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A gas reacts with a solid that is present in large chunks. Then the reaction is run again with the solid pulverized. How does the increase in the surface area of the solid affect the rate of its reaction with the gas? Explain.

Short Answer

Expert verified
Increasing the surface area of the solid speeds up the reaction rate because it allows more gas molecules to interact with the solid simultaneously.

Step by step solution

01

Identify the Reaction Components

Understand the components involved in the reaction: a gas and a solid. Note that the solid is initially in the form of large chunks.
02

Understand Surface Area

Recognize that surface area plays a crucial role in reactions involving solids and gases. The surface area is the measure of how much exposed area a solid object has.
03

Describe the Scenario with Large Chunks

In the initial scenario where the solid is in large chunks, the surface area exposed to the gas is limited. This means fewer particles from the gas can interact with the solid's surface at a given time.
04

Analyze the Pulverized Solid

When the solid is pulverized, its total surface area increases significantly. Pulverizing a solid breaks it into many smaller pieces, exposing more surface area for the gas to interact with.
05

Explain the Effect on Reaction Rate

An increase in surface area allows more gas particles to come into contact with the solid particles simultaneously. This increases the frequency of collisions between reactant molecules, thus speeding up the reaction rate.
06

Formulate the Conclusion

Therefore, pulverizing the solid increases its surface area, which in turn increases the rate of the reaction with the gas.

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Key Concepts

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

Gas-Solid Reactions
In chemistry, a gas-solid reaction involves a gas reacting with a solid substance. The rate at which this reaction occurs can be influenced by several factors, including temperature, pressure, and the physical state of the solid. When large chunks of a solid are used, the reaction rate may be slower due to the limited surface area. Pulverizing the solid into smaller pieces increases the surface area, allowing more gas molecules to interact with the solid surface simultaneously. This increased contact area accelerates the reaction, making gas-solid reactions sensitive to changes in surface area.
Chemical Reaction Rate
The chemical reaction rate indicates how fast a reaction proceeds. It is determined by the frequency of collisions between reactant molecules. In the case of gas-solid reactions, the rate can be affected by how well the gas molecules can access the solid surface. The more frequent and energetic these collisions are, the faster the reaction will occur. Key factors influencing the reaction rate include:
  • The concentration of reactants
  • The temperature of the system
  • The presence of catalysts
  • The surface area of the solid reactant
Increasing the surface area by pulverizing the solid results in more collisions and a higher reaction rate.
Surface Area in Chemistry
Surface area is a critical concept in chemistry when it comes to reactivity. It is the total area of the exposed surface of a substance. For solid reactants, increasing the surface area by breaking it into smaller pieces enhances their reactivity. More surface area means more sites available for chemical reactions to take place, leading to a higher frequency of collisions. This concept is crucial in processes such as catalysis, adsorption, and gas-solid reactions. To illustrate, a powdered solid will react much faster than a large chunk because the gas molecules have more surface to interact with.
Collision Theory
Collision theory explains how and why reactions occur at the molecular level. According to this theory, for a reaction to happen, reactant molecules must collide with sufficient energy and proper orientation. The energy required for a collision to result in a reaction is called the activation energy. In gas-solid reactions, increasing the surface area of the solid enhances the number of fruitful collisions between gas molecules and the solid surface. This means that pulverizing the solid leads to more frequent and effective collisions, thereby increasing the reaction rate.
Therefore, the principles of collision theory align with the observation that increasing the surface area of a solid reactant can significantly speed up the chemical reaction rate.

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Most popular questions from this chapter

Define the half-life of a reaction. Explain on the molecular level why the half-life of a first-order reaction is constant.

In a kinctics experiment, a chemist places crystals of iodine in a closed reaction vessel, introduces a given quantity of \(\mathrm{H}_{2}\) gas, and obtains data to calculate the rate of HI formation. In a second experiment, she uses the same amounts of iodine and hydrogen but first warms the flask to \(130^{\circ} \mathrm{C}\), a temperature above the sublimation point of iodine. In which of these experiments does the reaction proceed at a higher rate? Explain.

Nitrification is a biological process in which \(\mathrm{NH}_{3}\) in wastewater is converted to \(\mathrm{NH}_{4}^{+}\) and then removed according to the following reaction: \(\mathrm{NH}_{4}^{+}+2 \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) The first-order rate constant is given as \(k_{1}=0.47 e^{0.095\left(T-15^{\circ} \mathrm{C}\right)}\) where \(k_{1}\) is in day \(^{-1}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). (a) If the initial concentration of \(\mathrm{NH}_{3}\) is \(3.0 \mathrm{~mol} / \mathrm{m}^{3},\) how long will it take to reduce the concentration to \(0.35 \mathrm{~mol} / \mathrm{m}^{3}\) in the spring \(\left(T=20^{\circ} \mathrm{C}\right) ?\) (b) In the winter \(\left(T=10^{\circ} \mathrm{C}\right) ?\) (c) Using your answer to part (a), what is the rate of \(\mathrm{O}_{2}\) consumption?

In a study of ammonia production, an industrial chemist discovers that the compound decomposes to its elements \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in a first-order process. She collects the following data: $$ \begin{array}{llll} \text { Time (s) } & 0 & 1.000 & 2.000 \\ {\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L})} & 4.000 & 3.986 & 3.974 \end{array} $$ (a) Use graphical methods to determine the rate constant. (b) What is the half-life for ammonia decomposition?

Reaction rate is expressed in terms of changes in concentration of reactants and products. Write a balanced equation for the reaction with this rate expression: $$ \text { Rate }=-\frac{\Delta\left[\mathrm{CH}_{4}\right]}{\Delta t}=-\frac{1}{2} \frac{\Delta\left[\mathrm{O}_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta[\mathrm{H}, \mathrm{O}]}{\Delta t}=\frac{\Delta\left[\mathrm{CO}_{2}\right]}{\Delta t} $$

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