Problem 14 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Bromine and chlorine react to produce bromine monochloride according to the equation \\[ \mathrm{Br}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{BrCl}(\mathrm{g}) \\] \(0.2 \mathrm{mol}\) of bromine gas and \(0.2 \mathrm{mol}\) of chlorine gas are introduced into a sealed flask with a volume of \(5.0 \mathrm{dm}^{3}\). Under the conditions of the experiment, \(K=36.0 .\) How much BrCl will be present at equilibrium? (Section 15.4 )

Step-by-Step Solution

Answer

0.34 mol of BrCl will be present at equilibrium.

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{Br}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{BrCl}(\mathrm{g}) \) is given by:\[K = \frac{{[\mathrm{BrCl}]^2}}{{[\mathrm{Br}_2][\mathrm{Cl}_2]}}\]Where \([\mathrm{BrCl}], [\mathrm{Br}_2],\) and \([\mathrm{Cl}_2] \) are the equilibrium concentrations of BrCl, Br₂, and Cl₂, respectively.

Step 2: Calculate Initial Concentrations

The initial moles of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) are both 0.2 mol, in a 5.0 dm³ flask:\[[\mathrm{Br}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ dm}^3} = 0.04 \text{ M}\]\[[\mathrm{Cl}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ dm}^3} = 0.04 \text{ M}\]

Step 3: Set Up the Change in Concentration

Let \( x \) be the change in concentration for BrCl formed. The changes in concentration for the reactants will be \(-x\) for \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \), and \(+2x\) for \( \mathrm{BrCl} \) since the stoichiometry in the balanced equation gives 2 moles of \( \mathrm{BrCl} \) for every mol of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) that reacts.

Step 4: Write Equilibrium Concentrations in Terms of x

At equilibrium, the concentrations are:\[[\mathrm{Br}_2] = 0.04 - x\]\[[\mathrm{Cl}_2] = 0.04 - x\]\[[\mathrm{BrCl}] = 2x\]

Step 5: Substitute into Equilibrium Expression

Substituting into the equilibrium expression:\[K = \frac{(2x)^2}{(0.04 - x)(0.04 - x)}\]Given \( K = 36.0 \):\[36 = \frac{4x^2}{(0.04 - x)^2}\]

Step 6: Solve for x

Let's solve the equation: \[4x^2 = 36(0.04 - x)^2\]Expanding:\[4x^2 = 36(0.0016 - 0.08x + x^2)\]\[4x^2 = 0.0576 - 2.88x + 36x^2\]Simplify and rearrange:\[0 = 32x^2 - 2.88x + 0.0576\]Solve this quadratic equation to find \(x\). Using the quadratic formula,\[x \approx 0.034\].

Step 7: Determine Equilibrium Concentration of BrCl

Since \([\mathrm{BrCl}]= 2x\), substitute found \( x \):\[[\mathrm{BrCl}] = 2(0.034) = 0.068 \text{ M}\]Convert concentration to moles in the 5.0 dm³ flask:\[\text{Moles of } \mathrm{BrCl} = 0.068 \times 5.0 = 0.34 \text{ mol}\].

Key Concepts

Equilibrium Constant

In any reversible chemical reaction, the equilibrium constant, often symbolized as \( K \), plays a crucial role in determining the concentration of reactants and products at equilibrium.

• The equilibrium constant is derived from the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of reactants raised to their stoichiometric coefficients.
• For the reaction \( \mathrm{Br}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{BrCl} \), the expression for \( K \) is given by \[ K = \frac{{[\mathrm{BrCl}]^2}}{{[\mathrm{Br}_2][\mathrm{Cl}_2]}}. \]
• The value of \( K \), which is \( 36.0 \) in this case, helps us predict how far the reaction will proceed to produce \( \mathrm{BrCl} \).

A large \( K \) value means the reaction favors the formation of products, showing that at equilibrium there will be a higher concentration of \( \mathrm{BrCl} \). Understanding the equilibrium constant thus allows us to quantitatively describe chemical equilibria.

Concentration Calculation

Calculating concentrations is essential for setting up the problem and analyzing the equilibrium state.

• Initially, we have 0.2 moles of \( \mathrm{Br}_2 \) and \( \mathrm{Cl}_2 \) each, introduced into a 5.0 dm³ flask. The initial concentration \( [\mathrm{Br}_2]_0 \) and \( [\mathrm{Cl}_2]_0 \) is calculated by dividing the moles by the volume of the flask:
• \[ [\mathrm{Br}_2]_0 = [\mathrm{Cl}_2]_0 = \frac{0.2 \text{ mol}}{5.0 \text{ L}} = 0.04 \text{ M}. \]
• This step sets the stage for the changes that occur as the reaction moves toward equilibrium.

Understanding how to calculate these initial conditions allows you to effectively predict the behavior of the system under study.

Reaction Stoichiometry

In chemical reactions, stoichiometry is vital for understanding how the amount of reactants relates to the formation of products.

• The balanced reaction \( \mathrm{Br}_{2} + \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{BrCl} \) indicates that one mole of \( \mathrm{Br}_{2} \) reacts with one mole of \( \mathrm{Cl}_{2} \) to produce two moles of \( \mathrm{BrCl} \).
• When approaching equilibrium, a change \( x \) occurs where \( \mathrm{Br}_{2} \) and \( \mathrm{Cl}_{2} \) concentrations decrease by \( x \), while \( \mathrm{BrCl} \) increases by \( 2x \).
• The stoichiometric coefficients guide us in setting up these change relationships as: \[ [\mathrm{Br}_2] = 0.04 - x, \]\[ [\mathrm{Cl}_2] = 0.04 - x, \]\[ [\mathrm{BrCl}] = 2x. \]

Understanding stoichiometry helps us predict how concentrations will change as the reaction proceeds.

Quadratic Equation

The analysis of equilibrium often leads to solving a quadratic equation, a mathematical tool that helps determine equilibrium concentrations.

• We use the equilibrium expression and the known value of \( K \) to set up a quadratic equation as follows:
• The relationship \[ K = \frac{(2x)^2}{(0.04 - x)^2} \] leads to \[ 4x^2 = 36(0.04 - x)^2. \]
• Expanding and rearranging gives a quadratic equation: \[ 0 = 32x^2 - 2.88x + 0.0576. \]
• The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is then used to find \( x \), the change in concentration.

Solving this equation gives \( x \approx 0.034 \), leading to the calculation of equilibrium amounts of \( \mathrm{BrCl} \). Mastering this technique is crucial for any chemistry student dealing with equilibria.