The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: H 2 O 2 (aq) + I - (aq) → H 2 O(l) + IO - (aq) (slow) IO - (aq) + H 2 O 2 (aq) → H 2 O(l) + O 2 (g) + I - (aq) (fast) (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

Hey there. Welcome back. So we're going to be writing a rate law for the given mechanism here. Okay, whoops. So notice that uh this mechanism, we have actually 1233 different steps, but in order to write a rate law, we're gonna go ahead and only look at the slow steps for now. Okay. So we're gonna go ahead and take a look at the slow step and we're going to say that the rate law here is going to equal to K. K. Times the concentration of the reactant, right? And their coefficients here are going to be their orders. So if we right, C three H six here, we only have 11 roll of it. So it's just gonna be to the first power and then times P. T. H. There's also only one all of it. So it's just gonna be to the first power. Now, this is not the answer. Okay, So this be cannot be the answer. Why? Well, we we only this would work from the slow step. As long as none of the reactant in our slow step are intermediates. But if you take a look, we have P. T. H. Here and then we have P th right here. Right? So an intermediate is going to be a product first. And then in the very next step as a reactant. So what happens with intermediates is that they actually cancel each other out? So this p th is going to cancel one of these. Right? So this one is going to be completely canceled. And then this one we're just going to have one left over, Right? Because there's two moles here. Okay, so now what happens is that this is not going to be part of the rate law? But is this going to be the answer? No. It's not going to be deep. It's not as simple as that because we've lost one of our reactions. We actually need to um and get another one because we were canceling out the intermediate from the first step. We're going to go ahead and take a look at the first step. Now, Now we're going to go ahead and actually take the reactions of the first step and they are going to be part of our rate law for this slow steps. So we have H2. So let's go ahead and rewrite this to have ke we still still have C three H six. It is not an intermediate. It doesn't cancel with anything out. We don't have P. Th but we're going to have a church to now. And it looks like we have P. T. Right now. Let's go ahead and take a look at Petey. So, remember, a catalyst is not part of a reaction. It's not going to be in the rate law because it doesn't actually participate in the reaction. It just simply um you know, increases the rate of it. Now, Petey here. Remember catalysts in the first step, they're going to be a reactant. And then in the very last step, they're going to be products. So we have two pt here and two pt here, which means these are catalysts. So no, we're not going to have Petey and our answer. Okay, So I'm gonna go ahead and just erase it. Okay, now, what about H two? So here we have one mole of H two, but that is not going to be our answer because we can't have one mole of H two here anymore because we actually canceled one of the products. So before we had a church to producing two moles of P T. H. But because we canceled one of them, we only basically can use a half of H two. So this becomes just the coefficient in front of H two becomes one half. Okay. Which means the order of H two here is going to be too one half. Okay, so this one is going to be to the first power and this one is going to be to the one half power. And that is finally our answer. Okay, so option A here is the correct answer. So again, you can use a slow step, as long as it's um titled a slow to write the rate law. As long as none of the reactant are intermediates. If they are, then we actually have to take a look at that reaction where the intermediate is coming from. So, because we did have an intermediate and one of our reactant got canceled here. We needed to take a look at that first step from where the cancelation came from. Okay? Because we only had one mole of P th left, because in the second step there was only one mole, so it can only cancel one. We had to say that we can only use half of H two, and we couldn't use PT because it is a catalyst and it's not part of the reaction. Right, folks, please let us know if you have any questions and thanks so much for watching.

Ch.14 - Chemical Kinetics

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Brown 15th Edition

Ch.14 - Chemical Kinetics

Problem 72c

Chapter 14, Problem 72c

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:
H₂O₂(aq) + I^-(aq) → H₂O(l) + IO^-(aq) (slow)
IO^-(aq) + H₂O₂(aq) → H₂O(l) + O₂(g) + I^-(aq) (fast)
(c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

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Identify the rate-determining step in the mechanism, which is the slow step: H_2O_2(aq) + I^-(aq) → H_2O(l) + IO^-(aq).

The rate law is determined by the rate-determining step, so write the rate law based on the reactants of this step.

Since the rate-determining step involves H_2O_2 and I^-, the rate law will be first order with respect to each of these reactants.

Express the rate law as: rate = k[H_2O_2][I^-], where k is the rate constant.

Note that the fast step does not affect the rate law, as it is not the rate-determining step.

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Key Concepts

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

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 as rate = k[A]^m[B]^n, where k is the rate constant, and m and n are the orders of the reaction with respect to reactants A and B. Understanding the rate law is crucial for predicting how changes in concentration affect the reaction rate.

Rate Determining Step

The rate determining step (RDS) is the slowest step in a reaction mechanism that dictates the overall reaction rate. In multi-step reactions, the RDS is critical because it limits how quickly the entire process can occur. Identifying the RDS allows chemists to simplify the analysis of complex reactions and focus on the most significant factors influencing the rate.

Catalysis

Catalysis involves the acceleration of a chemical reaction by a substance called a catalyst, which is not consumed in the reaction. In this case, iodide ion acts as a catalyst for the decomposition of hydrogen peroxide. Understanding how catalysts work and their effect on reaction mechanisms is essential for predicting reaction rates and designing efficient chemical processes.

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Related Practice

Textbook Question

Consider the following energy profile.

Textbook Question

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:

H₂O₂(aq) + I^-(aq) → H₂O(l) + IO^-(aq) (slow)

IO^-(aq) + H₂O₂(aq) → H₂O(l) + O₂(g) + I^-(aq) (fast)

(a) Write the chemical equation for the overall process.

Textbook Question

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism:

H₂O₂(aq) + I^-(aq) → H₂O(l) + IO^-(aq) (slow)

IO^-(aq) + H₂O₂(aq) → H₂O(l) + O₂(g) + I^-(aq) (fast)

(b) Identify the intermediate, if any, in the mechanism.

Textbook Question

The reaction 2 NO1g2 + Cl21g2¡2 NOCl1g2 was performed and the following data were obtained under conditions of constant 3Cl24:

(a) Is the following mechanism consistent with the data? NO1g2 + Cl21g2ΔNOCl21g2 1fast2 NOCl21g2 + NO1g2¡2 NOCl1g2 1slow2

Textbook Question

You have studied the gas-phase oxidation of HBr by O₂: 4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g)

You find the reaction to be first order with respect to HBr and first order with respect to O₂. You propose the following mechanism:

HBr(g) + O₂(g) → HOOBr(g)

HOOBr(g) + HBr(g) → 2 HOBr(g)

HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)

(a) Confirm that the elementary reactions add to give the overall reaction.

Textbook Question

You have studied the gas-phase oxidation of HBr by O₂: 4 HBr(g) + O₂(g) → 2 H₂O(g) + 2 Br₂(g)

You find the reaction to be first order with respect to HBr and first order with respect to O₂. You propose the following mechanism:

HBr(g) + O₂(g) → HOOBr(g)

HOOBr(g) + HBr(g) → 2 HOBr(g)

HOBr(g) + HBr(g) → H₂O(g) + Br₂(g)

(b) Based on the experimentally determined rate law, which step is rate determining?