# Using an ICE Table, how do we calculate #K_c# for the following reaction?

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At #"373 K"# , #"0.1 mols"# of #N_2O_4# is heated in a one liter flask and at equilibrium the amount of nitrogen dioxide is #"0.12 mols"# .

#N_2O_4 rightleftharpoons 2NO_2#

At

##### 1 Answer

For starters, we (of course) assume *ideal gases* (these are both gases at room temperature).

Then, we recall the definition of the **concentration equilibrium constant** for a two-component equilibrium:

#K_c = ([B]^(nu_B))/([A]^(nu_A))# ,

for the equilibrium

Even though we are looking at two gases, we are given the **s** and the total vessel's **volume**, so we can still get initial and equilibrium concentrations as usual.

However, since the vessel is not going to change size, and ideal gases are assumed to fill the vessel completely and evenly, let us simply work in

Normally, we would write this:

#N_2O_4(g) rightleftharpoons 2NO_2(g)#

#"I"" "" "0.1" "" "" "0#

#"C"" "" "- x" "+2x#

#"E"" " 0.1 - x" "" "2x#

noting that the consumption of

But we know what

#K_c = ([NO_2]^(2))/([N_2O_4])#

#= ((2x)^2" mols"^cancel(2)"/" cancel("1") "L"^cancel(2))/((0.1 - x)cancel"mols""/" cancel"1 L")#

#= (4x^2)/(0.1 - x)# #"M"#

#= (4(0.06)^2)/(0.1 - 0.06)# #"M"#

However,

#color(blue)(K_c = 0.36)#