91Ó°ÊÓ

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

Step by step solution

01

Analyse the reaction and given rate-law expression

The reaction given is: \[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \] The rate law is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\). The overall order of the reaction is \(1 + 2 = 3\). It is not consistent with the reaction as written because the stoichiometry does not match the rate law.

02

Evaluate the first mechanism

The first proposed mechanism is: \[ \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] This mechanism has two steps, but the molecularity of the first one is inconsistent with the experimental rate law (it is bimolecular in both \(\mathrm{NO}\) and \(\mathrm{H}_{2}\)). Therefore, this mechanism is not consistent with the experimental rate law.

03

Evaluate the second mechanism

The second mechanism is: \[ 2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \] Let the intermediates be: \[\mathrm{I}_{1} = \mathrm{N}_{2} \mathrm{O}_{2}(g)\] Now, write the rate law for each step: \[ \text {rate}_{1} = k_{1}[\mathrm{NO}]^{2} \\ \text {rate}_{2} = k_{2}[\mathrm{H}_{2}][\mathrm{I}_{1}] \\ \text {rate}_{3} = k_{3}[\mathrm{H}_{2}][\mathrm{N}_{2} \mathrm{O}] \] Since the first step is a fast equilibrium, we can write: \[[\mathrm{I}_{1}] \approx \frac{k_{1}}{k_{-1}}[\mathrm{NO}]^{2}\] Also, to show that the mechanism is consistent, we have to consider that \(\text {rate}_{2} \approx \text {rate}_{3}\). Therefore, the experimental rate law is given by: \[ \text {rate} = k_{2}[\mathrm{H}_{2}] \cdot \frac{k_{1}}{k_{-1}}[\mathrm{NO}]^{2} \\ \text {rate} = k[\mathrm{H}_{2}][\mathrm{NO}]^{2} \\ \] Where \(k = \frac{k_{1}k_{2}}{k_{-1}}\). The second mechanism matches the experimental rate-law expression, which is first order in \(\mathrm{H}_{2}\) and second order in \(\mathrm{NO}\). Therefore, the second mechanism is consistent with the experimental rate-law expression.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rate Law Expression

Understanding the rate law expression is crucial for grasping how chemical reactions take place. The rate law shows how the rate of a reaction is related to the concentrations of reactants. It's written in the form: \[\text{rate} = k[\text{A}]^{m}[\text{B}]^{n} \cdots\], where \( k \) is the rate constant, \( [\text{A}] \) and \( [\text{B}] \) represent the molar concentrations of reactants A and B, and \( m \) and \( n \) are the reaction orders with respect to A and B.
In the provided exercise, the rate law was found to be first order in hydrogen (\( \mathrm{H}_{2} \)) and second order in nitric oxide (\( \mathrm{NO} \)), indicating the reaction rate depends linearly on the concentration of hydrogen and quadratically on the concentration of nitric oxide. To ensure that students can easily apply this concept, it's important to emphasize that the rate law can only be determined experimentally and may not directly correspond to the stoichiometry of the balanced chemical equation.

Reaction Mechanism

A reaction mechanism is a sequence of elementary steps that describe the pathway from reactants to products. Each step involves a transition state and potentially the formation of reactive intermediates. Understanding the mechanism helps explain the macroscopic phenomena observed, such as the rate law.
In our textbook example, the proposed mechanisms break down the overall reaction into simpler, more revealing stages. The first suggested mechanism did not align with the experimental rate law, as it didn't reflect the dependence of the reaction rate on the concentration of reactants. The second mechanism, however, demonstrated consistency with the rate law by introducing a fast initial step followed by subsequent steps that matched the rate dependencies observed experimentally. When discussing mechanisms with students, it's essential to clarify that a mechanism must agree with both the rate law's experimental data and the elementary steps, including their molecularity.

Reaction Order

The reaction order tells us how the rate of the reaction depends on the concentration of each reactant. It is determined by adding up the exponents in the rate law expression; for instance, a reaction that is first order in \( \mathrm{A} \) and second order in \( \mathrm{B} \) would be third order overall.
In our exercise, the reaction order provides insight into the dynamics of the chemical process. An interesting point to convey is that reaction orders are not always integers; they can be fractions or even zero, indicating varying levels of influence by each reactant on the reaction rate. Moreover, it’s necessary to stress that the overall reaction order can't be inferred from the stoichiometry and must be determined experimentally.

Molecular Kinetics

Molecular kinetics delves into the details of how molecules interact to proceed with chemical reactions. It looks at the reaction on a molecular level, considering the movement and collisions of the particles involved. Factors like temperature, pressure, and the presence of a catalyst can play a significant role in a reaction's kinetics.
On a molecular level, for the given reaction involving nitric oxide and hydrogen, the kinetics can help explain why certain mechanisms are favored over others. For instance, a fast initial step that quickly reaches equilibrium can significantly affect the overall reaction rate, as observed in the second, acceptable mechanism of our exercise. Teaching molecular kinetics involves linking the microscopic interactions between molecules to the macroscopic observables like reaction rate and the rate law expression.