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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 \(E_a^{\text{reverse}} = 290 \text{ kJ/mol}\).

Step by step solution

01

(Step 1: Sketch the Energy Profile)

To sketch the energy profile for the reaction, we will first draw the energy levels for the reactants and products. The reactants have higher energy than the products, with the difference being \(\Delta E\). Then, we will draw the "hill" representing the activation energy barrier.
02

(Step 2: Label Ea and ΔE)

Now that we have the energy profile, let's label the activation energy (\(E_a\)) and change in energy (\(\Delta E\)). The activation energy is the difference between the energy of the reactants and the highest point on the energy profile (the "hill"), while \(\Delta E\) is the difference between the energy of the reactants and the energy of the products.
03

(Step 3: Use The Relationship Between Activation Energies and Change in Energy)

To find the activation energy for the reverse reaction, we can use the following relationship: \( E_a^{\text{reverse}} = E_a^{\text{forward}} + \Delta E \) We are given the activation energy and change in energy for the forward reaction: \( E_a^{\text{forward}} = 154 \text{ kJ/mol} \) and \( \Delta E = 136 \text{ kJ/mol} \)
04

(Step 4: Calculate the Activation Energy for the Reverse Reaction)

Now, we can plug in the values for the activation energy and change in energy for the forward reaction into the relationship: \( E_a^{\text{reverse}} = 154 \text{ kJ/mol} + 136 \text{ kJ/mol} \) \( E_a^{\text{reverse}} = 290 \text{ kJ/mol} \) So, the activation energy for the reverse reaction is 290 kJ/mol.

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Key Concepts

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

Elementary Process
An elementary process is a single step in a chemical reaction that involves a direct change from reactants to products.
In an elementary process, molecules either collide or decompose to form products. This process is significant because it dictates how molecules interact and ultimately determines the speed and outcome of the reaction. For instance, the simplification of the decomposition of dinitrogen pentoxide (N2O5) into nitrogen dioxide (NO2) and nitrate radical (NO3) represents such a process.

Understanding this concept helps us to analyze complex reactions by breaking them down into more manageable steps, which facilitates the calculation of reaction rates and helps us to construct more detailed mechanisms for how reactions occur on a molecular level.
Energy Profile
An energy profile is a graphical representation showing how the energy of a system changes as a chemical reaction proceeds from reactants to products.
On an energy diagram, the y-axis represents energy, while the x-axis shows the progress of the reaction. The initial energy level of the reactants and the final energy level of the products are connected by a curve, typically resembling a hill, which illustrates the energy barrier that must be overcome for the reaction to proceed, known as the activation energy (Ea). The difference in energy between the reactants and the products is depicted as the enthalpy change (∆E) of the reaction. Sketching an energy profile allows us to visualize the energy landscape of a reaction more easily.
Enthalpy Change
Enthalpy change, designated as ∆H or ∆E in some contexts, refers to the heat exchange in a chemical reaction under constant pressure.
It indicates whether a reaction is exothermic (releases energy) or endothermic (absorbs energy). If ∆E is negative, the reaction is exothermic, whereas a positive ∆E suggests an endothermic reaction. For the given elementary process, the enthalpy change is 136 kJ/mol, which means there is a release of energy as N2O5 decomposes to NO2 and NO3, indicative of an exothermic process.

Knowing the enthalpy change of a reaction helps predict the spontaneity and feasibility of a reaction under different conditions and is an essential parameter for making energy calculations in thermodynamics.
Reverse Reaction Activation Energy
The activation energy for the reverse reaction, often denoted as Eareverse, is not simply the opposite of the activation energy for the forward reaction (Eaforward).
Instead, it can be calculated by adding the enthalpy change (∆E) of the forward reaction to its activation energy. For the decomposition of N2O5, the high activation energy for the reverse reaction (290 kJ/mol) reflects the energy required to recombine NO2 and NO3 back into N2O5. This concept is crucial in understanding chemical kinetics as it dictates the speed at which the reverse reaction will occur and also influences equilibrium dynamics in reversible reactions.

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Most popular questions from this chapter

As described in Exercise 14.41, the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the partial pressure of this substance after \(60 \mathrm{~s}\) ? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

(a) What is a catalyst? (b) What is the difference between a homogeneous and a heterogeneous catalyst? (c) Do catalysts affect the overall enthalpy change for a reaction, the activation energy, or both?

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \longrightarrow \operatorname{HOOBr}(g) \\\ \mathrm{HOOBr}(g)+\operatorname{HBr}(g) & \longrightarrow 2 \operatorname{HOBr}(g) \\ \mathrm{HOBr}(g)+\operatorname{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{~nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{~nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100\) ?

(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? (d) What is meant by the term rate-determining step?

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