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Arrhenius proposed that each reaction has an energy threshold that must be reached for the particles to react. The kinetic theory of gases proposes that the average kinetic energy of the particles is proportional to the absolute temperature. How do these concepts relate to the effect of temperature on rate?

Short Answer

Expert verified
Higher temperatures increase particles' kinetic energy, leading to more particles surpassing the activation energy, which in turn increases the reaction rate.

Step by step solution

01

Understanding the Energy Threshold

Arrhenius suggested that each chemical reaction has a specific energy threshold, known as the activation energy, that must be overcome for the reactants to convert into products.
02

Kinetic Theory of Gases

The kinetic theory of gases states that the average kinetic energy of gas particles is directly proportional to the absolute temperature. This means as the temperature increases, the average kinetic energy of the particles also increases.
03

Relating Temperature and Kinetic Energy

Since temperature increases the average kinetic energy of the particles, higher temperatures lead to more particles having sufficient energy to surpass the activation energy threshold.
04

Impact on Reaction Rate

As more particles have sufficient energy to reach the activation energy threshold, the number of successful collisions increases, thereby increasing the reaction rate.

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Key Concepts

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

Arrhenius Equation
The Arrhenius Equation is fundamental in understanding how temperature affects the rate of a chemical reaction. This equation is represented as follows:

\[ k = A e^{-\frac{E_a}{RT}} \]
where:
  • \( k \) is the rate constant,
  • \( A \) is the pre-exponential factor, also known as the frequency factor,
  • \( E_a \) is the activation energy,
  • \( R \) is the gas constant,
  • \( T \) is the absolute temperature in Kelvin.

The equation essentially tells us that the rate constant \( k \) increases exponentially with an increase in temperature \( T \). This results in a higher reaction rate as temperature rises. The exponential term \( e^{-\frac{E_a}{RT}} \) shows that as temperature increases, the exponent becomes less negative, leading to a larger value for \( k \). This directly links the temperature and reaction rate through the activation energy.
Activation Energy
Activation energy \( E_a \) is the minimum amount of energy needed for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to transform into products. This concept was introduced by Arrhenius.

Imagine reactants climbing a hill. They need a certain amount of energy to get over the top. Once they have enough energy, they can slide down the other side, completing the reaction.
  • If the particles do not have sufficient energy, they won't react when they collide.
  • Higher activation energy means fewer particles will have the required energy at a given temperature.
  • Lower activation energy means more particles can surpass this barrier with the same amount of energy.

By increasing the temperature, the number of particles with energy equal to or greater than the activation energy increases. This translates into a higher reaction rate, as more collisions result in successful reactions.
Kinetic Theory of Gases
The Kinetic Theory of Gases explains how gas particles move and interact. According to this theory, the average kinetic energy of gas particles is directly proportional to the absolute temperature.

Here are some key points to understand:
  • Particles in a gas are in constant, random motion.
  • The temperature of the gas is a measure of the average kinetic energy of the particles.
  • As temperature increases, the speed and kinetic energy of the gas particles also increase.

When considering chemical reactions, the higher average kinetic energy at elevated temperatures means that more particles will have sufficient energy to overcome the activation energy barrier set for the reaction.
Therefore, higher temperatures lead to more collisions having the required energy to result in a reaction, subsequently increasing the rate of that reaction. This interplay between kinetic energy and temperature underscores why reactions speed up with added heat.

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Most popular questions from this chapter

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.

Carbon disulfide, a poisonous, flammable liquid, is an excellent solvent for phosphorus, sulfur, and some other nonmetals. A kinetic study of its gaseous decomposition gave these data: $$ \begin{array}{ccc} \text { Experiment } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \end{array} & \begin{array}{c} \text { Initial }\left[\mathrm{CS}_{2}\right] \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 2.7 \times 10^{-7} & 0.100 \\ 2 & 2.2 \times 10^{-7} & 0.080 \\ 3 & 1.5 \times 10^{-7} & 0.055 \\ 4 & 1.2 \times 10^{-7} & 0.044 \end{array} $$ (a) Write the rate law for the decomposition of \(\mathrm{CS}_{2}\). (b) Calculate the average value of the rate constant.

The growth of Pseudomonas bacteria is modeled as a first-order process with \(k=0.035 \mathrm{~min}^{-1}\) at \(37^{\circ} \mathrm{C}\). The initial \(P\) seudomonas population density is \(1.0 \times 10^{3}\) cells/L. (a) What is the population density after \(2 \mathrm{~h}\) ? (b) What is the time required for the population to go from \(1.0 \times 10^{3}\) to \(2.0 \times 10^{3}\) cells/L?

For the reaction \(\mathrm{A}(g)+\mathrm{B}(g)+\mathrm{C}(g) \longrightarrow \mathrm{D}(g)\) the following data were obtained at constant temperature: $$ \begin{array}{ccccc} \text { Expt } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [C] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 6.25 \times 10^{-3} & 0.0500 & 0.0500 & 0.0100 \\ 2 & 1.25 \times 10^{-2} & 0.1000 & 0.0500 & 0.0100 \\ 3 & 5.00 \times 10^{-2} & 0.1000 & 0.1000 & 0.0100 \\ 4 & 6.25 \times 10^{-3} & 0.0500 & 0.0500 & 0.0200 \end{array} $$ (a) What is the order with respect to each reactant? (b) Write the rate law. (c) Calculate \(k\) (using the data from Expt 1 ). (d) Using the value of \(k\) calculated in part (c), calculate the rate when \([\mathrm{A}]=\) \(0.2000 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.1500 \mathrm{~mol} / \mathrm{L},\) and \([\mathrm{C}]=0.0300 \mathrm{~mol} / \mathrm{L}\).

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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