91Ó°ÊÓ

The equilibrium constant, denoted as \( K_{\mathrm{c}} \), is a fundamental concept in chemical equilibrium. It provides a measure of the extent to which a reaction occurs at equilibrium. By definition, \( K_{\mathrm{c}} \) is calculated from the ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients. For the reaction \( \mathrm{Br}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2\mathrm{BrF}(\mathrm{g}) \), the equilibrium constant expression is:\[ K_{\mathrm{c}} = \frac{[\mathrm{BrF}]^2}{[\mathrm{Br}_2][\mathrm{F}_2]} \]It is important to remember that \( K_{\mathrm{c}} \) is unitless and specific to a given reaction at a particular temperature.

• The equilibrium constant helps predict the direction of a reaction – if \( K_{\mathrm{c}} \) is much greater than 1, the products are favored.
• Conversely, if \( K_{\mathrm{c}} \) is much less than 1, the reactants are favored.

In the given problem, \( K_{\mathrm{c}} \) is 55.3, meaning that at equilibrium, the formation of BrF is significantly favored over the residual reactants.

Stoichiometry, the quantitative relationship between reactants and products in a chemical reaction, is an essential tool for solving equilibrium problems. This concept hinges on the law of conservation of mass, which dictates that matter is neither created nor destroyed in a chemical reaction. It enables us to relate changes in the amounts of different substances throughout the reaction.For the mentioned reaction, BrF is produced from Br\(_2\) and F\(_2\) in a 2:1:1 ratio. This means that for every mole of Br\(_2\) or F\(_2\) consumed, two moles of BrF are formed:

• Initial concentrations of both \([\mathrm{Br}_2] \) and \([\mathrm{F}_2] \) are \( 0.220 \mathrm{~mol/L} \).
• At equilibrium, their concentrations decrease by \( x \) while \([\mathrm{BrF}] \) increases by \( 2x \).

Understanding these stoichiometric relationships helps us visualize how quantities of substances change as the reaction progresses toward equilibrium. By assigning a variable \( x \) to these changes, we set up the equations necessary to solve for the equilibrium concentrations.

Chemistry often involves solving quadratic equations, especially when dealing with equilibrium scenarios. A quadratic equation arises when the concentration changes lead to a second-degree polynomial, such as in the example problem.The equation we get is:\[ 55.3 = \frac{4x^2}{(0.220 - x)^2} \]This equation simplifies to:\[ 51.3x^2 - 24.332x + 2.67652 = 0 \]To solve this, the quadratic formula is used, which is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 51.3 \), \( b = -24.332 \), and \( c = 2.67652 \). The plus-minus sign indicates two potential solutions, but in equilibrium problems, the solution that makes physical sense (usually the positive value) is chosen.

• This method allows us to solve for \( x \), indicating the change in concentration directly attributable to the reaction reaching equilibrium.
• In our problem, calculating with the formula yields \( x \approx 0.054 \) mol/L, indicating the change required to find the equilibrium concentrations of the reactive species.