Problem 14 - Chemistry The Central Science

Question

The equilibrium constant for the dissociation of molecular iodine, \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g),\) at 800 \(\mathrm{K}\) is \(K_{c}=3.1 \times 10^{-5} .\) (a) Which species predominates at equilibrium \(\mathrm{I}_{2}\) or \(\mathrm{I}\) ? (b) Assuming both forward and reverse reactions are elementary processes, which reaction has the larger rate constant, the forward or the reverse reaction?

Step-by-Step Solution

Answer

(a) I2 predominates at equilibrium. (b) The reverse reaction has the larger rate constant.

Step 1: (a) Determine predominant species

Let's first write the balanced equilibrium equation: \[I_2(g) \rightleftharpoons 2 I(g)\] Let's denote the equilibrium concentrations as: - [I2] = equilibrium concentration of I2 - [I] = equilibrium concentration of I The equilibrium constant Kc is given by: \[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\] We are given Kc = 3.1 x 10^(-5). Now we need to determine the predominant species (I2 or I) at equilibrium. If [I2] > [I], then Kc < 1; If [I2] < [I], then Kc > 1. In our case, Kc < 1, since Kc = 3.1 x 10^(-5). Thus, [I2] > [I], and I2 is the predominant species at equilibrium. Answer to (a): I2 predominates at equilibrium.

Step 2: (b) Compare rate constants of forward and reverse reactions

Since both forward and reverse reactions are elementary processes, we can use the expressions for the rate constants kf (forward reaction) and kr (reverse reaction). The relationship between the equilibrium constant Kc, the rate constants kf and kr is given by: \[K_c = \frac{k_f}{k_r}\] We need to compare kf and kr to determine which reaction has the larger rate constant. If Kc < 1, then kf < kr; If Kc > 1, then kf > kr. As Kc = 3.1 x 10^(-5) < 1, the forward reaction has a smaller rate constant compared to the reverse reaction. Answer to (b): The reverse reaction has the larger rate constant.

Key Concepts

Equilibrium Constant

In chemical reactions, especially those that can reverse, it's essential to know about the equilibrium constant, symbolized as \(K_c\). This value gives us a snapshot of a reaction's state at equilibrium, meaning that the forward and reverse reactions occur at the same rate. This doesn't mean the concentrations are equal, but rather stable over time.
For a given reaction like the dissociation of molecular iodine, \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \), \(K_c\) is represented by:
\[ K_c = \frac{[\text{I}]^2}{[\text{I}_2]} \]This formula tells us the ratio of the concentration of the products to the reactants.

A \(K_c\) value less than 1 indicates that, at equilibrium, the reaction favors the reactants. This means the concentration of \(\text{I}_2\) is higher than that of \(\text{I}\). For example, with \(K_c = 3.1 \times 10^{-5}\), iodine molecules \(\text{I}_2\) are more prevalent than \(\text{I}\) atoms. Understanding \(K_c\) helps predict which species is abundant, guiding chemists in harnessing or controlling reactions effectively.

Reaction Rate

The concept of reaction rate is pivotal in understanding how a chemical reaction progresses over time.
While the equilibrium constant \(K_c\) shows where the equilibrium lies, reaction rates focus on how fast a reaction reaches equilibrium.
In the dissociation of molecular iodine, the reaction can either proceed forward (\(\text{I}_2 \to 2\text{I}\)) or reverse (\(2\text{I} \to \text{I}_2\)). Each direction has a specific rate constant: \(k_f\) for the forward reaction and \(k_r\) for the reverse.

The relationship between these rates and the equilibrium constant is given by:
\[ K_c = \frac{k_f}{k_r} \]Knowing \(K_c < 1\) implies that the reaction rate constant for the forward reaction \(k_f\) is less than that for the reverse \(k_r\). Thus, while the forward process is slower, the reverse happens faster, ensuring more reactant presence at equilibrium.

Molecular Iodine Dissociation

Molecular iodine dissociation is a fascinating process where iodine molecules \(\text{I}_2\) break apart into individual iodine atoms \(\text{I}\). This is represented by the equation:
\( \text{I}_2(g) \rightleftharpoons 2\text{I}(g) \) This reversible reaction is key in several chemical and biological processes.

At 800 K, this reaction has a small equilibrium constant \(K_c\) of \(3.1 \times 10^{-5}\), highlighting that \(\text{I}_2\), the reactant, is more prevalent than \(\text{I}\), the product, at equilibrium.
The understanding of this reaction helps in areas like atmospheric chemistry, where iodine plays a crucial role in environmental processes. By controlling temperature and pressure, chemists can influence the position of equilibrium and the balance between \(\text{I}_2\) and \(\text{I}\), making it useful in both industrial and research settings.