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An organic compound A can decompose by either of two kinetically controlled pathways to form products B or C (see Exercis 14.79). The activation energy for the formation of \(B\) is greater than that for the formation of \(\mathrm{C}\). Will the ratio \([\mathrm{B}] /[\mathrm{C}]\) increase or decrease as the temperature is increased? Explain your answer.

Short Answer

Expert verified
The ratio \[B\]/\[C\] will increase as the temperature is increased because the reaction pathway to B has a higher activation energy, which causes its rate to increase more with temperature than the rate of formation of C.

Step by step solution

01

Understand Activation Energy

Activation energy is the minimum amount of energy that is required to initiate a chemical reaction. The rate of a reaction is dependent on the activation energy, with higher activation energies leading to slower reactions, because fewer molecules have the energy required to overcome the activation barrier.
02

Apply the Arrhenius Equation

The Arrhenius equation relates the rate of a chemical reaction to the temperature and activation energy. According to this equation, as the temperature increases, the rate of reaction increases more for reactions with higher activation energies.
03

Analyze the Temperature Dependence

Since the formation of B has a higher activation energy than that of C, as the temperature increases, the reaction rate for the formation of B will increase more compared to the reaction rate for the formation of C.
04

Determine the Ratio Change

Therefore, with an increase in temperature, the ratio [B]/[C] will increase because the formation rate of B will increase more significantly than that of C due to B's higher activation energy.

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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 a fundamental formula used to describe the effect of temperature on the rate of a chemical reaction. It is given by the mathematical relationship \( k = Ae^{-\frac{E_{a}}{RT}} \) where \( k \) is the reaction rate constant, \( A \) is the pre-exponential factor, \( E_{a} \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

The pre-exponential factor \( A \) reflects the frequency of collisions and the orientation of reactant molecules. The exponential term \( e^{-\frac{E_{a}}{RT}} \) represents the fraction of molecules that possess enough energy, or the minimum energy threshold - referred to as activation energy - to undergo the reaction.

Understanding the equation offers insight into how raising the temperature could potentially lead to an exponential increase in the rate of reaction, especially for reactions with high activation energies. This is because even a small increase in temperature can result in a significant decrease in the exponential term, thus increasing the rate constant \( k \) quite drastically. The Arrhenius equation is crucial for chemists to predict how different temperatures will affect the kinetics of a reaction.
Chemical Kinetics
Chemical kinetics involves the study of reaction rates and the pathways by which reactions occur. The main goal is to understand and describe the factors that influence the speed of chemical reactions. Factors such as concentration of reactants, surface area, catalysts, and temperature all play roles in how rapidly reactions proceed.

One core component of chemical kinetics is the concept of activation energy, which is the energy that must be overcome for a reaction to occur. It acts as an energy barrier: the higher the barrier, the slower the reaction. Catalysts work by lowering this barrier, which speeds up the reaction without being consumed in the process.

Kinetics also deals with reaction mechanisms, which describe the specific steps involved in transforming reactants to products. Each step has its own activation energy and rate, contributing to the overall reaction rate. Chemical kinetics is a vital part of developing new reactions in industries, pharmaceuticals, and environmental processes, ensuring that reactions are efficient and cost-effective.
Temperature Dependence in Reactions
The temperature of a system has a profound effect on chemical reactions, as depicted by the Arrhenius equation. At higher temperatures, molecules move faster and collide with more energy, which increases the likelihood of overcoming the activation energy barrier for the reaction to occur.

In the context of the example problem, if compound A has two pathways to form products B and C with different activation energies, temperature plays a pivotal role in determining the selectivity of the product. For the pathway with higher activation energy, a temperature increase would markedly boost the rate of formation, according to the exponential nature of the Arrhenius equation.

This temperature dependence can be visualized on a reaction energy profile, where lower temperatures may favor the formation of one product over another. Conversely, at higher temperatures, the product with the larger activation energy (B in this case) may be formed more readily. Consequently, in industrial processes, temperatures are carefully controlled to optimize yields and selectivity of desired products, demonstrating the importance of temperature in chemical kinetics and industry.

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Most popular questions from this chapter

(a) Nitrogen dioxide, \(\mathrm{NO}_{2}\), decomposes at \(6.5\) \(\mathrm{mmol} \cdot \mathrm{L}^{-1} \cdot \mathrm{s}^{-1}\) by the reaction \(2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\). Determine the rate of formation of \(\mathrm{O}_{2}\). (b) What is the unique rate of the reaction?

The rate law of the reaction \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is Rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]\), and the mechanism that has been proposed is Step \(1 \mathrm{NO}+\mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}\) Step \(2 \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Step \(3 \mathrm{~N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Which step in the mechanism is likely to be rate determining? Explain your answer. (b) Sketch a reaction profile for the overall reaction, which is known to be exothermic. Label the activation energies of each step and the overall reaction enthalpy.

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

The half-life of a substance taking part in a third-order reaction \(\mathrm{A} \rightarrow\) products is inversely proportional to the square of the initial concentration of A. How can this half-life be used to predict the time needed for the concentration to fall to (a) onehalf; (b) one-fourth; (c) one- sixteenth of its initial value?

Determine the rate constant for each of the following firstorder reactions, in each case expressed for the rate of loss of \(A\) : (a) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of \(\mathrm{A}\) decreases to one-half its initial value in \(1000 . \mathrm{s} ;\) (b) \(\mathrm{A} \longrightarrow \mathrm{B}\), given that the concentration of A decreases from \(0.67 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.53 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) in \(25 \mathrm{~s}\); (c) \(2 \mathrm{~A} \longrightarrow \mathrm{B}+\mathrm{C}\), given that \([\mathrm{A}]_{0}=0.153 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) and that after \(115 \mathrm{~s}\) the concentration of \(B\) rises to \(0.034 \mathrm{~mol} \cdot \mathrm{L}^{-1}\).

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