Question

The equilibrium constant for the dissolution of AgI(s) in water is 1.5×10-16. AgI(s) Ag +(aq) + I-(aq) A solution is prepared by diluting 2.0 mol AgI, 3.0×10-8 mol Ag+, and 3.0×10-8 mol I- to a volume of 1.0 L. Calculate the reaction quotient and determine whether a reaction occurs. A.Q = K; no reaction occurs. B.Q < K; Ag+ and I- combine to form AgI(s) until equilibrium is established. C.Q > K; AgI(s) dissolves until equilibrium is established. D.Q < K; AgI(s) dissolves until equilibrium is established. E.Q > K; Ag+ and I- combine to form AgI(s) until equilibrium is established.

Answer 1

For the given dissociation of AgI_(s) in water,

AgI_(s) = Ag⁺_(aq) + I^-_(aq)

The equilibrium constant Kc = 1.5×10^-16

[Ag⁺] = 3.0×10^-8 mol

[I^-] = 3.0×10^-8 mol

[AgI] = 2 mol

The reaction quotient Qc can be calculated as

Qc = [Ag⁺] [I^-] / [AgI]

Qc = 3.0×10^-8 * 3.0×10^-8 / 2

Qc = 4.5×10^-16

Now comparing between Kc and Qc, we find that

Qc > Kc

When Qc > Kc, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. Hence Ag⁺ and I^- combine to form AgI(s) until equilibrium is established.