/*! 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 20 The decomposition of nitrogen di... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. The rate constant for this reaction is \(3.40 \mathrm{L} / \mathrm{mol} \cdot \mathrm{min} .\) Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}\).

Short Answer

Expert verified
The time needed is approximately 0.0490 minutes.

Step by step solution

01

Identify the Reaction Order

Since the reaction is second-order, the rate law can be expressed using the second-order rate equation: \[\text{rate} = k[A]^2\] where \(k\) is the rate constant and \([A]\) is the concentration of the reactant.
02

Use the Integrated Rate Law for Second-Order Reactions

The integrated rate law for a second-order reaction is given by: \[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \] where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.
03

Substitute Known Values

The initial concentration \([A]_0\) is 2.00 mol/L, the concentration at time \(t\) \([A]_t\) is 1.50 mol/L, and the rate constant \(k\) is 3.40 L/mol·min. Substitute these values into the integrated rate law:\[ \frac{1}{1.50} = 3.40 \times t + \frac{1}{2.00} \].
04

Solve for Time \(t\)

Rearrange and solve the equation for \(t\):1. Calculate \( \frac{1}{1.50} = 0.6667 \) mol/L.2. Calculate \( \frac{1}{2.00} = 0.500 \) mol/L.3. Substitute these into the equation: \[0.6667 = 3.40 \times t + 0.500 \].4. Subtract 0.500 from both sides to solve for \(3.40t\): \[0.1667 = 3.40t\].5. Finally, divide by 3.40: \[t = \frac{0.1667}{3.40} \approx 0.0490 \text{ min}\].
05

Result Interpretation

The time required for the concentration of NO2 to decrease from 2.00 mol/L to 1.50 mol/L is approximately 0.0490 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.

Understanding Integrated Rate Laws
Integrated rate laws for chemical reactions are powerful tools that allow us to connect the concentration of a reactant to time. In second-order reactions, like the decomposition of nitrogen dioxide, this relationship shows how the concentration decreases as the reaction proceeds. The integrated rate law for a second-order reaction is given by the formula:\[ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \]where \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant. When you use this equation, you transform the non-linear rate process into a linear one, making it much easier to calculate the time required for a specific change in concentration. By rearranging this formula, chemists can predict how long a reaction will take under given conditions.
  • Second-order reactions depend on the square of the concentration of one reactant.
  • The formula directly helps in determining the time required for concentration changes.
  • This approach is particularly useful in controlling reaction conditions in industrial settings.
Exploring Reaction Kinetics
Reaction kinetics is the study of the rates at which chemical reactions proceed and the factors that affect these rates. For a reaction to occur, molecules must collide with sufficient energy and proper orientation. In second-order reactions, the rate depends on the concentration of the reactants squared, which is why the rate law is expressed as \( \text{rate} = k[A]^2 \). This tells us the speed and the overall mechanism of the reaction.
Several factors influence reaction kinetics, including:
  • The concentration of reactants: Higher concentrations generally increase the reaction rate.
  • Temperature: Increasing temperature typically increases reaction rates by providing more energy for collisions.
  • Catalysts: These substances can speed up reactions by lowering the energy required to reach the transition state.
Understanding these factors allows chemists to manipulate reactions to obtain desired outcomes, whether it be faster reaction completion or controlling the yield of a reaction.
Calculating the Rate Constant
The rate constant, \(k\), is a crucial part of the rate equation that helps us determine how fast a reaction proceeds. For a second-order reaction, we use the following form:\[ \text{rate} = k[A]^2 \]The units of \(k\) in second-order reactions are typically \( \text{L/mol} \cdot \text{time} \), and it is calculated or given for use in predicting reactions. Calculating \(k\) involves measuring the rate of reaction at various concentrations and then using these measurements in the rate law equation.
In the example exercise, \(k = 3.40 \text{ L/mol} \cdot \text{min} \), and this constant allows the determination of how much time is needed for the concentration of a reactant to fall to a desired level. Without an accurate \(k\), it would be impossible to predict the reaction's progress with precision.
  • The higher the rate constant, the faster the reaction occurs.
  • Rate constants also reflect the nature of the reaction, including temperature and pressure conditions.
  • In practical applications, knowing \(k\) informs decisions on reactant amounts and reaction time management.

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 substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L Iwhere L is an electron-pair donor such as \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\) ) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90,\) p. \(6927,1968 .\) A detailed study of the kinetics of the reaction led to the following mechanism: Slow: \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast: \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of 2 Doubling the concentration of L had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M},\) what is the concentration of the product after 5.0 minutes?

At temperatures below \(500 \mathrm{K}\), the reaction between carbon monoxide and nitrogen dioxide $$ \mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g}) $$ has the following rate equation: Rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\) Which of the three mechanisms suggested here best agrees with the experimentally observed rate equation? Mechanism 1 \(\quad\) single, elementary step $$ \mathrm{NO}_{2}+\mathrm{CO} \rightarrow \mathrm{CO}_{2}+\mathrm{NO} $$ Mechanism \(2 \quad\) Two steps Slow $$ \mathrm{NO}_{2}+\mathrm{NO}_{2} \rightarrow \mathrm{NO}_{3}+\mathrm{NO} $$ Fast $$ \mathrm{NO}_{3}+\mathrm{CO} \rightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} $$ Mechanism 3 Two steps Slow $$ \mathrm{NO}_{2} \rightarrow \mathrm{NO}+\mathrm{O} $$ Fast $$ \mathrm{CO}+\mathrm{O} \rightarrow \mathrm{CO}_{2} $$

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \begin{array}{c} \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] \text { where } k=0.17 / \mathrm{hr} \end{array} $$ (a) What is the rate of decomposition when \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.010 \mathrm{M} ?\) (b) What is the half-life of the reaction? (c) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a flask is \(0.050 \mathrm{atm},\) what is the pressure of all gases (i.e., the total pressure) in the flask after the reaction has proceeded for one half-life?

The compound \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) decomposes in a firstorder reaction to elemental Xe with a half-life of 30\. minutes. If you place \(7.50 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) in a flask, how long must you wait until only \(0.25 \mathrm{mg}\) of \(\mathrm{Xe}\left(\mathrm{CF}_{3}\right)_{2}\) remains?

To which species should an enzyme bind best: the substrate, transition state, or product of a reaction?

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.