chemistry
posted by Anonymous .
Calculate each of these equilibrium concentrations based on the reaction below.
2 NO(g) + O2(g) 2 NO2(g) K = 1.71 1012
(a) [NO] = 0.0048 M; [O2] = 0.000057 M; [NO2] = ?
(b) [NO] = 0.0026 M; [O2] = 0.000023 M; [NO2] = ?

K= [NO2]^2/([NO2]^2 * [O2])
conc NO2= sqrt(K*NO^2 * O2)
= sqrt(1.71E12*.0048^2*.000057)
= sqrt( 2250) moles/liter
check my work, it is easy to make a math error typo on a keyboard.
This is based on the very high forward reaction K you gave. I don't remember this reaction being anywhere close to this K value, if I guessed, I would guess 1E5 for K or thereabouts at room temp. 
I just checked it, so based on that formula, the other solution should come out to be something like....
sqrt(1.71E12*0.0026^2*0.000023)
=265 moles?