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Chapter 15: Problem 63

The rate constant \((k)\) for a reaction is measured as a function of tem- perature. A plot of ln \(k\) versus 1\(/ T(\) in \(\mathrm{K})\) is linear and has a slope of \(-7445 \mathrm{K}\) . Calculate the activation energy for the reaction.

Short Answer

Expert verified
The activation energy for the reaction is 61,925.93 J/mol.

Step by step solution

01

Understanding the Arrhenius Equation

Recognize that the slope in the plot of ln(k) versus 1/T comes from the Arrhenius equation in the form of ln(k) = (-Ea/R)(1/T) + ln(A), where Ea is the activation energy, R is the gas constant (8.314 J/mol*K), and A is the pre-exponential factor.
02

Slope Identification

Identify the slope of the line from the plot of ln(k) against 1/T. According to the given information, the slope is -7445 K.
03

Activation Energy Calculation

Calculate the activation energy (Ea) using the slope (which is equal to -Ea/R). Rearrange the equation and solve for Ea: Ea = -slope * R.
04

Plug Values into Formula

Substitute the values of the slope and R into the equation to find the activation energy: Ea = (-(-7445 K)) * (8.314 J/mol*K).

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Key Concepts

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

Arrhenius Equation
Understanding the Arrhenius equation is crucial when studying the rate at which chemical reactions occur. This equation links the rate constant \(k\) of a reaction to the temperature and activation energy, providing insight into how temperature affects reaction rates. Written as \(\text{ln}(k) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \text{ln}(A)\), the Arrhenius equation comprises of the activation energy \(E_a\), the gas constant \(R\), temperature \(T\), and the pre-exponential factor \(A\). The pre-exponential factor represents the frequency of collisions with the correct orientation, which could lead to a reaction if energy requirements are met.

Rate Constant
The rate constant \(k\) is a proportionality constant that defines the relationship between the reactant concentration and the rate of the reaction in the rate law equation. It is a crucial part of the rate equation that summarizes the overall rate at which a chemical process occurs. The value of \(k\) is determined experimentally and varies with temperature, as shown in the Arrhenius equation. The fact that \(k\) has a direct exponential dependence on temperature is an important character to understand reaction kinetics.
Reaction Kinetics
Reaction kinetics deals with the rates of chemical processes and the factors affecting these rates. It is a cornerstone of chemical kinetics - a field concerned with understanding the speed and mechanism of reactions. The rates are influenced by various parameters such as concentration, catalysts, and temperature. By analyzing the speed at which reactants turn into products, one can determine the kinetic parameters such as the activation energy and rate constant, which are fundamental in predicting the behavior of chemical reactions under different conditions.
Temperature Dependence
The temperature dependence of reaction rates is one of the key tenets of reaction kinetics. According to the Arrhenius equation, a higher temperature often means an increased rate constant \(k\), which translates to a faster reaction. This is because higher temperatures increase the kinetic energy of the molecules, leading to more frequent and effective collisions. As the temperature rises, more molecules possess the minimum amount of energy required to overcome the activation barrier, known as the activation energy \(E_a\), enabling the chemical transformation.

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Most popular questions from this chapter

Suppose that the reaction \(\mathrm{A} \longrightarrow\) products is exothermic and has an ac- tivation barrier of 75 \(\mathrm{kJ} / \mathrm{mol} .\) Sketch an energy diagram showing the en- ergy of the reaction as a function of the progress of the reaction. Draw a second energy curve showing the effect of a catalyst.

Why are reaction rates important (both practically and theoretically)?

The rate constant \((k)\) for a reaction is measured as a function of tem- perature. A plot of ln \(k\) versus 1\(/ T(\) in \(\mathrm{K})\) is linear and has a slope of \(-1.01 \times 10^{4} \mathrm{K}\) . Calculare the activation energy for the reaction.

Write integrated rate laws for zero-order, first-order, and second-order reactions of the form \(A \longrightarrow\) products.

Explain the difference between homogeneous catalysis and heteroge- neous catalysis.
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