Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 3

\(2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) The data below was gathered for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at 310 \(\mathrm{K}\) via the equation above. \(\begin{array}{|c|c|}\hline \text { Time(s) } & {\left[\mathrm{N}_{2} \mathbf{O}_{5}\right](M)} \\ \hline 0 & {0.250} \\ \hline 500 . & {0.190} \\\ \hline 1000 . & {0.145} \\ \hline 2000 . & {0.085} \\ \hline\end{array}\) (a) How does the rate of appearance of NO_{2} \text { compare to the rate of } disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? Justify your answer. (b) The reaction is determined to be first order overall. On the axes below, create a graph of some function of concentration vs. time that will produce a straight line. Label and scale your axes appropriately. (c) (i) What is the rate constant for this reaction? Include units. (ii) What would the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) be at \(t=1500 \mathrm{s} ?\) (iii) What is the half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\) ? (d) Would the addition of a catalyst increase, decrease, or have no effect on the following variables? Justify your answers. (i) Rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}\) (ii) Magnitude of the rate constant (iii) Half-life of \(\mathrm{N}_{2} \mathrm{O}_{5}\)

Short Answer

Expert verified
The rate of appearance of \(NO_{2}\) is twice the rate of disappearance of \(N_{2}O_{5}\). The graph will have \(ln[N_{2}O_{5}]\) vs time plotted. The rate constant and future concentration can be calculated from the slope of the plot. Half-life can be estimated using \(t_{1/2} = \frac{0.693}{k}\). A catalyst would increase the rate of disappearance of \(N_{2}O_{5}\) and the magnitude of the rate constant, but would not affect the half-life.

Step by step solution

01

Relationship between appearance and disappearance rates

The reaction equation \(2N_{2}O_{5}(g) \rightarrow 4NO_{2}(g)+O_{2}(g)\) gives a ratio between \(N_{2}O_{5}\) and \(NO_{2}\) of 2:4. This means that for every two molecules of \(N_{2}O_{5}\) decomposed, four molecules of \(NO_{2}\) are produced. Therefore, the rate of appearance of \(NO_{2}\) is twice the rate of disappearance of \(N_{2}O_{5}\).
02

Plotting the function of concentration vs. time

Since the reaction is first order, the function to be plotted for a straight line graph would be \(ln[N_{2}O_{5}]\) vs. time. \(N_{2}O_{5}\) concentration on the Y-axis (in a logarithmic scale) and Time on the X-axis (in seconds).
03

Calculation of the rate constant and future concentration

The rate constant \(k\) can be obtained from the slope of the line. The first-order rate equation will be \(ln[N_{2}O_{5}]=(-kt)+ln[N_{2}O_{5}]_{0}\). Plugging the values from any two data points into this equation will allow solving for \(k\). Given \(k\), the concentration of \(N_{2}O_{5}\) at a future time (t=1500s) can be calculated using the same equation.
04

Determining the half-life

The half-life (\(t_{1/2}\)) for a first-order reaction is given by \(t_{1/2} = \frac{0.693}{k}\). After calculating the rate constant k, it can be used to compute the half-life.
05

The effects of a catalyst

A catalyst speeds up the rate of a reaction. Therefore, it would increase the rate of disappearance of \(N_{2}O_{5}\), increase the magnitude of the rate constant \(k\), but have no effect on the half-life (\(t_{1/2}\)) because half-life for first-order reactions only depends on the rate constant and not on the concentration.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Which of the following pairs of ions would make the best buffer with a pH between 6 and 7? \(K_{\mathrm{a}}\) for \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}=1.75 \times 10^{-5}\) \(K_{\mathrm{a}}\) for \(\mathrm{HPO}_{4}^{2-}=4.8 \times 10^{-13}\) (A) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}\) (B) \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) (C) \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) (D) \(\mathrm{NaOH}\) and \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

$$\begin{array}{|c|c|c|}\hline & {\text { Initial pH }} & {\text { PH after NaOH }} \\ \hline \text { Acid 1 } & {3.0} & {3.5} \\ \hline \text { Acid 2 } & {3.0} & {5.0} \\ \hline\end{array}$$ Two different acids with identical pH are placed in separate beakers. Identical portions of NaOH are added to each beaker, and the resulting pH is indicated in the table above. What can be determined about the strength of each acid? (A) Acid 1 is a strong acid and acid 2 is a weak acid because acid 1 resists change in pH more effectively. (B) Acid 1 is a strong acid and acid 2 is a weak acid because the NaOH is more effective at neutralizing acid 2. (C) Acid 1 is a weak acid and acid 2 is a strong acid because the concentration of the weak acid must be significantly greater to have the same pH as the strong acid. (D) Acid 1 is a weak acid and acid 2 is a strong acid because the concentration of the hydrogen ions will be greater in acid 2 after the NaOH addition.

An unknown substance is found to have a high melting point. In addition, it is a poor conductor of electricity and does not dissolve in water. The substance most likely contains (A) ionic bonding (B) nonpolar covalent bonding (C) covalent network bonding (D) metallic bonding

Consider the Lewis structures for the following molecules: $$\begin{equation} \mathrm{CO}_{2}, \mathrm{CO}_{3}^{2-}, \mathrm{NO}_{2}^{-}, \text {and } \mathrm{NO}_{3}^{-} \end{equation}$$ Which molecule would have the smallest bond angle between terminal atoms? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{CO}_{3}^{2-}\) (C) \(\mathrm{NO}_{2}^{-}\) (D) \(\mathrm{NO}_{3}^{-}\)

A sample of \(\mathrm{H}_{2} \mathrm{S}\) gas is placed in an evacuated, sealed container and heated until the following decomposition reaction occurs at \(1000 \mathrm{K} :\) \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \qquad K_{\mathrm{c}}=1.0 \times 10^{-6}\) (A) \(K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{S}\right]^{2}}\) (B) \(K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2} \mathrm{S}\right]^{2}}{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{S}_{2}\right]}\) (C) \(K_{\mathrm{c}}=\frac{2\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]}{2\left[\mathrm{H}_{2} \mathrm{S}\right]}\) (D) \(K_{\mathrm{c}}=\frac{2\left[\mathrm{H}_{2} \mathrm{S}\right]}{2\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

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.

Sign-up for free