Posted by Linda on Wednesday, May 2, 2007 at 5:49pm.
Well I almost understand this question but still got stuck, I will show you the question and my answer so I need someone to tell me if I did it correctly or not and how I should fix it.
at 55degrees the K for the reaction: 2NO2(g)<=>N2O4 is 1.15
calc the concentration of N2O4 present in equilibrium with 0.50 mole of NO2
I first used the ICE chart then calculated
1.15 = x/(0.52x)^2
O.52x~ ).5
1.15 = x/(0.5)^2
1.15(0.25) = x
x= 0.2875
2(0.2875)  0.575
Therefore N2O4 = 0.575M?
I don't agree that you can assume .52x is approximately .5. Your final answer shows that it is not.
1.15 = x/(0.52x)^2
(0.52x)^2= x/1.15
.25 2x +4x^2= .696x
put the terms together, and use the quadratic equation for x.
then would my answer look like this
0.252.696x + 4x^2 = 0
and I would also like to understand how did you get 0.696?
I don't know how I got .696 . It should have been 1/1.15=.869
Now that you have the quadratic,
0.252.896x + 4x^2 = 0
Use the quadratic equation
x= (2.896 + sqrt (8.234) )/8
x= .618
or x=.104
Now obviously, the first is impossible (why?), so the concentration the product is .104, and the concentration of the reactant is .5.208
check my calc work

CHEM!  godluck maxawe, Wednesday, April 1, 2015 at 6:35am
0.2619

CHEM!  godluck maxawe, Wednesday, April 1, 2015 at 6:45am
1.15=x/[2{0.5x}]^2 u get x=0.9546 and x=0.2619 corrt anxw ix last 1