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

keq=(2x)^2/ (.06-x)(.03-x)=33

4x^2=33(.0018+x^2-.09x)

-29x^2+.09&33 x- .0018*33=0

ok,use the quadratic eqauation, solve for x.
2x is the conc of HI
.06-x is the conc of H2
.03-x is the conc of I2

To find the equilibrium concentrations, we need to use the given equilibrium constant (Keq) and the initial concentrations of the reactants. Here's how you can calculate the equilibrium concentrations:

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

Step 2: Identify the equilibrium concentrations:
[H2] = 0.0600 M (initial concentration of H2)
[I2] = 0.0300 M (initial concentration of I2)

Step 3: Use the given equilibrium constant (Keq) to set up an expression for the equilibrium concentrations:
Keq = [HI]² / ([H2] * [I2])

Step 4: Substitute the values into the equation:
33 = [HI]² / (0.0600 M * 0.0300 M)

Step 5: Solve for [HI]:
[HI]² = 33 * (0.0600 M * 0.0300 M)
[HI]² = 0.0594

Taking the square root of both sides (remembering to consider both the positive and negative square roots):
[HI] = ± √0.0594

There are two possible values for [HI]: +√0.0594 and -√0.0594. However, concentrations cannot be negative, so we can discard the negative root.

Therefore, the equilibrium concentration of [HI] is approximately +√0.0594.

Step 6: Calculate the equilibrium concentrations of [H2] and [I2]:
[H2] = 0.0600 M - (2 * 0.0594 M)
[H2] = 0.0600 M - 0.1188 M
[H2] ≈ -0.0588 M (Negative value is not possible, so [H2] ≈ 0 M)

[I2] = 0.0300 M - 0.0594 M
[I2] ≈ -0.0294 M (Negative value is not possible, so [I2] ≈ 0 M)

Therefore, the equilibrium concentrations are approximately:
[H2] ≈ 0 M
[I2] ≈ 0 M
[HI] ≈ +√0.0594 M

To solve this question, we need to use the equilibrium expression and set up an ICE (Initial, Change, Equilibrium) table.

The equilibrium expression for the given reaction is:

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

We are given that Keq = 33, [H2] = 0.0600 M, and [I2] = 0.0300 M.

Let's assume the change in concentration of HI is "x" M. Then the change in concentration for H2 and I2 will both be "-x" M since the stoichiometric coefficients for both reactants in the balanced equation are 1.

Setting up the ICE table:

H2 + I2 -> 2HI
Initial: 0.0600 + 0.0300 -> 0
Change: -x -x -> +2x
Equilibrium: 0.0600 - x + 0.0300 - x -> 2x

Using the equilibrium expression, Keq = [HI]^2 / ([H2] * [I2]), we can substitute the equilibrium concentrations:

33 = (2x)^2 / [(0.0600 - x)(0.0300 - x)]

Now, we can solve for "x" and find the equilibrium concentrations.

33 = (4x^2) / [(0.0600 - x)(0.0300 - x)]

Multiplying both sides by (0.0600 - x)(0.0300 - x):

33(0.0600 - x)(0.0300 - x) = 4x^2

0.198 - 0.09x - 0.099x + 0.03x^2 = 4x^2

0.03x^2 + 4x^2 - 0.189x + 0.198 = 0

Combining like terms:

4.03x^2 - 0.189x + 0.198 = 0

Using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

a = 4.03, b = -0.189, c = 0.198

Calculating the discriminant:

D = b^2 - 4ac = (-0.189)^2 - 4(4.03)(0.198) = 0.001236

Since the discriminant is positive, we have two real and distinct roots.

Using the quadratic formula:

x = (-(-0.189) ± √(0.189^2 - 4(4.03)(0.198))) / (2(4.03))

x = (0.189 ± √(0.001216024)) / (8.06)

x ≈ 0.01866 or x ≈ -0.000808

Since the negative value for "x" is not physically meaningful (concentration cannot be negative), we discard it.

Therefore, the equilibrium concentration of HI is approximately 0.0373 M (0.01866 M × 2).

Substituting the value of "x" into the expressions for [H2] and [I2]:

[H2] = 0.0600 - x = 0.0600 - 0.01866 ≈ 0.04134 M

[I2] = 0.0300 - x = 0.0300 - 0.01866 ≈ 0.01134 M

Thus, the equilibrium concentrations are approximately:

[H2] ≈ 0.0413 M
[I2] ≈ 0.0113 M
[HI] ≈ 0.0373 M