Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C if the initial concentrations are [N2O4]= 0.0130 M and [NO2]= 0.0300 M. The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C. I've worked through this problem a few times and still don't get the right answer. Can anyone show me the work??

Answer 1

N28469) 2248219) kc = 4.64x10-3 initial concentration of CM204] = 0.0130 m [MO2] = 0.0380m 2 Sm using ICE table N20419) Z 2N82 (9) 0.0300 0.0130 Ilm) c(m) -x +2x Elm) (0.0130-x) 10.0300 +2x) ke= = 4.64 x 10-3 [mron] (0.0300 + 2x)2 (0.0130-x) = 4.64X10-3 (0.0300) + (2x)?+210.0300) (2x) = 4.64x10-3 (0.0130-x) cs gelied uxht 0.12 X = 6.032x10-5-4.64X10-3x | CamScanner

ux² +0.12% + 4.64 X10-3x + 9x10-4_6.032x10-5=0 4x² +0.124x + 8.40 X10-9=o x² + 0.031 +2.1x18" =o it is a quadratic equation (ax²+bx+c=0) a=1, b = 0.031, c= 2.1810-4 -bt J 6²-yac 29 4 -0.031 + $10.031)? 4(1)(20141004) 2x 1 x= -0.031 + 0.011 -0.0lm 2 so, equilibrium concentration of U [Mzou] = (0.0130-x) = 0.0130 - (-0.01) = 0.0130 +0.01 [rrou] = 0.023m TMO - (0.0300+2x) = 0.0300 +2 (-000W = 0.0300 -0.02 Scanned with CamScanner [102] = 0.0lm