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Chapter 10: Problem 213

Rate of a reaction can be expressed by Arrhenius 2 equation as, \(k=\mathrm{Ae}^{\mathrm{E} / \mathrm{RT}}\) In this equation, E represents (a) the energy above which all the colliding molecules will react (b) the energy below which colliding molecules will not react (c) the total energy of the reacting molecules at a tem perature, T (d) the fraction of molecules with energy greater than the activation energy of the reaction

Short Answer

Expert verified
(b) the energy below which colliding molecules will not react

Step by step solution

01

Introduction to Arrhenius Equation

The Arrhenius equation is given as \( k = A \cdot e^{-E/RT} \). Here, \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. The equation describes how the rate constant \( k \) changes with temperature.
02

Understanding Activation Energy (E)

The activation energy, \( E \), is a critical component of the Arrhenius Equation. It is the minimum energy required for the reactants to undergo a reaction. Colliding molecules must have energy equal to or greater than this to overcome the energy barrier and form the products.
03

Analyzing Answer Choices

(a) and (b) both refer to energy conditions for reacting molecules - (a) implies energy above which reaction is possible, while (b) implies energy below which reaction isn鈥檛 possible, both indicating activation energy. (c) refers to total energy at a given temperature, unrelated to activation barrier. (d) talks about the fraction of molecules with energy greater than activation energy, which is a result but not a definition of \( E \).
04

Choosing the Correct Option

Based on the understanding of activation energy in the Arrhenius Equation, the correct choice is (b): \( E \) represents the energy below which the colliding molecules will not react. This defines the threshold energy that molecules need in order to successfully react.

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Key Concepts

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

Activation Energy
Activation energy is the key threshold that molecules must overcome for a chemical reaction to occur. In the Arrhenius Equation, it is denoted as \( E \). Here's why activation energy is so crucial:
  • It represents the minimum energy required for reactants to be converted into products.
  • Molecules with energy below this threshold are unable to react, as they do not possess sufficient power to break bonds and form new ones.
  • This concept explains why some reactions are slow or don't occur at all under certain conditions.
To think of it simply, if you're trying to roll a ball over a hill, the activation energy is the amount of push needed to get it to the top of the hill. Once at the peak (overcoming the activation barrier), the reaction can proceed to form products.
Reaction Rate
A key result of overcoming the activation energy is the increase in the reaction rate. Reaction rate, denoted by the rate constant \( k \) in the Arrhenius Equation, tells us how fast a reaction occurs once it gets started.
  • The rate at which a reaction proceeds is directly related to both the activation energy and the temperature.
  • Lower activation energy typically means a higher reaction rate because more molecules have the necessary energy to react.
  • A higher \( k \) value indicates a faster reaction.
The balance between activation energy and temperature plays a pivotal role in determining how quickly reactants are transformed into products. Think of it as a race track where the activation energy sets the obstacles and \( k \) records the speed at which racers (molecules) finish the race.
Temperature Dependence
Temperature is a major factor influencing how a chemical reaction unfolds, particularly in the context of the Arrhenius Equation. Here鈥檚 how temperature comes into play:
  • As temperature increases, the kinetic energy of molecules also rises.
  • This results in more frequent and more energetic collisions, meaning more molecules can surpass the activation energy barrier.
  • Thus, higher temperatures typically lead to quicker reactions by increasing the rate constant \( k \).
This dependence is key for understanding various natural processes and industrial applications, where controlling temperature helps manage reaction rates effectively. Consider this: just like making water boil faster by turning up the heat, raising the temperature speeds up the reaction.

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Most popular questions from this chapter

\(75 \%\) of a first-order reaction was completed in 32 min. When was \(50 \%\) of the reaction completed? (a) \(24 \mathrm{~min}\) (b) \(16 \mathrm{~min}\) (c) \(8 \mathrm{~min}\) (d) \(64 \mathrm{~min}\)

Consider the chemical reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) The rate of this reaction can be expressed in terms of time derivatives of concentration of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) or \(\mathrm{NH}_{3}(\mathrm{~g})\). Identify the correct relationship amongst the rate expressions. (a) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (b) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (c) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-1 / 3 \mathrm{~d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=2 \mathrm{~d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\) (d) rate \(=-\mathrm{d}\left[\mathrm{N}_{2}\right] / \mathrm{dt}=-\mathrm{d}\left[\mathrm{H}_{2}\right] / \mathrm{dt}=\mathrm{d}\left[\mathrm{NH}_{3}\right] / \mathrm{dt}\)

For an endothermic reaction, where \(\Delta \mathrm{H}\) represents the enthalpy of the reaction in \(\mathrm{kJ} /\) mole, the minimum value for the energy of activation will be (a) less than \(\Delta \mathrm{H}\) (b) zero (c) more than \(\Delta \mathrm{H}\) (d) equal to \(\Delta \mathrm{H}\)

The rate of reaction was found to be equal to its rate constant at any concentration of the reactant. The order of the reaction is (a) zero-order (b) first-order (c) second-order (d) third-order

In the following reaction, how is the rate of appear ance of the underlined product related to the rate of disappearance of the underlined reactant? \(\mathrm{BrO}_{3}^{-}(\mathrm{aq})+5 \mathrm{Br}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow 3 \mathrm{Br}_{2}(1)\) \(+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (b) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (c) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\) (d) \(\frac{\mathrm{d}\left[\mathrm{Br}_{2}\right]}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}[\mathrm{Br}]}{\mathrm{dt}}\)
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