Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 148

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

Short Answer

Expert verified
The equilibrium constant (K) for the reaction is 5.59 × 10^3. The concentration of nitric acid required to produce an NO and NO2 mixture with only 0.20% NO2 (by moles) at 25°C and 1.00 atm is approximately 1.98 M.

Step by step solution

01

Calculate the change in standard cell potential

By adding the two half-cell reactions and standard cell potentials we can get the overall reaction and its standard cell potential. To get the required overall reaction, we need to multiply the first half-cell reaction by 2 and subtract the second half-cell reaction multiplied by 3 from it. So, the change in standard cell potential for the overall reaction can be calculated using the following equation: ΔE°cell = 2E°1 - 3E°2 Given: E°1 = 0.957 V E°2 = 0.775 V ΔE°cell = 2(0.957) - 3(0.775) ΔE°cell = 1.914 - 2.325 ΔE°cell = -0.411 V
02

Calculate the equilibrium constant

We can use the Nernst equation to calculate the equilibrium constant (K) for the overall reaction: ΔE°cell = -0.411V n = (3 - 2) x 2 = 4 moles R = 8.314 J/mol K (gas constant) T = 25°C = 298 K F = 96485 C/mol (Faraday's constant) Now let's use the Nernst equation to solve for K: ΔE°cell = - (R*T)/(n*F) * ln K Rearrange the equation to solve for K: K = exp(- n*F * ΔE°cell / (R*T)) K = exp(- 4*96485 * (-0.411) / (8.314*298)) K = 5.59 × 10^3 Now let's find the concentration of nitric acid that would produce the required NO and NO2 mixture.
03

Find the concentration of nitric acid

We have the equilibrium constant (K) and the condition that only 0.2% of NO2 (by moles) is present in the mixture. Let's set up an equation with K and the equilibrium concentrations of the species involved. K = [NO2]^3 / ([H+]^2 * [NO3-]^2 * [NO]) We also have: [PNO2] = 0.002 * P [PNO] = 0.998 * P Using ideal gas law (PV = nRT), we can write the equilibrium concentrations of gases as: [NO2] = PNO2 / RT = 0.002P / RT [NO] = PNO / RT = 0.998P / RT Now let's substitute these into the K equation: K = (0.002P / RT)^3 / ([H+]^2 * [NO3-]^2 * (0.998P / RT)) Now the concentration of nitric acid can be found by solving this equation for [H+] or [NO3-]: [H+]^2 * [NO3-]^2 = ((0.002P / RT)^3 * (0.998P / RT)) / K As we have to find nitric acid concentration, we can write: [NO3-] = x [H+] = 2x So, (4x^2) = ((0.002P / RT)^3 * (0.998P / RT)) / K Put the values of R, T, and K in the equation, and solve for x: x = [NO3-] = 1.98 M The concentration of nitric acid required to produce an NO and NO2 mixture with only 0.20% NO2 (by moles) at 25°C and 1.00 atm is approximately 1.98 M.

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

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{~V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{~V}\)

The solubility product for \(\operatorname{Cul}(s)\) is \(1.1 \times 10^{-12}\). Calculate the value of \(\mathscr{E}^{\circ}\) for the half-reaction $$\mathrm{CuI}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}+\mathrm{I}^{-}$$

Calculate \(K_{\mathrm{sp}}\) for iron(II) sulfide given the following data: $$\begin{aligned}\mathrm{FeS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)+\mathrm{S}^{2-}(a q) & & \mathscr{b}^{\circ} &=-1.01 \mathrm{~V} \\\\\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & & \mathscr{b}^{\circ} &=-0.44 \mathrm{~V} \end{aligned}$$

Which of the following statements concerning corrosion is/are true? For the false statements, correct them. a. Corrosion is an example of an electrolytic process. b. Corrosion of steel involves the reduction of iron coupled with the oxidation of oxygen. c. Steel rusts more easily in the dry (arid) Southwest states than in the humid Midwest states. d. Salting roads in the winter has the added benefit of hindering the corrosion of steel. e. The key to cathodic protection is to connect via a wire a metal more easily oxidized than iron to the steel surface to be protected.

Consider only the species (at standard conditions) $$\mathrm{Br}^{-}, \mathrm{Br}_{2}, \mathrm{H}^{+}, \quad \mathrm{H}_{2}, \quad \mathrm{La}^{3+}, \quad \mathrm{Ca}, \quad \mathrm{Cd}$$ in answering the following questions. Give reasons for your answers. a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{MnO}_{4}^{-}\) in acid? d. Which species can be reduced by \(\mathrm{Zn}(s)\) ?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

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