] * [I2]).

Step 4: Substitute the given initial concentrations into the equation.

33 = [HI]^2 / (0.0600 * 0.0300).

Step 5: Rearrange the equation and solve for [HI].

([HI]^2) = 33 * (0.0600 * 0.0300).

[HI]^2 = 0.0594.

Take the square root of both sides:

[HI] = √0.0594.

Calculating this value, [HI] ≈ 0.243 M.

Step 6: Determine the equilibrium concentrations of H2 and I2. Since the reaction has a stoichiometric ratio of 1:1:2, the equilibrium concentration of H2 will also be 0.0600 M, and the equilibrium concentration of I2 will be 0.0300 M (since none of them react with each other).

Hence, the equilibrium concentrations are:
[H2] = 0.0600 M,
[I2] = 0.0300 M, and
[HI] ≈ 0.243 M.