The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism. Step 1: O 3 (g) ⇌ O 2 (g) + O(g) (fast) Step 2: O(g) + O 3 (g) → 2 O 2 (slow) (d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Hello everyone today. We are being asked to consider a formal reaction with a two step mechanism. So we have step one and we have step to note that step one is the slow step and this is going to come into importance later on. We are then asked if the rate law would change if the reaction occurred in a single step. If yes, what would the new rate law be? So the first thing you wanna do is you want to account for what this overall reaction would look like. So we're gonna look at what the overall reaction look like. First. By doing that, we take our step one, which is that two moles of N. 02 that form one, N. 03 plus one nitric oxide. And we're going to go and add it to the second step which is N. 03 plus carbon monoxide to form N. 02 and C. 02, we are then going to get rid of any molecules on opposite sides, the same molecules and opposite sides are going to subtract from each other and the same molecules on the same side will add. So we can note that for N. 02, we have one on the left here for the first reaction. The first step and then we have one on the right for the second reaction. So we're then going to be left with only one N. 02 at the on the first step on the left, we also note that no three is present on opposite sides of both step one and two. And so since there's only one of each, they both go away and thus we are left with the overall reaction. So the overall it's going to be no two plus C. O. Or carbon monoxide to form in oh and carbon dioxide. And so the rate law of the original. So we'll say rate law of the original reaction is going to be based off of the slow step. And we already set the slow step. Is that two N. 02 molecules. And so the rate is going to be equal to some constant K. Multiplied by our two molecules being used in 02. And since there's two, we're going to use an exponent of two. So this is the rate law of the original one. So if there was only one step we would then consider the overall reaction. And so the overall reaction uses N. 02 as well as ceo. So that rate law. So rate law for one step would be that the rate is equal to some constant K. Times N. 02 times C. O. And that is going to be our final answer. I hope this helps. And until next time

Ch.14 - Chemical Kinetics

All textbooks

Brown 14th Edition

Ch.14 - Chemical Kinetics

Problem 109d

Chapter 14, Problem 109d

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.
Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)
Step 2: O(g) + O₃(g) → 2 O₂ (slow)
(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Verified step by step guidance

Step 1: Identify the rate-determining step. In this case, the rate-determining step is the slowest step, which is Step 2: O(g) + O3(g) → 2 O2.

Step 2: Write the rate law based on the rate-determining step. The rate law for a reaction is generally of the form rate = k[A]^m[B]^n, where A and B are reactants, m and n are their respective orders, and k is the rate constant. For the given reaction, the rate law would be rate = k[O][O3].

Step 3: Consider what would happen if the reaction occurred in a single step. If the reaction occurred in a single step, it would be O3(g) → 2 O2(g).

Step 4: Write the rate law for the single-step reaction. The rate law for this reaction would be rate = k[O3], because there is only one reactant in the rate-determining step.

Step 5: Compare the two rate laws. The rate law for the two-step reaction is rate = k[O][O3], and the rate law for the single-step reaction is rate = k[O3]. Therefore, the rate law would change if the reaction occurred in a single step.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Reaction Mechanism

A reaction mechanism describes the step-by-step sequence of elementary reactions by which overall chemical change occurs. Understanding the individual steps, including their rates and intermediates, is crucial for predicting the behavior of the overall reaction. In this case, the two-step mechanism for ozone decomposition illustrates how the rate of the reaction can be influenced by the slowest step, known as the rate-determining step.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is determined experimentally and can vary depending on the reaction mechanism. For the two-step mechanism provided, the rate law is influenced by the slow step, while a single-step reaction would yield a different rate law based on the stoichiometry of that single step.

Elementary Reactions

Elementary reactions are single-step processes that occur in a chemical reaction, each with its own rate constant. The overall reaction can consist of multiple elementary steps, and the rate law for the overall reaction is derived from the elementary steps. In the context of the ozone decomposition, recognizing whether the reaction occurs in one or multiple steps is essential for determining how the rate law is formulated.

Recommended video:

Guided course

03:06

Reaction Mechanism Overview

Related Practice

Textbook Question

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism:

Cl(g) + O₃(g) → ClO(g) + O₂(g)

ClO(g) + O(g) → Cl(g) + O₂(g)

(b) What is the catalyst in the reaction?

621

views

Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(a) Write the balanced equation for the overall reaction.

Textbook Question

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.

Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)

Step 2: O(g) + O₃(g) → 2 O₂ (slow)

(b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.)

866

views

Textbook Question

The following mechanism has been proposed for the gasphasereaction of chloroform 1CHCl32 and chlorine:Step 1: Cl21g2 Δ k1 k - 1 2 Cl1g2 1fast2Step 2: Cl1g2 + CHCl31g2 ¡k2 HCl1g2 + CCl31g2 1slow2Step 3: Cl1g2 + CCl31g2 ¡k3 CCl4 1fast2(a) What is the overall reaction?

575

views

Textbook Question

The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine:

Step 1: Cl₂(g) k₁⇌ k_-1 2 Cl(g) (fast)

Step 2: Cl(g) + CHCl₃(g) k₂→ HCl(g) + CCl₃(g) (slow)

Step 3: Cl(g0 + CCl3(g) k₃→ CCl₄ (fast)

(e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.Step 1: O3(g) ⇌ O2(g) + O(g) (fast)Step 2: O(g) + O3(g) → 2 O2 (slow)(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

Verified video answer for a similar problem:

Key Concepts

Reaction Mechanism

Rate Law

Elementary Reactions

The gas-phase decomposition of ozone is thought to occur by the following two-step mechanism.
Step 1: O₃(g) ⇌ O₂(g) + O(g) (fast)
Step 2: O(g) + O₃(g) → 2 O₂ (slow)
(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?