I'll show my explanation in steps for your understanding
1. First set balance the equation
#2A + 2B = 2AB#
2. Now write the equilibrium constant for the reaction
#color(blue)(K_c = ([AB])/([A] [B]))#
3. As #color(blue)(K_c)# is being determined the temperature matters and thus volume also matters.So the conversion of moles to molarity is necessary
#M = "moles"/L#
4. As the volume of the container is #"2dm"^3#
Conversionof #dm^3# to litres is necessary
#1dm^3 = 1L#
#therefore# #2dm^3 = 2L#
5. Calculate the molarities
Conc. of A = #"0.2moles"/(2L) = 0.1M#
Conc. of B = #"0.4mole"/(2L) = 0.2M#
Conc. of A at equilibrium #"0.1moles"/(2L) = 0.05M#
5. Set up an ICE table
#color(blue)("For problems related to equilibrium") #
#color(blue) ("concentrations,you need to set up an ICE table.")#
#color(white)(mmmmml)"2A" color(white)(lm)+ color(white)(mmmm)"2B"color(white)(llll) → color(white)(ml) 2AB#
#"I/mol:"color(white)(mll)0.1Mcolor(white)(mmmmml)0.2Mcolor(white)(mmmmml)0#
#"C/mol:"color(white)(m)-xcolor(white)(mmmmmm)-xcolor(white)(mmmm)+2x#
#"E/mol:"color(white)(lm)0.1M-xcolor(white)(mml)0.2M-xcolor(white)(lmmml)2x#
From the given information we know that the equilibrium amount of A is 0.05M. Thus #0.1M - x = 0.05M#
Solve for #x#
#0.1M - x = 0.05M#
#x = 0.1M-0.05M#
#x = 0.05M#
6. Solve for the equilibrium amounts of B and AB
#AB = 2x = 2 xx 0.05M = 0.1M#
#B = 0.2M - 0.05M = 0.15M#
And we know the equilibrium amounts of A
7. Now solve for #K_c#
#K_c = "products"/"reactants"#
#K_c = ([ 0.1M])/([0.05M][0.15M])#
#K_c =13.333#
If you want #color(blue)("three sig figs")#
#K_c = 13.4#
You can also calculate the #Q_c# to see to which direction will the reaction proceed
#color(blue)(Q_c = ([AB]^2)/([A]^2[B]^2))#
#Q_c = 177.777777778#
As #color(purple)(K_c < Q_c) #
Reaction will proceed towards left
