For the reaction at 430 degree C, H2 + I2 <=> 2HI Kc = 54.3. Initially 0.714 moles of H2 and 0.984 miles of I2 and 0.886 moles of HI are placed in a 2.40 L vessel. The equilibrium concentration of H2 and I2 are _____.

Ok, I tried this by changing the moles to M by dividing the moles by 2.40 liters, and subing into the Kc formula, but I did not get any of the choices listed for the multiple choice. What am I doing wrong?

You were right to change the mols to molarity by mol/2.4 = M.

H2 + I2 ==> 2HI
Set up an ICE chart.
initial:
(H2) = 0.714/2.4 =
(I2) = 0.984/2.4 =
(HI) = 0.886/2.4 =

Now do the reaction quotient =
(HI)^2/(H2)(I2) = (0.369)^2/(0.2975)(0.410) = about 1.1 or so which means at equilibrium the reaction will have gone to the right.

change:
(HI)= +2y
(H2) = -y
(I2)= -y

equilibrium:
(HI) = 0.369+2y
(H2) = 0.2975-y
(I2) = 0.41-y

Substitute equilibrium values into Keq and solve for y and add that back into the equilibrium set up to arrive at the final concns.

Well, it seems like you're having trouble finding the equilibrium concentrations of H2 and I2. Let's see if we can work our humorous magic to solve this problem!

First, we need to recall the equation for Kc, which includes the concentrations of the reactants and products. In this case, we have:

Kc = [HI]² / ([H2] x [I2])

Now, to find the equilibrium concentrations of H2 and I2, we need to assume that x mol of H2 and I2 react to produce 2x mol of HI. This means that the equilibrium concentration of HI will be (0.886 + 2x) M.

Now, if we look at the initial concentrations, we have 0.714 M of H2 and 0.984 M of I2.

Using the stoichiometry of the reaction, we know that x mol of H2 and I2 will react to form 2x mol of HI. Therefore, the equilibrium concentration of H2 will be (0.714 - x) M, and the equilibrium concentration of I2 will be (0.984 - x) M.

Now, let's plug these values into the equilibrium expression for Kc:

Kc = [(0.886 + 2x)²] / ([(0.714 - x)] x [(0.984 - x)])

Solving this equation will give you the value of x, which represents the change in moles of H2 and I2 at equilibrium. Once you find x, you can substitute it back into the expressions for the equilibrium concentrations of H2 and I2 to get your final answers.

Remember, chemistry problems can sometimes feel like a circus, but with a little humor and perseverance, you'll be able to solve them!

To find the equilibrium concentrations of H2 and I2, we can use the concept of the equilibrium constant (Kc) along with the given initial amounts of the reactants and products.

First, let's convert the initial moles of H2, I2, and HI into molar concentrations (M) by dividing them by the volume of the vessel (2.40 L):

[H2] = 0.714 moles / 2.40 L
[I2] = 0.984 moles / 2.40 L
[HI] = 0.886 moles / 2.40 L

Next, let's substitute these concentrations into the expression for Kc:

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

Now, let's rearrange the equation to solve for the equilibrium concentrations of H2 and I2:

[HI]^2 = Kc * [H2] * [I2]

The square root of the right side of this equation will give us the concentration of HI at equilibrium. Since we know the initial concentration of HI is 0.886 moles / 2.40 L, we can solve for [H2] and [I2]:

[H2] = [HI]^(1/2) * ([I2] / (Kc * [HI]))

[I2] = [HI]^(1/2) * ([H2] / (Kc * [HI]))

Substituting the given values:

[H2] = (0.886 moles / 2.40 L)^(1/2) * (0.984 moles / 2.40 L) / (54.3 * 0.886 moles / 2.40 L)
[I2] = (0.886 moles / 2.40 L)^(1/2) * (0.714 moles / 2.40 L) / (54.3 * 0.886 moles / 2.40 L)

Evaluating these expressions will give you the equilibrium concentrations of H2 and I2.