I have a question on this question would I forgo the sqaure root on this on because there is no sqaure except on the products side which is 2HI? I am lost because I set up ICE then got to: 54=(4X-0.09)^2/x*x

Kc=[HI]^2/[x][x]

At a particular temperature, Kc = 54 for the reaction H2(g) + I2 (g) <=> 2 HI(g).
One mole of HI is placed in a 3.0-L container. What would be the equilibrium concentration of HI?
Choose one answer.

a. 0.035

b. 0.071

c. 0.29

d. 0.30

e. 0.33

CC--I went back this afternoon and worked that problem out showing you where you got off the track and finishing up except at the very end. Here is a link that will get you there. Post back here if you have any trouble. Check my arithmetic on that work.

http://www.jiskha.com/display.cgi?id=1269374970

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

54=(2x-0.3)^2/[X][X]

THEN I GET:

54=4X-0.09/[X]^2

SO SQAURE ROOT IT THEN I GET

7.348*(x-x) =2X-0.3
7.4348X-7.348X=2X-0.3

tHEN i GET STUCK BECAUSE i DON'T KNWO IF THE VARIABLES x-x is suppose to be here? because when I calculate my answers don't add up to any shown so I know I made a mistake somewhere.

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

54=(2x-0.3)^2/[X][X]

THEN I GET:

54=4X-0.09/[X]^2

SO SQAURE ROOT IT THEN I GET

7.348*(x-x) =2X-0.3
7.4348X-7.348X=2X-0.3

tHEN i GET STUCK BECAUSE i DON'T KNWO IF THE VARIABLES x-x is suppose to be here? because when I calculate my answers don't add up to any shown so I know I made a mistake somewhere.

To solve this problem, we need to use the concept of equilibrium and the expression for the equilibrium constant. We'll start by setting up an ICE table to determine the equilibrium concentrations.

Given:
- Initial concentration of HI (H2I) = 1 mole
- Volume of the container = 3.0 L

Let's assume the equilibrium concentration of HI is x, which means the concentrations of H2 and I2 will also be x because the stoichiometric coefficient for all three species is 1.

The ICE table is as follows:

H2(g) + I2 (g) <=> 2 HI(g)
Initial: x x 1
Change: -x -x +2x
Equilibrium: 1-x 1-x 1 + 2x

Now, let's substitute these equilibrium concentrations into the Kc expression:

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

Kc = (1 + 2x)^2 / (1 - x)(1 - x)

Since we are given that Kc = 54, we can write the equation:

54 = (1 + 2x)^2 / (1 - x)(1 - x)

Now, let's solve this equation.

Take the square root of both sides:

√54 = 1 + 2x / (1 - x)(1 - x)

Simplify:

√54 = (1 + 2x) / (1 - x)^2

Now, let's cross-multiply:

√54 * (1 - x)^2 = 1 + 2x

Expand and simplify:

54(1 - 2x + x^2) = 1 + 2x

54 - 108x + 54x^2 = 1 + 2x

Rearrange and simplify:

54x^2 - 110x + 53 = 0

Now, we have a quadratic equation. To solve for x, we can either factor or use the quadratic formula. In this case, factoring is more difficult, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values from the quadratic equation:

x = (-(-110) ± √((-110)^2 - 4(54)(53))) / (2(54))

Simplifying:

x = (110 ± √(12100 - 11664)) / 108

x = (110 ± √436) / 108

Now, we can calculate the two possible values of x:

x = (110 + √436) / 108 ≈ 0.331

or

x = (110 - √436) / 108 ≈ 0.162

Since we are looking for the equilibrium concentration of HI, we choose the positive value:

x ≈ 0.331

Therefore, the equilibrium concentration of HI is approximately 0.331 M.

The closest answer choice is e. 0.33, so that would be the correct option.