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Chapter 14: Problem 79

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2} .\) The reaction proceeds according to: $$ \begin{array}{l}{\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)}\end{array}$$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(S O_{2}\) by \(\mathrm{O}_{2}\) to give \(S O_{3} .(\mathbf{b})\) Do we consider \(N O_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? { ( d ) } Is this an example of homogeneous catalysis or heterogeneous catalysis?

Short Answer

Expert verified
(a) The overall reaction for the oxidation of SO₂ to SO₃ is: \(2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)\) (b) NO₂ acts as a catalyst in this reaction, as it is utilized in the first reaction and regenerated in the second. (c) NO is an intermediate in this set of reactions, as it is produced in the first reaction and consumed in the second. (d) This is an example of homogeneous catalysis, as all species involved are in the gas phase.

Step by step solution

01

(a) Finding the overall reaction

First, let's balance the two given reactions: 1. NO₂(g) + SO₂(g) → NO(g) + SO₃(g) (already balanced) 2. 2NO(g) + O₂(g) → 2NO₂(g) (already balanced) Now we can sum the reactions. In order to cancel the NO(g) and NO₂(g) species, we need two of the first reaction to balance the second reaction: 2(NO₂(g) + SO₂(g) → NO(g) + SO₃(g)) 2NO(g) + O₂(g) → 2NO₂(g) Summing the reactions, we get: 2NO₂(g) + 2SO₂(g) + 2NO(g) + O₂(g) → 2NO(g) + 2SO₃(g) + 2NO₂(g) Now we can cancel the species present on both sides: 2SO₂(g) + O₂(g) → 2SO₃(g) This is the overall reaction for the oxidation of SO₂ to SO₃.
02

(b) Identifying NOâ‚‚ as a catalyst or an intermediate

A catalyst is a substance that increases the rate of a chemical reaction but remains unchanged at the end of the reaction. Looking at the overall reaction, we can see that NOâ‚‚ does not appear in the final reaction. Since it is utilized in the first reaction and regenerated in the second, it is a catalyst in this reaction.
03

(c) Identifying NO as a catalyst or an intermediate

An intermediate is a reaction species that is produced during a reaction and is then consumed in a subsequent reaction. NO is produced in the first reaction and consumed in the second reaction, so it is an intermediate in this set of reactions.
04

(d) Classifying the type of catalysis

Homogeneous catalysis occurs when the catalyst and the reactants are in the same phase (e.g., both are gases, liquids, or solids), while heterogeneous catalysis occurs when the catalyst and the reactants are in different phases. In this case, all species are gases, so this is an example of homogeneous catalysis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Catalyst
In chemistry, a catalyst is a substance that speeds up a chemical reaction without being consumed in the process. Think of it as a helper that makes things happen faster but doesn't get used up itself. In the given reaction, NO₂ is a catalyst. It helps in converting SO₂ to SO₃ more quickly.

Catalysts work by lowering the activation energy required for a reaction. This means less energy is needed for the reaction to occur, making it faster. An important feature of catalysts is that they do not change the final amount of products; they only change how fast the products are made.
  • Catalysts are often used in industries to make chemical processes more efficient.
  • They are essential in reactions where speed is crucial.
NOâ‚‚ participates in the reaction but is regenerated, making it an ideal example of a catalyst. It enters in one form, participates, and exits unchanged so that it can continue to facilitate more reactions.
Chemical Reaction
A chemical reaction involves the rearrangement of atoms to change substances into different materials. In the exercise, we see two reactions taking place. The first reaction involves NO₂ and SO₂ transforming into NO and SO₃. The second reaction converts these intermediates back to NO₂ using additional O₂.

Chemical reactions can be represented by equations where reactants are transformed into products. Both reactions in this exercise are balanced, meaning the number of each type of atom on the reactant side equals the number on the product side.
  • Reactions must be balanced to obey the law of conservation of mass.
  • Reactants are the starting substances, while products are the new substances formed.
In the problem, understanding how to balance and combine these reactions helps in determining the overall reaction, which is integral to the goal of transforming SO₂ to SO₃ using the aid of a catalyst.
Oxidation Reaction
An oxidation reaction involves the loss of electrons by a molecule, atom, or ion. In simpler terms, it's a chemical process where a substance gains oxygen or loses hydrogen. In the context of the given exercise, SO₂ is oxidized to form SO₃. This means SO₂ gains oxygen in this process.

Oxidation reactions are often coupled with reduction reactions, where another substance gains electrons. This combined process is called a redox reaction. Redox reactions are vital in many biological and industrial chemical processes.
  • Oxidation increases the oxygen content in a compound.
  • It usually leads to the formation of oxides, like SO₃ in this reaction.
The importance of oxidation reactions cannot be understated, as they are crucial in processes ranging from energy production in cells to metal rusting. In the exercise, the final overall reaction shows this oxidation happening with Oâ‚‚, facilitated by the presence of NOâ‚‚ as a catalyst to speed up the transformation.

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Most popular questions from this chapter

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$\begin{array}{c}{\mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g)} \\ {\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g)}\end{array}$$ \(\begin{array}{l}{\text { (a) Write the balanced equation for the overall reaction. }} \\ {\text { (b) Identify any intermediates in the mechanism. (c) If }}\end{array}\) the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{array}{c}{\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \text { (slow) }} \\ {\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{g})+\mathrm{I}^{-}(a q) \text { (fast) }}\end{array} $$ \(\begin{array}{l}{\text { (a) Write the chemical equation for the overall process. }} \\ {\text { (b) Identify the intermediate, if any, in the mechanism. }} \\ {\text { (c) Assuming that the first step of the mechanism is rate }} \\ {\text { determining, predict the rate law for the overall process. }}\end{array}\)

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: \(\begin{array}{l}{\text { (a) } 2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)} \\\ {\text { (c) } 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\text { (d) } \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)}\end{array}\)

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\) . When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\) . (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of 1.8\(?\)

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c),\) has an activation energy of 86.8 \(\mathrm{kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) A g \(\mathrm{KOH}\) in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reac-solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving 0.335 g KOH in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Similarly, 1.453 \(\mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form 250.0 \(\mathrm{mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\) .
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