The hydrogen gas produced is collected over water at 25.0 oC. The volume of the gas is 7.80 L, and the total pressure in the gas
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First, we must write the <u>balanced chemical equation</u> for the process:
I₂ (g) ⇌ 2I (g)
The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because <u>the concentrations of the reactants and products remain constant over time</u>. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.
<u>The equilibrium constant (Kc) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products.</u> Its value in a chemical reaction <u>depends on the temperature</u>, so it must always be specified.
So, the <u>equilibrium constant </u>for the above reaction is,
Kc =
The initial concentration of I₂ is:
[I₂]₀ = → [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 - <em>initial</em>
0.0200 M - x 2x <em>equilibrium</em>
As the reaction doesn't go to completion, at equilibrium an amount <em>x</em> dissociates of the initial concentration of I₂ to produce <em>2x</em> of I (because of the stequiometric coefficients).
At equilibrium [I₂] = 0.0200 M - x and [I] = 2x. <u>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.</u>
If kc = 3.80ₓ10⁻⁵ we have,
Kc =
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,
→ <u>x = 4.36ₓ10⁻⁴ M</u>
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