Question

For the reaction \(\mathrm{I}_{2}+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g), K_{c}=280 \mathrm{at}\) \(150^{\circ} \mathrm{C}\). Suppose that \(0.500 \mathrm{~mol}\) IBr in a 2.00-L flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}^{2} \mathrm{I}_{2}\), and \(\mathrm{Br}_{2}\) ?

Short answer

The equilibrium concentrations of IBr, I₂, and Br₂ at 150°C are 0.1276 M, 0.0612 M, and 0.0612 M, respectively.

Step-by-step solution

Write down the given information and variables

We are given the following information: - Reaction: \(\mathrm{I}_{2} + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g)\) - Equilibrium constant: \(K_{c} = 280\) - Temperature: \(150^{\circ}\text{C}\) - Initial moles of IBr: \(0.500~\text{mol}\) - Volume of flask: \(2.00~\text{L}\) We need to find the equilibrium concentrations of \(\mathrm{IBr}\), \(\mathrm{I}_{2}\), and \(\mathrm{Br}_{2}\).

Write down the initial concentrations

We can start by finding the initial concentrations of the chemical species. The moles of IBr can be used to find its initial concentration: Initial concentration of IBr = \(\frac{\text{moles of IBr}}{\text{volume of flask}} = \frac{0.500}{2} = 0.250 \, \text{M}\) Since no other information about \(\mathrm{I}_{2}\) and \(\mathrm{Br}_{2}\) is given, we assume their initial concentrations are zero.

Write down the equilibrium expressions

Let's denote the change in the concentration of \(\mathrm{I}_2\) and \(\mathrm{Br}_2\) as x. So, the concentration of \(\mathrm{IBr}\) will decrease by \(2x\) as two moles of IBr are consumed for each mole of I2 and Br2 formed. At equilibrium, we have: \[ [\mathrm{I}_{2}] = x \] \[ [\mathrm{Br}_{2}] = x \] \[ [\mathrm{IBr}] = 0.250 - 2x \]

Write the equilibrium constant expression

The equilibrium constant expression for this reaction, based on the given equilibrium constant and stoichiometry, can be written as: \[ K_c = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_{2}][\mathrm{Br}_{2}]} \]

Substitute the equilibrium expressions and solve

Substitute the equilibrium constants and solve for x (concentration): \[ 280 = \frac{(0.250 - 2x)^2}{x^2} \] Rearrange the equation and solve for x: \[ x^2 = \frac{(0.250 - 2x)^2}{280} \] Solve for x either by factoring or using a quadratic formula. In this case, x would equal approximately \(0.0612\,\text{M}\).

Calculate the equilibrium concentrations

Now that we have x, we can find the equilibrium concentrations of each species: \[ [\mathrm{I}_{2}] = x = 0.0612 \,\text{M} \] \[ [\mathrm{Br}_{2}] = x = 0.0612 \,\text{M} \] \[ [\mathrm{IBr}] = 0.250 - 2x = 0.250 - 2(0.0612) = 0.1276 \,\text{M} \] So, the equilibrium concentrations of \(\mathrm{IBr}\), \(\mathrm{I}_{2}\), and \(\mathrm{Br}_{2}\) are \(0.1276\,\text{M}\), \(0.0612\,\text{M}\), and \(0.0612\,\text{M}\), respectively.

Key concepts

Equilibrium Constant

Understanding the concept of the equilibrium constant is essential in predicting how reactions behave at equilibrium. The equilibrium constant, denoted as \(K_c\), is a numerical value that quantifies the ratio of the concentrations of products to reactants at equilibrium. This ratio is influenced by the stoichiometry of the reaction.For the reaction \(\mathrm{I}_{2} + \mathrm{Br}_{2} \rightleftharpoons 2 \mathrm{IBr} \), the equilibrium constant expression is:

• \(K_c = \frac{[\mathrm{IBr}]^2}{[\mathrm{I}_{2}][\mathrm{Br}_{2}]}\)

The \(K_c\) value remains constant at a particular temperature and provides useful details:

• If \(K_c > 1\), products are favored at equilibrium.
• If \(K_c < 1\), reactants are favored at equilibrium.

In our exercise, a \(K_c\) of 280 at \(150^{\circ}\mathrm{C}\) indicates a product-favored equilibrium, meaning more \(\mathrm{IBr}\) will be present at equilibrium compared to \(\mathrm{I}_{2}\) and \(\mathrm{Br}_{2}\). This knowledge helps guide calculations to find the concentrations of the various chemical species involved.

Chemical Equilibrium

Chemical equilibrium is a state in which the concentrations of reactants and products remain constant over time, indicating a balance in the forward and reverse reactions. When a reaction reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
Equilibrium in a closed system does not mean the reactants and products are of equal concentrations; instead, it signifies no net change in the system. In the reaction \(\text{I}_2 + \text{Br}_2 \rightleftharpoons 2\text{IBr}\), equilibrium does not imply equal amounts of iodine and bromine compared to iodine bromide.

• Reactants and products coexist.
• The system's conditions (pressure, temperature, etc.) remain steady unless disturbed.

Various factors can shift this balance, such as temperature or pressure changes. For example, in our given reaction, maintaining a temperature of \(150^{\circ}\mathrm{C}\) ensures that the equilibrium constant \(K_c\) remains 280, stabilizing concentrations over time.

Reaction Stoichiometry

Stoichiometry involves using mole ratios from balanced chemical equations to calculate the amounts of reactants or products. Our exercise demonstrated this concept with the reaction:
\(\text{I}_2 + \text{Br}_2 \rightleftharpoons 2\text{IBr}\).
Here are key stoichiometric considerations:

• Each mole of \(\text{I}_2\) and \(\text{Br}_2\) produces two moles of \(\text{IBr}\).
• When \(x\) moles of \(\text{I}_2\) and \(\text{Br}_2\) form, \(2x\) moles of \(\text{IBr}\) are consumed.

Therefore, changes in concentration are based on these stoichiometric relationships:

• \([\text{I}_2]\) and \([\text{Br}_2]\) increase by \(x\).
• \([\text{IBr}]\) decreases by \(2x\).

By starting with known quantities, such as 0.500 mol of IBr in a 2.00 L flask, these mole ratios help find equilibrium concentrations of each species involved, giving the full picture of the reaction dynamics.