The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4} \mathrm{s}^{-1}\) at a certain temperature. \begin{equation} \begin{array}{l}{\text { a. What is the half-life for this reaction? }} \\\ {\text { b. How long will it take for the concentration of } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { to decrease to }} \\ {25 \% \text { of its initial concentration? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. If the initial concentration of } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { is } 1.00 \mathrm{M}, \text { how long will it take }} \\\ {\text { for the concentration to decrease to } 0.78 \mathrm{M} \text { ? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { d. If the initial concentration of } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { is } 0.150 \mathrm{M}, \text { what is the concen- }} \\\ {\text { tration of } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { after } 2.00 \times 10^{2} \text { s? After } 5.00 \times 10^{2} \text { s? }}\end{array} \end{equation}

Step by step solution

01

Understanding First Order Reactions

First order reactions have the property that the rate is directly proportional to the concentration of the reactant. The half-life of a first order reaction is independent of the initial concentration and can be determined using the formula: \( t_{1/2} = \frac{0.693}{k} \) where \( k \) is the rate constant.

02

Calculating the Half-life

Using the rate constant \( k = 1.42 \times 10^{-4} \, s^{-1} \), calculate the half-life using the formula \( t_{1/2} = \frac{0.693}{k} \).

03

Determining Time for Concentration to Decrease to 25%

For first-order reactions, the relationship between time \( t \), initial concentration \( [A]_0 \), and remaining concentration \( [A] \) at time \( t \) is given by: \( \ln{\frac{[A]_0}{[A]}} = kt \). To find the time it takes for the concentration to decrease to 25% of its initial value, set \( [A] = \frac{[A]_0}{4} \).

04

Calculating Time to Reach Specific Concentrations

Use established relationship to calculate the time taken for the concentration to decrease to \( 0.78 \, M \) from an initial concentration of \( 1.00 \, M \), and then find the time to reach \( 0.78 \, M \). Use the initial concentration of \( 1.00 \, M \) for part c and \( 0.150 \, M \) for part d.

05

Determining Concentration after Specific Times

Using the formula from Step 3, calculate the concentration of \( \mathrm{SO}_2\mathrm{Cl}_2 \) after \( 2.00 \times 10^2 \) seconds and after \( 5.00 \times 10^2 \) seconds with the initial concentration of \( 0.150 \, M \). Solve for \( [A] \) at each time point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics

Understanding chemical kinetics is fundamental to studying how chemical reactions occur. It focuses on reaction rates, which are the changes in the concentration of reactants or products over time. For reactions like the decomposition of \text{SO\(_{2}\)Cl\(_{2}\)}, familiar to our original exercise, kinetics offers insight into how quickly the reaction proceeds under specific conditions, such as temperature and pressure.

Particularly relevant to our discussion, kinetics explores how different reaction orders—such as first-order, second-order, or zero-order—affect the rate. In a first-order reaction, like the one given, the rate is directly proportional to the concentration of a single reactant. In practical terms, this means that if you double the concentration of the reactant, the rate of reaction doubles as well.

Reaction Rate Constant

The reaction rate constant, denoted as 'k', is a crucial aspect of the reaction rate equation for a given chemical reaction. It's unique to each reaction at a given temperature, representing the proportionality constant in the rate law. The value of 'k' tells us how quickly a reaction proceeds. For instance, in the original exercise, the rate constant for the decomposition of \text{SO\(_{2}\)Cl\(_{2}\)} is given as \text{1.42 \times 10\(^{-4}\) s\(^{-1}\)}, which reflects the rate at which \text{SO\(_{2}\)Cl\(_{2}\)} is decomposing per second at a certain temperature.

In the context of a first-order reaction, the magnitude of 'k' directly affects the half-life of the reaction (how long it takes for half of the reactant to be consumed) and is used to estimate how long it will take for the reaction to reach a certain concentration, as demonstrated in the steps for calculating time.

Half-Life of Reaction

The half-life of a chemical reaction is the time required for the concentration of a reactant to decrease to half of its initial value. For a first-order reaction, such as our \text{SO\(_{2}\)Cl\(_{2}\)} decomposition, the half-life is constant, meaning it does not depend on the initial concentration of the reactant. This is dramatically different from higher-order reactions, where half-life varies with initial concentration.

The formula for the half-life of a first-order reaction is elegantly simple: \text{t\(_{1/2}\) = \frac{0.693}{k}}. We see in our step-by-step solution that applying this formula tells us how quickly the concentration of \text{SO\(_{2}\)Cl\(_{2}\)} is halved, a key time frame for understanding the reaction's timescale.

Concentration-Time Relationship

The relationship between the concentration of reactants or products and time is foundational in chemical kinetics. For first-order reactions, like our textbook example, this relationship is represented by the following equation: \text{\text{ln}(\frac{[A]\(_{0}\)}{[A]}) = kt}. Here, \text{[A]\(_{0}\)} is the initial concentration, \text{[A]} is the concentration at time \text{t}, and \text{k} is the rate constant.

Using this relationship allows us to solve various problems related to the concentration at specific time points, as in parts b, c, and d of the original exercise. It's powerful because it connects the observed chemical behavior (the reactant concentration at various times) to an easily applicable mathematical model, which is why it is a central part of many kinetics studies.