Question

Two mechanisms are proposed for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ Mechanism \(1: \quad \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3}\) $$ \begin{aligned} & \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2} \\ \text { Mechanism 2: } & \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ & \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} \end{aligned} $$ Show that each of these mechanisms is consistent with the observed rate law: rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]\).

Short answer

$$ Answer: Yes, both Mechanism 1 and Mechanism 2 produce rate laws that match the observed rate law, making them consistent.

Step-by-step solution

Write the elementary reactions of Mechanism 1

The elementary reactions for Mechanism 1 are: $$ \mathrm{NO}+\mathrm{O}_{2} \rightleftharpoons \mathrm{NO}_{3}(1) $$ $$ \mathrm{NO}_{3}+\mathrm{NO} \longrightarrow 2 \mathrm{NO}_{2}(2) $$

Find the rate law for each elementary reaction in Mechanism 1

For the first elementary reaction (1), the rate law would be: $$ \text{rate}_1 = k_1[\mathrm{NO}][\mathrm{O}_{2}] - k_{-1}[\mathrm{NO}_{3}] $$ For the second elementary reaction (2), the rate law would be: $$ \text{rate}_2 = k_{2}[\mathrm{NO}_{3}][\mathrm{NO}] $$

Determine the overall rate law for Mechanism 1

Since the first reaction (1) is in equilibrium, the rate of the forward and reverse reactions are equal, which means: $$ k_1[\mathrm{NO}][\mathrm{O}_{2}] = k_{-1}[\mathrm{NO}_{3}] $$ Now, solving for \([\mathrm{NO}_{3}]\), we get: $$ [\mathrm{NO}_{3}] = \frac{k_1}{k_{-1}} [\mathrm{NO}][\mathrm{O}_{2}] $$ Substituting this expression back into the rate law for the second reaction (2), we get: $$ \text{rate} = \text{rate}_2 = k_{2}\left(\frac{k_1}{k_{-1}} [\mathrm{NO}][\mathrm{O}_{2}]\right)[\mathrm{NO}] $$ After simplifying, the overall rate law for Mechanism 1 becomes: $$ \text{rate} = (\frac{k_1 k_2}{k_{-1}})[\mathrm{NO}]^{2}[\mathrm{O}_{2}] $$ where \(k = \frac{k_1 k_2}{k_{-1}}\). #Mechanism 2#

Write the elementary reactions of Mechanism 2

The elementary reactions for Mechanism 2 are: $$ \mathrm{NO}+\mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} (3) $$ $$ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2} (4) $$

Find the rate law for each elementary reaction in Mechanism 2

For the first elementary reaction (3), the rate law would be: $$ \text{rate}_3 = k_3[\mathrm{NO}]^2 - k_{-3}[\mathrm{N}_{2}\mathrm{O}_{2}] $$ For the second elementary reaction (4), the rate law would be: $$ \text{rate}_4 = k_{4}[\mathrm{N}_{2} \mathrm{O}_{2}][\mathrm{O}_{2}] $$

Determine the overall rate law for Mechanism 2

Since the reaction (3) is in equilibrium, the rate of the forward and reverse reactions are equal, which means: $$ k_3[\mathrm{NO}]^2 = k_{-3}[\mathrm{N}_{2}\mathrm{O}_{2}] $$ Now, solving for \([\mathrm{N}_{2}\mathrm{O}_{2}]\), we get: $$ [\mathrm{N}_{2}\mathrm{O}_{2}] = \frac{k_3}{k_{-3}} [\mathrm{NO}]^2 $$ Substituting this expression back into the rate law for the second reaction (4), we get: $$ \text{rate} = \text{rate}_4 = k_{4}\left(\frac{k_3}{k_{-3}} [\mathrm{NO}]^2\right)[\mathrm{O}_{2}] $$ After simplifying, the overall rate law for Mechanism 2 becomes: $$ \text{rate} = (\frac{k_3 k_4}{k_{-3}})[\mathrm{NO}]^{2}[\mathrm{O}_{2}] $$ where \(k = \frac{k_3 k_4}{k_{-3}}\). Since both Mechanisms 1 and 2 produce rate laws in the form of \(k[\mathrm{NO}]^{2}[\mathrm{O}_{2}]\), they are both consistent with the observed rate law.

Key concepts

Understanding Rate Law

The rate law is a mathematical equation representing the relationship between the concentration of reactants and the rate of a chemical reaction. It's like a recipe that tells you how changing the amount of ingredients—the reactants—affects the speed of cooking—the reaction rate.

For a reaction with the general form \(aA + bB \rightarrow products\), the rate law could look like \(rate = k[A]^m[B]^n\). Here, \(k\) is the rate constant, governing the reaction speed, and \(m\) and \(n\) are the reaction orders that tell how the rate changes with concentrations. These orders are not necessarily the same as the stoichiometric coefficients \(a\) and \(b\); they are determined experimentally.

To illustrate, let's consider the rate law from the exercise we're discussing: \(rate = k[NO]^2[O_2]\). This implies that if you double the concentration of \(NO\), the reaction rate quadruples because \(2^2 = 4\). However, doubling the concentration of \(O_2\) only doubles the rate. Understanding this concept is crucial for predicting how a reaction behaves under different conditions and designing experiments in the lab.

Elementary Reactions and their Symphony

Elementary reactions are the simple steps that make up a complex reaction mechanism. Think of them as individual dance moves that come together to form an entire performance. Each elementary step has its own rate law based on molecularity—the number of molecules involved in the step.

Steps involving one molecule are unimolecular, with a rate law of the form \(rate = k[A]\). For two molecules, it's bimolecular, resembling \(rate = k[A][B]\) or \(rate = k[A]^2\) if both molecules are the same. And so on, for more molecules.

In our example, mechanism 1 features a bimolecular step followed by another bimolecular step—hence the squared concentration of \(NO\) in the rate law. Mechanism 2 starts with two \(NO\) molecules coming together, followed by their product reacting with \(O_2\), which gives a similar rate law. By determining the rate laws of elementary steps, chemists can propose mechanisms that agree with experimental observations, just as we've seen with the two proposed mechanisms for the reaction of \(NO\) and \(O_2\).

Chemical Kinetics: The Pace of Chemical Reactions

Chemical kinetics is the study of the rates of chemical processes and the factors affecting them. It's like being a timekeeper and strategist for chemical reactions. By understanding kinetics, you gain insight into how fast a reaction occurs under certain conditions, what the optimal conditions for a desired reaction rate are, and how to control reactions in industrial and lab settings.

Key factors that influence reaction rates include the presence of a catalyst, temperature, reactant concentrations, and the physical state of the reactants. Much like a coach's game plans, kinetics can be complex and nuanced, requiring an understanding of energy barriers, also known as activation energy, and transition states—the high-energy states reactants must pass through to become products.

The investigation into the reaction of \(NO\) with \(O_2\) is one such example of kinetics in action. By dissecting the reaction into potential elementary steps—mechanisms 1 and 2—and examining their compatibility with the observed rate law, we gain a better understanding of not just which steps might be taking place, but also how various conditions could alter the reaction's pace.