Question

At \(1,500 \mathrm{~K}\), iodine molecules break apart into iodine atoms. Write the equilibrium equation between these two species in two different ways.

Short answer

The equilibrium equations are \(I_2 (g) \rightleftharpoons 2I (g)\) and \(2I (g) \rightleftharpoons I_2 (g)\).

Step-by-step solution

Understand the Reaction

The process describes iodine molecules ( I_2 ) dissociating into iodine atoms ( I ). This reaction is at equilibrium.

Write the Equilibrium Reaction Equation

Represent the equilibrium reaction as:\[I_2 (g) \rightleftharpoons 2I (g)\]

Set up the Equilibrium Constant Expression

The equilibrium constant (K_c) is expressed in terms of the concentrations of products and reactants:\[K_c = \frac{[I]^2}{[I_2]}\]

Write the Equilibrium Reaction with 'Reactants' First

Another way to express the equation is to start with iodine atoms combining to form iodine molecules:\[2I (g) \rightleftharpoons I_2 (g)\]

Set up the Alternative Equilibrium Constant Expression

In this form, the equilibrium constant (K_c') would be:\[K_c' = \frac{[I_2]}{[I]^2}\]

Key concepts

Dissociation Reaction

In chemistry, a dissociation reaction occurs when a compound breaks down into smaller constituents such as ions or atoms. In the case of iodine at high temperatures, iodine molecules (\(I_2\)) undergo a dissociation reaction. When heated to 1,500 K, they separate into individual iodine atoms (\(I\)). This process is essential for understanding how complex structures can break down under certain conditions.

Dissociation reactions play a crucial role in various chemical processes. They can happen:

• When a substance dissolves in a solvent
• Under the influence of heat

Understanding these reactions helps chemists predict how substances will behave in different environments. That's why the study of dissociation in iodine is not only about this single element but also applicable in broader contexts of chemistry.

Iodine Equilibrium

Iodine equilibrium refers to the balance achieved between iodine molecules and iodine atoms in a chemical reaction. At a given temperature, such as the 1,500 K in the exercise, both forms coexist in stable proportions. This state is reached when the rate of iodine molecule formation is equal to the rate of dissociation into iodine atoms.

When considering iodine at equilibrium, we write the reaction equation as follows:

• For dissociating iodine: \[I_2 (g) \rightleftharpoons 2I (g)\]
• For iodine atoms forming molecules: \[2I (g) \rightleftharpoons I_2 (g)\]

These equations show the reversible nature of the reaction. It can proceed in both directions, depending on the conditions present. As a result, iodine equilibrium provides insight into how changing conditions such as temperature or pressure can affect chemical reactions.

Equilibrium Constant Expression

The equilibrium constant expression is a mathematical representation of the ratio of concentrations of products and reactants at equilibrium. For any reaction at equilibrium, the equilibrium constant (\(K_c\)) provides valuable information about the position of equilibrium.

In the case of iodine dissociation, where iodine molecules break down into iodine atoms, we represent the equilibrium constant as: \[K_c = \frac{[I]^2}{[I_2]}\]This expression tells us that the square of the concentration of iodine atoms divided by the concentration of iodine molecules equals the equilibrium constant.

Conversely, if we express the reaction by starting with iodine atoms forming iodine molecules, the equilibrium constant expression becomes:\[K_c' = \frac{[I_2]}{[I]^2}\]Depending on how the reaction is written, the equilibrium constant can give different insights into how concentrations shift during the reaction.

Understanding these expressions helps anticipate the reaction's behavior in various conditions, making it a foundational concept in equilibrium chemistry.