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Chapter 14: Problem 37

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\), is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). At \(600 \mathrm{~K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{~s}\). What is the rate constant at this temperature? (b) At \(320^{\circ} \mathrm{C}\) the rate constant is \(2.2 \times 10^{-5} \mathrm{~s}^{-1}\). What is the half-life at this temperature?

Short Answer

Expert verified
The rate constant at 600 K is approximately \(3.01 \times 10^{-6} \mathrm{~s}^{-1}\). The half-life at 320掳C is approximately \(3.14 \times 10^{4} \mathrm{~s}\).

Step by step solution

01

(Part a) Find the rate constant at 600 K

Given the half-life \(t_{1/2} = 2.3 \times 10^{5}\mathrm{~s}\) and the formula for the half-life of a first-order reaction \(t_{1/2} = \frac{0.693}{k}\). Now let's solve for the rate constant \(k\): 1. Use the half-life formula: \(t_{1/2} = \frac{0.693}{k}\). 2. Input the given value of \(t_{1/2} = 2.3 \times 10^{5}\mathrm{~s}\) to the formula: \(2.3 \times 10^{5}\mathrm{~s} = \frac{0.693}{k}\). 3. To find the value of \(k\), multiply both sides by \(k\): \(k \times (2.3 \times 10^{5}\mathrm{~s}) = 0.693\). 4. Finally, divide both sides by \((2.3 \times 10^{5}\mathrm{~s})\): \(k = \frac{0.693}{2.3 \times 10^{5}\mathrm{~s}}\). 5. Calculate the value of \(k\): \(k \approx 3.01 \times 10^{-6} \mathrm{~s}^{-1}\). The rate constant at 600 K is approximately \(3.01 \times 10^{-6} \mathrm{~s}^{-1}\).
02

(Part b) Find the half-life at 320掳C

Given the rate constant \(k = 2.2 \times 10^{-5} \mathrm{~s}^{-1}\) at 320掳C, let's find the half-life \(t_{1/2}\) using the formula for the half-life of a first-order reaction \(t_{1/2} = \frac{0.693}{k}\): 1. Use the half-life formula: \(t_{1/2} = \frac{0.693}{k}\). 2. Input the given value of \(k = 2.2 \times 10^{-5} \mathrm{~s}^{-1}\) to the formula: \(t_{1/2} = \frac{0.693}{2.2 \times 10^{-5} \mathrm{~s}^{-1}}\). 3. Calculate the value of \(t_{1/2}\): \(t_{1/2} \approx 3.14 \times 10^{4} \mathrm{~s}\). The half-life at 320掳C is approximately \(3.14 \times 10^{4} \mathrm{~s}\).

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Key Concepts

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

Rate constant
Understanding the rate constant is essential when dealing with first-order reactions. The rate constant, denoted as \( k \), offers a way to quantify the speed of a chemical reaction. In first-order reactions, the rate is directly proportional to the concentration of one reactant. Ideal conditions, temperature, and the nature of reactants all affect the value of the rate constant.
For a first-order reaction, the formula \( t_{1/2} = \frac{0.693}{k} \) relates the half-life of the reaction to the rate constant. This expression shows that the half-life is inversely proportional to \( k \). If the rate constant is known, predicting how quick a chemical undergoes change becomes easier.
In the exercise provided, we determined the rate constant at a certain temperature (600 K) using this formula. By knowing the half-life, we plugged it into the equation, rearranged, and solved for \( k \). This calculation helps us understand the inherent speed of the decomposition process at that temperature.
Half-life calculation
Calculating the half-life of a reaction provides critical insights into how long it takes for a substance to reduce to half its original amount. This is particularly important in first-order reactions, like radioactive decay or the decomposition reaction described in the exercise.
The formula \( t_{1/2} = \frac{0.693}{k} \) allows one to calculate the half-life if the rate constant is known. When the half-life is constant, as in first-order reactions, the calculations are straightforward. For example, if you know \( k \), you simply substitute it into the formula to find the half-life. This provides a predictable measure of how fast a reaction proceeds over time.
In part b of the exercise, we applied this concept by using a given rate constant of \( 2.2 \times 10^{-5} \text{s}^{-1} \) at 320掳C. Plugging this into the formula gave us a resulting half-life, indicating how quickly half of the original substance decomposes at that condition.
Decomposition reaction
Decomposition reactions are where a single compound breaks down into two or more simpler substances. These reactions are fundamental in chemistry due to their common occurrence and the insight they provide into reaction mechanisms. They are often endothermic, requiring energy input to occur, such as heat, electricity, or light.
In the specific decomposition of \( \text{SO}_2\text{Cl}_2 \) mentioned in the exercise, the compound decomposes into \( \text{SO}_2 \) and \( \text{Cl}_2 \). This is a first-order reaction, meaning the rate depends solely on the concentration of \( \text{SO}_2\text{Cl}_2 \).
Understanding decomposition reactions in terms of kinetics allows chemists to control and predict the reaction conditions and the time scales involved. This is extremely useful in industrial and laboratory settings where precise control of chemical processes is crucial.

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Most popular questions from this chapter

(a) In which of the following reactions would you expect the orientation factor to be least important in leading to reaction? \(\mathrm{NO}+\mathrm{O} \longrightarrow \mathrm{NO}_{2}\) or \(\mathrm{H}+\mathrm{Cl} \longrightarrow \mathrm{HCl}\) ? (b) How does the kinetic-molecular theory help us understand the temperature dependence of chemical reactions?

The rate of the reaction \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)\) was measured at several temperatures, and the following data were collected: $$ \begin{array}{ll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{M}^{-1} \mathrm{~s}^{-1}\right) \\ \hline 15 & 0.0521 \\ 25 & 0.101 \\ 35 & 0.184 \\ 45 & 0.332 \end{array} $$ Using these data, graph \(\ln k\) versus \(1 / T\). Using your graph, determine the value of \(E_{g}\)

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q) .\) A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval, in units of \(\mathrm{M} / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min}\), what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q)\) (slow) \(\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \quad\) (fast) (a) Write the rate law for each of the elementary reactions of the mechanism. (b) Write the chemical equation for the overall process. (c) Identify the intermediate, if any, in the mechanism. (d) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

The isomerization of methyl isonitrile \(\left(\mathrm{CH}_{3} \mathrm{NC}\right)\) to acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) was studied in the gas phase at \(215^{\circ} \mathrm{C}\), and the following data were obtained: $$ \begin{array}{ll} \hline \text { Time (s) } & \text { [CH }_{3} \text { NC] (M) } \\ \hline 0 & 0.0165 \\ 2,000 & 0.0110 \\ 5,000 & 0.00591 \\ 8,000 & 0.00314 \\ 12,000 & 0.00137 \\ 15,000 & 0.00074 \\ \hline \end{array} $$ (a) Calculate theaverage rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Graph \(\left[\mathrm{CH}_{3} \mathrm{NC}\right]\) versus time, and determine the instantaneous rates in \(\mathrm{M} / \mathrm{s}\) at \(t=5000 \mathrm{~s}\) and \(t=8000 \mathrm{~s}\).
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