Problem 52 - Chemistry The Central Science In Si Units

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 \(\mathrm{BrCl}\) ?

Step-by-Step Solution

Answer

Equilibrium concentrations are \( [\mathrm{Br}_2] = 0.0353 \), \( [\mathrm{Cl}_2] = 0.1353 \), \( [\mathrm{BrCl}] = 0.096 \) mol/L.

Step 1: Write the Initial Concentrations

We need to calculate the initial concentrations of each reactant. The initial moles given are 0.25 mol for \(\mathrm{Br}_2\) and 0.55 mol for \(\mathrm{Cl}_2\), and they are placed in a 3.0 L container.The initial concentration of \(\mathrm{Br}_2\) is \(\frac{0.25}{3.0} = 0.0833\, \mathrm{mol/L}\).The initial concentration of \(\mathrm{Cl}_2\) is \(\frac{0.55}{3.0} = 0.1833\, \mathrm{mol/L}\).The initial concentration of \(\mathrm{BrCl}\) is 0 mol/L since it is not present initially.

Step 2: Define the Change in Concentrations

Define changes in concentrations as the reaction proceeds to equilibrium. Let \(x\) represent the amount of \(\mathrm{Br}_2\) and \(\mathrm{Cl}_2\) that reacts to form \(2\mathrm{BrCl}\).The changes in concentration will be:- For \(\mathrm{Br}_2\): \(-x\)- For \(\mathrm{Cl}_2\): \(-x\)- For \(\mathrm{BrCl}\): \(+2x\) (since 2 moles of \(\mathrm{BrCl}\) are produced per mole of \(\mathrm{Br}_2\) and \(\mathrm{Cl}_2\) that reacts).

Step 3: Write the Equilibrium Concentrations

At equilibrium, the concentrations will be:- \(\mathrm{Br}_2\) = \(0.0833 - x\)- \(\mathrm{Cl}_2\) = \(0.1833 - x\)- \(\mathrm{BrCl}\) = \(2x\)

Step 4: Set Up the Equilibrium Expression

Using the equilibrium constant \(K_c\), write down the equilibrium expression: \[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]Substitute the equilibrium concentrations into this expression:\[ 7.0 = \frac{(2x)^2}{(0.0833 - x)(0.1833 - x)} \]

Step 5: Solve for x

Solve the equation using algebra. First, simplify:\[ 7.0 = \frac{4x^2}{(0.0833 - x)(0.1833 - x)} \]Cross-multiply to clear the fraction: \[ 7(0.0833 - x)(0.1833 - x) = 4x^2 \]Expand and simplify:\[ 7(0.015259 - 0.0833x - 0.1833x + x^2) = 4x^2 \]\[ 0.106813 - 0.9338x + 7x^2 = 4x^2 \]Move all terms to one side:\[ 3x^2 - 0.9338x + 0.106813 = 0 \]Solve this quadratic equation for \(x\). Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -0.9338\), and \(c = 0.106813\).

Step 6: Identify the Valid x Value

Calculate the roots of the quadratic equation:Using the quadratic formula:\[ x = \frac{0.9338 \pm \sqrt{(-0.9338)^2 - 4 \times 3 \times 0.106813}}{6} \]Solving gives two possible \(x\) values. Select the physically valid \(x\) where concentrations remain positive, which should be around \(x = 0.048\).

