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

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), decomposes by first- order kinetics, and \(k=2.81 \times 10^{-3} \mathrm{~min}^{-1}\) at a certain temperature. (a) Determine the half-life for the reaction. (b) Determine the time needed for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(10 \%\) of its initial concentration. (c) If \(14.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is sealed in a \(2500 .-\mathrm{L}\) reaction vessel and heated to the specified temperature, what mass will remain after \(1.5 \mathrm{~h}\) ?

In the brewing of beer, ethanal, which smells like green apples, is an intermediate in the formation of ethanol. Ethanal decomposes in the following first-order reaction: \(\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{CO}(\mathrm{g})\). At an elevated temperature the rate constant for the decomposition is \(1.5 \times 10^{-3} \mathrm{~s}^{-1}\). What concentration of ethanal, which had an initial concentration of \(0.120 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), remains \(20.0 \mathrm{~min}\) after the start of its decomposition at this temperature?

The pre-equilibrium and the steady-state approximations are two different approaches to deriving a rate law from a proposed mechanism. For the following mechanism, determine the rate law (a) by the pre-equilibrium approximation and (b) by the steady-state approximation. (c) Under what conditions do the two methods give the same answer? (d) What will the rate laws become at high concentrations of \(\mathrm{Br}^{-}\)? $$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}_{2}^{+} \text {(fast equilibrium) } \\ &\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O} \text { (slow) } \end{aligned} $$

The decomposition of A has the rate law Rate \(=k[\mathrm{~A}]^{a}\). Show that for this reaction the ratio \(t_{1 / 2} / t_{3 / 4}\), where \(t_{1 / 2}\) is the halflife and \(t_{3 / 4}\) is the time for the concentration of A to decrease to \(\frac{3}{4}\) of its initial concentration, can be written as a function of \(a\) alone and can therefore be used to make a quick assessment of the order of the reaction in A.

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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