Problem 42 - Chemistry and Chemical Reactivity

Question

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \mathrm{COBr}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K_{c}\) is 0.190 at \(73^{\circ} \mathrm{C}\). Suppose you place \(0.500 \mathrm{mol}\) of \(\mathrm{COBr}_{2}\) in a 2.00 -L. 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 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 \(\mathrm{COBr}_{2}\) that decomposed?

Step-by-Step Solution

Answer

Adding CO shifts equilibrium to the left, decreasing decomposition of COBr₂.

Step 1: Calculate Initial Concentrations

To solve the problem, first calculate the initial concentration of \( \mathrm{COBr}_2 \) initially placed in the flask. Since 0.500 mol of \( \mathrm{COBr}_2 \) is placed in a 2.00-L flask, its initial concentration is \( \frac{0.500 \text{ mol}}{2.00 \text{ L}} = 0.250 \text{ M} \). The initial concentrations of both \( \mathrm{CO} \) and \( \mathrm{Br}_2 \) are 0 M, as they are products of the reaction.

Step 2: Set Up the Initial ICE Table

An ICE (Initial, Change, Equilibrium) table helps to organize initial concentrations, changes, and those at equilibrium.\[\begin{array}{c|c|c|c} & \mathrm{COBr}_2 & \mathrm{CO} & \mathrm{Br}_2 \\hline\text{Initial (M)} & 0.250 & 0 & 0 \\hline\text{Change (M)} & -x & +x & +x \\hline\text{Equilibrium (M)} & 0.250-x & x & x \\end{array}\]We assume \(x\) mol/L of \( \mathrm{COBr}_2 \) decomposes.

Step 3: Apply Equilibrium Constant Expression

Use the equilibrium constant \( K_c \) expression:\[ K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\mathrm{COBr}_2]} = 0.190 \]Substitute the values from the ICE table:\[ 0.190 = \frac{x^2}{0.250-x} \]

Step 4: Solve Quadratic Equation

Solve the equilibrium equation for \( x \):\[ 0.190(0.250 - x) = x^2 \]\[ 0.0475 - 0.190x = x^2 \]Rearrange into a standard quadratic form:\[ x^2 + 0.190x - 0.0475 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 0.190, c = -0.0475 \). Calculate \( x \) (assume the positive root):\[ x \approx 0.200 \text{ M} \]

Step 5: Determine Equilibrium Concentrations Before Adding CO

Calculate the equilibrium concentrations:- \([\mathrm{COBr}_2] = 0.250 - 0.200 = 0.050 \text{ M} \)- \([\mathrm{CO}] = 0.200 \text{ M} \)- \([\mathrm{Br}_2] = 0.200 \text{ M} \)

Step 6: Add CO to the System

Add 2.00 mol of \( \mathrm{CO} \) to the system. The new total \( \mathrm{CO} \) concentration is:\[ [\mathrm{CO}] = 0.200 + \frac{2.00}{2.00} = 0.200 + 1.000 = 1.200 \text{ M} \] The reaction will shift towards the left to reach new equilibrium.

Step 7: Set Up New ICE Table After Adding CO

ICE table for the new scenario:\[\begin{array}{c|c|c|c} & \mathrm{COBr}_2 & \mathrm{CO} & \mathrm{Br}_2 \\hline\text{Initial (M)} & 0.050 & 1.200 & 0.200 \\hline\text{Change (M)} & +x & -x & -x \\hline\text{Equilibrium (M)} & 0.050+x & 1.200-x & 0.200-x \\end{array}\]

Step 8: Solve for New Equilibrium Concentrations

Set up the equilibrium expression again:\[ 0.190 = \frac{(1.200-x)(0.200-x)}{0.050+x} \]Simplify and solve this equation for \( x \).This will require using algebraic methods to find \( x \). Assume that the changes are small compared to the initial concentration of \( \mathrm{CO} \), then approximate \( x \). This often requires the assumption simplifies calculations.

