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Ch.14 - Chemical Kinetics
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All textbooksBrown 14th EditionCh.14 - Chemical KineticsProblem 80b

Chapter 14, Problem 80b

The addition of NO accelerates the decomposition of N2O, possibly by the following mechanism: NO1g2 + N2O1g2¡N21g2 + NO21g2 2 NO21g2¡2 NO1g2 + O21g2 (b) Is NO serving as a catalyst or an intermediate in this reaction?

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welcome back everyone below. We need to consider the given mechanism and determine whether our nitrate gas is acting as a catalyst or an intermediate for the given mechanism. Now, before we begin to determine that we want to recall our definition first for a catalyst. We should recall that a catalyst is going to be consumed in one step of our mechanism, but then it's going to be produced in another step as a product. Next, we want to recall our definition for an intermediate, and we should recall that an intermediate is going to be produced as a product in one step first, but then in another step it's going to be consumed as a reactant. And so let's go ahead and analyze how our nitrate gas is acting in this mechanism. So looking at step one of our mechanism, we have the expanded form where two moles of our nitrite gas are reacting with one another to form nitrate gas as a product, and nitrogen monoxide as a gaseous product as well. So we should recognize that our nitrate gas here is being produced of the product. So let's specifically right that nitrate is produced as a product. And moving on to step two of our mechanism, we have our nitrate gas as a reactant now reacting with carbon monoxide gas to form nitrite gas and carbon dioxide gas. And so we should recognize here that our nitrate gas is consumed As a reactant in step two. And so based on these observations, we can say that therefore our nitrate gas acts as an intermediate because in Step one, it acted or it was produced as a product, but in Step two it acted as a reactant where it was consumed to form our products. And so our final answer here is going to be this statement clarifying that nitrate gas acts as an intermediate. This completes this example. If you have any questions, leave them down below, and I will see everyone in the next practice video.
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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?

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(c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

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

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Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina 1Al2O32 or silica 1SiO22. (b) How does the surface area affect the rate of reaction?

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The enzyme urease catalyzes the reaction of urea, 1NH2CONH22, with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of 4.15 * 10-5 s-1 at 100 C. In the presence of the enzyme in water, the reaction proceeds with a rate constant of 3.4 * 104 s-1 at 21 C. (c) In actuality, what would you expect for the rate of the catalyzed reaction at 100 C as compared to that at 21 C?

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The activation energy of an uncatalyzed reaction is 95 kJ/mol. The addition of a catalyst lowers the activation energy to 55 kJ/mol. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) 25 C

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