Question

Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ \(K\) is 0.190 at \(73^{\circ} \mathrm{C} .\) If you place \(0.500 \mathrm{mol}\) of \(\mathrm{COBr}_{2}\) in a 2.00-L flask and heat it to \(73^{\circ}\) C, what are the equilibrium concentrations of \(\mathrm{COBr}_{2}, \mathrm{CO},\) and \(\mathrm{Br}_{2} ?\) What percentage of the original COBr, decomposed at this temperature?

Short answer

Equilibrium concentrations: [COBr2] = 0.214 M, [CO] = 0.036 M, [Br2] = 0.036 M. About 14.4% of COBr2 decomposed.

Step-by-step solution

Initial Concentration Calculation

First, calculate the initial concentration of \( \mathrm{COBr}_{2} \) in the flask. The initial moles of \( \mathrm{COBr}_{2} \) is 0.500 mol, and the volume of the flask is 2.00 L. Use the formula \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \):\[ [\mathrm{COBr}_{2}]_0 = \frac{0.500\, \mathrm{mol}}{2.00\, \mathrm{L}} = 0.250\, \mathrm{M}\]The initial concentrations of \( \mathrm{CO} \) and \( \mathrm{Br}_{2} \) are 0 M because they have not been formed yet.

Define Change in Concentration

Set up an ICE (Initial, Change, Equilibrium) table to understand how concentrations change. Let \( x \) be the change in concentration of \( \mathrm{COBr}_{2} \) that decomposes:- Initial concentrations: \([\mathrm{COBr}_{2}] = 0.250\), \([\mathrm{CO}] = 0\), \([\mathrm{Br}_{2}] = 0\)- Change: \([\mathrm{COBr}_{2}] = -x\), \([\mathrm{CO}] = +x\), \([\mathrm{Br}_{2}] = +x\)- Equilibrium: \([\mathrm{COBr}_{2}] = 0.250 - x\), \([\mathrm{CO}] = x\), \([\mathrm{Br}_{2}] = x\)

Write the Equilibrium Expression

Use the expression for the equilibrium constant \( K \):\[ K = \frac{[\mathrm{CO}][\mathrm{Br}_{2}]}{[\mathrm{COBr}_{2}]} = \frac{x \times x}{0.250 - x} = \frac{x^2}{0.250 - x} \]We know that \( K = 0.190 \), so we can set up the equation:\[ 0.190 = \frac{x^2}{0.250 - x} \]

Solve for x

Solve the quadratic equation \( 0.190 = \frac{x^2}{0.250 - x} \).Firstly, rearrange and express it as a quadratic equation:\[ x^2 = 0.190 (0.250 - x) \]\[ x^2 = 0.0475 - 0.190x \]\[ x^2 + 0.190x - 0.0475 = 0 \]Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = 0.190 \), and \( c = -0.0475 \) to find \( x \):\[ x = \frac{-0.190 \pm \sqrt{(0.190)^2 - 4 \times 1 \times (-0.0475)}}{2 \times 1} \]\[ x = \frac{-0.190 \pm \sqrt{0.0361 + 0.19}}{2} \]\[ x = \frac{-0.190 \pm 0.262}{2} \]Taking the positive root, \( x \approx 0.036 \).

Calculate Equilibrium Concentrations

With \( x = 0.036 \), substitute back into the equilibrium concentrations:- \([\mathrm{CO}] = x = 0.036\, \mathrm{M} \)- \([\mathrm{Br}_{2}] = x = 0.036\, \mathrm{M}\)- \([\mathrm{COBr}_{2}] = 0.250 - x = 0.214\, \mathrm{M}\)

Calculate Percentage Decomposed

Find the percentage of \( \mathrm{COBr}_{2} \) that decomposed:\[\text{Percentage decomposed} = \left( \frac{x}{[\mathrm{COBr}_{2}]_0} \right) \times 100 = \left( \frac{0.036}{0.250} \right) \times 100 \approx 14.4\%\]

Key concepts

Equilibrium Constant

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. The equilibrium constant \( K \) is a numerical value that describes the ratio of product concentrations to reactant concentrations at equilibrium for a particular reaction. It provides insight into the position of equilibrium:

• If \( K \) > 1, the products are favored.
• If \( K \) < 1, the reactants are favored.
• If \( K \) = 1, neither side is favored predominantly.

Knowing the equilibrium constant helps us understand how much a reaction occurs and can predict the concentrations of substances once the reaction is at equilibrium. For the decomposition of carbonyl bromide in this problem, \( K \) has a value of 0.190, indicating that the reactants are favored at 73°C.

ICE Table

An ICE table is a useful tool for tracking concentration changes in a chemical reaction. The acronym "ICE" stands for Initial, Change, and Equilibrium:

• Initial: Concentrations of all substances at the start of the reaction.
• Change: The change in concentration as the system reaches equilibrium, often expressed in terms of \( x \), a variable that represents the amount that reacts or is formed.
• Equilibrium: The final concentrations of substances when the reaction is at equilibrium.

In this example, we start with an initial concentration of \([\text{COBr}_2]=0.250\, \text{M}\) and no \([\text{CO}]\) or \([\text{Br}_2]\) because they have not formed yet. As the reaction proceeds, \( x \) moles of \( \text{COBr}_2 \) decompose, forming \( x \) moles of both \( \text{CO} \) and \( \text{Br}_2 \). At equilibrium, the reaction has balanced out these concentrations accordingly.

Concentration Calculation

Calculating concentrations of substances at equilibrium involves setting up the equilibrium expression using the equilibrium constant. For our reaction, the expression is:\[K = \frac{[\text{CO}][\text{Br}_2]}{[\text{COBr}_2]}\]By substituting the changes \(x\) into this expression using the ICE table, we get:\[0.190 = \frac{x^2}{0.250 - x}\]Solving this equation involves rearranging to form a quadratic equation and using the quadratic formula. This gives us the equilibrium concentrations:

• \([\text{CO}] = x = 0.036\, \text{M}\)
• \([\text{Br}_2] = x = 0.036\, \text{M}\)
• \([\text{COBr}_2] = 0.250 - x = 0.214\, \text{M}\)

These values describe how the concentrations adjust during the reaction until the equilibrium condition is reached.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more components. In the case of carbonyl bromide decomposing, it splits into carbon monoxide and bromine:\[\text{COBr}_2(\text{g}) \rightleftharpoons \text{CO}(\text{g}) + \text{Br}_2(\text{g})\]Such reactions are significant in understanding chemical processes and industrial applications where breaking down compounds is necessary. The equilibrium constant indicates whether the decomposition is significant or not under specific conditions—here, a relatively low \( K \) value of 0.190 suggests that at 73°C, only a limited fraction of \( \text{COBr}_2 \) has decomposed, about 14.4%. Understanding these principles helps us manage reactions in laboratories and real-world scenarios effectively.