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Chapter 14: Problem 58

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right.\) ) and overall \(\Delta E\) are \(154 \mathrm{~kJ} / \mathrm{mol}\) and \(136 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Short Answer

Expert verified
The activation energy for the reverse reaction is \(290 \mathrm{~kJ/mol}\).

Step by step solution

01

(a) Sketching the energy profile

For the elementary process: \(\mathrm{N_{2}O_{5}(g)} \longrightarrow \mathrm{NO_{2}(g)} + \mathrm{NO_{3}(g)}\), we are given the activation energy (Ea) as \(154 \mathrm{~kJ/mol}\) and overall change in energy (螖贰) as \(136 \mathrm{~kJ/mol}\). To sketch the energy profile: 1. Label the x-axis as "Reaction Progress" and the y-axis as "Potential Energy". 2. Draw an initial energy level representing the reactants and label it as \(\mathrm{N_{2}O_{5}(g)}\). 3. Draw a higher energy level representing the activated complex (transition state) and label it as "TS". 4. Draw a final energy level, representing the products, and label it as \(\mathrm{NO_{2}(g)} + \mathrm{NO_{3}(g)}\). 5. Connect these energy levels with a curve, going from the reactants to the transition state, and then to the products. 6. Indicate the activation energy (Ea) by drawing a vertical arrow from the initial energy level (reactants) to the energy level of the transition state (TS), and label it as "Ea = 154 kJ/mol". 7. Indicate the overall change in energy (螖贰) by drawing a vertical arrow from the initial energy level (reactants) to the final energy level (products), and label it as "螖贰 = 136 kJ/mol".
02

(b) Finding the activation energy for the reverse reaction

Given the activation energy \(154 \mathrm{~kJ/mol}\) and overall change in energy \(136 \mathrm{~kJ/mol}\) for the forward reaction, we can find the activation energy for the reverse reaction (called Ea_reverse) by using the following relationship: \[E_a + 螖贰 = E_{a, \text{reverse}}\] Now we substitute the given values and solve for Ea_reverse: \[E_{a, \text{reverse}} = 154 \mathrm{~kJ/mol} + 136 \mathrm{~kJ/mol}\] \[E_{a, \text{reverse}} = 290 \mathrm{~kJ/mol}\] So the activation energy for the reverse reaction is \(290 \mathrm{~kJ/mol}\).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the speed or rate at which a chemical reaction occurs and the factors that affect this rate. It seeks to understand the steps by which a reaction proceeds, known as the reaction mechanism, and how the rate can be modified. When considering the elementary process N2O5 decomposing into NO2 and NO3, kinetics provides insight into the rate at which this reaction occurs and the energy barriers involved.

In studying kinetics, we often discuss terms such as reaction rate, order of reaction, and rate constants, all of which combine to give a mathematical description of a chemical reaction's speed. The activation energy (Ea), which is the minimum energy that reacting molecules must have in order to undergo a chemical change, plays a significant role in this. It directly affects the reaction rate: a higher Ea generally means a slower reaction because fewer molecules will possess the necessary energy to surpass the energy barrier.
Elementary Process
An elementary process is a single step in a reaction mechanism that describes the changes from reactants to products in one discrete action. These processes are the simplest forms of chemical reactions and do not break down into simpler steps. They are integral to understanding the overall mechanism of a complex reaction.
For instance, the decomposition of N2O5 is an example of an elementary process, as it proceeds in one single step to form NO2 and NO3. An essential aspect to highlight in elementary reactions is that the molecularity, or the number of reactant particles involved, correlates with the reaction order. In the given decomposition reaction, it's unimolecular, involving one molecule of N2O5.
Energy Profile Diagram
An energy profile diagram visually represents the energy changes that occur during the course of a chemical reaction. It plots the potential energy of the reacting system against the progress of the reaction, illustrating the energy of reactants, products, and the transition state.

The diagram features a reaction coordinate axis, usually the x-axis, and a potential energy axis, the y-axis. Starting with reactants at an initial energy level, the curve rises to the topmost point, which corresponds to the transition state - the point at which the system has reached its highest potential energy and the activation energy (Ea). The difference in energy between this state and the reactants quantifies the activation energy needed for the reaction to proceed. Then, the curve descends to the energy level of the products, detailing the overall energy change (螖贰), which can be exothermic (releasing energy) or endothermic (absorbing energy).
Transition State Theory
Transition state theory is a concept that describes the high-energy states through which reactants pass on their way to becoming products. According to this theory, there exists a transient configuration called the transition state or activated complex. This state is not isolable but crucial for the reaction, as it represents the point of maximum potential energy before the reaction descends into product formation.

It proposes that the rate of reaction is proportional to the concentration of this transition state, which is a fleeting combination of reactants in a potential energy maximum. The activation energy (Ea) represents the energy barrier to achieving this state. Once reached, the transition state can either regress back to reactants (especially if the energy is taken away) or progress to form the products, with the overall energy change (螖贰) depicting whether the process absorbed or released energy.

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Most popular questions from this chapter

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly known as table sugar, reacts in dilute acid solutions to form two simpler sugars, glucose and fructose, both of which have the formula \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} .\) At \(23{ }^{\circ} \mathrm{C}\) and in \(0.5 \mathrm{M} \mathrm{HCl}\), the following data were obtained for the disappearance of sucrose: $$ \begin{array}{rl} \hline \text { Time }(\mathrm{min}) & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](M)} \\ \hline 0 & 0.316 \\ 39 & 0.274 \\ 80 & 0.238 \\ 140 & 0.190 \\ 210 & 0.146 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to \(\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right] ?(\mathbf{b})\) What is the rate constant? (c) Using this rate constant, calculate the concentration of sucrose at 39,80,140 , and 210 min if the initial sucrose concentration was \(0.316 \mathrm{M}\) and the reaction was zero order in sucrose.

The reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first order rate constant: $$ \begin{array}{rl} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2} \text { (atm) } \\ \hline 0 & 1.000 \\ 2,500 & 0.947 \\ 5,000 & 0.895 \\ 7,500 & 0.848 \\ 10,000 & 0.803 \\ \hline \end{array} $$

Indicate whether each statement is true or false. If it is false, rewrite it so that it is true. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.

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 \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- \(125,\) which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the halflives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00 -mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00 -mg sample of each isotope remains after 4 days?
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