Question

For the equilibrium $$ \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0\). If \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) and \(0.25 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a \(1.0\) -L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2}\), and \(\mathrm{BrCl} ?\)

Short answer

The equilibrium concentrations of \(\mathrm{Br}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{BrCl}\) are approximately \(0.136\, \mathrm{mol/L}\), \(0.136\, \mathrm{mol/L}\) and \(0.228\, \mathrm{mol/L}\), respectively.

Step-by-step solution

Write the balanced chemical equation and the expression for \(K_c\)

First, let's write down the balanced chemical equation and the expression for \(K_c\) for the given reaction: \[ \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g) \] The expression for \(K_c\) for this equilibrium is: \[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]

Set up the ICE table

Write down the initial concentration of the reactants, the change in their concentrations, and the equilibrium concentrations: \[ \begin{array}{c|c|c|c} & [\mathrm{Br}_2] & [\mathrm{Cl}_2] & [\mathrm{BrCl}] \\ \hline \text{Initial} & 0.25 & 0.25 & 0.00 \\ \text{Change} & -x & -x & +2x \\ \text{Equilibrium} & 0.25-x & 0.25-x & 2x \end{array} \] Here, x is the amount of \(\mathrm{Br}_2\) and \(\mathrm{Cl}_2\) that react to form \(\mathrm{BrCl}\) at equilibrium.

Substitute the equilibrium concentrations into the \(K_c\) expression

Using the concentrations from the ICE table, substitute the equilibrium concentrations into the \(K_c\) expression: \[ 7.0 = \frac{(2x)^2}{(0.25-x)(0.25-x)} \]

Solve for x

Now, we need to solve the quadratic equation for x that we have obtained: \[ 7.0 = \frac{4x^2}{(0.25-x)^2} \] We can cross-multiply and expand the equation to obtain: \[ 4x^2 = 7(0.25 - x)^2 \] Solve for x (Note: You can use either the Quadratic Formula, factoring, or any other appropriate method to find the value of x). \[ x \approx 0.114 \]

Find the equilibrium concentrations

After finding the value of x, plug it back into the equilibrium concentrations in the ICE table: \[ [\mathrm{Br}_2] = 0.25 - x \approx 0.136\,\mathrm{mol/L} \] \[ [\mathrm{Cl}_2] = 0.25 - x \approx 0.136\,\mathrm{mol/L} \] \[ [\mathrm{BrCl}] = 2x \approx 0.228\,\mathrm{mol/L} \] So, the equilibrium concentrations of \(\mathrm{Br}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{BrCl}\) are approximately \(0.136\, \mathrm{mol/L}\), \(0.136\, \mathrm{mol/L}\) and \(0.228\, \mathrm{mol/L}\), respectively.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as K or Kc when referring to concentrations, is a fundamental concept in the study of chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at a state of equilibrium for a reversible reaction. For the reaction \[\[\begin{align*}\mathrm{Br}_{2}(g) + \mathrm{Cl}_{2}(g) &\rightleftharpoons 2 \mathrm{BrCl}(g),\end{align*}\]\] the equilibrium constant expression is:\[\[\begin{align*}K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]}\end{align*}\]\]The value of Kc (7.0 in this exercise) remains constant at a given temperature, providing valuable insights into the extent of the reaction and the proportion of reactants and products present at equilibrium.

Understanding the equilibrium constant is crucial for predicting the direction of the reaction and for calculating the equilibrium concentrations of each species involved in a reaction.

ICE Table

An ICE table is an organized way to calculate changes in the concentration of reactants and products during chemical reactions. ICE stands for Initial, Change, and Equilibrium, representing the stages of the reaction. In the exercise, we start with known initial concentrations of the reactants \[\[\begin{align*}&[\mathrm{Br}_2] = 0.25\end{align*}\]\] and \[\[\begin{align*}&[\mathrm{Cl}_2] = 0.25\end{align*}\]\] moles per liter, while the initial concentration of the product \[\[\begin{align*}&[\mathrm{BrCl}] = 0\end{align*}\]\] is zero. As the reaction proceeds, these concentrations change by a certain amount (denoted as 'x'), eventually leading to the equilibrium state where \[\[\begin{align*}&[\mathrm{Br}_2] = 0.25 - x,\end{align*}\]\] \[\[\begin{align*}&[\mathrm{Cl}_2] = 0.25 - x,\end{align*}\]\] and \[\[\begin{align*}&[\mathrm{BrCl}] = 2x\end{align*}\]\] are the concentrations at equilibrium. This methodical approach assists in visually tracing the progression of the reaction and significantly simplifies the computations involved.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products in a chemical reaction at the state of equilibrium. These concentrations become constant over time, indicating that a dynamic balance has been reached between the forward and reverse reactions, and the net change is zero. In the given exercise, solving for 'x' allows us to determine that the equilibrium concentrations are:\[\[\begin{align*}[\mathrm{Br}_2] &\approx 0.136\end{align*}\]\] moles per liter, \[\[\begin{align*}[\mathrm{Cl}_2] &\approx 0.136\end{align*}\]\] moles per liter, and \[\[\begin{align*}[\mathrm{BrCl}] &\approx 0.228\end{align*}\]\] moles per liter. Knowledge of these concentrations is crucial for understanding the composition of the system at equilibrium. It is a key concept that finds extensive use in fields such as chemistry, biology, environmental science, and engineering.

Grasping the concept of equilibrium concentrations is important for students aiming to comprehend the dynamic nature of chemical systems and for those tackling more complex applications, such as in pharmaceutical and industrial processes.

Reaction Quotient

The reaction quotient, denoted as Q, is closely related to the equilibrium constant but applies to a system where reactants and products have not yet reached equilibrium. Its expression is identical to that of the equilibrium constant, but it uses the current concentrations of the reactants and products. For the reaction under consideration, the reaction quotient is given by:\[\[\begin{align*}Q = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]}\end{align*}\]\]Comparing Q to Kc can predict the direction in which the reaction will proceed to achieve equilibrium. If Q < Kc, the forward reaction is favored, and more product will form. Conversely, if Q > Kc, the reverse reaction is favored, reducing the amount of product. This concept is an essential tool for understanding and predicting reaction dynamics and can guide adjustments in reaction conditions to achieve the desired outcome in industrial and laboratory settings.

By mastering the concept of the reaction quotient, students can make educated predictions about chemical reactions outside of equilibrium and learn how to steer them in the desired direction.