#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) #
