Question

The Keq for the decomposition of hydrogen iodide to form hydrogen and iodine as follows

2HI(g)
↔
H2(g) + I2(g)

is 1.6 x 10–4. If the concentration at equilibrium of H2 and I2 are each 2 x 10–3 M, what is [HI]?

Answer 1

The equilibrium constant (Keq) is given for the reversible reactions. The concentration of hydrogen iodide will be 1.58 M.

What is Keq?

The equilibrium constant (Keq) is the proportion of the ratios of the concentration of the products to the reactants.

The balanced chemical reaction is shown as:

2HI  ⇔ H₂ + I₂

The equilibrium expression for the reaction can be given as:

Keq = [H₂] [I₂] / [HI]²

Given,

Equilibrium constant = 1.6 × 10⁻⁴

Concentration of dihydrogen = 2  × 10⁻³ M

Concentration of iodine = 2 × 10⁻³ M

The concentration of hydrogen iodide is calculated as:

1.6 × 10⁻⁴ = [ 2  × 10⁻³ M][2  × 10⁻³ M] ÷ [HI]²

[HI]² = 4  × 10⁻⁶ ÷ 1.6 × 10⁻⁴

= 2.5 M

[HI] = 1.58 M

Therefore, 1.58 M is the concentration of hydrogen iodide.

Learn more about the equilibrium constant here:

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