Problem: Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equationN2O4(g) ⇌ 2NO2(g).If at equilibrium N2O4 is 28.0% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions?A) 0.44B) 2.3C) 0.31D) 0.78E) 0.11

We’re being asked to calculate the equilibrium constant of HA if a 0.0538 M solution is 3.57% ionized.

Recall that the percent ionization is given by:

$\boxed{\mathbf{\%}\mathbf{ }\mathbf{dissociation}\mathbf{ }\mathbf{=}\frac{\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{initial}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{final}}}{\mathbf{[}\mathbf{A}{\mathbf{]}}_{\mathbf{initial}}}\mathbf{\times }\mathbf{100}}$

The dissociation is as follows:

N₂O₄(g) ⇌ 2NO₂(g)

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and K_a expression.

The K_c expression for is:

$\boxed{{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\mathbf{=}\frac{{\mathbf{\left[}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right]}}}$

Note that each concentration is raised by the stoichiometric coefficient: [NO₂] is raised to 2 and [N₂O₄] is raised to 1.