Chemical Equilibrium

$W+X\rightleftharpoons Y+Z$Initial concentrations:W starts at $2 x, X$ starts at $x$, and both Y and Z start at 0 .At equilibrium:Let the amount of Y and Z formed be $y$. So, at equilibrium:$\;W=2 x-y$$\;X=x-y$$\;Y=y$$\;Z=y$Relationship between Y and X at equilibrium:At equilibrium, the amount of Y is four times the amount of X . This gives us:$[Y]_{e q}=4[X]_{e q}$So, $y=4(x-y)$Solve for $y$ :$y=4 x-4 y$$y+4 y=4 x$$5 y=4 x$$y=\;\frac{4}{5}x$Write $K_c$ expression and substitute values:The equilibrium constant is:$K_c=\;\frac{[Y][Z]}{[W][X]}$Plug in the equilibrium values:$K_c=\;\frac{\left(\frac{4}{5}x\right)\left(\frac{4}{5}x\right)}{\left(2x-\;\frac{4}{5}x\right)\left(x-\;\frac{4}{5}x\right)}$Simplify:Numerator: $\left(\;\frac{4}{5}x\right)^2=\;\frac{16}{25}x^2$Denominator:$\;2 x-\;\frac{4}{5}x=\;\frac{10 x-4 x}{5}=\;\frac{6 x}{5}$$\;x-\;\frac{4}{5}x=\;\frac{5 x-4 x}{5}=\;\frac{x}{5}$So denominator: $\;\frac{6 x}{5} \cdot \;\frac{x}{5}=\;\frac{6 x^2}{25}$Therefore, $K_c=\;\frac{\frac{16x^2}{25}}{\frac{6x^2}{25}}=\;\frac{16}{6}=2.666$