First, we must write the balanced chemical equation for the process:
I₂ (g) ⇌ 2I (g)
The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.
The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.
So, the equilibrium constant for the above reaction is,
Kc = [tex]\frac{[I]^{2} }{[I_{2} ]}[/tex]
The initial concentration of I₂ is:
[I₂]₀ = [tex]\frac{0.0456 mol}{2.28 L}[/tex] → [I₂]₀ = 0.0200 M
To facilitate the visualization of the reaction, we can write it in the following way,
I₂ ⇌ 2I
0.0200 M - initial
0.0200 M - x 2x equilibrium
As the reaction doesn't go to completion, at equilibrium an amount x dissociates of the initial concentration of I₂ to produce 2x of I (because of the stequiometric coefficients).
At equilibrium [I₂] = 0.0200 M - x and [I] = 2x. By substituting this in the above equation for kc we will be able to find the value of x and then calculate the concentrations of both gases in equilibrium.
If kc = 3.80ₓ10⁻⁵ we have,
Kc = [tex]\frac{[I]^{2} }{[I_{2} ]} = \frac{(2x)^{2} }{0.0200 - x} = 3.80x10^{-5}[/tex]
Since the dissociation constant has such a low value, we can assume that the degree of dissociation of I₂ is very low, therefore x ≈ 0 and,
[tex]\frac{(2x)^{2} }{0.0200 - x} = \frac{4x^{2} }{0.0200} = 3.80x10^{-5}[/tex]
→ x = 4.36ₓ10⁻⁴ M
Then, the concentrations of the gases I₂ and I at equilibrium are,
[I₂] = 0.0200 M - x = 0.0200 M - 4.36ₓ10⁻⁴ M → [I₂] = 0.0196 M
[I] = 2x = 2 x 4.36ₓ10⁻⁴ M → [I] = 8.72ₓ10⁻⁴ M
