Question

At high temperatures, dinitrogen tetroxide gas decomposes to nitrogen dioxide gas. At 500 ^oC, a sealed vessel containing some dinitrogen tetroxide gas was allowed to reach equilibrium. At equilibrium, the mixture contained 3.92 M dinitrogen tetroxide gas. Calculate the equilibrium concentration of nitrogen dioxide if K_C at this temperature is 0.559.

Answer 1

The decomposition reaction of dinitrogen tetroxide gas to nitrogen dioxide gas is

N₂O_4(g) ----------> 2NO_2(g)  

i.e form the decomposition of 1 mole of N₂O_4(g) , we get 2 moles of NO_2(g)  

Now given-

the equilibrium constant value(Kc) for the reaction = 0.559.

Now we know Equilibrium constant is the ratio between product of concentration of products, each raised to the power of their number of moles involved to product of concentration of reactants, each raised to the power of their number of moles involved, at equilibrium.

i.e for the above reaction-

Kc = [NO_2(g)]² / [N₂O_4(g)]

Now lets say initially we have x+3.92 M of [N₂O_4(g) present. And xM of it reacted to from 2x M of NO_2(g) . So Now to calculate these concentration values at equilibrium lets form the ICE table-

Reaction	[N2O4(g)] ------>	2[NO2(g) ]
Initial	x+3.92 M	0
Change	-x	+2x
Equilibrium	3.92 M	2x M

Now putting these values in the formula-

Kc = [NO_2(g)]² / [N₂O_4(g)]

0.559 = [2x]² / [3.92]

0.559 * [3.92] = 4x²  

2.191 = 4x²  

2.191/4 = x²  

x² = 0.54775‬

x = 0.74

That means at equilibrium, we have

concentration of nitrogen dioxide = [NO_2(g) ] = 2x = 2 * 0.74 = 1.48 M