Question

\(\mathrm{I}_{2}(a q)+\mathrm{I}^{-}(a q) \rightleftharpoons \mathrm{I}_{3}^{-}(a q) .\) We started with 1 mole of \(\mathrm{I}_{2}\) and \(0.5\) mole of \(\mathrm{I}^{-}\) in one litre flask. After equilibrium is reached, excess of \(\mathrm{AgNO}_{3}\) gave \(0.25\) mole of yellow precipitate. Equilibrium constant is : (a) \(1.33\) (b) \(2.66\) (c) \(2.0\) (d) \(3.0\)

Short answer

After plugging in the values and solving, the equilibrium constant (K) is 3.0, making option (d) the correct answer.

Step-by-step solution

Write the balanced chemical equation

Write the balanced chemical reaction: \(I_{2}(aq) + I^{-}(aq) \rightleftharpoons I_{3}^{-}(aq)\).

Determine the number of moles of product formed

Determine the number of moles of the product, \(I_{3}^{-}\), formed by the reaction of excess \(AgNO_{3}\) with the triiodide ions, resulting in 0.25 moles of yellow precipitate.

Calculate the change in moles of reactants

Since each \(I_{3}^{-}\) ion is composed of one \(I^{-}\) ion and one \(I_{2}\) molecule, 0.25 moles of \(I_{3}^{-}\) will consume 0.25 moles of \(I_{2}\) and 0.25 moles of \(I^{-}\).

Set up the equilibrium expression

Set up the equilibrium expression: \([I_{3}^{-}] = \frac{[I_{2}]_{initial} - x}{V} \cdot \frac{[I^{-}]_{initial} - x}{V}\), where \(x\) is the moles of \(I_{2}\) and \(I^{-}\) that reacted and \(V\) is the volume in liters.

Plug in the known values

Plug in the known values, where initial moles of \(I_{2}\) was 1, initial moles of \(I^{-}\) was 0.5, volume \(V = 1\) L, and \(x = 0.25\) moles to solve for the equilibrium constant \(K\).

Solve for the equilibrium constant (K)

Solve the equilibrium equation \(K = [I_{3}^{-}] / ([I_{2}] \cdot [I^{-}])\) using the concentrations after the reaction to find the value of the equilibrium constant.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is pivotal to grasping the behavior of reactions in a closed system. It occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. This doesn't mean the reactions stop, but rather that they proceed at the same rate in both directions, achieving a balance.

In the exercise, we're dealing with the equilibrium between triiodide, iodine, and iodide ions in an aqueous solution. When equilibrium is reached, knowing the initial quantities of reactants and the amount of product formed allows us to calculate the equilibrium constant, which is a quantitative measure of the reaction's position at equilibrium. High values of the equilibrium constant indicate a reaction heavily in favor of products, while low values suggest a reaction favoring reactants.

Reaction Stoichiometry

Reaction stoichiometry involves quantitatively analyzing the relationships between reactants and products in a chemical reaction. It's governed by the balanced chemical equation, which serves as a recipe for the number of moles of each substance involved. The mole concept, another core topic, plays a critical role here as it allows chemists to count molecules in a way similar to counting eggs by the dozen.

For calculating the equilibrium constant, stoichiometry helps us understand that one mole of iodine reacts with one mole of iodide to produce one mole of triiodide. This one-to-one-to-one ratio simplifies the calculation, ensuring that each mole of triiodide formed will reduce the moles of iodine and iodide by one mole each. Knowing this stoichiometric relationship is essential for setting up the equilibrium expression correctly.

Mole Concept

The mole concept is an indispensable part of chemistry that allows us to work with the submicroscopic world of atoms, molecules, and ions. A mole is defined as the amount of substance containing as many elementary entities (such as atoms, molecules, or ions) as there are atoms in 12 grams of pure carbon-12. The number of entities per mole is Avogadro's number, which is approximately 6.022 x 10^23.

In the textbook exercise, we initially have 1 mole of iodine (\( I_2 \)) and 0.5 moles of iodide ions (\(I^-\)), which helps us calculate how the equilibrium shifts. By applying the mole concept, we can equate the change in concentration to the formation of triiodide ions, enabling us to compute the equilibrium constant. It also clarifies that a 0.25 mole yield of precipitate reflects a direct consumption of 0.25 moles of iodine and iodide ions, important for the subsequent steps of the calculation.