CHEM!
posted by Linda on .
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

0.2619

1.15=x/[2{0.5x}]^2 u get x=0.9546 and x=0.2619 corrt anxw ix last 1