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Chapter 14: Problem 69

The reaction \(2 \mathrm{NO}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{NOCl}\) has the rate law: rate of reaction \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right] .\) Propose a twostep mechanism for this reaction consisting of a fast reversible first step, followed by a slow step.

Short Answer

Expert verified
The proposed mechanism consists of two steps: \n1. Fast Step: \(2NO + Cl_{2} \rightleftharpoons N_{2}O_{2}Cl_{2}\) \n2. Slow Step: \(NO + N_{2}O_{2}Cl_{2} \longrightarrow 2NOCl\). Combining, we get back the original reaction.

Step by step solution

01

Writing the Fast and Reversible Step

We begin by proposing a fast and reversible step. Two copies of NO could combine with a molecule of Cl2 to form an unstable intermediate. The reaction is given as: \(2NO + Cl_{2} \rightleftharpoons N_{2}O_{2}Cl_{2}\)
02

Writing the Slow Step

Next, we propose a slow step. Because the fast step is in equilibrium, the unstable intermediate \(N_{2}O_{2}Cl_{2}\) would slightly dissociate back into its reactants. However, a molecule of NO could collide with it, causing it to react further and reducing the overall activation energy. Hence, the slow step is formulated as: \(NO + N_{2}O_{2}Cl_{2} \longrightarrow 2NOCl\)
03

Combine the Steps

Finally, observe the results when both steps are combined. The intermediate compound \(N_{2}O_{2}Cl_{2}\) used in the first and second step will cancel out, resulting in the net reaction. We end up with the original reaction: \(2NO + Cl_{2} \longrightarrow 2NOCl\)

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Key Concepts

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

Rate Law
Understanding the rate law of a chemical reaction is essential for predicting how fast a reaction occurs. In this instance, the given rate law is:
  • Rate = k[NO]2[Cl2]
This equation tells us that the reaction rate depends on the concentration of nitric oxide (NO) squared and chlorine (Cl2). A higher concentration of these reactants generally means a faster reaction rate.

The rate constant \( k \) is specific to each reaction and temperature. If you increase the concentrations of NO and Cl2, the rate will increase. This quadratic relation with NO means NO's presence has a significant impact on the reaction speed.

In the proposed mechanism, the rate law helps identify that the slow step determines the rate, involving one NO and the intermediate derived from NO and Cl2. This aligns perfectly with our proposed second slow step in the mechanism.
Chemical Kinetics
Chemical kinetics involves studying the rate at which reactions occur and the steps involved in those reactions. This can include multiple shifts from reactants to intermediate substances and then to products. For this reaction, we implemented a two-step mechanism:

  • Fast, Reversible Step: This involves an initial rapid formation of an intermediate compound (not present in the final product) which is both created and broken down quickly until equilibrium is reached.
  • Slow Step: This is the rate-determining step, meaning it controls the speed of the overall reaction. Here, the slow step involves NO reacting with the intermediate to form the final product NOCl.
In our example, a slow second step suggests a bottleneck, dictating the overall rate, and coincides with the rate law expression. This step-wise analysis shows how intermediates play a crucial role, even if they don鈥檛 appear in the final reaction.
Activation Energy
Activation energy is the minimum energy required for a reaction to proceed. It's like an energy barrier that must be overcome for reactants to transform into products. In our example, the slow step has a higher activation energy, making it the rate-determining step.

Reducing activation energy can speed up a chemical reaction. Catalysts or specific reaction conditions like temperature can lower this energy threshold, increasing the rate. The proposed mechanism reflects this by suggesting the collision between NO and the intermediate is critical to overcoming the activation energy.

When looking at energy profiles of reactions, the peak represents the activation energy. The first fast step in our mechanism would have energy barriers quickly accessible, whereas the second has a more significant barrier, acting as a checkpoint for the reaction's pace. Understanding this energy barrier helps to control reaction rates in practical applications.

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Most popular questions from this chapter

According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

The reaction \(A \longrightarrow\) products is first order in A. (a) If \(1.60 \mathrm{g} \mathrm{A}\) is allowed to decompose for 38 min, the mass of A remaining undecomposed is found to be 0.40 g. What is the half-life, \(t_{1 / 2}\), of this reaction? (b) Starting with \(1.60 \mathrm{g} \mathrm{A},\) what is the mass of \(\mathrm{A}\) remaining undecomposed after \(1.00 \mathrm{h} ?\)

