H2+I2->2HI

If 3 moles of H2, I2, and HI are in a 3L flask, what will be the equilibrium amount of each. Kc=5

The answer is H2= I2= .71M, HI= 1.584M(given by teacher)

Just need to know how to do the work.

Do you mean the problem to sound the way it does. Are these the EQUILIBRIUM concentrations? I don't think so. Otherwise, K = 1

(H2) = 3 mols/3 L = ??
(I2) = same
(HI) = same
Write the equilibrium constant expression, plug in the numbers, and solve

I think you meant to say that 3 mols HI, H2, and I2 were placed in a 3 L container and allowed to come to equilibrium. Calculate the concentrations of each.

Write Kc = (HI)^2/(H2)(I2)
Concentrations at the beginning are 1 M for each.
At equilibrium,
(HI) = 1+2x
(H2) = 1-x
(I2) = 1-x
Plug into Kc expression and solve the quadratic for x, then evaluate 1-x, 1-x, and 1+2x.
Post your work if you get stuck.

To determine the equilibrium amounts of H2, I2, and HI in the given reaction H2 + I2 -> 2HI, we first need to understand how to use the equilibrium expression and the given value of Kc.

The equilibrium expression for this reaction is: Kc = [HI]^2 / ([H2] * [I2])

Given that Kc = 5, and the initial number of moles of each reactant and product are 3 moles, we can set up the following equation:

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

Let's represent the equilibrium amount of HI as x, and the equilibrium amounts of H2 and I2 as y. Therefore, we have the following:

[H2] = 3 - y
[I2] = 3 - y
[HI] = x

Now, substitute these values into the equilibrium expression:

5 = (x^2) / ((3 - y) * (3 - y))

If we rearrange the equation, we get:

5 * ((3 - y) * (3 - y)) = x^2

15 - 10y + 2y^2 = x^2 (Equation 1)

Now, we need to account for the stoichiometry of the reaction: 1 mole of H2 reacts with 1 mole of I2 to form 2 moles of HI. Since we started with 3 moles of H2 and I2, and we know that y moles of H2 are reacted, the moles of I2 reacted will also be y. Therefore, the moles of HI formed will be 2y.

The total moles of HI at equilibrium will be the sum of HI formed and the initial moles of HI, which is 3 moles. So, we can say:

[HI] = 2y + 3

Now, substitute this expression for [HI] back into Equation 1:

15 - 10y + 2y^2 = (2y + 3)^2

Solve this quadratic equation to find the value of y (the equilibrium amount of H2 and I2).

Once you find the value of y, substitute it back into the equations [H2] = 3 - y and [I2] = 3 - y to get the equilibrium amounts of H2 and I2, respectively.

Similarly, substitute the value of y into the equation [HI] = 2y + 3 to determine the equilibrium amount of HI.

By plugging in the correct values into these equations, you should obtain the answer given by your teacher: H2 = I2 = 0.71M and HI = 1.584M.