/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 The activation energy for the re... [FREE SOLUTION] | 91Ó°ÊÓ

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The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is 125 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\) . What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Short Answer

Expert verified
The activation energy for the reverse reaction is \(-91 \, \mathrm{kJ/mol}\).

Step by step solution

01

Identify the given values

For this exercise, we are given the activation energy for the forward reaction (125 kJ/mol), the change in energy of the reaction (-216 kJ/mol), and we are asked to find the activation energy for the reverse reaction.
02

Use the formula

To find the activation energy for the reverse reaction, we should use the formula that relates the activation energy of the forward and reverse reactions and the change in energy of the reaction: Activation energy of reverse reaction (Ea_reverse) = Activation energy of forward reaction (Ea_forward) + Change in energy of reaction (∆E) Plug in the given values into the formula: Ea_reverse = Ea_forward + ∆E Ea_reverse = \(125 \, \mathrm{kJ/mol} - 216 \, \mathrm{kJ/mol}\)
03

Calculate the activation energy for the reverse reaction

To find the activation energy for the reverse reaction, simply perform the subtraction in the equation: Ea_reverse = \(-91 \, \mathrm{kJ/mol}\) The activation energy for the reverse reaction is -91 kJ/mol. This negative value indicates that the activation energy barrier for the reverse reaction is actually lower than that of the forward reaction, making the reverse reaction more favorable to occur under certain conditions.

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Key Concepts

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

Reaction Thermodynamics
In chemical reactions, thermodynamics helps us understand how energy is transformed. It focuses on whether a reaction is energy-releasing or energy-absorbing. When discussing reaction thermodynamics, we often refer to the change in energy of a reaction, symbolized as \( \Delta E \). This value shows us how much energy is absorbed or released as reactants turn into products.

For example, in the reaction \( \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \), we are given a \( \Delta E \) of \(-216 \mathrm{kJ/mol} \). This negative value means the reaction releases energy; it's exothermic. Particularly, 216 kJ are released for every mole of reaction that occurs.

Understanding \( \Delta E \) is crucial because it tells us about the energy landscape of the reaction. It serves as a guide to predict whether a reaction will proceed energetically, as reactions that release energy (like this one) are typically more spontaneous.
Chemical Kinetics
Chemical kinetics revolves around the rate at which a chemical reaction occurs and the factors influencing this rate. It's all about understanding how quickly reactants are transformed into products.

A key factor impacting reaction rate is activation energy \( (E_a) \). Activation energy is the minimum energy that colliding molecules need to start a reaction. It acts like an energy hill that must be climbed.
  • A lower activation energy means reactants can more easily form products, making the reaction faster.
  • A higher activation energy means the reaction is slower because it's harder for molecules to achieve the necessary energy to react.
For the reaction \( \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \), the activation energy is 125 kJ/mol. This quantifies the energy barrier that must be overcome for the reaction to proceed. The reaction's rate will depend significantly on this parameter.

By understanding chemical kinetics, scientists can better control reactions, optimizing conditions to make them faster or slower depending on the desired outcome.
Energy Change of Reaction
The energy change of a reaction, indicated by \( \Delta E \), tells us the difference in energy between reactants and products. It is a key component in both reaction thermodynamics and chemical kinetics.

For any given reaction, such as \( \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \), knowing \( \Delta E \) helps us understand the thermodynamic favorability. This particular reaction has a \( \Delta E \) of \(-216 \mathrm{kJ/mol} \), indicating an energy release.
  • When \( \Delta E \) is negative, the reaction tends to release energy, possibly making it spontaneous.
  • When \( \Delta E \) is positive, energy is absorbed, suggesting the reaction might need an input of energy to occur.
The energy change also plays a role in the activation energy for the reverse reaction. In our example, we calculated the activation energy for the reverse process as \(-91 \mathrm{kJ/mol} \), meaning that the reverse reaction has a lower energy barrier to overcome. This can sometimes make the reverse reaction more feasible under differing conditions. Understanding the energy change of a reaction thus allows us to predict the energy demands and overall feasibility of chemical processes.

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Most popular questions from this chapter

The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at 273 \(\mathrm{K}\) is \(2.3 \times 10^{-12} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) , and the activation energy is 111 \(\mathrm{kJ} / \mathrm{mol}\) . How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of 2.5 \(\mathrm{atm}\) to 1.5 \(\mathrm{atm}\) at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{array}{c}{\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}} \\ {\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}\end{array} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

A certain reaction has the form $$ \mathrm{aA} \longrightarrow $$ At a particular temperature, concentration versus time data were collected. A plot of 1\(/[\mathrm{A}]\) versus time (in seconds) gave a straight line with a slope of \(6.90 \times 10^{-2} .\) What is the differential rate law for this reaction? What is the integrated rate law for this reaction? What is the value of the rate constant for this reaction? If \([\mathrm{A}]_{0}\) for this reaction is \(0.100 M,\) what is the first half-life (in seconds)? If the original concentration (at \(t=0 )\) is \(0.100 M,\) what is the second half-life (in seconds)?

The reaction $$ \mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C} $$ is known to be zero order in A and to have a rate constant of \(5.0 \times 10^{-2} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . An experiment was run at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.0 \times 10^{-3} \mathrm{M} .\) a. Write the integrated rate law for this reaction. b. Calculate the half-life for the reaction. c. Calculate the concentration of \(\mathrm{B}\) after \(5.0 \times 10^{-3} \mathrm{s}\) has elapsed assuming \([\mathrm{B}]_{0}=0\)

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section 12.7 is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section 12.7\(?\)

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