/*! 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 26 The decomposition of \(\mathrm{S... [FREE SOLUTION] | 91Ó°ÊÓ

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The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2},\) and the reaction has a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a 1.0 -L. flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

Short Answer

Expert verified
It takes approximately 1021 minutes.

Step by step solution

01

Understanding First-Order Kinetics

For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The equation that describes a first-order reaction is the integrated rate law: \[[A] = [A]_0 e^{-kt}\]where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time.
02

Calculate the Rate Constant

We know the reaction is first order and the half-life \(t_{1/2}\) for a first-order reaction is:\[t_{1/2} = \frac{0.693}{k}\]Given \(t_{1/2} = 245\) minutes, we can solve for the rate constant \(k\):\[k = \frac{0.693}{245} \approx 0.00283\, ext{min}^{-1}\]
03

Apply the Integrated Rate Law

Using the integrated rate law equation, we can solve for \(t\). The initial concentration \([A]_0\) is \(3.6 \times 10^{-3}\, ext{mol/L}\), and \([A]\) is \(2.00 \times 10^{-4}\, ext{mol/L}\):\[2.00 \times 10^{-4} = 3.6 \times 10^{-3} \times e^{-0.00283t}\]To isolate \(e^{-0.00283t}\), divide both sides by \(3.6 \times 10^{-3}\):\[e^{-0.00283t} = \frac{2.00 \times 10^{-4}}{3.6 \times 10^{-3}} \approx 0.0556\]
04

Solve for Time \(t\)

To solve for \(t\), take the natural logarithm of both sides:\[-0.00283t = \ln(0.0556)\]Calculate \(\ln(0.0556)\), which is approximately \(-2.89\). Then:\[-0.00283t = -2.89\]Solving for \(t\):\[t = \frac{-2.89}{-0.00283} \approx 1021\, ext{minutes}\]
05

Final Calculation and Result

Therefore, it will take approximately 1021 minutes for the concentration of \( ext{SO}_2 ext{Cl}_2\) to decrease from \(3.6 \times 10^{-3}\, ext{mol/L}\) to \(2.00 \times 10^{-4}\, ext{mol/L}\). This is the time required for this first-order reaction under the given conditions.

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Key Concepts

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

Half-Life Calculation
In first-order reaction kinetics, the concept of half-life is a key element, which pertains to the time required for the concentration of a reactant to decrease to half of its initial value. This is especially important because, for first-order reactions, the half-life is constant and does not depend on the initial concentration of the reactant. This characteristic allows chemists to predict how quickly a reaction proceeds based on the half-life value.

For a first-order reaction, you can calculate the half-life (\(t_{1/2}\)) using the equation:
  • \(t_{1/2} = \frac{0.693}{k}\)
Here, \(k\) is the rate constant. The constant value 0.693 comes from the natural logarithm of 2. In the example problem, where the half-life of SO\(_2\)Cl\(_2\) is given as 245 minutes, this formula can be used to find the rate constant. By rearranging the formula, you solve for \(k\):
  • \(k = \frac{0.693}{245}\)
Integrated Rate Law
The integrated rate law for a first-order reaction elegantly describes how the concentration of a reactant decreases over time. This formula is crucial for determining the amount of reactant remaining after a given period has elapsed. By integrating the differential rate law for first-order reactions, we arrive at the formula:
  • \([A] = [A]_0 e^{-kt}\)
In this equation, \([A]_0\) represents the initial concentration, while \([A]\) is the concentration at time \(t\). The term \(e^{-kt}\) reflects the exponential decay nature of first-order reactions. With the problem's given values, the initial concentration of \(3.6 \times 10^{-3}\) mol/L eventually reduces to \(2.00 \times 10^{-4}\) mol/L.

This formula allows for flexible calculations depending on the unknown variable. Whether solving for final concentration, initial concentration, rate constant, or time elapsed, rearranging the equation reveals any necessary parameter with precision. The ease of applying this equation is a significant reason why integrated rate laws are fundamental in chemical kinetics.
Rate Constant
The rate constant, symbolized as \(k\), is an intrinsic factor of a chemical reaction signifying how fast the reaction occurs. Importantly, it functions independently of the reactant concentrations, specifically in first-order reactions. Instead, its value depends on the nature of the reaction and the conditions under which it takes place, such as temperature and the presence of a catalyst.

The units of the rate constant in first-order kinetics are s\(^{-1}\) or min\(^{-1}\), which illustrate that the reaction rate is proportional to the concentration of a single reactant. To find \(k\), you can rearrange the half-life equation for a first-order reaction. With a known half-life, like in our example (245 minutes), we use:
  • \(k = \frac{0.693}{t_{1/2}}\)
This provides the rate constant as approximately 0.00283 min\(^{-1}\). Understanding \(k\) enables predictions about how fast the concentration of reactants decreases and is vital for comparing reaction speeds between different reactions or conditions.

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Most popular questions from this chapter

Data for the following reaction are given in the table. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})$$ $$\begin{array}{clll}\text { Experiment } & \text { [NO] }(\mathrm{M}) & {\left[\mathrm{Br}_{2}\right](\mathrm{M})} &\begin{array}{l}\text { Initial Rate } \\\\(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s})\end{array} \\\\\hline 1 & 1.0 \times 10^{-2} & 2.0 \times 10^{-2} & 2.4 \times 10^{-2} \\\2 & 4.0 \times 10^{-2} & 2.0 \times 10^{-2} & 0.384 \\\3 & 1.0 \times 10^{-2} & 5.0 \times 10^{-2} & 6.0 \times 10^{-2} \\\\\hline\end{array}$$ What is the order of the reaction with respect to [NO] and \(\left[\mathrm{Br}_{2}\right],\) and what is the overall order of the reaction?

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ with a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0-L. flask, what is the partial pressure of each reactant and product after 245 minutes? What is the partial pressure of each reactant after 12 hours?

Hypofluorous acid, HOF, is very unstable, decomposing in a first-order reaction to give HF and \(\mathrm{O}_{2},\) with a half-life of \(30 .\) minutes at room temperature: $$\mathrm{HOF}(\mathrm{g}) \rightarrow \mathrm{HF}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$$ If the partial pressure of HOF in a \(1.00-\mathrm{L}\). flask is initially \(1.00 \times 10^{2} \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C},\) what are the total pressure in the flask and the partial pressure of HOF after exactly 30 minutes? After 45 minutes?

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.

The reaction of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{CO}(\mathrm{g})\) is thought to occur in two steps: Step 1: Slow \(\mathrm{NO}_{2}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(\mathrm{g})\) Step 2: Fast \(\mathrm{NO}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\) (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.

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