Step 7: Calculate the Equilibrium Concentrations

Substitute \(x = 0.048\) into the equilibrium concentration expressions:- \(\mathrm{Br}_2\) = \(0.0833 - 0.048 = 0.0353\, \mathrm{mol/L}\)- \(\mathrm{Cl}_2\) = \(0.1833 - 0.048 = 0.1353\, \mathrm{mol/L}\)- \(\mathrm{BrCl}\) = \(2 \times 0.048 = 0.096\, \mathrm{mol/L}\)

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K_c\), is a fundamental concept in chemical equilibrium. At a given temperature, it is a fixed value that relates the concentrations of products and reactants at equilibrium in a chemical reaction. For a reaction such as \(\mathrm{Br}_{2}(g) + \mathrm{Cl}_{2}(g) \rightleftharpoons 2\mathrm{BrCl}(g),\)the equilibrium constant expression is:\[K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]}.\]The equilibrium constant provides insight into the position of equilibrium:

• If \(K_c\) is much greater than 1, the reaction favors the formation of products.
• If \(K_c\) is much less than 1, the reaction favors the formation of reactants.

In our exercise, \(K_c = 7.0\), indicating that at \(400 \mathrm{~K}\), the products are favored, suggesting significant formation of \(\mathrm{BrCl}\) at equilibrium.

Initial Concentration

Initial concentration refers to the concentration of reactants or products in a chemical reaction before the reaction has reached equilibrium. These are calculated using the formula:\[\text{Initial concentration} = \frac{\text{moles of substance}}{\text{volume of container}}.\]In this problem, the initial moles and volumes are given:

• \(0.25 \mathrm{~mol}\) of \(\mathrm{Br}_2\)
• \(0.55 \mathrm{~mol}\) of \(\mathrm{Cl}_2\)

Both are in a \(3.0 \mathrm{~L}\) container.- Initial concentration of \(\mathrm{Br}_2\): \(\frac{0.25}{3.0} = 0.0833\, \mathrm{mol/L}\)- Initial concentration of \(\mathrm{Cl}_2\): \(\frac{0.55}{3.0} = 0.1833\, \mathrm{mol/L}\)- \(\mathrm{BrCl}\) starts at \(0\, \mathrm{mol/L}\) since it is not initially present.
Knowing these helps in setting up the chemical reaction and determining how the reactants convert to products.

Quadratic Formula

The quadratic formula is a standard mathematical tool used to solve quadratic equations of the form:\[ax^2 + bx + c = 0.\]The solutions for \(x\) are derived using:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]In the context of chemical equilibrium, quadratic equations often arise when substituting equilibrium concentrations into the equilibrium constant expression. For the given problem:
The quadratic equation derived is:\[3x^2 - 0.9338x + 0.106813 = 0.\]Here, \(a = 3\), \(b = -0.9338\), \(c = 0.106813\). Solving this yields possible \(x\) values, representing reacted or produced concentrations. You choose the \(x\) that results in non-negative concentrations, keeping the solution physically meaningful.

Reaction Quotient

The reaction quotient \(Q_c\) is similar to the equilibrium constant but applied at any point in the reaction, not just at equilibrium. It compares the state of a reaction at any moment to its equilibrium state. The formula follows the form of the equilibrium constant expression:\[Q_c = \frac{[\text{products reactive coefficients]]}{[\text{reactants reactive coefficients]}}.\]

• If \(Q_c < K_c\), the reaction proceeds forward to reach equilibrium.
• If \(Q_c = K_c\), the reaction is already at equilibrium.
• If \(Q_c > K_c\), the reaction will shift backwards to reach equilibrium.

Conceptually, \(Q_c\) helps predict which direction the reaction will proceed to achieve equilibrium conditions.

Le Chatelier's Principle

Le Chatelier's Principle is a key concept in understanding how equilibrium shifts in response to changes in conditions. It states that if a change is made to the conditions of a system at equilibrium, the system adjusts to partly counteract that change, aiming to restore equilibrium. Changes include:

• Concentration: Adding or removing substances shifts the equilibrium to re-balance product and reactant concentrations.
• Pressure: For gaseous reactions, increasing pressure by reducing volume causes a shift towards the side with fewer moles of gas.
• Temperature: Raising temperature favors the endothermic direction, while lowering favors the exothermic.

Le Chatelier's Principle is immensely useful for predicting how a system will react to external changes, making it crucial for practical applications in chemical reactions.