Bookwork 5.39

[Meagan Kimbrell 1I]

5.39) In an experiment, 0.020 mol NO2 was introduced into a flask of volume 1.00 L and the reaction 2NO2(g) <--> N2O4(g) 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 am struggling with part (a). I have set up K = x/(0.020-x)^2= 6.1 x 10^-3. Then, I began solving for x with x = (6.1 x 10^-3)x^2 - 1.0002x + 2.44 x 10^-6. But, I am using the quadratic equation to find x, but I cannot get the correct answer. I am guessing I might have the wrong equation for x. Any help would be greatly appreciated!

[Jamie Lin 1K]

I think you left out the stoichiometric coefficient in your ice table. The K expression should be K=x/(0.02-2x)^2.