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Ch.14 - Chemical Kinetics
All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 112c
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Chapter 14, Problem 112c

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:
Step 1: (CH3)3AuPH3 k1 k -1 (CH3)3Au + PH3 (fast)
Step 2: (CH3)3Au k2 C2H6 + (CH3)Au (slow)
Step 3: (CH3)Au + PH3 ¡k3 1(CH3)AuPH3 (fast)
(c) What is the molecularity of each of the elementary steps?

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Key Concepts

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

Molecularity

Molecularity refers to the number of reactant molecules involved in an elementary reaction step. It can be classified as unimolecular (one molecule), bimolecular (two molecules), or termolecular (three molecules). Understanding molecularity is crucial for determining the rate law of a reaction and predicting how changes in concentration affect the reaction rate.
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Elementary Steps

Elementary steps are individual reactions that occur in a multi-step reaction mechanism. Each step represents a single molecular event, and the overall reaction is the sum of these steps. Analyzing elementary steps helps in understanding the pathway of the reaction and the role of intermediates, which are species formed during the reaction but not present in the final products.
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Reaction Mechanism Overview

Rate Determining Step

The rate determining step (RDS) is the slowest step in a reaction mechanism that dictates the overall reaction rate. It is crucial to identify the RDS because it has the highest activation energy and thus controls how quickly the reaction proceeds. In the provided mechanism, the slow step is where (CH₃)₃Au decomposes, making it essential for calculating the reaction kinetics.
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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.)

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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)

(d) If instead the reaction occurred in a single step, would the rate law change? If so, what would it be?

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Textbook Question
The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine: Step 1: Cl21g2 Δ k1 k - 1 2 Cl1g2 1fast2 Step 2: Cl1g2 + CHCl31g2 ¡k2 HCl1g2 + CCl31g2 1slow2 Step 3: Cl1g2 + CCl31g2 ¡k3 CCl4 1fast2 (a) What is the overall reaction?
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Textbook Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:

Step 1: (CH3)3AuPH3 k1 k -1 (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au k2 C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 ¡k3 1(CH3)AuPH3 (fast)

(e) What is the rate law predicted by this mechanism?

529
views
Textbook Question

In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for the decomposition of (CH3)3AuPH3:

Step 1: (CH3)3AuPH3 k1 k -1 (CH3)3Au + PH3 (fast)

Step 2: (CH3)3Au k2 C2H6 + (CH3)Au (slow)

Step 3: (CH3)Au + PH3 ¡k3 1(CH3)AuPH3 (fast)

(f) What would be the effect on the reaction rate of adding PH3 to the solution of (CH3)3AuPH3?

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Textbook Question

Platinum nanoparticles of diameter 2 nm are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (b) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere 14pr22 and assuming that the 'footprint' of one Pt atom can be estimated from its atomic diameter of 2.8 A .

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