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Chapter 13: Problem 104

Consider a car fitted with a catalytic converter. The first 5 minutes or so after it is started are the most polluting. Why?

Short Answer

Expert verified
The first few minutes after starting are the most polluting because the catalytic converter hasn't reached its optimal operating temperature, thus being less efficient at converting harmful gases into less harmful substances. Moreover, during the initial phase, the engine operates at a rich fuel to air mixture, producing more emissions.

Step by step solution

01

Understand Catalytic Converters

A catalytic converter is a device designed to reduce harmful emissions from a car's engine by converting them into less harmful substances. It uses catalysts such as platinum, palladium and rhodium to initiate reactions converting toxic gases: carbon monoxide (a poisonous gas) to carbon dioxide, nitrogen oxides (which contribute to smog and acid rain) to nitrogen and oxygen, and unburned hydrocarbons (which contribute to smog) to carbon dioxide and water.
02

Role of temperature

The chemical reactions that occur within the catalytic converter require heat to happen efficiently. That's why they are located near the engine to utilize its heat. Under ideal conditions, the converter temperature should be between 400 and 800 degrees Celsius for the reactions to occur proficiently.
03

The initial phase after starting the car

When a car is first started, the catalytic converter is still cold (ambient temperature) and doesn't work efficiently. The necessary reactions do not occur adequately due to the lower temperature. During this phase (usually the first 5-10 minutes of operation), the car ends up releasing more pollutants.
04

How engines contribute

At the same time, when an engine is cold, it operates at a rich fuel-air mixture, which means it burns more fuel than necessary producing more emissions. Until the engine reaches its optimal operating temperature and the air-fuel mixture leans, excess unburned fuel may pass through the engine and enter the exhaust stream, exacerbating the emission problem.

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Most popular questions from this chapter

The rate law for the decomposition of ozone to molecular oxygen \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\) is rate \(=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}\) The mechanism proposed for this process is \(\mathrm{O}_{3} \stackrel{k_{1}}{k_{-1}} \mathrm{O}+\mathrm{O}_{2}\) \(\mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2}\) Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

Write the equations relating the half-life of a secondorder reaction to the rate constant. How does it differ from the equation for a first-order reaction?

Distinguish between homogeneous catalysis and heterogeneous catalysis. Describe three important industrial processes that utilize heterogeneous catalysis.

The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 M\)

Determine the molecularity and write the rate law for each of the following elementary steps: (a) \(\mathrm{X} \longrightarrow\) products (b) \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products (c) \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products (d) \(\mathrm{X}+\mathrm{X} \longrightarrow\) products (e) \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products
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