You have studied the gas-phase oxidation of HBr by O 2 : 4 HBr(g) + O 2 (g) → 2 H 2 O(g) + 2 Br 2 (g) You find the reaction to be first order with respect to HBr and first order with respect to O 2 . You propose the following mechanism: HBr(g) + O 2 (g) → HOOBr(g) HOOBr(g) + HBr(g) → 2 HOBr(g) HOBr(g) + HBr(g) → H 2 O(g) + Br 2 (g) (a) Confirm that the elementary reactions add to give the overall reaction.

Hello everyone today. We are being given a series of steps over an overall reaction and then asked if these two reactions add to give the overall reaction. So in order to even find out what that means, we have to remember two rules. So the first one, is that the same species on the same side? So same species, same side they are going to add. And the second role is that the same species on opposite sides. So opposite sides, you're going to subtract. So same species, same side, you add them, same species on opposite sides, you subtract. So if you write out our step one, we have N 02 in the gas form, adding to N 02 in the gas form to yield an gasses as well as you know, in the gasses, we're then going to react that N. 03 in the gaseous form with carbon monoxide gas to yield N. 02 in the gas form as well as C. 02 in the gas form as well. And so using the same rules are going to get rid of some ions as well as add them together. So, off the bat, we note that N 02 is visible for step one on the left and on step two on the right. And so we also see that N. 03 on the right of step one is also present on the left of Step two. Leaving us with an overall reaction of N. 02 in the gas form of course, plus our C. O. In the gas form yielding us, you know, or nitric oxide, and the gas form, as well as C. +02 in the gas form. And so from the question stem, we have the overall reaction right here, and we can see that yes, they and up to the overall reaction. So yes, the two steps add up to the overall reaction. I hope this helped, and until next time.

Ch.14 - Chemical Kinetics

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Brown 14th Edition

Ch.14 - Chemical Kinetics

Problem 74a

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Chapter 14, Problem 74a

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.

Verified step by step guidance

insert step 1> Identify the given elementary reactions and the overall reaction.

insert step 2> Write down each elementary reaction: 1) HBr(g) + O_2(g) → HOOBr(g), 2) HOOBr(g) + HBr(g) → 2 HOBr(g), 3) HOBr(g) + HBr(g) → H_2O(g) + Br_2(g).

insert step 3> Add the reactants and products of the elementary reactions to see if they match the overall reaction: 4 HBr(g) + O_2(g) → 2 H_2O(g) + 2 Br_2(g).

insert step 4> Cancel out any intermediates that appear on both sides of the combined equation.

insert step 5> Verify that the remaining reactants and products match the overall reaction, confirming the mechanism is consistent with the overall reaction.

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Key Concepts

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

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves balancing chemical equations to ensure that the number of atoms for each element is conserved. In this context, confirming that the elementary reactions add up to the overall reaction requires checking that the total number of each type of atom on the reactant side equals that on the product side.

Reaction Mechanism

A reaction mechanism is a step-by-step description of the pathway from reactants to products in a chemical reaction. It includes elementary steps, which are individual reactions that occur in sequence. Understanding the proposed mechanism helps in analyzing how the overall reaction occurs and how the rate laws are derived from these elementary steps.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For this reaction, it is stated to be first order with respect to both HBr and O2, indicating that the rate is directly proportional to the concentration of each reactant. This concept is crucial for understanding how the proposed mechanism aligns with the observed reaction order.

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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.

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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)

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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

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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?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important?

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