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Calculate the equilibrium constant (Kc) for the reaction between iodine (I2) and potassium iodide (KI) in an aqueous solution, given that at equilibrium, the concentration of iodine is 0.008 M and the concentration of potassium iodide is 0.15 M. The balanced equation for the reaction is given below:I2 (aq) + 2KI (aq) ⇌ 2K+ (aq) + 2I- (aq)

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To calculate the equilibrium constant  Kc  for the reaction, we first need to determine the concentrations of all species at equilibrium. The balanced equation for the reaction is:I2  aq  + 2KI  aq   2K+  aq  + 2I-  aq At equilibrium, the concentration of iodine  I2  is given as 0.008 M, and the concentration of potassium iodide  KI  is given as 0.15 M. Since the stoichiometry of the reaction shows that 1 mole of I2 reacts with 2 moles of KI, we can determine the change in concentration for each species.Change in concentration of I2 = -0.008 M  consumed Change in concentration of KI = -2 * 0.008 M = -0.016 M  consumed Change in concentration of K+ = 2 * 0.008 M = 0.016 M  produced Change in concentration of I- = 2 * 0.008 M = 0.016 M  produced Now, we can determine the equilibrium concentrations of each species:[I2] = 0.008 M  initial  - 0.008 M  change  = 0 M[KI] = 0.15 M  initial  - 0.016 M  change  = 0.134 M[K+] = 0 M  initial  + 0.016 M  change  = 0.016 M[I-] = 0 M  initial  + 0.016 M  change  = 0.016 MNow we can calculate the equilibrium constant  Kc  using the expression:Kc = [K+]^2 * [I-]^2 /  [I2] * [KI]^2 Plugging in the equilibrium concentrations:Kc =  0.016 ^2 *  0.016 ^2 /  0 * 0.134^2 Since the concentration of I2 is 0, the denominator becomes 0, and we cannot divide by 0. This indicates that the reaction has proceeded to completion, and there is no equilibrium between the reactants and products. In this case, we cannot calculate an equilibrium constant  Kc  for the reaction.

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