Step 9: Find New Percent Decomposition of COBr2

Calculate the percentage decomposition of \( \mathrm{COBr}_2 \) using:\[ \left( \frac{\text{amount decomposed}}{\text{initial amount}} \right) \times 100\%\]Initially, decomposition was 0.200. After adding CO, the degree of decomposition may have decreased based on the new value of \(x\), compare the change.

Key Concepts

Le Chatelier's Principle

Have you ever wondered how a chemical equilibrium reacts to changes in the system? This is where Le Chatelier's Principle becomes incredibly useful. Leitner’s Principle helps us predict the direction in which an equilibrium shifts when the conditions are altered. For instance, if the concentration of a reactant or product in an equilibrium system is changed, the system will react by shifting its position to counteract this change.

In our example, when additional CO is added to the equilibrium mixture, the system experiences an increase in CO concentration. Le Chatelier's Principle indicates that the reaction will attempt to decrease this stress by shifting the equilibrium position to the left, forming more reactants (COBr extsubscript{2}).

• Systems will shift in the opposite direction of the change.
• For a concentration increase, the reaction moves towards the side that will consume the added component.
• The goal is always to restore equilibrium balance.

Understanding this principle provides critical insight into predicting how changes will affect equilibrium.

Reaction Quotient

The reaction quotient (Q extsubscript{c}) is similar to the equilibrium constant but applied to any state of a reaction, not necessarily at equilibrium. It's a snapshot of the system at a given moment. To find the reaction quotient, use the same formula as for the equilibrium constant, substituting the concentrations at that moment.

Mathematically, for the reaction \[ \mathrm{COBr}_2(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{g}) \], the reaction quotient is expressed as:

\[ Q_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\mathrm{COBr}_2]} \]

When adding additional CO, Q extsubscript{c} changes as the concentrations shift. By comparing Q extsubscript{c} to K extsubscript{c}, you can predict the system’s behavior:

• If Q extsubscript{c} < K extsubscript{c}, the system will shift to the right (towards products).
• If Q extsubscript{c} = K extsubscript{c}, the system is at equilibrium.
• If Q extsubscript{c} > K extsubscript{c}, the system will shift to the left (towards reactants).

The reaction quotient offers a "real-time" measure of where the reaction lies relative to equilibrium.

Equilibrium Constant

The equilibrium constant (K extsubscript{c}) at a given temperature describes the ratio of product to reactant concentrations at equilibrium for a reversible reaction. It is a fundamental quantity that remains constant for a specific reaction at a specific temperature.

In our reaction of decomposing COBr extsubscript{2}, K extsubscript{c} is given as 0.190 at 73°C. This means that at equilibrium, the concentration of products raised to their stoichiometric coefficients divided by reactants, also raised to their coefficients, equals 0.190. The expression is:

\[ K_c = \frac{[\mathrm{CO}][\mathrm{Br}_2]}{[\mathrm{COBr}_2]} \]

The equilibrium constant provides a snapshot of the balance between reactants and products in equilibrium. Knowing K extsubscript{c} helps predict how the concentrations will change when the system is disturbed, such as when additional CO is introduced.

Concentration Changes

Concentration changes are a common way to disturb chemical equilibrium. When you add or remove a substance, the equilibrium position shifts to counteract these changes, according to Le Chatelier's Principle.

In the scenario of adding 2.00 mol of CO to the system, the concentration of CO initially increases significantly. This shifts the equilibrium position to the left in response, leading to a decrease in the concentration of CO over time as more COBr extsubscript{2} and Br extsubscript{2} form to re-establish equilibrium.

Key points to understand:

• Deviations in concentration prompt the system to re-establish the existing K extsubscript{c} .
• The magnitude of concentration changes determines how far the equilibrium shifts.
• Over time, the conditions return to a new state of equilibrium with adjusted concentrations.

These insights into concentration changes are pivotal for predicting and understanding equilibrium dynamics.