In a hydrocarbon solution, the gold compound
1CH323AuPH3 decompos...

Question

In a hydrocarbon solution, the gold compound
1CH323AuPH3 decomposes into ethane 1C2H62 and a
different gold compound, 1CH32AuPH3. The following
mechanism has been proposed for the decomposition of
1CH323AuPH3:
Step 1: 1CH323 AuPH3 Δ k1 k -1 1CH323Au + PH3 1fast2
Step 2: 1CH323 Au ¡k2 C2H6 + 1CH32Au 1slow2
Step 3: 1CH32Au + PH3 ¡k3 1CH32AuPH3 1fast2
(e) What is the rate law predicted by
this mechanism?

Accepted Answer

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.

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