Question

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

Short answer

The half-life for this reaction is approximately \(2.56~\text{s}\). After \(5.12~\text{s}\), the concentration of \(\mathrm{I}_{2}(g)\) remaining is approximately \(0.016~\text{M}\).

Step-by-step solution

Part (a) - Calculate the half-life of the reaction

For first-order reactions, the half-life is given by the equation: \[t_{1/2} = \frac{0.693}{k}\] where \(t_{1/2}\) is the half-life and \(k\) is the rate constant. In this case, the rate constant is given as \(0.271~\text{s}^{-1}\). Plugging the values into the equation, we get: \[t_{1/2} = \frac{0.693}{0.271~\text{s}^{-1}}\] Calculating the half-life, we find: \[t_{1/2} \approx 2.56~\text{s}\] The half-life for this reaction is approximately \(2.56~\text{s}\).

Part (b) - Calculate the concentration of \(\text{I}_{2}\) after \(5.12~\text{s}\)

To determine the concentration of \(\text{I}_{2}\) after a certain time, we use the integrated rate law for first-order reactions: \[\ln{\frac{[\text{I}_{2}]_{t}}{[\text{I}_{2}]_{0}}} = -kt\] where \([\text{I}_{2}]_{t}\) is the concentration at time \(t\), \([\text{I}_{2}]_{0}\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time elapsed. In this case, the initial concentration is \(0.050~\text{M}\), the rate constant is \(0.271~\text{s}^{-1}\), and the time elapsed is \(5.12~\text{s}\). Plugging these values into the equation, we get: \[\ln{\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}}} = -0.271~\text{s}^{-1}(5.12~\text{s})\] Now, we just need to solve for \([\text{I}_{2}]_{t}\): \[\ln{\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}}} \approx -1.387\] \[\frac{[\text{I}_{2}]_{t}}{0.050~\text{M}} = e^{-1.387}\] \[ [\text{I}_{2}]_{t} \approx 0.050~\text{M} \cdot e^{-1.387}\] Calculating the concentration, we find: \[[\text{I}_{2}]_{t} \approx 0.016~\text{M}\] After \(5.12~\text{s}\), the concentration of \(\mathrm{I}_{2}(g)\) remaining is approximately \(0.016~\text{M}\).

Key concepts

First-Order Reaction

In chemical kinetics, first-order reactions are a fundamental type that describes how the rate of a reaction is directly proportional to the concentration of a single reactant. This means that as the concentration of the reactant decreases, so does the rate of the reaction. These reactions are commonly represented by the following general form:

• A single substance decomposes into products: A → Products
• The rate law of the reaction: rate = k[A]

In this setup, - **A** stands for the reactant, - **k** is the rate constant, - **[A]** is the concentration of the reactant. The simplicity of first-order reactions makes them particularly useful for understanding the principles of reaction kinetics.

Half-Life Calculation

The concept of half-life is an essential aspect of first-order reactions and kinetics. It refers to the time required for half of the reactant to be converted into products. This property is unique because, for first-order reactions, the half-life remains constant irrespective of the initial concentration of the reactant.The formula for calculating the half-life, given a first-order reaction, is:\[t_{1/2} = \frac{0.693}{k}\]where:- **\(t_{1/2}\)** is the half-life,- **\(k\)** is the rate constant.For instance, when dealing with a reaction involving molecular iodine dissociating, the use of this formula allows for quick determination of how long it takes for half of the iodine to convert, providing insights into the reaction's behavior over time.

Rate Constant

In the context of reaction kinetics, the rate constant, denoted as **k**, plays a crucial role in understanding the speed of a chemical reaction. In a first-order reaction, this constant makes the connection between the rate of the reaction and the concentration of the reactant through the rate law:\[ ext{rate} = k[A]\]The value of **k** is influenced by:- **Temperature**: Higher temperatures usually increase the rate constant.- **Nature of reactants**: Different chemical species react at different rates.In the exercise example of iodine dissociation at a temperature of \(625 \mathrm{~K}\), the rate constant \(k\) was given as \(0.271 \mathrm{~s}^{-1}\). Understanding this helps chemists and students predict how fast reactions will proceed under various conditions.

Integrated Rate Law

The integrated rate law provides a mathematical description of how the concentration of reactants changes over time in a chemical reaction. For a first-order reaction, the integrated rate law is:\[\ln{\left(\frac{[A]_t}{[A]_0}\right)} = -kt\]Here,- **\([A]_t\)** is the concentration of the reactant at time **t**,- **\([A]_0\)** is the initial concentration,- **k** is the rate constant,- **t** is the elapsed time.This formula is very useful for calculating the concentration of a reactant left after a certain period, which helps in understanding the progress of the reaction. For example, using the iodine dissociation problem, it can predict the remaining concentration after a given time, showing the decay of the iodine molecules efficiently over time with precise calculations.