what is the equilibrium concentration of NO2 in mol/L?

Question

The reaction

N2O4(g) ↔ 2NO2(g)

is an equilibrium reaction at some temperature with an equilibrium constant K = 4.00 x 10−7.
If the intial condition is pure N2O4(g) at a concentration of 2.20 mol/L what is the equilibrium concentration of NO2 in mol/L ?

Answer 1

N2O4 <--> 2NO2

K = [NO2]^2/[N2O4]

x= equib conc of NO2

2.20 - x = [N2O4] at equilib

so 4.00 *10^-7 = x^2/(2.2-x) ==> x^2 + 8.8*10-7x - 8.8*10-7=0

x = -b +- sqrt(b2 - 4ac)/2a <==> omit the imaginary so only +
= (-8.8 *10^-7 + sqrt(4*8.8*10-7))/2 = 0.0009 M

note in sqrt I didn't need to calculate b^2 since it's too small compared to 4ac, in other words b^2-4ac = -4ac

Answer 2

N2O4(g) = 2NO2(g) enable "X" equivalent the concentrationt of N2O4 that reacts. Then "2X" will equivalent the concentration of NO2 that varieties. by using fact the fee of ok is very small, we are in a position to assume that the quantity that reacts is negligible (and could be ignored) from the equilibrium expression. ok= [NO2]^2 / [N2O4] 4 X 10^-7 = [2X]^2 / a million.seventy two 6.88 X10^-7 = 4 X^2 a million.seventy two X 10^-7 = X^2 X= (a million.seventy two X10-7)^0.5 X= 4.15 X 10^-4 answer: The concentration of NO2 replaced into represented as "2X", subsequently [NO2] = (2)(4.15 X10^-4) = 8.30 X 10^-4

Answer 3

uhhhhh 5. Just a guess.