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In a series of experiments on the decomposition of dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), rate constants were determined at two different temperatures. At \(35^{\circ} \mathrm{C}\), the rate constant was \(1.4 \times\) \(10^{-4} / \mathrm{s}\); at \(45^{\circ} \mathrm{C}\), the rate constant was \(5.0 \times 10^{-4} / \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant at \(55^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Use the Arrhenius equation to find activation energy and extrapolate rate constant.

Step by step solution

01

Introduce the Arrhenius Equation

To find the activation energy, we use the Arrhenius equation:\[k = A e^{-\frac{E_a}{RT}}\]where:- \(k\) is the rate constant,- \(A\) is the frequency factor,- \(E_a\) is the activation energy,- \(R\) is the universal gas constant \(8.314 \, \text{J/molâ‹…K}\),- \(T\) is the temperature in Kelvin.
02

Use the Two-Point Form of the Arrhenius Equation

Rearranging and taking the natural logarithm of the Arrhenius equation, we have:\[\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where:- \(k_1\) and \(k_2\) are the rate constants at temperatures \(T_1 = 35^{\circ}\text{C} = 308\, \text{K}\) and \(T_2 = 45^{\circ}\text{C} = 318\, \text{K}\).

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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 mathematical expression that shows the dependence of the rate constant on temperature and activation energy. It is given by the formula: \[ k = A e^{-\frac{E_a}{RT}} \]Here, - \( k \) represents the rate constant, describing how fast a reaction proceeds, - \( A \) is the frequency factor, which accounts for the frequency of collisions leading to a reaction,- \( E_a \) is the activation energy, indicating the minimum energy that reacting molecules must possess,- \( R \) is the universal gas constant, with a value of \( 8.314 \, \text{J/molâ‹…K} \),- \( T \) is the absolute temperature in Kelvin.
In the context of chemical kinetics, understanding this equation helps predict how changes in temperature affect the speed of chemical reactions. The equation shows that as temperature increases, the rate constant \( k \) usually increases, leading to a faster reaction rate. To determine reaction rates at different temperatures, one often rearranges the equation and takes its natural logarithm, simplifying the calculation of activation energy.
Activation Energy
Activation Energy, denoted as \( E_a \), plays a crucial role in chemical reactions. It is the amount of energy required to initiate a reaction. Think of it as an energy barrier that reactant molecules need to overcome to transform into products.
A higher activation energy means that fewer molecules will have sufficient energy to react, making the reaction slower. Conversely, a lower activation energy implies that more molecules can surpass the energy barrier, leading to a faster reaction.
Finding the activation energy can be done using the two-point form of the Arrhenius equation: \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]Here, \( k_1 \) and \( k_2 \) are the rate constants at two different temperatures, \( T_1 \) and \( T_2 \). By knowing the rate constants and temperatures, we can solve for \( E_a \) using this relationship. It helps us understand why certain reactions might be sluggish or rapid at different temperatures.
Rate Constants
Rate Constants, often symbolized as \( k \), are vital parameters in chemical kinetics that directly correlate to the speed of a reaction. Essentially, it is a proportionality constant in the rate law equation that connects the reaction rate with the concentrations of the reactants.
The rate constant is dependent on temperature, as illustrated by the Arrhenius equation. Higher temperatures typically result in higher rate constants, signifying that reactions occur quickly.
For instance, in the reaction involving the decomposition of dinitrogen pentoxide, we observe different rate constants at varying temperatures. At \( 35^{\circ} \text{C} \), the rate constant is \( 1.4 \times 10^{-4} / \text{s} \), while at \( 45^{\circ} \text{C} \), it increases to \( 5.0 \times 10^{-4} / \text{s} \), reflecting the reaction’s enhanced pace with rising temperature. Understanding rate constants is fundamental for predicting reaction rates and controlling chemical processes.

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Most popular questions from this chapter

The decomposition of ozone is believed to occur in two steps: \- \(\mathrm{O}_{3} \rightleftharpoons \mathrm{O}_{2}+\mathrm{O} \quad\) (elementary reaction) $$\mathrm{O}_{3}+\mathrm{O} \longrightarrow 2 \mathrm{O}_{2}$$ (elementary reaction) Identify any reaction intermediate. What is the overall reaction?

Given the following mechanism for a chemical reaction: $$\begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} & \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{IO}^{-} \\ \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{IO}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+\mathrm{I}^{-} \end{aligned}$$ a. Write the overall reaction. b. Identify the catalyst and the reaction intermediate. C. With the information given in this problem, can you write the pate law? Fxnlaiv

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Draw and label the potential-energy curve for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; \Delta H=57 \mathrm{~kJ}$$ The activation energy for the reverse reaction is \(23 \mathrm{~kJ} .\) Note \(\Delta H\) and \(E_{a}\) on the diagram. What is the activation energy for the forward reaction? For which reaction (forward or reverse) will the reaction rate be most sensitive to a temperature increase? Explain.

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