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

The following data are obtained for the initial rates of reaction in the reaction \(A+2B+C \longrightarrow 2 D+E.\) $$\begin{array}{lllll} \hline & \text { Initial } & \text { Initial } & & \\ \text { Expt } & \text { [A], M } & \text { [B],M } & \text { [C], M } & \text { Initial Rate } \\ \hline 1 & 1.40 & 1.40 & 1.00 & R_{1} \\ 2 & 0.70 & 1.40 & 1.00 & R_{2}=\frac{1}{2} \times R_{1} \\ 3 & 0.70 & 0.70 & 1.00 & R_{3}=\frac{1}{4} \times R_{2} \\ 4 & 1.40 & 1.40 & 0.50 & R_{4}=16 \times R_{3} \\ 5 & 0.70 & 0.70 & 0.50 & R_{5}=? \\ \hline \end{array}$$ (a) What are the reaction orders with respect to A, B, and C? (b) What is the value of \(R_{5}\) in terms of \(R_{1} ?\)

The object is to study the kinetics of the reaction between peroxodisulfate and iodide ions. $$\begin{aligned} &\text { (a) } \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \end{aligned}$$ The \(I_{3}^{-}\) formed in reaction (a) is actually a complex of iodine, \(\mathrm{I}_{2},\) and iodide ion, \(\mathrm{I}^{-}\). Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) also present in the reaction mixture, reacts with \(\mathrm{I}_{3}^{-}\) just as fast as it is formed. $$\text { (b) } 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{I}^{-}(\mathrm{aq})$$ When all of the thiosulfate ion present initially has been consumed by reaction (b), a third reaction occurs between \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) and starch, which is also present in the reaction mixture. $$\text { (c) } \mathrm{I}_{3}^{-}(\mathrm{aq})+\operatorname{starch} \longrightarrow \text { blue complex }$$ The rate of reaction (a) is inversely related to the time required for the blue color of the starch-iodine complex to appear. That is, the faster reaction (a) proceeds, the more quickly the thiosulfate ion is consumed in reaction (b), and the sooner the blue color appears in reaction (c). One of the photographs shows the initial colorless solution and an electronic timer set at \(t=0 ;\) the other photograph shows the very first appearance of the blue complex (after 49.89 s). Tables I and II list some actual student data obtained in this study. $$\begin{array}{l} \hline\text { TABLE I } \\ \text { Reaction conditions at } 24^{\circ} \mathrm{C}: 25.0 \mathrm{mL} \text { of the } \\ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8}(\text { aq) listed, } 25.0 \mathrm{mL} \text { of the } \mathrm{KI}(\mathrm{aq}) \\ \text { listed, } 10.0 \mathrm{mL} \text { of } 0.010 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}), \text { and } 5.0 \mathrm{mL} \\ \text { starch solution are mixed. The time is that of the } \\ \text { first appearance of the starch-iodine complex. } \\ \hline & \text { Initial Concentrations, } \mathrm{M} \\ \hline \text { Experiment } & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8} & \mathrm{KI} & \text { Time, s } \\ \hline 1 & 0.20 & 0.20 & 21 \\ 2 & 0.10 & 0.20 & 42 \\ 3 & 0.050 & 0.20 & 81 \\ 4 & 0.20 & 0.10 & 42 \\ 5 & 0.20 & 0.050 & 79 \\ \hline \end{array}$$ $$\begin{array}{l} \hline \text { TABLE II } \\ \text { Reaction conditions: those listed in Table I for } \\ \text { Experiment } 4, \text { but at the temperatures listed. } \\ \hline \text { Experiment } & \text { Temperature, }^{\circ} \mathrm{C} & \text { Time, } \mathrm{s} \\ \hline 6 & 3 & 189 \\ 7 & 13 & 88 \\ 8 & 24 & 42 \\ 9 & 33 & 21 \\ \hline \end{array}$$ (a) Use the data in Table I to establish the order of reaction (a) with respect to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and to I \(^{-}\). What is the overall reaction order? [Hint: How are the times required for the blue complex to appear related to the actual rates of reaction? (b) Calculate the initial rate of reaction in Experiment 1 expressed in \(\mathrm{M} \mathrm{s}^{-1} .\) [Hint: You must take into account the dilution that occurs when the various solutions are mixed, as well as the reaction stoichiometry indicated by equations \((a),(b), \text { and }(c) .]\) (c) Calculate the value of the rate constant, \(k,\) based on experiments 1 and 2 (d) Calculate the rate constant, \(k\), for the four different temperatures in Table II. (e) Determine the activation energy, \(E_{\mathrm{a}}\), of the peroxodisulfate- iodide ion reaction. (f) The following mechanism has been proposed for reaction (a). The first step is slow, and the others are fast. $$\begin{array}{c} \mathrm{I}^{-}+\mathrm{S}_{2} \mathrm{O}_{8}^{2-} \longrightarrow \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \\ \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}^{+} \\ \mathrm{I}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} \\ \mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} \end{array}$$ Show that this mechanism is consistent with both the stoichiometry and the rate law of reaction (a). Explain why it is reasonable to expect the first step in the mechanism to be slower than the others.
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