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The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)-\rightarrow 2 \mathrm{NOCl}(g)\) obeys the rate law, rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]\). The following mechanism has been proposed for this reaction: $$ \begin{aligned} \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) & \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \cdots & 2 \mathrm{NOCl}(g) \end{aligned} $$ (a) What would the rate law be if the first step were rate determining? (b) Based on the observed rate law, what can we conclude about the relative rates of the two steps?

Short Answer

Expert verified
The rate law considering the first step as rate-determining would be \(\text{rate} = k_1[\mathrm{NO}][\mathrm{Cl}_{2}]\). Comparing this with the observed rate law, \(\text{rate} = k[\mathrm{NO}]^{2}[\mathrm{Cl}_{2}]\), we notice a difference in the concentration terms of \(\mathrm{NO}\). Based on this discrepancy, we can infer that the second step is likely to be the rate-determining step in the mechanism, with a comparatively slower rate than the first step.

Step by step solution

01

(a) Finding the rate law considering the first step as rate determining

To find the rate law considering the first step as rate-determining, we should apply the rate law only to the first reaction. The first reaction is \[ \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{NOCl}_{2}(g) . \] If this step is rate-determining, then the rate law for this step would be: \[ \text{rate} = k_1[\mathrm{NO}][\mathrm{Cl}_{2}] , \] where \(k_1\) is the rate constant for the first step.
02

(b) Comparing observed rate law with the rate law obtained in (a)

Now, we will compare the rate law we got in (a) with the given observed rate law, which is: \[\text{rate} = k[\mathrm{NO}]^{2}[\mathrm{Cl}_{2}] .\] When comparing the two rate laws, we notice that the rate law obtained in (a) lacks the squared concentration of \(\mathrm{NO}\) and the observed rate law has the squared concentration of \(\mathrm{NO}\). This difference in rate laws suggests that the first step cannot be the rate-determining step because it doesn't match the observed rate law. Therefore, we can conclude that the second step is likely to be the rate-determining step. By comparing the observed rate law with the proposed reaction mechanism, we can conclude that the second step is comparatively slower than the first step, as it is more likely to be the rate-determining step.

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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 is crucial for grasping how chemical reactions occur. Essentially, the rate law expresses the relationship between the concentration of reactants and the rate of the reaction.

The rate law for a reaction is often determined experimentally and is usually of the form \[ \text{rate} = k[\text{Reactant}_1]^{m}[\text{Reactant}_2]^{n} \], where \( k \) is the rate constant and \( m \) and \( n \) are the reaction orders for the respective reactants. These orders are not generally equal to the stoichiometric coefficients in the balanced equation for the reaction.

When you encounter an exercise involving rate laws, it's vital to compare the proposed reaction mechanism with the observed rate law. If the rate law derived from the proposed mechanism matches the observed rate law, the mechanism is plausible. However, if there is a discrepancy, like a difference in the reaction orders, it might suggest that the proposed mechanism does not accurately describe the steps of the reaction.
Reaction Mechanism
Diving deeper, the reaction mechanism tells the story of how reactants are transformed into products at the molecular level. It’s a step-by-step description that shows the sequences of elementary reactions that take place.

When considering a mechanism, chemists break it down into elementary steps—each step has its own rate law. For an overall reaction, such as \(2 \mathrm{NO}(g) + \mathrm{Cl}_2(g) \rightarrow 2 \mathrm{NOCl}(g)\), a proposed mechanism might involve several species that do not appear in the overall balanced equation. These are called intermediates.

By analyzing each step, we can deduce which step is the slowest, also known as the rate-determining step, which has a significant effect on the overall reaction rate. It's like a bottleneck in a production line—no matter how fast the earlier steps are, if one step is slow, it limits the overall rate of production. Similarly, in a reaction, the slowest elementary step determines the reaction rate.
Rate-Determining Step
The rate-determining step is akin to the slowest runner in a relay race—it sets the pace for the whole reaction. It’s the slowest step in the proposed reaction mechanism, and it controls the speed at which the overall reaction proceeds.

In our given exercise, we compared the rate law from the rate-determining step in the proposed mechanism to the experimentally determined rate law. The discrepancy between the two helped us identify that the first step couldn't be the rate-determining step because it failed to match the complexity of the observed rate law that included a squared concentration of \(\mathrm{NO}\).

This process is fundamental in chemical kinetics because understanding which step is rate-determining in a reaction allows chemists to tweak reaction conditions to achieve desired rates or to understand better why certain reactions happen the way they do. Remember, the rate-determining step will always give you insight into the kinetics of the reaction, even if the overall mechanism is still under investigation.

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Most popular questions from this chapter

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & \text { Rate Constant (s }^{-1} \text { ) } \\\ \hline 300 & 3.2 \times 10^{-11} \\ 320 & 1.0 \times 10^{-9} \\ 340 & 3.0 \times 10^{-8} \\ 355 & 2.4 \times 10^{-7} \\ \hline \end{array} $$ From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})\). The following data are collected: \begin{tabular}{lccccc} \hline Time (s) & 0 & 40 & 80 & 120 & 160 \\ \hline Moles of A & \(0.100\) & \(0.067\) & \(0.045\) & \(0.030\) & \(0.020\) \\ \hline \end{tabular} (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each \(40-\mathrm{s}\) interval, in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant. The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has a molar absorptivity constant of \(5.60 \times 10^{3} \mathrm{~cm}^{-1} \mathrm{M}^{-1}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) within \(30.0\) min. Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). 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 \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ?

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