In a hydrocarbon solution, the gold compound (CH 3 ) 3 AuPH 3 decomposes into ethane (C 2 H 6 ) and a different gold compound, (CH 3 )AuPH 3 . The following mechanism has been proposed for the decomposition of (CH 3 ) 3 AuPH 3 : Step 1: (CH 3 ) 3 AuPH 3 k1 k -1 (CH 3 ) 3 Au + PH 3 (fast) Step 2: (CH 3 ) 3 Au k2 C 2 H 6 + (CH 3 )Au (slow) Step 3: (CH 3 )Au + PH 3 ¡k3 1(CH 3 )AuPH 3 (fast) (e) What is the rate law predicted by this mechanism?

Welcome back everyone in this example, we have the reaction where three moles of a reacts with one mole of B. To produce two moles of C and one mole of D. As products were told to observe this reaction, which occurs in a three step mechanism where we have our first step as an equilibrium fast step. So we'll label that here, and we recognize that this is an equilibrium fast step because we have an equilibrium arrow separating our reactant and be from our product F. Here. Moving on to step two. This occurs at a slow step here, and we should recall that our slow step is going to be our rate determining step, and this is going to be what we would base our final answer for or on to complete this prompt here. Now, moving on to step three of our mechanism, we have G plus A, which forms C and D. S products in a fast up. And what we should recognize based on our mechanism is that we have F being produced as a product in step one, where it's consumed as a reactant in step two, and then G is also produced as a product in step two, but then consumed as a reactant in step three. And so we would confirm that F and G both are produced In one step, as products and then consumed in another step. And so we would recall that therefore F and G are intermediates. And sorry, so just so it's visible, we would say F and G are intermediates here. So this is just something to make note of here, and looking at F being our first intermediate, we see that in step one, it's our product in our fast equilibrium step. So we want to write out a rate expression to represent this equilibrium fast step for the formation of our product being F. So we would say that for step one We have the rate constant. K. for step one. For the forward reaction of our concentration of the reactant a times our concentration of reactant B, Which are both first order reactant with regards to step one. Because they have coefficients of one here is going to be equal because we have an equilibrium here to the rate constant of the reverse reaction times the concentration of F. Now being a reacting. Since it would be considered an equilibrium reverse reaction. Where F is now in the reverse being a reactant instead of a product here. Now we want to go ahead and isolate this rate expression for our equilibrium fast step for our concentration of our product here being F. And so in order to do so we would divide both sides by the reverse reaction of our rate constant. So we would divide both sides. Bye. K. to the -1. So this would now leave us with our formation for our product or the concentration of our product, F being equal to our rate constant. For the forward reaction of step one multiplied by our concentration of our reactant. A and our concentration of reactant B. Divided by our rate constant for the reverse reaction. And we can place this in parentheses here. So now that we've isolated for the relationship of the formation of the concentration of our product F. We want to go ahead and move on to step two. Of our mechanism, which we again stated is our rate determining slow step here, and this is going to be the basis of our final answer. So for step two, as we stated, is a slow step. We can say that our rate or a rate law for step two is going to be that the rate is equal to The rate constant K. for step two. So we'll have a subscript of two here, which is then multiplied by the concentration of the reactant involved in step two. And as we can see we have F. And a being our reactant in step two. So this is multiplied by the concentration of F. Which is then multiplied by the concentration of a. Which are both First order with respect to Step two because they have coefficients of one and now that we know what our concentration for F can be substituted as which would be by the above relationship we've outlined here, we're going to take this relationship that we've outlined and we're going to substitute this and we'll just draw the arrow closer. So we'll substitute this for our concentration of F. Here for our rate determining rate law here. So what we can say is that therefore, and we'll just use the color purple. Actually our rate should now equal. Our numerator is here will take so we have K one times K. Two in the numerator, Divided by our denominator being the rate constant of the reverse reaction in step one, so K -1. So this quotient is multiplied by our concentration of a times the concentration of B. So this would complete substituting this value for our concentration of F. Here and all we would be left over with is our concentration of reacting again. And so we would include that here. Below. And now we want to simplify this rate expression so we can simplify this so that we have our rate set equal two instead of three expressions of our rate constant because we know that this is all an expression of just the same value. We would just simplify this to just K. And so our rate constant K. We can say is equal to we have eight times a. Which would give us a squared. I'm sorry, eight times A. Here gives us our A. Squared which is then multiplied by our concentration of reactant B. And so this would represent our simplified rate law as our final answer to complete this example. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below. And I will see everyone in the next practice video.

Ch.14 - Chemical Kinetics

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Brown 14th Edition

Ch.14 - Chemical Kinetics

Problem 112e

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Chapter 14, Problem 112e

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?

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Step 1: Identify the slow step in the reaction mechanism. In this case, it is Step 2: (CH3)3Au k2 C2H6 + (CH3)Au. The rate law of the reaction is determined by the slowest step, also known as the rate-determining step.

Step 2: Write the rate law for the slow step. The rate law is written as the rate equals the rate constant times the concentration of the reactants. In this case, the reactant is (CH3)3Au, so the rate law for this step is rate = k2[(CH3)3Au].

Step 3: Check if the reactants in the rate law are the products of the previous steps. If they are, we need to substitute them with the reactants of the previous steps. In this case, (CH3)3Au is the product of Step 1.

Step 4: Substitute the reactant in the rate law with the reactants of the previous step. The reactants of Step 1 are (CH3)3AuPH3. So, the rate law becomes rate = k2[(CH3)3AuPH3].

Step 5: This is the final rate law predicted by the mechanism. It shows that the rate of the reaction is directly proportional to the concentration of (CH3)3AuPH3.

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Key Concepts

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

Reaction Mechanism

A reaction mechanism is a step-by-step description of the pathway through which reactants are converted into products. It outlines individual elementary steps, each with its own rate, and helps in understanding how the overall reaction occurs. In this case, the mechanism involves three steps, with the slowest step determining the rate of the overall reaction.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It is typically formulated based on the slowest step of the reaction mechanism, where the rate is proportional to the concentrations of the reactants raised to their respective powers. For the given mechanism, identifying the slow step is crucial to derive the correct rate law.

Elementary Steps

Elementary steps are individual reactions that occur in a reaction mechanism. Each step can be characterized by its own rate constant and molecularity, which indicates the number of molecules involved in the step. Understanding these steps is essential for determining the overall rate law, as the slowest elementary step dictates the rate of the entire reaction.

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Reaction Mechanism Overview

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Related Practice

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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Open Question

Consider the hypothetical reaction 2 A + B → 2 C + D. The following two-step mechanism is proposed for the reaction: Step 1: A + B → C + X Step 2: A + X → C + D. X is an unstable intermediate. (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

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)

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Textbook Question

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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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. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0-nm nanoparticle.

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