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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) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism?

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose When the concentration of invertase is \(4.2 \times 10^{-7} M\) and the concentration of sucrose is \(0.0077 M\), invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s} .\) When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme- substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

The temperature dependence of the rate constant for the reaction is tabulated as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & k\left(\mathbf{M}^{-1} \mathbf{s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \end{array} $$ Calculate \(E_{g}\) and \(A\).

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.
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