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The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: $$ \begin{array}{ll} \hline \text { Temperature (K) } & \text { Rate Constant (s }^{-1} \text { ) } \\\ \hline 300 & 3.2 \times 10^{-11} \\ 320 & 1.0 \times 10^{-9} \\ 340 & 3.0 \times 10^{-8} \\ 355 & 2.4 \times 10^{-7} \\ \hline \end{array} $$ From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

Short Answer

Expert verified
The activation energy for the reaction of the organic compound with water is approximately 53.38 kJ/mol.

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation is given by: \(k = A\exp(\frac{-E_a}{RT})\), where k is the rate constant, A is the pre-exponential (or frequency) factor, E_a is the activation energy, R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin.
02

Find the natural logarithm of the Arrhenius equation

We will find the natural logarithm of the Arrhenius equation to make it easier to handle. \(\ln{k} = \ln{A} - \frac{E_a}{RT}\)
03

Find the difference between two temperatures

We will choose two different temperatures from the given data and find the difference between their natural logarithm values. Let's take temperature T1 = 300 K and T2 = 355 K, with their respective rate constants k1 = 3.2 x 10^{-11} s^{-1} and k2 = 2.4 x 10^{-7} s^{-1}. \(\ln{\frac{k_2}{k_1}} = - \frac{E_a}{R}(\frac{1}{T_2} - \frac{1}{T_1})\)
04

Calculate the activation energy

Using the values from Step 3, calculate the activation energy E_a. \(\ln{\frac{2.4 \times 10^{-7}}{3.2 \times 10^{-11}}} = - \frac{E_a}{8.314}(\frac{1}{355} - \frac{1}{300})\) Now calculate the activation energy E_a: \(E_a = - 8.314 \times \ln{\frac{2.4 \times 10^{-7}}{3.2 \times 10^{-11}}} \times (\frac{1}{355} - \frac{1}{300})^{-1}\) \(E_a \approx 53378.78\, \mathrm{J/mol} \)
05

Convert the activation energy to kJ/mol

Now, we need to convert the activation energy from J/mol to kJ/mol. \(E_a= \frac{53378.78\,J/mol}{1000}\) \(E_a \approx 53.38\, \mathrm{kJ/mol} \) The activation energy for the reaction of the organic compound with water is approximately 53.38 kJ/mol.

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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 in chemical kinetics to describe how reaction rates vary with temperature. It is given by the expression: \( k = A\exp\left(\frac{-E_a}{RT}\right) \). Here, \( k \) is the rate constant of the reaction, which tells us how fast the reaction proceeds.

The term \( A \), known as the pre-exponential factor, represents the frequency of collisions leading to product formation. Meanwhile, \( E_a \) signifies the activation energy, which is the minimum energy required for the reaction to occur. The constant \( R \) symbolizes the gas constant, which has a value of 8.314 J/mol K. Finally, \( T \) stands for the absolute temperature in Kelvin, highlighting the temperature's role in reaction kinetics.

The equation emphasizes that a reaction's rate increases exponentially with an increase in temperature. This makes intuitive sense as higher temperatures generally provide more thermal energy to overcome activation energy barriers, thereby accelerating the reaction.
Rate constant
The rate constant \( k \) is a critical parameter in chemical kinetic studies that quantifies the speed at which a reaction occurs. It has different units depending on the order of the reaction. For example, in this case, it is a first-order reaction, and thus \( k \) has units of \( s^{-1} \).

The value of the rate constant is influenced by several factors:
  • Temperature: An increase in temperature typically results in an increase in the rate constant.
  • Nature of reactants: Certain substances react more readily than others.
  • Catalysts: These can significantly alter the rate constant by providing an alternate pathway with a lower activation energy.
When looking at how \( k \) changes with temperature, as demonstrated in the exercise data, we can see the effect of these factors and why understanding the rate constant is essential in predicting reaction behavior.
Temperature dependence
The temperature dependence of reaction rates is one of the essential elements in the study of kinetics. According to the Arrhenius equation, the rate constant \( k \) is exponentially dependent on temperature. This means even minor changes in temperature can lead to significant variations in \( k \).

In a practical sense, this can be visualized using an Arrhenius plot, where \( \ln k \) is plotted against \( 1/T \). Such plots usually yield a straight line whose slope is related to the activation energy \( E_a \). A steeper slope indicates higher energy barriers to the reaction, implying that the reaction is more sensitive to changes in temperature.

This temperature dependence also explains why some reactions, which may be incredibly slow at room temperature, proceed rapidly when heated or conducted at elevated temperatures. It gives chemists a way to control reaction rates through temperature adjustment.
Kinetics
Kinetics is the branch of chemistry concerned with understanding the rates of chemical reactions and the steps by which they occur – which is why it’s crucial for analyzing chemical processes.

Several concepts are pivotal in kinetics, including:
  • Reaction mechanisms: These are step-by-step sequences of elementary reactions showing the path from reactants to products.
  • Molecularity: Refers to the number of molecules participating in an elementary reaction step.
  • Reaction order: Describes how the concentration of reactants influences the reaction rate.
For chemists, understanding kinetics doesn't just inform about how fast a reaction proceeds but also provides insights into the mechanism and details about the potential energy landscape over which the reaction traverses. It's essential for optimizing conditions in industrial processes, designing new materials, and even understanding biological systems.

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Most popular questions from this chapter

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the rates of reactions. (c) What is the difference between average rate and instantaneous rate?

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

(a) What is meant by the term elementary reaction? (b) What is the difference between a unimolecular and a bimolecular elementary reaction? (c) What is a reaction mechanism?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00\) L of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a 10.0-L container. (Assume that the products do not dissolve in chloroform.)

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1}\). Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L}\). (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0\) min? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half- life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C}\) ?

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