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April 20, 2014

April 20, 2014

Posted by **Linda** on Wednesday, May 2, 2007 at 5:49pm.

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/(0.5-2x)^2

O.5-2x~ ).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 dont agree that you can assume .5-2x is approximately .5. Your final answer shows that it is not.

1.15 = x/(0.5-2x)^2

(0.5-2x)^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

0.25-2.696x + 4x^2 = 0

and I would also like to understand how did you get 0.696?

I dont know how I got .696 . It should have been 1/1.15=.869

Now that you have the quadratic,

0.25-2.896x + 4x^2 = 0

Use the quadratic equation

x= (2.896 +- sqrt (8.23-4) )/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

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