/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 The decomposition of \(\mathrm{S... [FREE SOLUTION] | 91Ó°ÊÓ

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The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ The rate constant for the reaction is \(2.8 \times 10^{-3} \mathrm{min}^{-1}\) at \(600 \mathrm{K} .\) If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.24 \times 10^{-3}\) mol/L, how long will it take for the concentration to drop to \(0.31 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Short Answer

Expert verified
It takes approximately 495 minutes.

Step by step solution

01

Understand the Problem

We need to determine the time required for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease from \(1.24 \times 10^{-3}\ \text{mol/L}\) to \(0.31 \times 10^{-3}\ \text{mol/L}\) for a first-order reaction with a given rate constant.
02

Recall the First-Order Kinetics Formula

For a first-order reaction, the formula to calculate time \(t\) based on initial concentration \([A]_0\), final concentration \([A]\), and rate constant \(k\) is: \[t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right)\]
03

Identify Given Values

From the problem, we identify \([A]_0 = 1.24 \times 10^{-3}\ \text{mol/L}\), \([A] = 0.31 \times 10^{-3}\ \text{mol/L}\), and \(k = 2.8 \times 10^{-3}\ \text{min}^{-1}\).
04

Calculate the Natural Logarithm Ratio

Calculate the ratio of initial to final concentration:\[ \frac{[A]_0}{[A]} = \frac{1.24 \times 10^{-3}}{0.31 \times 10^{-3}} = 4.0 \]Then find the logarithm:\[ \ln(4.0) \approx 1.386 \]
05

Solve for Time using the Formula

Substitute the values into the formula:\[ t = \frac{1}{2.8 \times 10^{-3}} \times 1.386 \]Calculate \(t\):\[ t \approx 495 \ \text{minutes} \]

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Key Concepts

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

Understanding the Rate Constant in First-Order Reactions
The rate constant, often denoted as \(k\), is a crucial element in understanding the speed of a chemical reaction. For first-order reactions, like the decomposition of \(\text{SO}_2\text{Cl}_2\), it provides the rate at which the concentration of a reactant decreases over time. In this exercise, the rate constant is given as \(2.8 \times 10^{-3}\, \text{min}^{-1}\). This means for every minute, a constant fraction of the original concentration decomposes. The unit \(\text{min}^{-1}\) suggests a reaction progress that is linear with time on a logarithmic scale. Understanding the rate constant allows chemists to predict how quickly a reaction will proceed under specific conditions. Key points to remember include that the rate constant is dependent on temperature, thus changing if the temperature varies.
The Role of the Natural Logarithm in Kinetics
When dealing with first-order reactions, the natural logarithm, \(\ln\), is central to calculating the time required for a change in concentration. This is because the kinetics of first-order reactions can be described using logarithmic relationships. In our case, to find out how long it takes for the concentration to change, we use \(\ln\) to determine the ratio of the initial and final concentrations. The reason \(\ln\) is used here is because it transforms the rapidly changing exponential decay into a linear format which makes manipulating and solving equations more straightforward. In this example, we calculated \( \ln(4.0) \approx 1.386 \), which represents the natural logarithm of the ratio of initial to final concentration, translating a relationship that might seem complex into a simpler arithmetic problem.
Applying the First-Order Kinetics Formula
The core concept to solving problems like this one lies in the first-order kinetics formula. For any first-order reaction, the time \(t\) can be determined using the formula \[ t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right) \], where \([A]_0\) and \([A]\) are the initial and final concentrations, respectively, and \(k\) is the rate constant. This formula allows us to connect the change in concentration with time directly. To use the formula effectively, ensure that all the concentrations, the rate constant, and time units are consistent. As we've seen in the solution, by substituting \([A]_0 = 1.24 \times 10^{-3} \) mol/L, \([A] = 0.31 \times 10^{-3} \) mol/L, and \(k = 2.8 \times 10^{-3} \ \text{min}^{-1}\), we can solve for \(t\), resulting in a time of approximately 495 minutes. This is a practical application of kinetic theory allowing prediction of reaction timelines.

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Most popular questions from this chapter

A three-step mechanism for the reaction of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and \(\mathrm{H}_{2} \mathrm{O}\) is proposed: Step 1 Slow $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{Br}^{-}$$ Step 2 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}$$ Step 3 Fast $$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{HBr}$$ (a) Write an equation for the overall reaction. (b) Which step is rate-determining? (c) What rate law is expected for this reaction?

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (b) \(\mathbf{O}(\mathrm{g})+\mathbf{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\)

The rate equation for the hydrolysis of sucrose to fructose and glucose $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})$$ is " \(-\Delta[\text { sucrose }] / \Delta t=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] .\) After \(2.57 \mathrm{h}\) at \(27^{\circ} \mathrm{C}\) the sucrose concentration decreased from \(0.0146 \mathrm{M}\) to \(0.0132 \mathrm{M} .\) Find the rate constant, \(k.\)

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1 Slow \(\quad \mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2 Fast \(\quad \mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Show that the elementary steps add up to give the overall, stoichiometric equation. (b) What is the molecularity of each step? (c) For this mechanism to be consistent with kinetic data, what must be the experimental rate equation? (d) Identify any intermediates in this reaction.

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

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