Question

The equilibrium constant, \(K_{\mathrm{c}}\), for the reaction $$ \mathrm{Br}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{BrF}(\mathrm{g}) $$ is 55.3 . Calculate what the equilibrium concentrations of all these gases are if the initial concentrations of bromine and fluorine were both \(0.220 \mathrm{~mol} / \mathrm{L}\). (Assume constantvolume conditions.)

Short answer

At equilibrium: \([\mathrm{Br}_2] = [\mathrm{F}_2] = 0.166\,\mathrm{~mol/L}\), \([\mathrm{BrF}] = 0.108\,\mathrm{~mol/L}\).

Step-by-step solution

Write the Equilibrium Expression

The equilibrium expression for the given reaction \( \mathrm{Br}_{2} + \mathrm{F}_{2} \rightleftharpoons 2\mathrm{BrF} \) is derived from the law of mass action. Thus, the expression for the equilibrium constant \( K_{\mathrm{c}} \) is: \[ K_{\mathrm{c}} = \frac{[\mathrm{BrF}]^2}{[\mathrm{Br}_2][\mathrm{F}_2]} \]

Define Initial and Change in Concentrations

Initially, the concentrations are: - \([\mathrm{Br}_2] = 0.220 \mathrm{~mol/L}\)- \([\mathrm{F}_2] = 0.220 \mathrm{~mol/L}\)- \([\mathrm{BrF}] = 0 \mathrm{~mol/L}\) Assume \( x \) moles/L of \( \mathrm{Br}_2 \) and \( \mathrm{F}_2 \) react to form \( 2x \) moles/L of \( \mathrm{BrF} \). Thus, the change in concentration is \( -x \) for \( \mathrm{Br}_2 \) and \( \mathrm{F}_2 \), and \( +2x \) for \( \mathrm{BrF} \).

Express Equilibrium Concentrations

The equilibrium concentrations can be expressed as:- \([\mathrm{Br}_2] = 0.220 - x\)- \([\mathrm{F}_2] = 0.220 - x\)- \([\mathrm{BrF}] = 2x\)

Substitute into Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium expression:\[ 55.3 = \frac{(2x)^2}{(0.220 - x)(0.220 - x)} \] This simplifies to:\[ 55.3 = \frac{4x^2}{(0.220 - x)^2} \]

Solve for x

We need to solve for \( x \) by solving the equation: \[ 55.3(0.220 - x)^2 = 4x^2 \] Expanding and solving this equation:1. \( 55.3(0.0484 - 0.44x + x^2) = 4x^2 \)2. \( 2.67652 - 24.332x + 55.3x^2 = 4x^2 \)3. Simplify to: \( 51.3x^2 - 24.332x + 2.67652 = 0 \)4. Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 51.3 \), \( b = -24.332 \), \( c = 2.67652 \) 5. Calculate to find \( x \approx 0.054 \).

Calculate Equilibrium Concentrations

Using the value of \( x = 0.054 \), calculate equilibrium concentrations:- \([\mathrm{Br}_2] = 0.220 - x = 0.220 - 0.054 = 0.166 \mathrm{~mol/L}\)- \([\mathrm{F}_2] = 0.220 - x = 0.220 - 0.054 = 0.166 \mathrm{~mol/L}\)- \([\mathrm{BrF}] = 2x = 2 \times 0.054 = 0.108 \mathrm{~mol/L}\)

Key concepts

Equilibrium Constant Calculation

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.

Reaction Stoichiometry

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.

Quadratic Equation in Chemistry

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.