Ch.15 - Chemical Kinetics

Tro - Chemistry: A Molecular Approach 6th Edition

Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217Not the one you use?Change textbook

Tro 6th Edition

Ch.15 - Chemical Kinetics

Problem 82

Chapter 15, Problem 82

Consider this overall reaction, which is experimentally observed to be second order in X and first order in Y: X + Y → XY. a. Does the reaction occur in a single step in which X and Y collide? b. Is this two-step mechanism valid? 2X →k1/k2 X2 (Fast) X2 + Y →k3 XY + X (Slow)

Name: Chemistry: A Molecular Approach
Author: Tro
ISBN: 9780137832217

Verified step by step guidance

The reaction is given as second order in X and first order in Y. This means the rate law is rate = k[X]^2[Y].

For a reaction to occur in a single step, the molecularity must match the reaction order. Here, the reaction order is third (2 in X and 1 in Y), which suggests a termolecular process. Termolecular reactions are rare due to the low probability of three molecules colliding simultaneously with the correct orientation and energy.

The mechanism consists of two steps: (1) 2X → X2 (fast) and (2) X2 + Y → XY + X (slow).

The slow step (step 2) is the rate-determining step. The rate law for this step is rate = k3[X2][Y].

From step 1, X2 is formed quickly and is in equilibrium with 2X. Therefore, [X2] can be expressed in terms of [X] using the equilibrium constant for the fast step. Substitute [X2] in the rate law of the slow step to see if it matches the observed rate law rate = k[X]^2[Y]. If it does, the mechanism is valid.>

Key Concepts

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

Reaction Order

Reaction order refers to the power to which the concentration of a reactant is raised in the rate law of a chemical reaction. It provides insight into the relationship between the concentration of reactants and the rate of the reaction. In this case, the reaction is second order in X and first order in Y, indicating that the rate depends quadratically on the concentration of X and linearly on the concentration of Y.

Mechanism of Reaction

The mechanism of a reaction describes the step-by-step sequence of elementary reactions by which overall chemical change occurs. Understanding the mechanism helps in determining how reactants are converted to products, including whether the reaction occurs in a single step or multiple steps. The proposed two-step mechanism suggests that the first step is fast and reversible, while the second step is slow and rate-determining.

Elementary Steps and Rate-Determining Step

Elementary steps are individual reactions that make up the overall reaction mechanism. The rate-determining step is the slowest step in the mechanism, which controls the overall reaction rate. In the provided mechanism, the second step is the slow step, making it the rate-determining step, which aligns with the observed reaction orders and helps validate the proposed mechanism.

Recommended video:

Guided course

01:15

Rate Law Determination

Related Practice

Textbook Question

Which of these two reactions would you expect to have the smaller orientation factor? Explain. a. O(g) + N2(g) → NO( g) + N(g) b. NO(g) + Cl2(g) → NOCl( g) + Cl(g)

1178

views

Textbook Question

Consider this three-step mechanism for a reaction:

Cl₂ (g) k₁⇌k₂ 2 Cl (g) Fast

Cl (g) + CHCl₃ (g) →k₃ HCl (g) + CCl₃ (g) Slow

Cl (g) + CCl₃ (g) →k₄ CCl₄ (g) Fast

a. What is the overall reaction?

625

views

Textbook Question

Consider these two gas-phase reactions: a. AA(g) + BB(g) → 2 AB(g) b. AB(g) + CD(g) → AC(g) + BD(g) If the reactions have identical activation barriers and are carried out under the same conditions, which one would you expect to have the faster rate?

1584

views

Textbook Question

Consider this three-step mechanism for a reaction:

Cl₂ (g) k₁⇌k₂ 2 Cl (g) Fast

Cl (g) + CHCl₃ (g) →k₃ HCl (g) + CCl₃ (g) Slow

Cl (g) + CCl₃ (g) →k₄ CCl₄ (g) Fast

c. What is the predicted rate law?

3695

views