Does a catalyst increase reaction rate by the same means as a rise in temperature does? Explain.

Short Answer

Expert verified
No, a catalyst lowers activation energy, while a rise in temperature increases kinetic energy and collision frequency.

Step by step solution

01

Understand the Role of a Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It achieves this by providing an alternative reaction pathway with a lower activation energy.
02

Understand the Effect of Temperature on Reaction Rate

Increasing the temperature of a system generally increases the reaction rate. This is because higher temperatures provide more kinetic energy to the reactants, increasing the number of collisions with sufficient energy to overcome the activation energy barrier.
03

Compare the Two Mechanisms

While both a catalyst and an increase in temperature raise the reaction rate, they achieve this differently. A catalyst lowers the activation energy, making it easier for the reaction to proceed at any given temperature. On the other hand, increasing temperature raises the kinetic energy of particles, which increases the frequency and energy of collisions.
04

Conclusion

Catalysts and temperature increases both speed up reactions, but they do so through different mechanisms. A catalyst changes the pathway and lowers activation energy, while temperature increase enhances kinetic energy and collision frequency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Activation Energy
Activation energy is a key concept in understanding chemical reactions. It refers to the minimum amount of energy that reactant molecules need to have in order to undergo a successful chemical reaction. You can think of activation energy as a barrier that reactants must overcome to form products.

Catalysts play a crucial role in reducing this activation energy. By providing an alternative pathway, catalysts make it easier for the reactants to transform into products. As a result, the reaction can proceed at a faster rate, even at lower temperatures. This is fundamentally different from how temperature affects reaction rates, which we will discuss next.
Reaction Rate
The reaction rate is a measure of how quickly reactants are converted into products in a chemical reaction. Several factors can affect the reaction rate, including the concentration of reactants, surface area, temperature, and the presence of a catalyst.

Catalysts specifically increase the reaction rate by lowering the activation energy, therefore, making it easier for the reactants to convert to products. This does not involve increasing the kinetic energy of the reactants, but rather making the path to the products more accessible.

In contrast, increasing the temperature boosts the kinetic energy of the molecules, increasing the number and force of successful collisions. Though both methods increase the reaction rate, they do so through different mechanisms.
Temperature Effect on Reactions
Temperature has a significant impact on the rate of a chemical reaction. When you increase the temperature, the kinetic energy of the reactants also increases. This means the molecules move faster and collide with each other more frequently and with greater energy.

According to the collision theory, more collisions with energy equal to or greater than the activation energy result in a higher reaction rate. Consequently, heating up a reaction mixture generally speeds up the reaction.

In summary, while both catalysts and temperature increases can accelerate chemical reactions, they operate differently. Catalysts lower the activation energy barrier, providing an easier pathway for the reaction. On the other hand, increasing temperature boosts the kinetic energy of molecules, leading to more frequent and energetic collisions that can overcome the activation energy barrier. Both strategies are valuable tools in controlling and optimizing chemical reaction rates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are studying the reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow 2 \mathrm{AB}(g)\) to determine its rate law. Assuming that you have a valid experimental procedure for obtaining \(\left[\mathrm{A}_{2}\right]\) and \(\left[\mathrm{B}_{2}\right]\) at various times, explain how you determine (a) the initial rate, (b) the reaction orders, and (c) the rate constant.

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

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both \(\mathrm{NOBr}\) and \(\mathrm{Br}_{2}\) are reddish brown: $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ Use the data below to answer the following: (a) Determine the average rate over the entire experiment. (b) Determine the average rate between 2.00 and \(4.00 \mathrm{~s}\). (c) Use graphical methods to estimate the initial reaction rate. (d) Use graphical methods to estimate the rate at \(7.00 \mathrm{~s}\). (e) At what time does the instantaneous rate equal the average rate over the entire experiment? $$ \begin{array}{cc} \text { Time (s) } & \text { [NOBr] (mol/L) } \\ \hline 0.00 & 0.0100 \\ 2.00 & 0.0071 \\ 4.00 & 0.0055 \\ 6.00 & 0.0045 \\ 8.00 & 0.0038 \\ 10.00 & 0.0033 \end{array} $$

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.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free