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The gas-phase reaction $$2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ has an activation energy of \(103 \mathrm{kJ} / \mathrm{mol},\) and the rate constant is \(0.0900 \min ^{-1}\) at \(328.0 \mathrm{K}\). Find the rate constant at \(318.0 \mathrm{K}.\)

Short Answer

Expert verified
The rate constant at 318.0 K is approximately 0.00185 min⁻¹.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation is used to calculate the rate constant of a reaction at a different temperature. It is given by:\[k = A \, e^{-E_a/RT}\]where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant \( (8.314 \, \text{J/mol K}) \), and \( T \) is the temperature in Kelvin.
02

Rearrange the Arrhenius Equation for Two Temperatures

We need to compare the rate constants at two different temperatures using the following rearrangement:\[\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where \( k_1 \) is the rate constant at the initial temperature \( T_1 = 328.0 \, \text{K} \), and \( k_2 \) is the rate constant at the new temperature \( T_2 = 318.0 \, \text{K} \).
03

Substitute Known Values

Substitute the given values into the equation:- \( E_a = 103,000 \, \text{J/mol} \) (converted from kJ/mol),- \( R = 8.314 \, \text{J/mol K} \),- \( T_1 = 328.0 \, \text{K} \),- \( T_2 = 318.0 \, \text{K} \),- \( k_1 = 0.0900 \, \text{min}^{-1} \).The equation becomes:\[\ln\left(\frac{k_2}{0.0900}\right) = -\frac{103,000}{8.314}\left(\frac{1}{318.0} - \frac{1}{328.0}\right)\]
04

Calculate the Rate Constant at New Temperature

Calculate the expression on the right:\[-\frac{103,000}{8.314} = -12391.14 \, \text{K}\]\[\frac{1}{318.0} - \frac{1}{328.0} = 0.000314 \, \text{K}^{-1}\]Then multiply these values:\[(-12391.14) \times (0.000314) = -3.889\]Now, find \( k_2 \):\[\ln\left(\frac{k_2}{0.0900}\right) = -3.889\]\[\frac{k_2}{0.0900} = e^{-3.889}\]Solve for \( k_2 \):\[k_2 = 0.0900 \times e^{-3.889} \]\[k_2 = 0.0900 \times 0.0205 \approx 0.00185 \, \text{min}^{-1}\]

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Key Concepts

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

Activation Energy
Activation energy is a crucial concept in chemical kinetics, playing a significant role in determining how fast a reaction occurs. It is the minimum amount of energy required for reactant molecules to collide and form products. In the context of the Arrhenius Equation, activation energy is denoted by \( E_a \) and it reflects the energy barrier that must be overcome for a reaction to proceed. - Higher activation energies imply more energy is needed for a reaction to proceed, making the reaction slower.- Lower activation energies mean less energy is required, potentially resulting in faster reactions.
For example, in the provided gas-phase reaction, the activation energy is given as \( 103 \, \text{kJ/mol} \). This value indicates the amount of energy needed to initiate the conversion of \( \text{N}_2\text{O}_5 \) to \( \text{NO}_2 \) and \( \text{O}_2 \). Understanding and calculating activation energy can help predict the rate at which a reaction will occur.
Rate Constant
The rate constant \( k \) is a pivotal parameter in the rate equation of a chemical reaction, defining how quickly a reaction proceeds at a given temperature. The magnitude of the rate constant depends on several factors including temperature, the nature of the reactants, and the activation energy.In the Arrhenius Equation, the rate constant is expressed as: \[ k = A \, e^{-E_a/RT} \]where \( A \) is the pre-exponential factor, which represents the frequency of collisions in the correct orientation. The rate constant given in the exercise is \( 0.0900 \, \text{min}^{-1} \) at \( 328.0 \, \text{K} \), highlighting that at this temperature, the reaction occurs at a certain rate. Understanding variations in the rate constant with changing temperatures is crucial for predicting how a reaction behaves under different conditions.
Temperature Dependence of Rates
Temperature plays a critical role in influencing the rates of chemical reactions. As the temperature increases, reactant molecules obtain more kinetic energy, leading to a higher frequency of successful collisions. This often results in an increased rate constant, meaning the reaction proceeds faster.
The Arrhenius Equation provides a mathematical model this temperature dependence:\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]This formula allows us to calculate a new rate constant \( k_2 \) at a different temperature \( T_2 \) using a known rate constant \( k_1 \) at temperature \( T_1 \). In the exercise provided, decreasing the temperature from \( 328.0 \, \text{K} \) to \( 318.0 \, \text{K} \) results in a new, slower rate constant of \( 0.00185 \, \text{min}^{-1} \). Understanding this relationship is key in controlling reaction speeds in industrial processes and laboratory settings.
Gas-phase Reaction Kinetics
Gas-phase reactions involve reactants in the gas state and are crucial in fields such as atmospheric chemistry and industrial processing. In gas-phase reaction kinetics, the behavior of reactions is influenced by variables like temperature, pressure, and volume.The provided exercise reflects a basic gas-phase reaction where \( \text{N}_2\text{O}_5 \) decomposes into \( \text{NO}_2 \) and \( \text{O}_2 \). Key aspects of gas-phase reactions include:- Collisions involving the reactant gas molecules must have adequate energy and correct orientation to form products.- Reaction rates are directly impacted by changes in temperature, as described by the Arrhenius Equation.Studying gas-phase reactions is critical for understanding naturally occurring processes like the formation of pollutants and synthetic applications in chemical manufacturing. With foundational knowledge in these kinetics principles, students and researchers can better predict and manipulate reaction outcomes.

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Most popular questions from this chapter

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NOBr}(\mathrm{g})\) (b) \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\)

To determine the concentration dependence of the rate of the reaction $$\mathrm{H}_{2} \mathrm{PO}_{3}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ you might measure \(\left[\mathrm{OH}^{-}\right]\) as a function of time using a pH meter. (To do so, you would set up conditions under which \(\left[\mathrm{H}_{2} \mathrm{PO}_{3}^{-}\right]\) remains constant by using a large excess of this reactant.) How would you prove a second-order rate dependence for \(\left[\mathrm{OH}^{-}\right] ?\)

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. Step 1: Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2: Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate \(=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of \(\mathrm{B}\) is halved, what happens to the reaction rate?

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(\mathrm{aq}) \rightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\)

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