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All Message Boards : Equilibruim Constant
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 Message 1 of 2 in Discussion 
From: MSN NicknameAlbert1145  (Original Message)Sent: 8/22/2008 9:43 AM
Sorry Steve for asking dumb questions. My brain is somewhere else these couple of days.

At a particular temperature, Kc = 54.3 for the reaction:
H2 + I2 <-> 2HI
If the initial concentrations of both H2 and I2 are 0.5M, what are the equilibrium concentrations of each gas involved?

My lecturer said to use the ICE table, so I figured
H2 I2 2HI
Initial 0.5 0.5 0.0
Change -x -x +2x
Equi. (0.5-x) (0.5-x) 2x

Kc = [HI]^2 / [H2][I2]
54.3 = (2x)^2 / (0.5-x)(0.5-x)

Is it right so far? How do I go on from here?


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 8/22/2008 3:12 PM
That is correct.  Now you just have to solve for x.  Often you must use the quadratic formula to do that, but in this case you just have to take the square root of both sides.

54.3    =              (2x)2             =       (2x)2      
                 (0.5 –x)(0.5 �?x)            (0.5 �?x)2

7.36885   =         2x     
                     0.5 �?x

2x   =   7.36885 (0.5 �?x)

Once you have x, you can get the equilibrium concentrations:

[H2]eq   =   0.5 �?x
[I2]eq   =   0.5 �?x
[HI]eq   =   2x

Once you get these values, do a check by plugging them back into the equilibrium constant expression, Kc  =  [HI]eq2 / [H2]eq[I2]eq, and see if you get 54.3 or something close to that.