/*! 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 49 The gas-phase decomposition of $... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 49

The gas-phase decomposition of $\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\( \)2 \mathrm{NO}(g)+\mathrm{O}_{2}(g),$ is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?(\mathbf{b})\) What is the rate constant? (c) Predict the reaction rates at the beginning of the reaction for initial concentrations of \(0.200 \mathrm{M}, 0.100 \mathrm{M},\) and $0.050 \mathrm{M} \mathrm{NO}_{2}$.

Short Answer

Expert verified
The reaction is first order with respect to the concentration of $\mathrm{NO}_{2}$ and the rate constant $k \approx 0.573\, \text{s}^{-1}$. The predicted reaction rates at the beginning of the reaction for initial concentrations of $0.200 \mathrm{M}, 0.100 \mathrm{M},$ and $0.050 \mathrm{M}\ \mathrm{NO}_{2}$ are $0.115\, \mathrm{M\,s}^{-1}, 0.0573\, \mathrm{M\,s}^{-1},$ and $0.0287\, \mathrm{M\,s}^{-1}$ respectively.

Step by step solution

01

Calculating the reaction rates

: Using the given concentration data, we can calculate the reaction rate at each time by finding the change in concentration per unit time. The reaction rate is given by: \[R = \frac{-\Delta\left[ \mathrm{NO}_{2} \right]}{\Delta t}\]
02

Plotting the data for different orders

: Once we have the reaction rates, we will plot the reaction order graphs for both first order and second order reactions using the given concentration data and the rate we calculated earlier. The first-order reaction plot will be a plot of \(\ln\left( \frac{[\mathrm{NO}_{2}]}{[\mathrm{NO}_{2}]_{0}}\right)\) vs time (t) while the second-order reaction plot will be a plot of \(\frac{1}{[\mathrm{NO}_{2}]} - \frac{1}{[\mathrm{NO}_{2}]_{0}}\) vs time (t). The one which gives a linear plot will indicate the order of the reaction.
03

Determining the reaction order

: By examining the plots, we will determine which plot is linear. If the first-order plot is linear, then the reaction is first order with respect to the concentration of \(\mathrm{NO}_{2}\). If the second-order plot is linear, then the reaction is second order with respect to the concentration of \(\mathrm{NO}_{2}\).
04

Calculating the rate constant

: Once the reaction order plot is determined, we can now calculate the rate constant (k). For first-order reactions, the slope of the linear plot will be equal to -k, while for second-order reactions, the slope of the linear plot will be equal to k.
05

Predicting reaction rates for various initial concentrations

: Using the rate constant (k) calculated in step 4, we can predict the reaction rates at the beginning of the reaction for initial concentrations of \(0.200 \mathrm{M}, 0.100 \mathrm{M},\) and \(0.050 \mathrm{M}\ \mathrm{NO}_{2}\). For first-order reactions, the rate equation is given by: \[R = k\left[ \mathrm{NO}_{2} \right]\] For second-order reactions, the rate equation is given by: \[R = k\left[ \mathrm{NO}_{2} \right]^2\] By substituting the respective concentrations and the rate constant (k) in the appropriate rate equation, we can find the reaction rates at the beginning of the reaction for the various initial concentrations.

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 91影视!

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 iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: \(\mathrm{OCl}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{OI}^{-}+\mathrm{Cl}^{-}\). This rapid reaction gives the following rate data: (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{OCI}^{-}\right]=2.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=5.0 \times 10^{-4} \mathrm{M}\).

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(l)+\mathrm{I}^{-}(\mathrm{alc})\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C}\). (b) \(\mathrm{A}\) solution of \(\mathrm{KOH}\) in ethanol is made up by dissolving \(0.335 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C}\) ? (c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? (d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(50^{\circ} \mathrm{C}\).

Based on their activation energies and energy changes and assuming that all collision factors are the same, which of the following reactions would be fastest and which would be slowest? (a) \(E_{a}=45 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{~kJ} / \mathrm{mol}\) (b) \(E_{a}=35 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{~kJ} / \mathrm{mol}\) (c) \(E_{a}=55 \mathrm{~kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{~kJ} / \mathrm{mol}\)

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\). are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2}\), and \(\mathrm{N}_{2}\) - (b) Using a resource such as Table \(8.4\), look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

The activation energy of an uncatalyzed reaction is \(95 \mathrm{~kJ} / \mathrm{mol}\). The addition of a catalyst lowers the activation energy to \(55 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C}\), (b) \(125^{\circ} \mathrm{C}\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation
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.