Question

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) and \(0.55 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-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 the species are - \( [\mathrm{Br}_2]_{eq} = 0.034 \, M \) - \( [\mathrm{Cl}_2]_{eq} = 0.134 \, M \) - \( [\mathrm{BrCl}]_{eq} = 0.098 \, M \).

Step-by-step solution

Write down the balanced chemical equation and the given information

First, let's write the balanced chemical equation: \[ \mathrm{Br_{2}(g) + Cl_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g)} \] Given information: - \( T = 400 \, K \) - \( K_c = 7.0 \) - \( [\mathrm{Br}_2]_0 = \frac{0.25 \, mol}{3.0 \, L} = 0.083 \, M \) - \( [\mathrm{Cl}_2]_0 = \frac{0.55 \, mol}{3.0 \, L} = 0.183 \, M \) - \( [\mathrm{BrCl}]_0 = 0 \, M \)

Write the equilibrium expression

The equilibrium expression for the given reaction can be written as: \[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]

Set up the ICE table

Next, let's use the ICE (Initial, Change, Equilibrium) table to find the changes in concentrations for each species: \(\begin{array}{c|c|c|c} & [\mathrm{Br}_2] & [\mathrm{Cl}_2] & [\mathrm{BrCl}] \\ \hline \mathrm{Initial} & 0.083 & 0.183 & 0 \\ \mathrm{Change} & -x & -x & +2x \\ \mathrm{Equilibrium} & 0.083-x & 0.183-x & 2x \end{array}\) Where x is the change in molarity for Br₂ and Cl₂, and 2x represents the change in molarity for BrCl.

Substitute equilibrium concentrations into the equilibrium expression and solve for x

Substituting the equilibrium concentrations into the equilibrium expression: \[ 7.0 = \frac{(2x)^2}{(0.083 - x)(0.183 - x)} \] We now have a quadratic equation to solve for x. This quadratic equation can be a bit challenging to solve by hand, so using a solver might be best.

Solve for x and find the equilibrium concentrations

After using a quadratic equation solver, we get \(x = 0.049\). Now, we can find the equilibrium concentrations: - \( [\mathrm{Br}_2]_{eq} = 0.083 - 0.049 = 0.034 M \) - \( [\mathrm{Cl}_2]_{eq} = 0.183 - 0.049 = 0.134 M \) - \( [\mathrm{BrCl}]_{eq} = 2(0.049) = 0.098 M \) So, the equilibrium concentrations of the species are: - \( [\mathrm{Br}_2]_{eq} = 0.034 \, M \) - \( [\mathrm{Cl}_2]_{eq} = 0.134 \, M \) - \( [\mathrm{BrCl}]_{eq} = 0.098 \, M \)

Key concepts

ICE Table

An ICE table helps manage the changes in concentrations for substances during a chemical reaction reaching equilibrium. ICE stands for Initial, Change, and Equilibrium, which are the stages of shifts in concentrations as the reaction progresses. By organizing the data for each chemical species, it simplifies the process to visualize how the system evolves:

• Initial: Represents the initial concentrations of the reactants and products before any reaction occurs.
• Change: Denotes how concentrations change; negative for reactants as they are used up, positive for products being formed.
• Equilibrium: Calculated after using the initial counts and the changes. Represents the concentrations at equilibrium.

This table is particularly useful when you start with a known amount of reactants and need to find out where equilibrium lies. By assigning variables to depict changes (often represented by 'x'), you can track how the concentration of each element or compound adjusts. Once this is set, it provides the groundwork needed to solve for unknown concentrations.

Equilibrium Expression

The equilibrium expression is a mathematical representation of a balanced chemical reaction at equilibrium. For reversible reactions like \[\mathrm{Br}_2(g) + \mathrm{Cl}_2(g) \rightleftharpoons 2 \mathrm{BrCl}(g)\],we represent the position of equilibrium with a constant, referred to as the equilibrium constant \(K_c\).

This expression is a key player in equilibrium calculations:

• Setup: The equilibrium expression is formulated by taking the products’ concentrations, raised to the power of their coefficients in the balanced chemical equation, divided by the reactants’ concentrations, similarly adjusted.
• Calculation: For the provided reaction, it would be \(K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]}\).
• Use: This expression with known equilibrium constant value helps in determining unknown concentrations when combined with the ICE table.

Understanding how to formulate and manipulate the equilibrium expression is crucial for predicting the position of equilibrium and the ultimate concentrations of species present.

Quadratic Equation

Solving chemical equilibria often involves mathematical equations. The quadratic equation is frequently encountered when dealing calculations involving the changes in concentrations as seen in our scenario.

When we substitute values from the ICE table into the equilibrium expression, we sometimes end up with a quadratic form:

• Quad Form: Looks like \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants derived from the problem's specific parameters.
• Solving: Can be solved using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Using this, you find the roots of the equation, representing possible values for variables like \(x\).
• Application: You would then use \(x\) to calculate equilibrium concentrations, considering physically plausible values (like positive concentrations).

In this context, the quadratic equation is an essential tool for finding the exact concentrations of substances in equilibrium, crucial for predicting the system's behavior.