N_2O_4(g) -> 2NO_2(g)
K_c = ([NO_2(g)]^2)/([N_2O_4(g)])
We need to find the equilibrium concentrations for N_2O_4(g) " and " 2NO_2(g); therefore we can use color(blue)("ICE table") ( I don't know how to format it into a table!):
Remember, the volume is equal to 1.000 L and therefore,
[N_2O_4(g)]_0 = n/V = (0.0300 mol)/(1.000 L) = 0.0300M
at equilibrium: [N_2O_4(g)] = n/V = (0.0236 mol)/(1.000 L) = 0.0236M
N_2O_4(g) " "->" " 2NO_2(g)
Initial: " " " " " ""0.0300M" " " " " " " " " " ""0M"
Change: " " " " " "color(blue)(- x)"M" " " " " " " " " " "color(blue)(+2x)"M"
Equilibrium: " " "0.0236M" " "" " " " " " " "+2x"M"
Therefore, at equilibrium, [N_2O_4(g)] = (0.0300 - x)M = 0.0236M
=> x= 0.0300 - 0.0236 = 0.0064M
[NO_2(g)] = +2xM = 2xx0.0064M = 0.0128M
K_c = ([NO_2(g)]^2)/([N_2O_4(g)]) = ((0.0128)^2)/0.0236 = 0.00694 =6.94xx10^(-3)
