Equilibrium Chemistry Homework: Find [HI] at 400°C

AI Thread Summary
At 400°C, the equilibrium constant Kc for the reaction H2 + I2 = 2HI is 64. The initial concentrations of H2 and I2 are both 0.75 M in a 4.0 L flask. The user attempts to calculate the equilibrium concentration of HI but arrives at an incorrect value of 6.0 M. To find the correct concentration, it's important to account for the stoichiometry of the reaction, which indicates that the concentrations of H2 and I2 will decrease as HI is produced. A balanced equation and stoichiometric ratios should be used to determine the final concentrations at equilibrium.
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Homework Statement


At 400 degrees Celcius, 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 Celcius.


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!
 
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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.
 
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