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Understanding the high-temperature behavior of nitrogen oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide \((\mathrm{NO})\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is second order with a rate constant of \(0.0796 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(737^{\circ} \mathrm{C}\) and \(0.0815 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(947{ }^{\circ} \mathrm{C}\). Calculate the activation energy for the reaction.

Short Answer

Expert verified
The activation energy for the decomposition reaction of nitric oxide (NO) to N₂ and O₂ is approximately -114.8 kJ/mol.

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation relates the rate constant (k) to temperature (T) and activation energy (Ea): \(k = Ae^{-Ea / RT}\) Where A is the pre-exponential factor (or frequency factor), R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin. For this exercise, we have the rate constants and temperatures given, so we'll need to convert these temperatures to Kelvin and then find the activation energy (Ea).
02

Convert the given temperatures to Kelvin

To convert Celsius to Kelvin, simply add 273.15 to the given temperatures: Temperature (T1): \(737^\circ \mathrm{C} = 737 + 273.15 = 1010.15\ \mathrm{K}\) Temperature (T2): \(947^\circ \mathrm{C} = 947 + 273.15 = 1220.15\ \mathrm{K}\)
03

Use the Arrhenius equation for both temperatures

We can use the Arrhenius equation for both temperatures T1 and T2: \(k_1 = Ae^{-Ea / (R \cdot T_1)}\) \(k_2 = Ae^{-Ea / (R \cdot T_2)}\) Where \(k_1\) is the rate constant at T1 and \(k_2\) is the rate constant at T2.
04

Divide the two Arrhenius equations

Divide the first equation by the second equation to eliminate the pre-exponential factor (A) and simplify: \(\frac{k_1}{k_2} = \frac{e^{-Ea / (R \cdot T_1)}}{e^{-Ea / (R \cdot T_2)}}\)
05

Rearrange the equation for Ea

Next, we can simplify by taking the natural logarithm of both sides and rearranging the equation to solve for Ea: \(ln(\frac{k_1}{k_2}) = \frac{Ea}{R} (\frac{1}{T_2} - \frac{1}{T_1})\) Ea = R * ln(\(\frac{k_1}{k_2}\)) * \(\frac{T_1 \cdot T_2}{T_2 - T_1}\)
06

Substitute the given rate constants and temperatures

Now, substitute the given rate constants and temperatures into the equation: Ea = 8.314 * ln(\(\frac{0.0796}{0.0815}\)) * \(\frac{1010.15 \cdot 1220.15}{1220.15 - 1010.15}\)
07

Calculate the activation energy

Finally, calculate the activation energy: Ea ≈ 8.314 * -0.0247 * 5568.82 Ea ≈ -114.8 kJ/mol The activation energy for the decomposition reaction of nitric oxide (NO) is approximately -114.8 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 chemistry to describe how the rate constant of a reaction changes with temperature. It is expressed as:\[ k = Ae^{-Ea / RT} \]where:
  • k is the rate constant.
  • A represents the pre-exponential factor or frequency factor, which indicates the frequency of collisions that lead to a reaction.
  • Ea is the activation energy, the minimum energy required for the reaction to occur.
  • R is the ideal gas constant, approximately 8.314 J/mol K.
  • T is the temperature in Kelvin.
This equation highlights the impact of temperature on the speed of chemical reactions. A higher temperature generally increases the rate constant, bringing about faster reactions.
Rate Constants
Rate constants are crucial to describing how quickly a reaction proceeds. They are specific to each reaction and depend on factors like temperature, pressure, and the presence of a catalyst.

The rate constant \(k\) is used together with the concentration of reactants in a rate law to predict the rate at which a reaction occurs. In the context of the Arrhenius equation, rate constants are linked to the activation energy and temperature.

For example, in the decomposition of nitric oxide, the rate constants at different temperatures help us understand how the reaction rate changes, assisting in calculations such as finding the activation energy.
Nitrogen Oxides Decomposition
The decomposition of nitrogen oxides, specifically nitric oxide ( O ), into nitrogen ( N ) and oxygen ( O ) gases is a crucial chemical process, often used as an example for studying second-order reactions.

This reaction's progress can significantly influence environmental conditions, since nitrogen oxides are pollutants found in vehicle emissions. Understanding the decomposition helps in devising methods to control these emissions, thus reducing pollution and protecting environmental health.

The rate law for this reaction is second order, meaning it depends on the concentration of nitric oxide squared, which is important for students to understand when performing kinetic analyses.
Temperature Conversion
Temperature conversion between Celsius and Kelvin is a straightforward but essential process in chemistry. To convert a temperature from Celsius to Kelvin, one simply adds 273.15 to the Celsius measurement. This is crucial in calculations involving the Arrhenius equation, where temperature must be in Kelvin.

For the problem of nitric oxide decomposition, converting the temperatures from Celsius to Kelvin allows accurate use of the Arrhenius equation, ensuring the right calculations for finding activation energy at different temperatures.

Learning how to switch effortlessly between these temperature scales is important for anyone studying chemistry, as many formulas require Kelvin.
Second Order Reaction
A second order reaction refers to a chemical reaction where the rate is proportional to the concentration of one reactant squared or the product of the concentrations of two reactants.

The rate law for a second order reaction is expressed as:\[ rate = k[A]^2 \]where:
  • k is the rate constant.
  • [A] represents the concentration of the reactant.
In the decomposition of nitric oxide, it is a second order in \( ext{NO} \), highlighting the importance of concentration changes in the rate of reactions. These reactions tend to slow down more noticeably as the reactant is consumed compared to first-order reactions.

Understanding the order of a reaction is fundamental for predicting how the reaction will proceed under different conditions and can be a powerful tool in studying complex chemical processes.

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Most popular questions from this chapter

Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) that is first order in A, zero order in B, and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when \([A]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([\mathrm{C}]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? \((f)\) By what factor does the rate change when the concentrations of all three reactants are cut in hal??

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

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: $$ \begin{array}{cc} \hline \text { Time (s) } & \text { Pressure } \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{kPa}) \\ \hline 0 & 101.3 \mathrm{kPa} \\ 2500 & 95.95 \mathrm{kPa} \\ 5000 & 90.69 \mathrm{kPa} \\ 7500 & 85.92 \mathrm{kPa} \\ 10,000 & 81.36 \mathrm{kPa} \\ \hline \end{array} $$

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28{ }^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2} \mathrm{~s} \mathrm{} 0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-}\)?

(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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