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(a) For a reaction with a given \(E_{a},\) how does an increase in \(T\) affect the rate? (b) For a reaction at a given \(T,\) how does a decrease in \(E_{\mathrm{a}}\) affect the rate?

Short Answer

Expert verified
An increase in temperature increases the reaction rate. A decrease in activation energy increases the reaction rate.

Step by step solution

01

Understand the Arrhenius Equation

The rate of a chemical reaction is given by the Arrhenius equation: \[ k = A e^{-E_a / RT} \]where: - \( k \) is the rate constant - \( A \) is the pre-exponential factor - \( E_a \) is the activation energy - \( R \) is the universal gas constant - \( T \) is the temperature in Kelvin.
02

Analyze the effect of increasing Temperature on Rate

As the temperature \( T \) increases, the exponential term \[ e^{-E_a / RT} \] becomes larger because the exponent \[ -\frac{E_a}{RT} \] becomes less negative. Since an increase in \( T \) reduces the magnitude of the negative exponent, the rate constant \( k \) increases. Therefore, the rate of reaction increases with an increase in temperature.
03

Analyze the effect of decreasing Activation Energy on Rate

For a given temperature \( T \), a decrease in activation energy \( E_a \) reduces the exponent \[ -\frac{E_a}{RT} \].A smaller \( E_a \) makes the negative exponent less negative, increasing the value of \[ e^{-E_a / RT} \]. As a result, the rate constant \( k \) increases, which increases the rate of the reaction.

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Key Concepts

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

activation energy
Activation energy, represented as \( E_{a} \), is an essential concept in chemical kinetics. It refers to the minimum energy required for a chemical reaction to occur. Think of it as the energy barrier that reactants need to overcome to transform into products.
If the activation energy is high, fewer molecules have sufficient energy to surpass this barrier. This results in a slower reaction rate. Conversely, if the activation energy is low, more molecules can overcome the barrier, leading to a faster reaction rate.
Reducing the activation energy, for instance through the use of catalysts, can significantly increase the rate of a reaction. Catalysts work by providing an alternative reaction pathway with a lower activation energy.
reaction rate
The rate of a reaction is a measure of how quickly reactants are converted into products over time. It's influenced by various factors, including temperature, activation energy, and the presence of catalysts.
The Arrhenius equation allows us to understand the relationship between these factors and the reaction rate:
\[ k = A e^{-E_a / RT} \]
In this equation, \( k \) is the rate constant, and it is directly related to the reaction rate. A higher \( k \) value means a faster reaction rate.
Let's break down how each factor affects the reaction rate:
  • Temperature: As temperature increases, the rate constant \( k \) increases, accelerating the reaction rate.
  • Activation Energy: A lower activation energy results in an increased value of \( k \,\) thus speeding up the reaction.
  • Pre-exponential Factor (\( A \)): This factor, sometimes related to the frequency of collisions, also influences \( k \). A higher \( A \) can lead to a faster reaction rate.
Overall, understanding these factors helps us control and optimize chemical reactions in various fields, from industrial processes to biochemical pathways.
temperature effect
The temperature effect on reaction rate is one of the most significant factors in chemical kinetics. According to the Arrhenius equation, the reaction rate increases exponentially with a rise in temperature.
When temperature \( T \) increases, the exponent \( -E_a / RT \) in the Arrhenius equation becomes less negative. This makes the exponential term \( e^{-E_a / RT} \) larger, leading to a higher rate constant \( k \).
Here's why temperature has such a profound effect:
  • Increased Molecular Collisions: Higher temperatures provide more kinetic energy to molecules. This results in more frequent and energetic collisions, increasing the likelihood of overcoming the activation energy barrier.
  • Faster Movement: As temperature rises, molecules move faster, making it easier for them to reach the transition state and proceed to the product side of the reaction.
In practical terms, increasing the temperature is a common method to speed up reactions. However, it's essential to control it carefully. Excessive temperatures can lead to undesirable side reactions or even damage sensitive reactants.

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Most popular questions from this chapter

Is it possible for more than one mechanism to be consistent with the rate law of a given reaction? Explain.

Sulfonation of benzene has the following mechanism: (1) \(2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}+\mathrm{SO}_{3} \quad[\) fast \(]\) (2) \(\mathrm{SO}_{3}+\mathrm{C}_{6} \mathrm{H}_{6} \longrightarrow \mathrm{H}\left(\mathrm{C}_{6} \mathrm{H}_{5}^{+}\right) \mathrm{SO}_{3}^{-} \quad\) [slow] (3) \(\mathrm{H}\left(\mathrm{C}_{6} \mathrm{H}_{5}^{+}\right) \mathrm{SO}_{3}^{-}+\mathrm{HSO}_{4}^{-} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}+\mathrm{H}_{2} \mathrm{SO}_{4} \quad\) [fast] (4) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \quad\) [fast \(]\) (a) Write an overall equation for the reaction. (b) Write the overall rate law in terms of the initial rate of the reaction.

A biochemist studying the breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a halflife of 12 yr. How long does it take DDT in a soil sample to decrease from 275 ppbm to \(10 .\) ppbm (parts per billion by mass)?

The rate constant of a reaction is \(4.50 \times 10^{-5} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(195^{\circ} \mathrm{C}\) and \(3.20 \times 10^{-3} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(258^{\circ} \mathrm{C}\). What is the activation energy of the reaction?

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.

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