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Sulfur dioxide is oxidized to sulfur trioxide in the following sequence of reactions: $$ \begin{aligned} &2 \mathrm{SO}_{2}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)+2 \mathrm{NO}(g) \\ &2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{aligned} $$ (a) Write the chemical equation for the overall reaction. (b) Identify any molecule that acts as a catalyst or intermediate in this reaction.

Short Answer

Expert verified
The overall reaction is \(2 \text{SO}_2 (g) + \text{O}_2 (g) \to 2 \text{SO}_3 (g)\). \(\text{NO}_2\) is the catalyst; \(\text{NO}\) is the intermediate.

Step by step solution

01

Analyze Individual Reactions

The given reactions are two-step sequences. In the first reaction, \(2 \text{SO}_2\) reacts with \(2 \text{NO}_2\) to produce \(2 \text{SO}_3\) and \(2 \text{NO}\). In the second reaction, \(2 \text{NO}\) combines with \(\text{O}_2\) to regenerate \(2 \text{NO}_2\). Both reactions involve the transformation of substances.
02

Identify Overall Reaction

To find the overall reaction, add up both reactions given: \[ \begin{aligned} & 2 \text{SO}_2 (g) + 2 \text{NO}_2 (g) \to 2 \text{SO}_3 (g) + 2 \text{NO} (g) \ & 2 \text{NO} (g) + \text{O}_2 (g) \to 2 \text{NO}_2 (g) \end{aligned} \] The \(\text{NO}\) produced in the first reaction is fully consumed in the second, and the \(2 \text{NO}_2\) is regenerated, leaving the final equation: \[2 \text{SO}_2 (g) + \text{O}_2 (g) \to 2 \text{SO}_3 (g)\]
03

Identify Catalyst and Intermediate

A catalyst is a substance that speeds up a reaction without being consumed, and an intermediate is a molecule that is produced and consumed during the reaction pathway. Here, \(\text{NO}_2\) is regenerated in the second step, indicating it acts as a catalyst. \(\text{NO}\) is formed in the first reaction and consumed in the second, indicating it serves as an intermediate.

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Key Concepts

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

Catalysis
Catalysis plays a crucial role in chemical reactions by speeding up the reaction process without being consumed. Consider catalysts like helpful guides—they assist but remain unchanged. In the sulfur dioxide oxidation to sulfur trioxide example, \( \text{NO}_2 \) acts as a catalyst. It facilitates the transformation of \( \text{SO}_2 \) into \( \text{SO}_3 \), while being regenerated at the end.
This is why catalysts are incredibly valuable in industrial processes. They increase efficiency, lower energy costs, and sometimes even allow reactions to occur under more manageable conditions. Understanding catalysis is vital for both chemistry students and professionals alike.
Reaction Intermediates
Reaction intermediates are molecules that form temporarily during a reaction sequence. Think of them as pit stops—critically important, but only temporarily present. In the oxidation of sulfur dioxide to sulfur trioxide, \( \text{NO} \) is the reaction intermediate.
It is produced in the first reaction and consumed in the second. This characteristic makes it essential in the sequence, even though it doesn’t appear in the overall reaction equation.
  • Intermediates are often short-lived.
  • They help complete the reaction pathway.
  • Their identification is key to understanding complex reactions.
Knowing how intermediates function helps in visualizing and managing each step in reaction mechanisms.
Oxidation Reactions
Oxidation reactions involve the transfer of electrons, leading to an increase in the oxidation state of a molecule. They're a fundamental part of countless chemical processes. In our example, sulfur dioxide \((\text{SO}_2)\) is oxidized to sulfur trioxide \((\text{SO}_3)\) by gaining an oxygen atom.
Key characteristics of oxidation reactions include:
  • The loss of electrons by a molecule, atom, or ion.
  • Often paired with reduction reactions, where another species gains the electrons.
  • Essential in processes like energy production and metabolic pathways.
Understanding oxidation reactions provides deeper insights into chemical changes and energy transformations that occur across numerous scientific and real-world applications.

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Most popular questions from this chapter

Chlorite is reduced by bromide in acidic solution according to the following balanced equation: \(\begin{aligned} \mathrm{ClO}_{2}^{-}(a q)+4 \mathrm{Br}^{-}(a q)+4 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+& 2 \mathrm{Br}_{2}(a q) \\ &+2 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned}\) (a) If \(\Delta\left[\mathrm{Br}_{2}\right] / \Delta t=4.8 \times 10^{-6} \mathrm{M} / \mathrm{s}\), what is the value of \(\Delta\left[\mathrm{ClO}_{2}^{-}\right] / \Delta t\) during the same time interval? (b) What is the average rate of consumption of \(\mathrm{Br}^{-}\) during the same time interval?

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