Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 13

The rate law for the reaction $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is given by rate \(=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right] .\) At \(25^{\circ} \mathrm{C},\) the rate constant is \(3.0 \times 10^{-4} / M \cdot\) s. Calculate the rate of the reaction at this temperature if \(\left[\mathrm{NH}_{4}^{+}\right]=0.26 M\) and \(\left[\mathrm{NO}_{2}^{-}\right]=0.080 M\)

Short Answer

Expert verified
The rate of the reaction is \(6.24 \times 10^{-6}\) M.s^-1.

Step by step solution

01

Understand the rate law

The rate of the reaction is calculated using the rate law. The rate law for this reaction is given by \( rate = k[\mathrm{NH}_{4}^{+}][\mathrm{NO}_{2}^{-}] \) where 'k' is the rate constant, '[NH4+]' and '[NO2-]' are the concentrations of the reactants NH4+ and NO2- respectively.
02

Substitute known values

At 25° C, the rate constant 'k' is given as 3.0 x 10^-4 / M.s. The concentrations of NH4+ and NO2- are also given, so these could be substituted in the above rate law. So we get, \( rate = (3.0 \times 10^{-4}/\text{M.s}) \times (\text{0.26 M}) \times (\text{0.080 M}) \).
03

Calculation

The units 'M' get cancelled out in the above expression and the value left to be calculated is: \( rate = 3.0 \times 10^{-4}/\text{s} \times \text{0.26} \times \text{0.080} \). Simple multiplicative calculation gives us the rate of the chemical reaction.

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

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right),\) commonly called table sugar, undergoes hydrolysis (reaction with water) to produce fructose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} $$ fructose \(\quad\) glucose This reaction is of considerable importance in the candy industry. First, fructose is sweeter than sucrose. Second, a mixture of fructose and glucose, called invert sugar, does not crystallize, so the candy containing this sugar would be chewy rather than brittle as candy containing sucrose crystals would be. (a) From the following data determine the order of the reaction. (b) How long does it take to hydrolyze 95 percent of sucrose? (c) Explain why the rate law does not include \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) even though water is a reactant. $$ \begin{array}{cc} \hline \text { Time (min) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]} \\ \hline 0 & 0.500 \\ 60.0 & 0.400 \\ 96.4 & 0.350 \\ 157.5 & 0.280 \\ \hline \end{array} $$

The rate constants of some reactions double with every 10 -degree rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

What are the units of the rate constant for a thirdorder reaction?

The following expression shows the dependence of the half-life of a reaction \(\left(t_{\frac{1}{2}}\right)\) on the initial reactant concentration [A] \(_{0}:\) $$ t_{\frac{1}{2}} \propto \frac{1}{[\mathrm{~A}]_{0}^{n-1}} $$ where \(n\) is the order of the reaction. Verify this dependence for zero-, first-, and second-order reactions.

Consider the reaction $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of \(0.066 M / \mathrm{s}\). (a) At what rate is \(\mathrm{NO}_{2}\) being formed? (b) At what rate is molecular oxygen reacting?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

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