Textbook 5.39

[Amanda Nguyen dis 2E]

In an experiment, 0.020 mol NO2 was introduced into a flask of volume 1.00 L and the reaction 2NO2 --> N2O4
was allowed to come to equilibrium at 298 K. (a) Using information in Table 5G.2, calculate the equilibrium concentrations of the two gases. (b) The volume of the flask is reduced to half its original value. Calculate the new equilibrium concentrations of the gases.

I found K to be 1.6x10^2 by finding the reciprocal of 6.1 x 10^-3. Using that and an ICE table I got an equation of

1.6 x 10^2 = x / (.02 - 2x)^2

However, when I solve for x I keep on getting the wrong answer. Is this equation the correct one to solve for x, or did I make a mistake before I got the equation?

[Shria G 2D]

Hi,
I had that same equation for part a when I solved it. After simplifying it, my quadratic equation was (6.4x10^2)x^2 - 13.8x + 0.064 = 0 and then I used the quadratic formula so I got 0.015 and 0.0068 as x but only 0.0068 would make sense for this problem. Does that help?