/*! 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 4 The activation energy for the de... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The activation energy for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) according to the equation $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ is \(102 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). At \(45.0^{\circ} \mathrm{C}\) the rate constant is \(5.0 \times 10^{-4} \mathrm{~s}^{-1} .\) What is the rate constant at \(65.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
Use the Arrhenius equation to find that \(k_2\) at 65 °C is approximately \(7.1 \times 10^{-4} \mathrm{s}^{-1}\).

Step by step solution

01

Understand the Arrhenius Equation

To solve this problem, we need to use the Arrhenius equation, which is given by \[ k = A e^{\frac{-E_a}{RT}} \]. Here, \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
02

Convert Celsius to Kelvin

First, convert the given temperatures from Celsius to Kelvin. The formula for conversion is \( T(K) = T(°C) + 273.15 \).At 45.0° C: \[ T_1 = 45.0 + 273.15 = 318.15 \text{ K} \] At 65.0° C: \[ T_2 = 65.0 + 273.15 = 338.15 \text{ K} \]
03

Use the Arrhenius Equation Form

We use the form of the Arrhenius equation that relates two temperatures and their respective rate constants: \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] where \(k_1\) is the known rate constant at \(T_1\) and \(k_2\) is the unknown rate constant at \(T_2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Activation Energy
Activation energy is a critical concept in understanding how chemical reactions occur. It is defined as the minimum amount of energy required to initiate a reaction. This energy barrier must be overcome for reactants to be transformed into products. Think of it as the hill that molecules need to climb over in order to react.
To visualize this:
  • An exothermic reaction, which releases energy overall, still requires an input of energy to start.
  • The activation energy is this initial energy 'hurdle'.
Using the Arrhenius equation, activation energy (\(E_a\)) can influence how quickly a reaction proceeds. A larger \(E_a\) typically means a slower reaction, as fewer molecules have enough energy to overcome the barrier.
In the given exercise, the activation energy for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is given as \(102 \text{ kJ} \cdot \text{mol}^{-1}\), which provides insight into the required energy needed to break the bonds of the \(\mathrm{N}_{2} \mathrm{O}_{5}\) molecules.
Rate Constant
The rate constant, denoted by \(k\), is a crucial part of the Arrhenius equation. It represents the speed of a reaction under specific conditions. In a chemical equation, \(k\) gives us an idea about the reaction's frequency or likelihood.
Its value depends on several factors:
  • Temperature: A higher temperature usually increases the rate constant, meaning the reaction speeds up.
  • Activation Energy: Lower activation energy can increase \(k\) since the reaction becomes easier to start.
In the exercise, at \(45.0^{\circ} \mathrm{C}\), the rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(5.0 \times 10^{-4} \ \mathrm{s}^{-1}\). Using the Arrhenius equation, you can predict the rate constant at another temperature (\(65.0^{\circ} \mathrm{C}\) in this case), demonstrating the temperature dependence of reaction rates.
Temperature Conversion
Temperature conversion is an essential step in using the Arrhenius equation properly, as this equation demands temperature input in Kelvin, not Celsius. To convert: simply add 273.15 to a Celsius value.
For example, converting \(45.0^{\circ} \mathrm{C}\) to Kelvin gives \(318.15 \text{ K}\), and \(65.0^{\circ} \mathrm{C}\) becomes \(338.15 \text{ K}\).
It's important to note:
  • Celsius and Kelvin scales have the same increment size, hence, only an offset.
  • This uniformity makes conversion straightforward without altering the nature of temperature differences or changes.
In scientific calculations, this accuracy is crucial because it ensures consistency when comparing values or inputting them into equations.
Gas Constant
The gas constant (\(R\)) is a universal constant crucial for calculations involving gases and energies in chemistry. It appears in multiple fundamental equations, like the ideal gas law and the Arrhenius equation.
Its value is generally \(8.314 \text{ J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}\), which factors in energy per mole per Kelvin. This makes it versatile for calculating energy changes when temperature shifts occur.
Considerations include:
  • Using the correct units for \(R\) based on context (e.g., when energy is presented in kilojoules).
  • Ensuring dimensional consistency, which helps solve the Arrhenius equation accurately.
In the problem, \(R\) helps relate activation energy, temperature, and the rate constant, providing an interconnected view of how these variables influence one another.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The reaction of carbon dioxide with hydroxide ion in aqueous solution described by $$ \mathrm{CO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ is postulated to occur according to the mechanism (1) \(\mathrm{CO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{HCO}_{\mathrm{s}}^{-}\) (slow) (2) \(\mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (fast) The rate law for the disappearance of \(\mathrm{CO}_{2}(a q)\) was found experimentally to be $$ \text { rate of reaction }=k\left[\mathrm{CO}_{2}\right]\left[\mathrm{OH}^{-}\right] $$ Is this mechanism consistent with the observed rate law? Justify your answer.

