Question

2CrO4 2−(aq)+2H+(aq)⇌Cr2O7 2−(aq)+H2O(l)Write the equation for the equilibrium constant (K) of the reaction studied in this exercise.

Answer 1

Answer:

$K = \frac {[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^{2} \cdot [H_{3}O^{+}]^{2}}$  

Explanation:  

The equilibrium constant for a given reversible aqueous reaction is defined by the product ratio of the concentrations between the products and reactants:      

                                           aA + bB ⇄ cC + dD

$K = \frac {[products]^{p}}{[reactants]^{r}} = \frac {[C]^{c} \cdot [D]^{d}}{[A]^{a} \cdot [B]^{b}}$  

where K: is the equilibrium constant, [C] and [D]: are the product concentrations, [A] and [B]: are the reactant concentrations and a,b,c,d: are the stoichiometric coefficients from the reaction.  

Therefore, based on the definition the equilibrium constant of our reaction is:    

2CrO₄²⁻(aq) + 2H₃O⁺(aq) ⇄ Cr₂O₇²⁻(aq) + 3H₂O(l)

$K = \frac {[Cr_{2}O_{7}^{2-}]}{[CrO_{4}^{2-}]^{2} \cdot [H_{3}O^{+}]^{2}}$  

Generally, the water concentration is omitted from the expressions.

I hope it helps you!