Question

Write the concentration equilibrium constant expression for this reaction. cr2o−27 (aq) + 6i− (aq) + 14h+ (aq) → 3i2 (s) + 2cr+3 (aq) + 7h2o (l)

Answer 1

The equilibrium constant of reaction, usually denoted as K, is a unit of ratio. The ratio involves concentrations of products to reactants. But you also have to incorporate their stoichiometric coefficients in the reaction as their respective exponents. Note that substances in their aqueous state are the ones that are included only in the expression. To properly show you how it's done, consider this equilibrium reaction:

aA (aq) + bB (l) ⇆ nN (aq)

Since only reactant A and product N are aqueous, the equilibrium constant for this reaction is:
K = [N]ⁿ/[A]ᵃ
where the [] brackets denotes concentration in molarity

Now, let's apply this to the given equation:
Cr₂O²⁻ (aq) + 6 I⁻ (aq) + 14 H⁺ (aq) → 3 I₂ (s) + 2 Cr³⁺ (aq) + 7 H₂O (l)
I think there is a typographical error because Cr₂O²⁻ has a negative 2 charge rather than -27. Remember that only substances in aqueous states are included in the K expression. Therefore, the expression for K is:

K = [Cr³⁺]² / [Cr₂O²⁻][I⁻]⁶[H⁺]¹⁴