The reaction $2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$ was performed and the following data were obtained under conditions of constant \(\left[\mathrm{Cl}_{2}\right]:\) (a) Is the following mechanism consistent with the data? $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) & \rightleftharpoons \mathrm{NOCl}_{2}(g) \quad(\text { fast }) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \longrightarrow 2 \mathrm{NOCl}(g) &(\text { slow }) \end{aligned} $$ (b) Does the linear plot guarantee that the overall rate law is second order?

Short Answer

Expert verified
The proposed reaction mechanism is consistent with the experimental data, as it shows a first-order dependence on NO concentration. The linear plot indicates a first-order overall rate law concerning NO but does not guarantee that the reaction is second-order overall. Our analysis of the mechanism shows that the reaction is first-order with respect to NO concentration.

Step by step solution

01

Determine the rate law for the slow step

The slow step in the proposed mechanism is: $$ \mathrm{NOCl}_{2}(g) + \mathrm{NO}(g) \longrightarrow 2 \mathrm{NOCl}(g) $$ Since the rate of the reaction depends on the slowest step, the rate law for this step can be written as: $$ \text{rate} = k[\mathrm{NOCl}_{2}][\mathrm{NO}] $$ Here, k is the rate constant for this step.
02

Use the steady-state approximation to find the overall rate law

We use the steady-state approximation to assume that the concentration of the intermediate, NOCl₂, remains constant throughout the reaction. This means that the rate of its formation equals the rate of its consumption. The intermediate is formed in the fast step and consumed in the slow step. Therefore, we can write: $$ k_{1}[\mathrm{NO}][\mathrm{Cl}_{2}] = k_{2}[\mathrm{NOCl}_{2}][\mathrm{NO}] $$ Here, \(k_{1}\) and \(k_{2}\) are the rate constants of the fast and slow steps, respectively. Now, we can solve for [\(\mathrm{NOCl}_{2}\)]: $$ [\mathrm{NOCl}_{2}] = \frac{k_{1}}{k_{2}}\frac{[\mathrm{NO}][\mathrm{Cl}_{2}]}{[\mathrm{NO}]} $$ By simplifying, we get: $$ [\mathrm{NOCl}_{2}] = \frac{k_{1}}{k_{2}}[\mathrm{Cl}_{2}] $$ Since Cl₂ has a constant concentration, \(\frac{k_{1}}{k_{2}}[\mathrm{Cl}_{2}]\) is also a constant. Let's call it K: $$ [\mathrm{NOCl}_{2}] = K $$
03

Compare the overall rate law with the experimental data

Now, we can substitute the expression for [\(\mathrm{NOCl}_{2}\)] in the rate law of the slow step: $$ \text{rate} = k[\mathrm{NOCl}_{2}][\mathrm{NO}] = k K [\mathrm{NO}] $$ This rate law indicates a first-order dependence on NO. As the concentration of Cl₂ is held constant, the observed experimental data is consistent with this mechanism.
04

Discuss whether the linear plot guarantees a second-order overall rate law

The linear plot initially given in the problem refers to a plot of the rate versus the concentration of NO. If the plot is linear, it indicates a first-order overall rate law concerning NO. However, this does not guarantee that the reaction is second-order. We can only conclude that the reaction is first-order with respect to NO concentration, as shown by our analysis of the proposed mechanism.

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Most popular questions from this chapter

The first-order rate constant for the decomposition of $\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g), \quad\( at \)\quad 70^{\circ} \mathrm{C}$ is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with $0.0250 \mathrm{~mol}\( of \)\mathrm{N}_{2} \mathrm{O}_{5}(g)\( in a volume of \)2.0 \mathrm{~L} .(\mathbf{a})\( How many moles of \)\mathrm{N}_{2} \mathrm{O}_{5}$ will remain after \(5.0 \mathrm{~min} ?\) (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to $0.010 \mathrm{~mol}\( ? (c) What is the half-life of \)\mathrm{N}_{2} \mathrm{O}_{5}$ at \(70{ }^{\circ} \mathrm{C}\) ?

The rate of the reaction $4 \mathrm{PH}_{3}(g) \rightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)\( was studied by charging \)\mathrm{PH}_{3}(g)$ into a constant-volume reaction vessel and measuring the total pressure.

(a) A certain first-order reaction has a rate constant of $2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}\(. What is the value of \)k$ at \(60^{\circ} \mathrm{C}\) if $E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{b})\( Another first-order reaction also has a rate constant of \)2.75 \times 10^{-2} \mathrm{~s}^{-1}\( at \)20^{\circ} \mathrm{C}$. What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if $E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?(\mathbf{c})$ What assumptions do you need to make in order to calculate answers for parts (a) and (b)?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a 1.00-cm sample cell, and the only colored species in the reaction has an extinction coefficient of $5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\( at \)520 \mathrm{nm} .$ (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 at $30.0 \mathrm{~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 \(0.100 ?\)

(a) Develop an equation for the half-life of a zero-order reaction. (b) Does the half-life of a zero-order reaction increase, decrease, or remain the same as the reaction proceeds?

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