Question

Equilibrium Help?? Medals and favorite, multiple choice.
At a high temperature, the equilibrium constant for the decomposition of hydrogen iodide is 65.0. If the initial concentration of HI is 1.60 M, what is the concentration of hydrogen at equilibrium?
2HI(g) <-> H2(g) + I2(g)

A. 1.42 M
B. .802 M
C. .240 M
D. .753 M

Thanks(:

Answer 1

Use the ICE box approach

Answer 2

Right, which takes the initial concentrations + the change which you reach the equilibrium....So is it saying that the equilibrium amount for hydrogen iodide is 65.0 or is it saying that K(eq) = 65? The initial concentrations for the others is 0, correct?

Answer 3

It's telling you the K is 65. The starting amount of HI is 1.65M, the starting concentrations of the products is zero

Answer 4

That's what I thought. I really don't know how to go find the change or the rest of the equilibrium amounts of the products...

Answer 5

You know the starting amounts. The reactant loses "2x" and each product gains "x", because of the stoichiometric coefficients. The equilibrium amounts are "1.65-2x", "x", and "x". Insert them into the Keq expression and solve for x

Answer 6

So k= (x)(x)/ (1.65-2x)^2
square root both 65= x/ 1.65-2x
65= 1/1.65-x
x=1.63??
Which is pretty close to answer A, so is that it?

Answer 7

Did you forget to square root the 65?

Answer 8

Oh yeah whoops!

Answer 9

I got .76..which corresponds with D. Is this correct?

Answer 10

Not having done the math, that sounds about right

Answer 11

Alright, thankyou so much for your help!(:

Answer 12

YVW