[zmasterflex]

Consider the following gas-phase equilibrium;
H2(g) + I2(g) <--> 2HI(g)
At a certain temperature, the equilibrium constant Kc is 4.0. Starting with equimolar quantities of H2 and I2 and no HI, when equilibrium was established, .2 moles of HI was present. How much H2 was used to start the reaction?

Initial concentration of H2 and I2 = "x" and final concentration of HI =.2

[.2]^2 / x^2= 4.0 gives x = .1 mole. The correct answer is .2 moles. The mistake I'm making is that I need to put something along with "x" in the denominator, "x" minus something. What? thanks

[Corribus]

Are you familiar with an ICE table?

[zmasterflex]

Yes. I tried that also but I get the same result.
Initial moles of H2 and I2 are "x", they both lose an amount "Y" and end up with an amount of "X-Y". The HI starts with zero and gains "2Y" and ends with .2 moles. Therefore "Y" = .1 and the equation reads; [.2]^2/[x-.1]^2 = 4.0

Solving we get .04 = 4(x^2 - .2x + .01) ---> x^2 - .2x + .01 = .01
Therefore x^2-.2x=0 and X = .2 which is the wrong answer. Unless the answer key has a mistake which happens sometimes..

[Shipwreck]

Initial concentration of H2 and I2 = "x" and final concentration of HI =.2

[.2]^2 / x^2= 4.0 gives x = .1 mole.

If the initial concentrations of H₂ and I₂ are x, how come you are using that variable in the equation for the equilibrium constant?

ICE table this bad boy one more time.

[Corribus]

In your original post you said 0.2 is the correct answer according to your guide. By your first method you also got 0.1, not 0.2, so you didn't appear to get the same result for your two methods.

[zmasterflex]

Yes. You are correct, .2 is the correct answer which is the result of using the ICE table. My mistake. I have the correct answer now.