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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’t 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

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

For the reaction \(A \longrightarrow\) products, the following data give \([\mathrm{A}]\) as a function of time \(t=0 \mathrm{s},[\mathrm{A}]=0.600 \mathrm{M};100 \mathrm{s}, 0.497 \mathrm{M} ; 200 \mathrm{s}, 0.413 \mathrm{M} ; 300 \mathrm{s}, 0.344 \mathrm{M} ; 400 \mathrm{s}\) \(0.285 \mathrm{M} ; 600 \mathrm{s}, 0.198 \mathrm{M} ; 1000 \mathrm{s}, 0.094 \mathrm{M}.\) (a) Show that the reaction is first order. (b) What is the value of the rate constant, \(k ?\) (c) What is \([\mathrm{A}]\) at \(t=750 \mathrm{s} ?\)

The rate equation for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) is found to be rate \(=k[\mathrm{A}][\mathrm{B}] .\) For this reaction, we can conclude that (a) the unit of \(k=s^{-1} ;\) (b) \(t_{1 / 2}\) is constant; (c) the value of \(k\) is independent of the values of \([\mathrm{A}]\) and \([\mathrm{B}] ;\) (d) the rate of formation of \(\mathrm{C}\) is twice the rate of disappearance of A.

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 \(R_{1} ?\)

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