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Chapter 13: Problem 55

The rate law for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ is given by rate \(=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] .\) (a) What is the order of the reaction? (b) A mechanism involving the following steps has been proposed for the reaction: $$ \begin{array}{c} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) \longrightarrow 2 \mathrm{NOCl}(g) \end{array} $$ If this mechanism is correct, what does it imply about the relative rates of these two steps?

Short Answer

Expert verified
The order of the reaction is 2. If the given mechanism is correct, it implies that the first step of the mechanism: NO(g) + Cl2(g) -> NOCl2(g) would be the rate-determining or the slow step of the reaction.

Step by step solution

01

Determine the order of the reaction

The order of the reaction is determined by the sum of the powers of the concentrations of the reactants in the rate law. In this case, the rate law is \(k[\mathrm{NO}][\mathrm{Cl}_2]\). Therefore, the order of the reaction is \(1 (for NO) + 1 (for Cl_2) = 2\).
02

Analyze the proposed mechanism

The given proposed mechanism involves two steps: 1) NO(g) + Cl2(g) -> NOCl2(g) and 2) NOCl2(g) + NO(g) -> 2 NOCl(g). The rate of a reaction is determined by its slowest step, which is known as the rate-determining step.
03

Relate the proposed mechanism to the rate law

The slowest step in the mechanism should align with the rate law provided. In this case, since the rate law is \(k[\mathrm{NO}][\mathrm{Cl}_2]\), this mechanism implies that the slow step of the reaction must be the first step: NO(g) + Cl2(g) -> NOCl2(g). This is because the first step includes both NO and Cl2 reactants, which are present in the observed rate law.

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Most popular questions from this chapter

For the reaction $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ the frequency factor \(A\) is \(8.7 \times 10^{12} \mathrm{~s}^{-1}\) and the activation energy is \(63 \mathrm{~kJ} / \mathrm{mol} .\) What is the rate constant for the reaction at \(75^{\circ} \mathrm{C} ?\)

Assume that the formation of nitrogen dioxide: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the condition described in (b), the half-life of the reaction has been estimated to be \(6.4 \times 10^{3}\) min. What would be the half-life if the initial concentration of NO were \(10 \mathrm{ppm} ?\)

Determine the molecularity and write the rate law for each of the following elementary steps: (a) \(\mathrm{X} \longrightarrow\) products (b) \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products (c) \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products (d) \(\mathrm{X}+\mathrm{X} \longrightarrow\) products (e) \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products

Consider the following elementary step: $$ \mathrm{X}+2 \mathrm{Y} \longrightarrow \mathrm{XY}_{2} $$ (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{XY}_{2}\) is \(3.8 \times 10^{-3} \mathrm{M} / \mathrm{s}\) and the initial concentrations of \(\mathrm{X}\) and \(\mathrm{Y}\) are \(0.26 \mathrm{M}\) and \(0.88 M\), what is the rate constant of the reaction?

The activation energy \(\left(E_{\mathrm{a}}\right)\) for the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=-164 \mathrm{~kJ} / \mathrm{mol} $$ is \(240 \mathrm{~kJ} / \mathrm{mol} .\) What is \(E_{\mathrm{a}}\) for the reverse reaction?
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