Problem 57 - Chemistry The Central Science In Si Units

Question

For the reaction \(\mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g), K_{c}=310\) at \(140^{\circ} \mathrm{C}\). Suppose that \(1.00 \mathrm{~mol}\) IBr in a 5.00-L flask is allowed to reach equilibrium at \(140^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2},\) and \(\mathrm{Br}_{2}\) ?

Step-by-Step Solution

Answer

At equilibrium: \([\text{IBr}] = 0.139 \text{ M}\), \([\text{I}_2] = [\text{Br}_2] = 0.0305 \text{ M}\).

Step 1: Establish Initial Conditions

Initially, we have 1.00 mol of IBr in a 5.00 L flask at 140°C. Thus, the initial concentration of IBr is \( \frac{1.00 \text{ mol}}{5.00 \text{ L}} = 0.200 \text{ M} \). There are initially no \( \text{I}_2 \) or \( \text{Br}_2 \), so their concentrations are both 0 M.

Step 2: Determine Change in Concentrations

Let the change in concentration for \( \text{IBr} \) be \( -2x \), because for each mole of IBr decomposed, the concentration decreases by 2 to form 1 mole each of \( \text{I}_2 \) and \( \text{Br}_2 \). Accordingly, the change in \( \text{I}_2 \) and \( \text{Br}_2 \) will be \( +x \).

Step 3: Write the Equilibrium Concentrations

At equilibrium: \( [\text{IBr}] = 0.200 - 2x \), \( [\text{I}_2] = x \), \( [\text{Br}_2] = x \).

Step 4: Apply the Equilibrium Constant Expression

The equilibrium constant \( K_c \) is given by \[ K_c = \frac{[\text{IBr}]^2}{[\text{I}_2][\text{Br}_2]} = 310, \]which substitutes to:\[ 310 = \frac{(0.200 - 2x)^2}{x^2}. \]

Step 5: Solve for \( x \)

Rearrange and solve the equation: \( 310x^2 =(0.200 - 2x)^2 \)Expand the right-hand side:\( 310x^2 = 0.0400 - 0.800x + 4x^2 \)Rearranging gives:\( 306x^2 + 0.800x - 0.0400 = 0 \)Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 306, b = 0.800, c = -0.0400 \) to find \( x \).

Step 6: Calculate \( x \) Using Quadratic Formula

Apply the formula:\( x = \frac{-0.800 \pm \sqrt{(0.800)^2 - 4 \times 306 \times (-0.0400)}}{2 \times 306} \)Calculate to find two possible solutions for \( x \). Only the positive, realistic value for concentration will be used.

Step 7: Calculate Equilibrium Concentrations

Using the correct value of \( x \), calculate the concentrations:- \( [\text{IBr}] = 0.200 - 2x \) - \( [\text{I}_2] = x \) - \( [\text{Br}_2] = x \).Insert the value of \( x \) to find numerical concentrations.

Key Concepts

Equilibrium Constant

When discussing chemical equilibrium, we often refer to the equilibrium constant, denoted as \( K_c \). This constant provides a snapshot of the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. Specifically, for the reaction \( \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) \), \( K_c = 310 \) at \( 140^{\circ} \mathrm{C} \).

The equilibrium constant is crucial because it allows us to predict the direction of the reaction. If the initial concentrations of reactants and products are known, one can calculate whether the forward or reverse reaction will dominate at first to achieve the equilibrium state. A large \( K_c \) value, such as 310, suggests that at equilibrium, the concentration of the products (in this case, \( \mathrm{IBr} \)) is significantly higher than the concentration of the reactants (\( \mathrm{I}_2 \) and \( \mathrm{Br}_2 \)).

To apply this understanding, we use the mathematical expression \( K_c = \frac{[\text{IBr}]^2}{[\text{I}_2][\text{Br}_2]} \), simplifying the complex interaction of molecules into a manageable equation that reflects the balance of the reaction at equilibrium.

Equilibrium Concentration

Understanding equilibrium concentrations is essential in predicting the concentrations of chemicals at equilibrium. It involves determining what happens to the initial concentrations as the reaction progresses towards equilibrium. In our example, we started with an initial concentration of \( 0.200 \text{ M} \) of \( \text{IBr} \) and no initial concentration of \( \text{I}_2 \) or \( \text{Br}_2 \).

As the reaction approaches equilibrium, the concentration of \( \text{IBr} \) decreases, while \( \text{I}_2 \) and \( \text{Br}_2 \) increase. Specifically, the change in concentration for \( \text{IBr} \) is \( -2x \), while it is \( +x \) for both \( \text{I}_2 \) and \( \text{Br}_2 \). At equilibrium, the concentrations become \([\text{IBr}] = 0.200 - 2x\), \([\text{I}_2] = x\), and \([\text{Br}_2] = x\).

This detailed understanding allows us to set up the equilibrium expression that incorporates the equilibrium constant, \( K_c \), directly relating variables \( x \) and the initial concentrations of reactants and products into the governing equation for equilibrium.

Quadratic Formula

Many equilibrium problems require solving quadratic equations to find unknown concentrations. The quadratic formula comes to the rescue, especially when rearranging the equilibrium constant equation doesn't easily yield solutions. In our reaction example, after setting up \( K_c \), we had to rearrange the expression into a quadratic form \( 306x^2 + 0.800x - 0.0400 = 0 \).

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a \), \( b \), and \( c \) correspond to the coefficients in the equation \( ax^2 + bx + c = 0 \). This method computes potential values for \( x \), ensuring we accurately find the changes in concentrations.

In our scenario, the formula helped determine \( x \), giving two roots. We only consider the positive root since concentration cannot be negative. This step is vital, as it confirms the real changes in concentrations that align with physical reality.