Question

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

Answer 1

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

Answer 2

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.

Answer 3

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.