Consider this three-step mechanism for a reaction:
Cl2( g) Δk1k2...

Question

Consider this three-step mechanism for a reaction: 
Cl2( g) Δk1k22 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?

Accepted Answer

Welcome back everyone in this example. We need to identify the rate law for the theoretical three step mechanism given below. So we want to first recall what a rate law is, recall that a rate law is an expression that relates the rate of a reaction to the concentration of its reactant. Their respective orders, which we should recall are the coefficients the Moeller coefficients to be specific and the rate constant K. Now let's make sure that each of these reactions are balanced. Looking at our first step, step one, we have an equilibrium fast because we have this equilibrium arrow here. So let's label that. And so we recognize that we have two moles of a on the left hand side of this reaction and just one more on the right hand side. So we're going to bounce this by placing a coefficient of two in front of the A. Looking at the second step of our mechanism, we see we have one mole of each of our re agents, so we don't need to add any coefficients. And then the third step we have one mole of B. A. On the reactant side, and two moles of B A. On the product side. So to mend this, we're going to place a coefficient of two in front of our B. A. Here on the reactant side. So now everything is balanced. We next want to write out a rate expression for our first step in this mechanism and again this is an equilibrium fast up. So to write out the rate law, we would say that our rate is equal to because this is an equilibrium reaction step. We want to have to rate constants one for one side of the reaction and one for the second side of the reaction. So we'll have K one and K two. And so our rate law is going to have K one being multiplied by the concentration of our first reactant, which is a two, where we have just one mole of that. And so we would raise it to its respective order as the exponent one, which we don't really have to write in because that should be understood. We would set this equal to the rate constant K2 for the right hand side of our reaction, where we have that being multiplied by the concentration of our second, technically reactant, which is going to be our two moles of a on the right hand side of this equilibrium expression. So we would have our concentration of a raised to its coefficient as an order being to which is why you can see it was important to balance this reaction. Now that we have this rate expression, we next want to identify any intermediates in our mechanism, recall that an intermediate Is being produced in one step and consumed as a reactant, another. And so looking at our mechanism, we have our two moles of a which are produced in our first step of our mechanism where it's being consumed as a reactant in step two of the mechanism. So for our first intermediate we would have A and then we can go ahead and just rule it out because we don't want to consider this in our rate expression for our answer for this problem. So moving on to any other intermediates, we can see that B. A. Is produced in step two of our mechanism as a product where it's being consumed as a reactant in our third step of our mechanism. So we can also rule it out from our rate expression as or for our rate from our rate law as an intermediate. So A and B. A. R. Intermediates. You would say next. We want to make note of the fact that we are told that the second step of this mechanism is a slow step. We want to recall that our slow step is going to be our rate determining step, meaning this is what we will base our rate law off of. And so what we want to do is add all of these reactions together to come up with an overall reaction that we can base our rate law off of. And what we would have is everything that's left after we canceled out our intermediates. So we would say that our rate law and sorry about that. So we would say that what we have left for our overall reaction is going to be A two Plus C. B gives us C plus B A two. So we're going to use this fourth reaction here to help us determine our rate law as well as our rate determining step. So going back to this rate determining step, we want to recognize that this is going to be our third rate constant K three, Whereas this would be K4. But what matters here is our third step here are slow step. So we're going to write a rate expression for our rate determining step. So rate determining step rate law We can say is going to be that our rate is equal to K3 times the concentration of our first reactant in step, two which is our intermediate A multiplied by the concentration of our second reactant in step two, which is our concentration of C. B. And these are both raised to their respective orders as coefficients of one, which we are understanding here. So we don't have to write that in Now. We also agreed that A is an intermediate. So technically we need to substitute something in place of A for our rate law here. And so we need to go back to that equilibrium expression that we wrote for step one of this mechanism to isolate it for something that we can substitute for our concentration of A. And so going back to step two. So we're gonna say, since A is intermediate, I'm sorry, not going back two steps, you're going back to step one we would go back to step one and we would say that our first rate constant K one multiplied by its concentration of the reactant. A two to its order one is equal to our rate constant K two or second rate constant K two times or times the concentration of the second reactant being our concentration of A. To its order being too since it had that coefficient of two that we added in to balance out the reaction. So now we want to simplify this expression to just get a alone and so to do so we're going to divide both sides by K two. What that's going to give us is K1 divided by K2 times our concentration of a. two is equal to our concentration of a. To the second order as a coefficient or sorry as an exponent. And so now we want to get rid of that exponents by taking both sides to a power of one half. And that will cancel that out on the right hand side. Which is going to give us that our concentration of our intermediate a is equal to the expression K1 divided by K two times our concentration of a. Two to the one half power. And this is what we will use to substitute substitute for a. So we'll say use this to substitute a in rate law. And so with that in mind we can now say that our rate, Our rate law for our rate determining slow step is now K times the and sorry, this should actually be a parenthesis here. So Kay three times our substitution for A which is K one over K two times the concentration of A two Which is all raised to the 1/2 power. Where this is done multiplied by the concentration of our second reactant from our rate determining slow step, C. B. Now we want to recall that K. Our rate constant is just expressed in three different ways as K one, K two and K three for all the three steps of our mechanism. But they really only represent one constant. So we would say K represents one constant rather let's be more specific and say K one K two. Oops, sorry K one K two And K three represents one constant. And so therefore we can say that K three multiplied by our expression where we say K one over K two is raised to the one half power is all equal to just our single rate constant. K. Since it's just all the same constant value. And so with that being understood since we've simplified it to that, we would say that for our final answer, our rate law is equal to the rate constant as just K multiplied by our Concentration of a two raised to the one half power, which is multiplied by our concentration of our second reactant, C B. And this would be our final answer to complete this example as our rate law. So this is our rate law for our three step mechanism. So I hope that everything I went through is clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

Consider this three-step mechanism for a reaction: Cl2( g) Δk1k22 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?

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