Question

Write the concentration equilibrium constant expression for this reaction. 2cui (s) + i2 (aq) → 2cu+2 (aq) + 4i− (aq)

Answer 1

2CuI (s) + I₂ (aq) → 2 Cu⁺² (aq) + 4 I⁻ (aq)

When writing an equilibrium expression, we use the following values:

A (aq) + 3B (aq) → 2C (aq) + 2D (aq)

The numbers were arbitrary molar equivalents and the uppercase letters are the molecules in the reaction. The species used in the equilibrium expression but all be in the same state, e.g., solid, liquid, aqeuous.

Kc = [C]²[D]² / [A][B]³

We write the formula by taking the concentration of the products, each to the power of their molar equivalent, and multiply them together. We then divide the products by the concentration of the reactants, also to the power of their molar equivalent.

Going back to the initial equation given, we can now write a Kc expression.

Kc = [Cu⁺²]²[I⁻]⁴ / [I₂]

It should be noted that the CuI (s) in the reaction was left out of the Kc expression. Pure solids and liquids are left out of the expression and only the aqueous species are included. The reason being that, in this case, solid CuI does not affect the amount of reactant at equilibrium. Therefore, we just leave the concentration for [CuI] = 1, and remove it from the expression.

Answer 2

Answer : The expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{[Cu^{2+}]^2[I^-]^4}{[I_2]}$

Explanation :

The given balanced equilibrium reaction is,

$2CuI(s)+I_2(aq)\rightleftharpoons 2Cu^{2+}(aq)+4I^-(aq)$

The general expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{\text{Concentration of products}}{\text{Concentration of reactants}}$

As we know that the concentration of solid is equal to 1.

So, the expression for equilibrium constant for this reaction will be,

$K_{eq}=\frac{[Cu^{2+}]^2[I^-]^4}{[I_2]}$