A reaction vessel contains 0.100 m cadmium ion, 0.500 m chloride and 0.250 m tetrachlorocadmate ion. which direction will the reaction proceed?

First by considering the following reaction:

Cd (aq) + 4Cl (aq) ⇄ CdCl4

and according to Kc (the constant of equilibrium) low:

when Kc = [Concentration of products]^ (n) / [ concentration of the reactants]^ (n)

when n is the stoichiometric coefficient for the products and reactants

and we have the equilibrium concentrations of:

Cl = 0.5m, Cd = 0.1 m, and CdCl4 = 0.250 m

So by substitution:

∴ Kc = [CdCl4] / ([Cd] [C]^4

= 0.250 / ((0.1) * (0.5) ^4)

= 40

So, ∵ Kc > 1 (that's means the concentration of the products is more than the concentration of the reactants)

-we should have the rate of right reaction equal the rate of left reaction in the equilibrium reaction. so the Kc value should be 1 but when it becomes > 1 that means we need to form more reactants So, that leads the reaction to go leftward to the direction of the reactants to make the equilibrium.

∴ The reaction goes (leftward) to form more reactants.