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 \(3.0-\mathrm{L}\) container at \(400 \mathrm{~K},\) what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{BrCl} ?\)

Short answer

At equilibrium, the concentrations of Br₂, Cl₂, and BrCl are \( 0.040 \ \frac{\text{mol}}{\text{L}} \), \( 0.140 \ \frac{\text{mol}}{\text{L}} \), and \( 0.086 \ \frac{\text{mol}}{\text{L}} \), respectively.

Step-by-step solution

Write the balanced equilibrium equation

The balanced equilibrium equation is as follows: \[ \mathrm{Br}_{2}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \ \mathrm{BrCl}(g) \]

Calculate initial concentrations

To find the initial concentrations (moles per liter) of Br₂ and Cl₂, divide the moles by the volume of the container (3.0 L): \[ [\mathrm{Br}_{2}]_0 = \frac{0.25 \ \text{mol}}{3.0 \ \text{L}} = 0.083 \frac{\text{mol}}{\text{L}} \] \[ [\mathrm{Cl}_{2}]_0 = \frac{0.55 \ \text{mol}}{3.0 \ \text{L}} = 0.183 \frac{\text{mol}}{\text{L}} \] Since no BrCl is initially present, its concentration is 0.

Set up the ICE table

To do this, list the initial concentrations, changes, and equilibrium concentrations for all substances: | | Br₂ | Cl₂ | BrCl | |---------|-----|-----|------| | Initial | 0.083 | 0.183 | 0 | | Change | -x | -x | +2x | | Equilibrium | 0.083-x | 0.183-x | 2x | Here, x represents the change in concentration of Br₂ and Cl₂ due to the reaction.

Use Kc expression

Write the equilibrium constant (Kc) expression based on the balanced equation: \[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]

Plug in equilibrium concentrations

Using the equilibrium concentrations from the ICE table, substitute them into the Kc expression: \[ 7.0 = \frac{(2x)^2}{(0.083 - x)(0.183 - x)} \]

Solve for x

To simplify the expression and find x, you can either use algebraic methods, like factoring or using the quadratic formula, or use a numerical method such as iterative or graphical methods. After solving: \[ x = 0.043 \]

Find equilibrium concentrations

Now that we have the value of x, substitute it back into the equilibrium concentrations in the ICE table: \[ [\mathrm{Br}_{2}] = 0.083 - x = 0.083 - 0.043 = 0.040 \frac{\text{mol}}{\text{L}} \] \[ [\mathrm{Cl}_{2}] = 0.183 - x = 0.183 - 0.043 = 0.140 \frac{\text{mol}}{\text{L}} \] \[ [\mathrm{BrCl}] = 2x = 2(0.043) = 0.086 \frac{\text{mol}}{\text{L}} \] At equilibrium, the concentrations of Br₂, Cl₂, and BrCl are 0.040, 0.140, and 0.086 mol/L, respectively.

Key concepts

Equilibrium Constant

The equilibrium constant (K) is a crucial concept in chemistry that quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction. It is a reflection of the extent to which a reaction will proceed until the system reaches a state of balance, where the rates of the forward and reverse reactions are equal.

For the reaction \[ \mathrm{Br}_{2}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons 2\operatorname{BrCl}(g), \]
the equilibrium constant at a certain temperature, represented as Kc, is calculated using the equation:
\[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]
Here, Kc has a value of 7.0 at 400 K, indicating the ratio of the concentration of BrCl squared to the product of the concentrations of Br2 and Cl2 when the reaction has reached equilibrium. A higher Kc value denotes a greater extent of reaction towards the product side. Understanding Kc helps predict the direction of the reaction and the relative amounts of reactants and products present at equilibrium.

ICE Table Method

The ICE table method is an invaluable tool used to organize and simplify the calculation of equilibrium concentrations in chemical reactions. ICE stands for Initial, Change, and Equilibrium, which are the stages we consider for each substance involved in the equilibrium.

In our example:

• Initial: The starting concentrations of Br2 and Cl2 are given, and BrCl is initially zero.
• Change: We represent the changes in concentration using a variable 'x', with Br2 and Cl2 decreasing by 'x' and BrCl increasing by '2x' due to the stoichiometry of the reaction.
• Equilibrium: We use these changes to express the equilibrium concentrations of all species in relation to 'x'.

By setting up an ICE table,
\[\begin{array}{c|ccc} & \mathrm{Br}_2 & \mathrm{Cl}_2 & \mathrm{BrCl} \ \hline\mathrm{Initial} & 0.083 & 0.183 & 0 \mathrm{Change} & -x & -x & +2x \mathrm{Equilibrium} & 0.083 - x & 0.183 - x & 2x \ \end{array}\]
we create a systematic way to approach the equilibrium calculations and assist in solving for the unknown variable 'x' that allows us to find the equilibrium concentrations.

Equilibrium Concentrations

Equilibrium concentrations are the concentrations of the reactants and products in a chemical reaction when the reaction has reached its equilibrium state. They are determined by the stoichiometry of the reaction, the initial concentrations, and the equilibrium constant (Kc).

In the reaction we're examining, once we have established Kc and the initial concentrations using an ICE table, we can solve for the unknown 'x', which represents the shift in molar concentrations of reactants and products as the system reaches equilibrium.
\[ x = 0.043 \frac{\text{mol}}{\text{L}} \]
Substituting 'x' back into the ICE table, we calculate the equilibrium concentrations:

• Br2: \( [\mathrm{Br}_{2}] = 0.083 - x = 0.040 \frac{\text{mol}}{\text{L}} \)
• Cl2: \( [\mathrm{Cl}_{2}] = 0.183 - x = 0.140 \frac{\text{mol}}{\text{L}} \)
• BrCl: \( [\mathrm{BrCl}] = 2x = 0.086 \frac{\text{mol}}{\text{L}} \)

The accurate determination of these equilibrium concentrations is essential for predicting how a system behaves at equilibrium and is fundamental for further applications, such as calculating reaction yields or optimizing industrial chemical processes.