Problem 123 - Chemistry The Molecular Science

Question

A student studies the equilibrium $$ \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) $$ at a high temperature. She finds that the total pressure at equilibrium is \(40 . \%\) greater than it was originally, when only \(\mathrm{I}_{2}\) was present at a pressure of \(1.00 \mathrm{~atm}\) in the same sealed container. Calculate \(K_{\mathrm{p}}\).

Step-by-Step Solution

Answer

\( K_p \approx 1.07 \)

Step 1: Initial setup of pressures

Initially, only \( \text{I}_2 \) is present at 1.00 atm. At equilibrium, the total pressure increases by 40%, so it becomes \( 1.00 \times 1.40 = 1.40 \) atm.

Step 2: Define change in pressures

Let the change in pressure due to the dissociation of \( \text{I}_2 \) be \( x \). Thus, the pressures of \( \text{I}_2 \) and \( \text{I} \) at equilibrium are \( 1.00 - x \) atm and \( 2x \) atm, respectively, because \( 2 \) moles of \( \text{I} \) form for every mole of \( \text{I}_2 \) dissociated.

Step 3: Expression for total equilibrium pressure

Total pressure at equilibrium will be the sum of \( \text{I}_2 \) and \( \text{I} \) pressures: \( (1.00 - x) + 2x = 1.40 \) atm.

Step 4: Solve for x

Simplifying the total pressure expression gives \( 1.00 + x = 1.40 \). Thus, \( x = 0.40 \).

Step 5: Calculate partial pressures at equilibrium

With \( x = 0.40 \), the partial pressures are: \( \text{I}_2 = 1.00 - 0.40 = 0.60 \) atm and \( \text{I} = 2 \times 0.40 = 0.80 \) atm.

Step 6: Calculate the equilibrium constant Kp

The equilibrium constant \( K_p \) is given by \( K_p = \left(\frac{P_{\text{I}_2}}{P_{\text{I}}^2}\right) \). Substituting the equilibrium pressures gives \( K_p = \frac{(0.80)^2}{0.60} = 1.067 \approx 1.07 \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products no longer change over time. This does not mean that the reactants and products are in equal amounts, but rather that their rates of formation and decomposition are equal. In other words, the forward and reverse reactions occur at the same rate.

In our example of the iodine dissociation reaction \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \), chemical equilibrium is reached when the rate at which \( \text{I}_2 \) molecules break down into iodine atoms equals the rate at which iodine atoms recombine to form \( \text{I}_2 \). This balance results in stable amounts of both \( \text{I}_2 \) and \( \text{I} \) at the equilibrium state.

• The system must be closed, meaning no substances enter or leave the container.
• Equilibrium can be disturbed by changing conditions such as pressure, volume, or temperature.
• Reaches a point where the Gibbs free energy is minimized.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. In a system with a mixture of gases, like the iodine dissociation, the total pressure is the sum of individual partial pressures of the gases.

In the dissociation reaction \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \), initially, only \( \text{I}_2 \) is present at 1.00 atm. When equilibrium is reached, the total pressure increased to 1.40 atm—40% greater than the original. To find individual partial pressures at equilibrium:

• Let the pressure change due to the reaction be \( x \).
• The partial pressure of \( \text{I}_2 \) is reduced to \( 1.00 - x \).
• The partial pressure for \( \text{I} \), which forms from the dissociation of \( \text{I}_2 \), becomes \( 2x \).

With \( x \) calculated as 0.40 atm, the partial pressures at equilibrium are determined as follows:
- \( \text{I}_2 = 0.60 \) atm.
- \( \text{I} = 0.80 \) atm.

Dissociation Reaction

A dissociation reaction is a process where a compound separates into two or more components. These reactions are especially common with gases and liquids and often involve breaking of specific chemical bonds.

The dissociation of iodine gas, \( \text{I}_2(g) \), into individual iodine atoms, \( \text{I}(g) \), is one such reaction. In this case, each molecule of \( \text{I}_2 \) dissociates into two iodine atoms, a process that affects the equilibrium position and the total pressure inside the reaction vessel.

• Dissociation reactions often require energy input to overcome the bond energies.
• The extent of dissociation can significantly influence the pressure relationships in the system.
• In this exercise, dissociation results in an increase in total pressure as new, additional molecules are formed.

Understanding how these reactions work helps to predict changes in pressure, composition, and even the equilibrium constant \( K_p \), which in this reaction was calculated to be 1.07.