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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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Most popular questions from this chapter

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

The reaction \(A+B \longrightarrow C+D\) is second order in \(A\) and zero order in B. The value of \(k\) is \(0.0103 \mathrm{M}^{-1} \mathrm{min}^{-1}.\) What is the rate of this reaction when \([\mathrm{A}]=0.116 \mathrm{M}\) and \([\mathrm{B}]=3.83 \mathrm{M} ?\)

The first-order reaction \(A \longrightarrow\) products has \(t_{1 / 2}=180 \mathrm{s}\) (a) What percent of a sample of A remains unreacted \(900 \mathrm{s}\) after a reaction has been started? (b) What is the rate of reaction when \([\mathrm{A}]=0.50 \mathrm{M} ?\)

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).
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