The reaction described by $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g) $$ has an experimentally determined rate law of $$ \text { rate of reaction }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] $$ Some proposed mechanisms for this reaction are given below: Mechanism a. (1) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{HI}(g) \quad\) (one-step reaction) Mechanism b. (1) \(\mathrm{I}_{2}(g) \frac{k_{\mathrm{l}}}{{ }_{k_{-1}}} 2 \mathrm{I}(g) \quad\) (fast equilibrium) (2) \(\mathrm{H}_{2}(g)+2 \mathrm{I}(g) \stackrel{k_{z}}{\longrightarrow} 2 \mathrm{HI}(g) \quad\) (slow) Mechanism c. (1) \(\mathrm{I}_{z}(g) \frac{k_{\mathrm{l}}}{\mathrm{k}_{-\mathrm{t}}} 2 \mathrm{I}(g) \quad\) (fast equilibrium) (2) \(\mathrm{I}(g)+\mathrm{H}_{2}(g) \frac{k_{2}}{{ }_{k_{-2}}} \mathrm{H}_{2} \mathrm{I}(g) \quad\) (fast equilibrium) (3) \(\mathrm{H}_{2} \mathrm{I}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HI}(g)+\mathrm{I}(g) \quad\) (slow) (a) Which of these mechanisms are consistent with the observed rate law? (b) In \(1967 \mathrm{~J} . \mathrm{H}\). Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I-I bond in an \(\mathrm{I}_{2}\) molecule. Which mechanism(s) are consistent with both the rate law and this additional observation? Justify your answer.

The reaction described by the equation $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightarrow 2 \mathrm{HBr}(g) $$ has a fractional dependence on bromine. The observed rate law is $$ \text { rate of reaction }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{Br}_{2}\right]^{1 / 2} $$ One proposed mechanism for this reaction is (2) \(\operatorname{Br}(g)+\mathrm{H}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \operatorname{HBr}(g)+\mathrm{H}(g) \quad\) (slow) (3) \(\mathrm{H}(g)+\mathrm{Br}_{2}(g) \stackrel{k_{3}}{\longrightarrow} \operatorname{HBr}(g)+\operatorname{Br}(g) \quad\) (fast) (a) Show that the observed rate law is consistent with the proposed mechanism. (b) What are the units of the rate constant for this reaction? (c) Is it possible that this reaction occurs in a single-step mechanism?

The rate of decomposition of acetaldehyde, \(\mathrm{CH}_{3} \mathrm{CHO}(g)\), into \(\mathrm{CH}_{4}(g)\) and \(\mathrm{CO}(g)\) in the presence of \(\mathrm{I}_{2}(g)\) at \(800 \mathrm{~K}\) follows the rate law $$ \text { rate of reaction }=k\left[\mathrm{CH}_{3} \mathrm{CHO}\right]\left[\mathrm{I}_{2}\right] $$ The decomposition is believed to occur by the following two-step mechanism: (1) \(\mathrm{CH}_{3} \mathrm{CHO}(g)+\mathrm{I}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)+\mathrm{CO}(g)\) (2) \(\mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) \rightarrow \mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g)\) (a) What is the catalyst for the reaction? (b) Which step in the proposed mechanism is most likely the rate-limiting step?

Cryosurgical procedures involve lowering the body temperature of the patient prior to surgery. Given that the activation energy for the beating of the heart muscle is about \(30 \mathrm{~kJ}\), estimate the pulse rate at \(22^{\circ} \mathrm{C}\). Assume the pulse rate at \(37^{\circ} \mathrm{C}\) (body temperature) to be 75 beats \(\cdot \min ^{-1}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.