Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 24

In the Haber process for the production of ammonia, $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Short Answer

Expert verified
The relationship between the rate of production of ammonia and the rate of consumption of hydrogen in the Haber process is \(\frac{\Delta[H_2]}{\Delta t}=-3\frac{\Delta[NH_3]}{2\Delta t}\).

Step by step solution

01

Write down the balanced chemical equation.

Write down the balanced chemical equation for the Haber process: \(N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)\)
02

Analyze the stoichiometry of the reaction.

In the balanced chemical equation, the stoichiometric coefficients tell us the ratio in which the reactants are consumed and the products are formed. From the equation, we can see that 1 mole of nitrogen (N₂) reacts with 3 moles of hydrogen (H₂) to produce 2 moles of ammonia (NH₃).
03

Express the rate of production and consumption in terms of moles.

To express the relationship between the rates of production and consumption in terms of moles, we will call the rate of consumption of hydrogen (-Δ[H₂]/Δt) and the rate of production of ammonia (Δ[NH₃]/Δt). The stoichiometry of the equation tells us that for each mole of N₂ reacting, 3 moles of H₂ are consumed, and 2 moles of NH₃ are formed.
04

Find the relationship between the rates of production and consumption.

Using the stoichiometry of the reaction, we can now find the relationship between the rates of production and consumption of H₂ and NH₃. Since 3 moles of H₂ are consumed for every 2 moles of NH₃ produced, we can write the relationship as: \(-\frac{\Delta[H_2]}{3}=\frac{\Delta[NH_3]}{2}\) Next, we solve for the rate of consumption of hydrogen: \(\frac{\Delta[H_2]}{\Delta t}=-3\frac{\Delta[NH_3]}{2\Delta t}\) Therefore, the relationship between the rate of production of ammonia and the rate of consumption of hydrogen in the Haber process is: \(\frac{\Delta[H_2]}{\Delta t}=-3\frac{\Delta[NH_3]}{2\Delta t}\)

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

The thiosulfate ion \(\left(S_{2} O_{3}^{2-}\right)\) is oxidized by iodine as follows: $$2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q)$$ In a certain experiment, \(7.05 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-} .\) Calculate the rate of production of iodide ion.

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{h}\) if the initial solution of \(\operatorname{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{g}\) InCl \((s)\) in dilute HCl to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{h} ?\)

Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.) $$2 \mathrm{Ce}^{4+}(a q)+\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+2 \mathrm{Hg}^{2+}(a q)$$ $$\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)$$

The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter \(11 .\) What are these two procedures? Explain how each method is used to determine rate laws.

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0,\) and 40.0 min for an experiment in which \([\mathrm{A}]_{0}=0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{min}\) b. \(30.0 \mathrm{min}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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