Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C if the initial concentrations are [N2O4]= 0.0210 M and [NO2]= 0.0340 M. The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

1) Calculate the equilibrium concentration of N2O4.

2) Calculate the equilibrium concentration of NO2.

I have worked through this problem several times and keep getting the wrong answer. I think I'm making a math error along the way. Could you show as much work as possible so I can see where I'm messing up? Thank you!

Answer 1

N2O4 < -----> 2NO2

Qc = [NO2]^2 / [N2O4]
     = (0.034)^2 / (0.0210)
     =0.055
Since Qc>Kc, reaction will shift to left

Let at equilibrium:
[N2O4] = 0.0210 + x
[NO2] = 0.0340 -2x

use:
Kc = [NO2]^2 / [N2O4]
4.64*10^-3 = (0.034 -2x )^2 / (0.0210+x)
4.64*10^-3*(0.0210+x) = (0.034-2x)^2
9.744*10^-5 + 4.64*10^-3*x = 1.156*10^-3 - 0.136*x + 4x^2
4x^2 - 0.14064 x +1.059*10^-3 = 0

Solving above equation we get,
x = 0.024 M
and x = 0.011 M

x can't ne 0.024 M because in that case [NO2] = 0.0340-2x will become negative
so,
x = 0.011 M

[N2O4] = 0.0210 + x = 0.0210 + 0.011 = 0.032 M
[NO2] = 0.0340 -2x = 0.0340 - 2x = 0.0340 - 2*0.011 = 0.012 M