Question

The dissociation of molecular iodine into iodine atoms is represented as i2(g) ⇌ 2i(g) at 1000 k, the equilibrium constant kc for the reaction is 3.80 × 10−5. suppose you start with 0.0456 mol of i2 in a 2.28−l flask at 1000 k. what are the concentrations of the gases at equilibrium? what is the equilibrium concentration of i2

Answer 1

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 = $\frac{[I]^{2} }{[I_{2} ]}$

The initial concentration of I₂ is:

[I₂]₀ = $\frac{0.0456 mol}{2.28 L}$ → [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 = $\frac{[I]^{2} }{[I_{2} ]} = \frac{(2x)^{2} }{0.0200 - x} = 3.80x10^{-5}$

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,

$\frac{(2x)^{2} }{0.0200 - x} = \frac{4x^{2} }{0.0200} = 3.80x10^{-5}$

→ 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