Question

\(\mathrm{I}_{2}+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} .\) This reaction is set up in aqueous medium. We start with \(1 \mathrm{~mol}\) of \(\mathrm{I}\), and \(0.5 \mathrm{~mol}\) of \(\mathrm{I}^{-}\) in \(1 \mathrm{~L}\) flask. After equilibrium is reached, excess of \(\mathrm{AgNO}_{3}\) gave \(0.25 \mathrm{~mol}\) of yellow precipitate. Equilibrium constant is (a) \(1.33\) (b) \(2.66\) (c) \(0.375\) (d) \(0.75\)

Short answer

The equilibrium constant K is 2.66, so the correct answer is (b) 2.66.

Step-by-step solution

Determine Limiting Reactant

With 1 mol of I2 and 0.5 mol of I-, I- is the limiting reactant because it requires 2 moles of I- to completely react with 1 mole of I2 to form I3-.

Calculate Initial Molarity of Reactants

Initial molarity of I2 is 1M (1 mol/L) since there is 1 mol in a 1 L flask. For I-, the initial molarity is 0.5M.

Determine Change in Concentration at Equilibrium

The addition of AgNO3 causes a yellow precipitate of AgI to form. The amount of precipitate (0.25 mol) gives the amount of I- at equilibrium, subtracted from the initial I- (0.5 mol) to determine the change in concentration of I-.

Calculate Equilibrium Concentration of I3- and I-

At equilibrium, 0.5 mol - 0.25 mol = 0.25 mol of I- is left, and since 1 I3- forms from 1 I2 and 1 I-, the equilibrium concentration of I3- is also 0.25M. For I2, it changes by -0.25M as well.

Use the ICE Table to Find the Equilibrium Constant K

Using the ICE table, write down the initial concentrations, changes, and equilibrium concentrations. Then, apply the reaction quotient, Q, using the equilibrium concentrations to solve for the equilibrium constant K.

Calculate K Using the Equation K = [I3-] / ([I2] * [I-]^2)

K can be calculated using the equilibrium concentration of I3- divided by the product of the equilibrium concentrations of I2 and I- squared. Plug in the values to get K

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is essential for studying chemical reactions. The limiting reactant is the substance in a chemical reaction that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot proceed further without it. For instance, in our exercise with the reaction \( \mathrm{I}_{2} + \mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} \), we start with 1 mole of \( \mathrm{I}_{2} \) and 0.5 moles of \( \mathrm{I}^{-} \) in a 1-liter flask. Here, \( \mathrm{I}^{-} \) is the limiting reactant because there is not enough \( \mathrm{I}^{-} \) to react with all of the \( \mathrm{I}_{2} \). This concept dictates the maximum amount of product that can be formed and plays a crucial role in stoichiometric calculations.

When identifying the limiting reactant, you compare the molar ratio of the reactants used in the reaction to their stoichiometric ratio in the balanced equation. The reactant that provides the smallest ratio (in this case, \( \mathrm{I}^{-} \)) is the limiting reagent. Knowing the limiting reactant allows you to calculate the theoretical yield of the product.

ICE Table Method

The ICE Table method stands for Initial, Change, Equilibrium and is an organized way to tackle equilibrium calculations. Initially, you list the molar concentrations of reactants and products. The 'Change' row records the changes in concentration as the reaction approaches equilibrium. This is denoted by a minus sign for reactants consumed and a plus sign for products formed. Lastly, 'Equilibrium' row indicates the final concentrations when the system has reached equilibrium.

For the given reaction \( \mathrm{I}_{2} + \mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} \), we create an ICE table to track changes in concentrations from start to equilibrium. The ICE table helps us identify the shifts occurring within the reaction and can guide us through the process to find the equilibrium constant, which is essential for understanding how the system behaves at equilibrium.

Molarity and Concentration

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The concept of molarity allows chemists to work with the amount of substance in a predictable way, as it relates to the volume of the solution. In the context of the given equilibrium problem, we calculate the initial molarity of \( \mathrm{I}_{2} \) by dividing the number of moles by the volume of the solution, giving us a concentration of 1M. Similarly, the concentration of \( \mathrm{I}^{-} \) is calculated as 0.5M. These concentrations are vital for calculating how much of each reactant is present at the start and how much product can be formed.

Reaction Quotient (Q)

The reaction quotient (Q) is a measure that predicts the direction in which a reaction will proceed to reach equilibrium. It has the same form as the equilibrium constant (K), but while K uses equilibrium concentrations, Q can be calculated using the initial concentrations of the reactants and products.

For the reaction \( \mathrm{I}_{2} + \mathrm{I}^{-} \rightleftharpoons \mathrm{I}_{3}^{-} \), the reaction quotient is expressed as Q = [\(\mathrm{I}_{3}^{-}\)] / ([\(\mathrm{I}_{2}\)] * [\(\mathrm{I}^{-}\)]^2). When Q is compared to K, it indicates whether the reaction is at equilibrium (Q = K), or which way it needs to shift to reach equilibrium (if Q > K or Q < K). In our example, we use the equilibrium concentrations to calculate Q, which in this case is equivalent to the equilibrium constant K because we're at equilibrium.