91Ó°ÊÓ

The concept of the equilibrium constant, denoted as \( K_c \), is crucial for understanding how reactions behave when they reach equilibrium. It represents the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients from the balanced equation. For the decomposition reaction of carbonyl bromide, we have the equilibrium expression:

• \( K_c = \frac{[CO][Br_2]}{[COBr_2]} \)

This formula tells us the relationship between the concentrations of carbon monoxide \([CO]\), bromine \([Br_2]\), and carbonyl bromide \([COBr_2]\) at equilibrium. A small \( K_c \) value, like 0.190 in this scenario, indicates that the reactants are favored over the products at equilibrium. Understanding \( K_c \) allows chemists to predict the composition of the reaction mixture once it has reached equilibrium, which is essential for various industrial and laboratory processes.

Concentration is a fundamental concept that helps determine how much of a substance is present in a given volume of solution. In this exercise, we're asked to calculate the initial concentration of the reactant \( COBr_2 \) placed in the flask. We use the formula:

• \([\text{COBr}_2]_0 = \frac{\text{moles}}{\text{volume}}\)

Given that there are 0.500 moles of \( COBr_2 \) in a 2.00 L flask, the calculation becomes:\[ [\mathrm{COBr}_2]_0 = \frac{0.500 \text{ mol}}{2.00 \text{ L}} = 0.250 \text{ M} \]This tells us that the initial concentration of \( COBr_2 \) is 0.250 M. Calculating concentrations in this way is vital for setting up equilibrium expressions and solving them, as seen in this decomposition reaction.

In chemical reactions, decomposition refers to the breakdown of a compound into simpler substances. Here, \( \text{COBr}_2 \) decomposes into \( \text{CO} \) and \( \text{Br}_2 \). In equilibrium calculations, we often set up an "ICE table," which stands for Initial, Change, and Equilibrium concentrations:

• Initially, only \( \text{COBr}_2 \) is present: \([\text{COBr}_2]_0 = 0.250 \text{ M}\)
• Change: Some \( \text{COBr}_2 \) decomposes into \( \text{CO} \) and \( \text{Br}_2 \); let \( x \) be the change in concentration.
• Equilibrium: The concentration of \( \text{COBr}_2 \) becomes \( 0.250 - x \), while \([\text{CO}] = x\) and \([\text{Br}_2] = x\).

This setup helps us find the equilibrium concentrations of all species involved in the reaction, by inserting these expressions into the equilibrium constant equation.

Many chemical equilibrium problems involve solving a quadratic equation, which arises when we substitute concentration terms involving \( x \) (the change in concentration) into the equilibrium constant expression. In our example:\[ K_c = \frac{x^2}{0.250 - x} \]This leads to the quadratic equation:\[ x^2 + 0.190x - 0.0475 = 0 \]The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is then used to solve for \( x \), where \( a = 1 \), \( b = 0.190 \), and \( c = -0.0475 \). Calculating \( x \) involves evaluating:\[ x = \frac{-0.190 \pm \sqrt{0.0361 + 0.190}}{2} \]We derive \( x = 0.033 \) M as the solution, representing the amount of \( \text{COBr}_2 \) that has decomposed. This method is a common tool in chemistry, allowing us to find concentrations that meet the conditions set by the equilibrium constant.