Problem 40 - Chemistry and Chemical Reactivity

Question

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K_{\mathrm{c}}\) is 0.190 at \(73^{\circ} \mathrm{C} .\) Suppose you place \(0.500 \mathrm{mol}\) of COBr, in a \(2.00-\) I. flask and heat it to \(73^{\circ} \mathrm{C}\) (see Study Question 17 ). After equilibrium has been achieved, you add an additional 2.00 mol of \(\mathrm{CO}\) (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of \(\mathrm{COBr}_{2}, \mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) (c) How has the addition of CO affected the percentage of COBr \(_{2}\) that decomposed?

Step-by-Step Solution

Answer

(a) Adding CO shifts equilibrium left, decreasing decomposition. (b) Calculate new equilibrium concentrations using the additional 1.038 M CO. (c) Addition of CO reduces the percentage of COBr2 decomposed.

Step 1: Write the Initial Reaction and Setup the ICE Table

The chemical reaction is given as \( \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \). Initially, we have \( [\operatorname{COBr}_2] = \frac{0.500\text{ mol}}{2.00\text{ L}} = 0.250\, \mathrm{M} \), and both products start at 0 M. When the system reaches equilibrium, let \( x \) be the change in concentration of \( \operatorname{COBr}_2 \) that decomposes, making the concentrations \( [\operatorname{COBr}_2] = 0.250 - x \), \([\operatorname{CO}] = x\), and \([\operatorname{Br}_2] = x\).

Step 2: Apply the Equilibrium Constant Expression

The equilibrium constant expression for the reaction is \( K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\operatorname{COBr}_2]} = 0.190 \). Substitute the equilibrium concentrations to get the equation \( \frac{x^2}{0.250-x} = 0.190 \). Solve for \( x \) to find the equilibrium concentrations before the CO addition.

Step 3: Solve for x at Initial Equilibrium

Solve the equation \( x^2 = 0.190 (0.250 - x) \) to find \( x \). This forms the quadratic equation \( x^2 + 0.190x - 0.0475 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), find \( x \). Once calculated, \( x \approx 0.038 \).

Step 4: Calculate Equilibrium Concentrations before CO Addition

Using \( x = 0.038 \), determine the equilibrium concentrations: \([\operatorname{COBr}_2] = 0.250 - 0.038 = 0.212\, \mathrm{M} \), \([\mathrm{CO}] = 0.038\, \mathrm{M} \), \([\mathrm{Br}_2] = 0.038\, \mathrm{M} \) before the addition of \( \mathrm{CO} \).

Step 5: Incorporate Additional CO and Re-calculate Equilibrium

Adding 2.00 mol of CO to the 2.00 L flask increases the concentration of CO by \( \frac{2.00\text{ mol}}{2.00\text{ L}} = 1.00\, \mathrm{M} \). New \([\mathrm{CO}] = 0.038 + 1.00 = 1.038\, \mathrm{M} \). New equilibrium concentrations are found by letting \( y \) be the amount of \( \operatorname{COBr}_2 \) that decomposes after CO addition and using \( K_c = \frac{(1.038-y)(0.038+y)}{0.212+y} = 0.190 \).

Step 6: Solve for y and Final Concentrations

Expand and solve the new equilibrium expression to find \( y \). Initially, the expression simplifies to a quadratic, which can be solved numerically or using an appropriate approximation technique for \( y \). Once \( y \) is found, recalculate: \([\operatorname{COBr}_2] = 0.212 + y, [\mathrm{CO}] = 1.038 - y, [\mathrm{Br}_2] = 0.038 + y\).

Step 7: Impact of CO Addition on COBr2 Decomposition

Calculate the initial and final percentage decomposition of \( COBr_2 \). Initial decomposition is \( \frac{0.038}{0.250} \times 100\% \). After adding CO, use \( y \) to determine the final amount decomposed relative to the new \([\operatorname{COBr}_2]\) at equilibrium and compare it to the initial decomposition.

Key Concepts

Le Chatelier's principle

Le Chatelier's principle is a key concept in understanding how chemical equilibria respond to changes in conditions. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. This could mean changes in concentration, temperature, or pressure. In our case, adding more carbon monoxide (CO) affects the equilibrium of the reaction \[ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \],which initially achieves equilibrium as per the description. When more CO is added, according to Le Chatelier's principle, the equilibrium will shift to the left to reduce the increased concentration of CO. This means more \( \operatorname{COBr}_2 \) is created until a new equilibrium is established. Understanding this principle helps predict how a reaction will proceed under different conditions without needing extensive calculations.

Equilibrium constant

The equilibrium constant \( K_c \) provides insight into the ratio of product concentrations to reactant concentrations at equilibrium for a given reaction. For a reaction like \( \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \), the equation is expressed as:\[ K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\operatorname{COBr}_2]} \].This constant remains unchanged with concentration changes because it is tied to temperature conditions (given as 73°C here). When more CO is introduced, while the concentrations shift to adjust, the ratio of these concentrations at a new equilibrium will still yield the same \( K_c \). This concept underpins how we define the 'balance' of reactions and predict outcomes based on initial conditions and changes applied through external factors.

ICE table

An ICE table is a helpful tool in chemistry for organizing information about the initial concentrations, changes, and equilibrium concentrations of reactants and products in a reaction. The term 'ICE' stands for Initial, Change, and Equilibrium. Using the reaction \[ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \],an ICE table helps visualize these concentration changes. Initially, we have 0.250 M \( \operatorname{COBr}_{2} \) and 0 M for both products. As the reaction progresses, let \( x \) represent the amount of \( \operatorname{COBr}_2 \) decomposing, causing the concentrations to shift to 0.250 - \( x \) for \( \operatorname{COBr}_2 \), and \( x \) for both CO and \( \text{Br}_2 \) at equilibrium. It's essential to understand this table not only for keeping track of values but also for solving complex equilibrium problems in chemistry where multiple variables shift.

Quadratic equations in chemistry

Quadratic equations often appear in chemistry when solving for equilibrium concentrations involving an unknown change, such as \( x \) in our example. In this context, setting up an equilibrium constant expression results in a quadratic equation of the form\[ x^2 + bx + c = 0 \].For the decomposition of \( \operatorname{COBr}_{2} \), after substituting into our equilibrium constant expression \( K_c = 0.190 \),the quadratic equation becomes \( x^2 + 0.190x - 0.0475 = 0 \). To solve, apply the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].Even for students not adept at algebra, understanding that these equations model the balance of chemical reactions is crucial. Often you'll find one solution is negative, which doesn't make sense physically as a concentration, so the positive solution is considered. Solving these equations allows for determining precise values of \( x \), hence the equilibrium concentrations, crucial for understanding how systems reach equilibrium.