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Chapter 14: Problem 41

(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 halflife 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
(a) At 600 K, the rate constant (k) is approximately \(3.01 \times 10^{-6} s^{-1}\). (b) The half-life at 320掳C (593.15 K) is approximately \(3.15 \times 10^{4} s\).

Step by step solution

01

(a) Finding the rate constant at 600 K

For a first-order reaction, the relationship between the half-life (t陆) and the rate constant (k) is given by: \[t_{陆} = \frac{0.693}{k}\] We are given t陆 at 600 K, which is \(2.3 \times 10^{5} s\). We can now solve the equation for the rate constant (k) at 600 K: \[k = \frac{0.693}{t_{陆}}\] Plug in the value of t陆: \[k = \frac{0.693}{2.3 \times 10^{5} s}\] Now, calculate the value of k: \[k 鈮� 3.01 \times 10^{-6} s^{-1}\]
02

(b) Finding the half-life at 320掳C

First, convert the given temperature from Celsius to Kelvin: \[T = 320掳C + 273.15 = 593.15 K\] At 320掳C (593.15 K), we are given the rate constant, k, which is \(2.2 \times 10^{-5} s^{-1}\). We can use the same equation as before to find the half-life (t陆) at this temperature: \[t_{陆} = \frac{0.693}{k}\] Plug in the value of k: \[t_{陆} = \frac{0.693}{2.2 \times 10^{-5} s^{-1}}\] Now, calculate the value of t陆: \[t_{陆} 鈮� 3.15 \times 10^{4} s\] So the half-life at 320掳C (593.15 K) is approximately \(3.15 \times 10^{4} s\).

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Key Concepts

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

Rate laws in chemistry
Rate laws in chemistry, often referred to as 'rate equations', are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. For a reaction where a substance A decomposes into products, the rate law can be expressed as the rate = k[A]n, where 'k' is the rate constant, [A] is the concentration of A, and 'n' is the order of the reaction with respect to A.

In first-order reactions, the rate of the reaction is directly proportional to the concentration of the reactant. This means that if the concentration of the reactant A doubles, so does the rate of the reaction. The rate law for a first-order reaction is simply rate = k[A], with 'n' equal to 1. This type of rate law indicates that the rate of reaction depends on the concentration of a single reactant to the first power.
Half-life of a reaction
The concept of half-life is critical in understanding both nuclear and chemical reactions. In chemical kinetics, the half-life of a reaction refers to the time it takes for half of the reactant to be consumed or for the concentration to decrease by half.

For first-order reactions, the half-life is constant and does not depend on the initial concentration of the reactant. It is given by the formula t陆 = 0.693/k, where 't陆' denotes the half-life and 'k' is the rate constant of the reaction. This formula exemplifies the principle that, regardless of the concentration, it will always take the same amount of time for half of the reactant to react. This is a distinctive feature of first-order kinetics.
Chemical kinetics
Chemical kinetics is the branch of chemistry that studies the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which reactions occur. It is not only important to know what products are formed in a reaction, but also how fast they are produced.

In the realm of chemical kinetics, scientists investigate how various conditions such as concentration, temperature, and catalysts influence the speed of chemical reactions. Using graphs and mathematical models, chemists can predict how the rate will change over time and under different circumstances. Kinetics is fundamental in the development of new chemical processes and the optimization of existing ones in various industries.
Temperature dependence of reaction rates
The rates of chemical reactions are greatly affected by temperature. Generally, increasing the temperature increases the rate at which the reaction occurs. This is because higher temperatures provide more energy to the reactant molecules, leading to more frequent and more energetic collisions which are more likely to overcome the activation energy barrier and result in a reaction.

The precise relationship between temperature and the rate of reaction is described by the Arrhenius Equation, which shows how the rate constant 'k' changes with temperature: k = A * exp(-Ea / (RT)), where 'Ea' is the activation energy of the reaction, 'R' is the universal gas constant, 'T' is the temperature in Kelvin, and 'A' is the frequency factor. As shown in the formula, even a small increase in temperature can lead to a significant increase in the rate constant, and thereby the reaction rate, illustrating how sensitive reaction rates are to changes in temperature.

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Most popular questions from this chapter

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M}\) ? (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M}\) ?

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\). are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\) - (b) Using a resource such as Table \(8.4\), look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{aligned} &\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) \(\mathrm{Do}\) we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous catalysis or heterogeneous catalysis?
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