/*! 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 40 The first-order rate constant fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The first-order rate constant for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}, 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\), at \(70^{\circ} \mathrm{C}\) is \(6.82 \times 10^{-3} \mathrm{~s}^{-1} .\) Suppose we start with \(0.0250 \mathrm{~mol}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) in a volume of \(2.0 \mathrm{~L} .\) (a) How many moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) will remain after \(5.0 \mathrm{~min}\) ? (b) How many minutes will it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to drop to \(0.010\) mol? (c) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(70^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) After 5 minutes, there are approximately 0.0129 moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) remaining. (b) It will take approximately 6.8 minutes for the quantity of \(\mathrm{N}_{2}\mathrm{O}_{5}\) to drop to 0.010 mol. (c) The half-life of \(\mathrm{N}_{2}\mathrm{O}_{5}\) at \(70^{\circ}\mathrm{C}\) is approximately 1.7 minutes.

Step by step solution

01

(a) Calculate the remaining moles of Nâ‚‚Oâ‚… after 5 minutes

To find the remaining moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) after 5 minutes, we can use the first-order integrated rate law: \[N_t = N_0e^{-kt}\] Where \(N_t\) is the moles at time \(t\), \(N_0\) is the initial moles, \(k\) is the rate constant, and \(t\) is the time in seconds. First, convert 5 minutes to seconds: \(5.0 \, \text{min} \times \frac{60 \, \text{s}}{1 \, \text{min}} = 300 \, \text{s}\) Next, plug the given values into the equation: \[N_t = (0.0250 \, \text{mol})e^{-(6.82 \times 10^{-3} \, \text{s}^{-1})(300 \, \text{s})}\] Calculate the remaining moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\): \[N_t = 0.0250 \, \text{mol} \times e^{-2.046} \approx 0.0129 \, \text{mol}\] So after 5 minutes, there are approximately 0.0129 moles of \(\mathrm{N}_{2}\mathrm{O}_{5}\) remaining.
02

(b) Calculate the time for Nâ‚‚Oâ‚… to drop to 0.010 mol

Rearrange the first-order integrated rate law to solve for \(t\): \[t = -\frac{\ln{\frac{N_t}{N_0}}}{k}\] Plug in the given values and the desired final quantity of 0.010 mol: \[t = -\frac{\ln{\frac{0.010 \, \text{mol}}{0.0250 \, \text{mol}}}}{6.82 \times 10^{-3} \, \text{s}^{-1}}\] Calculate the time in seconds, then convert to minutes: \[t \approx 406.9 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \approx 6.8 \, \text{min}\] So it will take approximately 6.8 minutes for the quantity of \(\mathrm{N}_{2}\mathrm{O}_{5}\) to drop to 0.010 mol.
03

(c) Find the half-life of N₂O₅ at 70°C

The half-life for a first-order reaction is given by: \[t_{1/2} = \frac{\ln{2}}{k}\] Plug in the given rate constant: \[t_{1/2} = \frac{\ln{2}}{6.82 \times 10^{-3} \, \text{s}^{-1}}\] Calculate the half-life in seconds, then convert to minutes: \[t_{1/2} \approx 101.7 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \approx 1.7 \, \text{min}\] The half-life of \(\mathrm{N}_{2}\mathrm{O}_{5}\) at \(70^{\circ}\mathrm{C}\) is approximately 1.7 minutes.

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.

First-order Reaction
In chemical kinetics, reactions are classified based on their order, which refers to how the rate of reaction is dependent on the concentration of reactants. A **first-order reaction** is one where the rate is directly proportional to the concentration of a single reactant. This means that if the concentration of the reactant is doubled, the reaction rate also doubles. In mathematical terms, the rate law for a first-order reaction can be expressed as: - \[ ext{Rate} = k[N_2O_5]\] where \(k\) is the rate constant, and \[N_2O_5\] is the concentration of reactant.
Such reactions often involve a single molecule breaking down into smaller molecules or atoms. For example, the decomposition of \(N_2O_5\) to \(NO_2\) and \(O_2\) as given in the exercise is a perfect example of a first-order reaction.
Rate Constant
The **rate constant** (**k**) is a crucial factor in the rate equation for any chemical reaction. For first-order reactions, it provides insights into how fast the reaction occurs independent of the reactant's initial concentration. The units of the rate constant for a first-order reaction are \( ext{s}^{-1}\)\.
💡 Key roles of the rate constant:
  • Indicates reaction speed: A larger value of \(k\) means the reaction proceeds more rapidly. A small \(k\) suggests a slower reaction process.
  • Unique for every reaction: It varies depending on the reaction and its conditions, such as temperature.
  • Essential for calculating other parameters: It's used in the integrated rate law to determine the concentration of reactants at any time.
In our exercise, the rate constant is \(6.82 imes 10^{-3} \, ext{s}^{-1}\), which helps in calculating both the remaining concentration of \(N_2O_5\) at any given time and its half-life.
Half-life
The **half-life** of a reaction is the time required for the concentration of a reactant to reduce to half its initial value. It is a useful measure because it describes how quickly a reaction progresses. For first-order reactions, the half-life is constant and can be calculated using the following formula: - \[ t_{1/2} = \frac{\ln{2}}{k} \] where \(\ln{2}\) is approximately equal to 0.693.
Features of half-life in first-order reactions:
  • Independent of initial concentration: This is a unique characteristic of first-order reactions. The half-life remains constant regardless of how much reactant you start with.
  • Can be used to predict how long it will take for the reactant to decrease to a certain level.
In our example, calculating the half-life of \({N_2O_5}\) using \(t_{1/2} = \frac{0.693}{6.82 \times 10^{-3} \, ext{s}^{-1}}\) gives approximately 1.7 minutes. This indicates that every 1.7 minutes, the amount of \(N_2O_5\) is halved.
Integrated Rate Law
The **integrated rate law** for first-order reactions is pivotal for understanding how the concentration of reactants decreases over time. It is derived from the basic rate law and provides a relation between the concentration of reactants and time. The formula can be expressed as: - \[[ ext{N}_t] = [ ext{N}_0] e^{-kt} \] where \[N_t\] is the concentration at time \(t\), \[N_0\] is the starting concentration, \(k\) is the rate constant, and \(e\) is the base of the natural logarithm.
Here’s why the integrated rate law is crucial:
  • Allows calculation of concentration: It helps us determine how much of a reactant is left after a certain period.
  • Time prediction for desired concentration: We can rearrange the formula to find out how long it takes for a reactant to reach a specific concentration.
  • Useful in experimental data analysis: Scientists can use it to plot and examine the progress of a reaction over time.
In the exercise, we use the integrated rate law to find how many moles of \(N_2O_5\) remain, or to determine how long it takes to reach a certain molarity after starting with \(0.0250 \, ext{mol}\).

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

(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?

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbonic acid with carbon dioxide and water. 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?

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00\) L of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant. The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has a molar absorptivity constant of \(5.60 \times 10^{3} \mathrm{~cm}^{-1} \mathrm{M}^{-1}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is \(0.605\) at the beginning of the reaction. (b) The absorbance falls to \(0.250\) within \(30.0\) min. Calculate the rate constant in units of \(\mathrm{s}^{-1} .\) (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftarrows}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g)\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4}\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

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.