91Ó°ÊÓ

In a chemical reaction at equilibrium, the equilibrium constant (\(K_c\)) is a crucial value that provides insight into the proportions of reactants and products. \(K_c\) is derived from the concentrations of the chemical species at equilibrium. It's specifically expressed for a reaction as the ratio of the product of the concentration of the products to the product of the concentration of the reactants. Each concentration is raised to the power of its coefficient in the balanced chemical equation.For the given reaction of iodine and bromine forming iodine monobromide:- The balanced equation is \[\mathrm{I}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{IBr}(g)\]- The equilibrium expression becomes: \[K_c = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_2][\mathrm{Br}_2]}\]The \(K_c\) value indicates how far the reaction proceeds before reaching equilibrium. In this problem, \(K_c\) is 108, suggesting a significant formation of \(\mathrm{IBr}\) over the reactants at equilibrium.

During a chemical reaction, the concentration of substances changes from the initial to the equilibrium state. Understanding these changes is fundamental to finding the equilibrium concentrations. Initially, we calculate the concentrations by dividing the moles of substances by the volume of the container.In our problem:

• Initial moles of both \(\mathrm{I}_2\) and \(\mathrm{Br}_2\) = 0.0019 mol
• Volume = 5.0 L

The initial concentrations are:\[[\mathrm{I}_2]_0 = \frac{0.0019}{5.0} = 0.00038 \ \text{mol/L}\]\[[\mathrm{Br}_2]_0 = \frac{0.0019}{5.0} = 0.00038 \ \text{mol/L}\]As the reaction proceeds, let \(x\) represent the change in concentration that occurs for \(\mathrm{I}_2\) and \(\mathrm{Br}_2\). This change also affects the formation of \(\mathrm{IBr}\), doubling its increment because for every mole of each reactant consumed, two moles of \(\mathrm{IBr}\) are formed.

Sometimes, solving for the equilibrium state requires the use of algebraic equations, such as quadratic equations. The relation between concentration changes and the equilibrium constant often leads to a quadratic equation.For this reaction:- After defining changes in concentration, the equation:\[108 = \frac{(2x)^2}{(0.00038-x)(0.00038-x)}\]is simplified to a standard quadratic equation form:\[4x^2 = 108 \times (0.00038 - x)^2\]Rearranging gives:\[4x^2 = 108 (0.00038^2 - 2 \times 0.00038 \times x + x^2)\]Solving this involves expanding and applying the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where coefficients \(a, b,\) and \(c\) are derived from rearranging the equation. This calculation finds the value of \(x\), representing how much the concentrations have changed to reach equilibrium.

Stoichiometry is the area in chemistry that pertains to the quantification of elements in chemical reactions. It plays a crucial role in understanding reactions by connecting the balanced equation's ratios to actual substance amounts.For the reaction:\(\mathrm{I}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{IBr}(g)\), stoichiometry dictates that 1 mole of \(\mathrm{I}_2\) reacts with 1 mole of \(\mathrm{Br}_2\) to produce 2 moles of \(\mathrm{IBr}\).In our calculation:

• Every change \(x\) reduced from \(\mathrm{I}_2\) and \(\mathrm{Br}_2\) produces 2x\(\mathrm{IBr}\)
• This relates the concentrations directly to their molecular quantities based on initial moles provided and volume
• It allows calculating equilibrium concentrations as: - \([\mathrm{I}_2]_{eq} = 0.00038 - x\) - \([\mathrm{Br}_2]_{eq} = 0.00038 - x\) - \([\mathrm{IBr}]_{eq} = 2x\)

Understanding stoichiometry is pivotal to predict how changes in one part of the reaction affect the entire system, especially in determining the equilibrium composition.