Finding equilibrium [HI] for H2 + I2 reaction at 400°C

[ColinTI89]

Homework Statement

At 400 degrees celsius, Kc=64 for the equilibrium H2 + I2 = 2HI. If 3.0 mol H2 and 3.0 mol of I2 are introduced into an empty 4.0 L flask, find the equilibrium concentration of HI at 400 degrees celsius.

Homework Equations

[HI]^2
--------

[I2]

The Attempt at a Solution

The answer given in the answer key is 1.2 M, which I am not getting. Since molarity is defined as mol/Liter, I calculate the molarity of the two given substances which is both 0.75 M. I set up the equation like this (sorry for the crappy quality):

...[HI]^2
64 = ----------
...[0.75] [0.75]

Multiplying 0.75 x 0.75 gives 0.5625 on the bottom. I then multiply both sides by 0.5625 to isolate [HI]^2. It now looks like this:

36=[HI]^2.

When I square root both sides I get 6.0, which doesn't sound correct anyways. Can anyone help? P.S. It's my first post on PhysicsForums!

 

[PhaseShifter]

Keep in mind that the same reaction that produces HI also uses up both I2 and H2. Those elements may initially have concentrations of 0.75 mol/L, but this will decrease as they are consumed by the reaction.

Try writing a balanced equation for the reaction, and use the stoichiometric ratios to determine the amount of each element left after you've produced x moles of HI.