Untitled - Kelly Walsh High School

Chemical Equilibria 215
Quick Tip
If we use our assumption in the equilibrium expression, it changes as indicated
below:
K 3.76 105 [4 x2 [2 x] ]
[0.500]
2
[2 x]
[0.500]
2
[I]
[0.500 x]
2
[I2 ]
We can now rearrange this equation and determine the value of x:
x 2.1679 103 3.76 10
(unrounded)
Å
5 (0.500)
4
To finish the problem, we must enter this value into the bottom line of our table:
I2 I
Initial 0.500 0
Change x 2 x
Equilibrium 0.500 x 2 x
Equilibrium 0.500 2.1679 103 2 (2.1679 103 )
The equilibrium concentration of I 2 0.498 M and the equilibrium concentration
of I is 4.34 10 3 M. At this point you should apply whatever method your
instructor wants you to use to determine the validity of the assumption.
If you enter your answers into the equilibrium expression, the result should be
near K.
If you do not or cannot use the assumption method, then it is necessary to do
the problem the “long way.” We will go back to this point.
K 3.76 105 [2 x] 2
[I]
[0.500 x]
2
[I2 ]
If we do not neglect the x in the denominator, we must rearrange this equation to:
4 x2 (0.500 x)(3.76 105 )
We rearrange this to:
4 x 2 3.76 10 5 x 1.88 10 5 0