91Ó°ÊÓ

An equilibrium constant, represented as \(K_c\), plays a critical role in determining the position of equilibrium in a chemical reaction. The equilibrium constant quantifies the ratio of concentrations of products to reactants at equilibrium. In the given reaction, iodine monobromide (IBr) decomposes into iodine \((\text{I}_2)\) and bromine \((\text{Br}_2)\).
When we describe this in terms of \(K_c\), we write the expression as:
- \(K_c = \frac{[\text{I}_2][\text{Br}_2]}{[\text{IBr}]^2}\)
This formula shows that the equilibrium constant is based on the concentrations of the products \([\text{I}_2]\) and \([\text{Br}_2]\) in the numerator, divided by the concentration of the reactant \([\text{IBr}]\) squared.
The given value of \(K_c\) is \(0.026\) at \(100^\circ \text{C}\). This tells us that at this temperature, the concentration of products is much lower compared to the reactants, signifying that the reaction strongly favors the reverse formation of IBr. Understanding \(K_c\) is essential because it provides insight into how a change in conditions influences equilibrium in reversible reactions.

In any equilibrium system, the concentration changes when the reaction is set to reach a state of equilibrium from its initial condition. For the decomposition of iodine monobromide, IBr, we start with initial concentrations:
- \([\text{IBr}]_0 = 0.010 \text{ M}\)- \([\text{I}_2]_0 = [\text{Br}_2]_0 = 0 \text{ M}\)
These values imply that initially, only iodine monobromide is present in the system, while iodine and bromine begin at zero concentrations.
As the system approaches equilibrium, these concentrations change. We can use \(x\) to represent the change in concentration. Since forming iodine and bromine requires the dissociation of IBr, each time 1 mole of I\(_2\) and 1 mole of Br\(_2\) is produced, 2 moles of IBr must decompose:

• The concentration change of \([\text{IBr}] = -2x\)
• The concentration change of both \([\text{I}_2]\) and \([\text{Br}_2]\) = \(x\)

Reaching equilibrium, we have:

• \([\text{IBr}] = 0.010 - 2x\)
• \([\text{I}_2] = x\)
• \([\text{Br}_2] = x\)

Solving for \(x\) provides a straightforward way of determining how much of each species exists at equilibrium, which is crucial for understanding chemical dynamics.

The decomposition of iodine monobromide (IBr) serves as a classic example of a reversible chemical reaction. Initially, a defined amount of IBr is present, and over time it decomposes into iodine \((\text{I}_2)\) and bromine \((\text{Br}_2)\). This reaction can be represented by the equation:
- \(2 \text{IBr}(g) \rightleftharpoons \text{I}_2(g) + \text{Br}_2(g)\)
At equilibrium, the total amount of reactants and products stop changing, even though the forward and reverse reactions continue to occur at equal rates. The equilibrium state minimizes free energy, striking a balance between the decomposition of IBr into iodine and bromine, and the reverse formation of IBr from these products.
Given the context of the reaction's \(K_c\) value and initial conditions, we can determine the moles of each species at equilibrium:

• Moles of \(\text{I}_2 = \text{Br}_2 = 0.0023 \text{ moles}\)
• Moles of remaining \(\text{IBr} = 0.0054 \text{ moles}\)

These calculations illustrate how chemical equilibrium involves continuous yet stable interconversion between reactants and products, orchestrated to maintain a state of dynamic stability.