How to find the equilibrium "k"

N2(g) + 3H2(g) ⇌ 2NH3(g)

In a given experiment, 1.84 moles of N2(g) and 4.52 moles of H2(g) were placed in a 4.63 L vessel.

[N2] (M) [H2] (M) [NH3] (M)
Initial 0.397 0.976 0
Change -x -3x +2x
Equilibrium 0.397-x 0.976-3x 2x

At a particular temperature, the equilibrium concentration of NH3 is found to be 0.424 M.
Given this information, determine the numerical value of K for this reaction, at this temperature?

(IVE TRIED A MILLION TIMES AND FOR SOME REASON CANT GET IT)

0.424 M = 2x
x = 0.212

[N2] = 0.397 - 0.212 = .185 M
[H2] = 0.976 - 3(0.212) = 0.340 M

K = [NH3]^2 / [N2][H2]^3 = (0.424 M)^2 / {(0.185 M)(0.340 M)^3} = 24.7 1/M^2

Ke = {NH3} ^2 / {N2} {H2}^3
Find the molarities. N2 = 1.84 / 4.63 = 0.397 M H2 = 4.52 / 4.36 = 1.04 M NH3 is 0.424M
Fill in the data. Ke = ( 4.2x10^-1)^2 / ( 3.87x10^-1) (1.04)^3
Ke = 1.01x10^-1 ??