Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 94

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

Short Answer

Expert verified
a. Elevated temperature increases reaction rate but decreases ammonia yield due to the exothermic nature of the reaction, as per Le Chatelier's principle. b. Removing ammonia prevents the reverse reaction and shifts the equilibrium towards the product, increasing the ammonia yield. c. Using a catalyst speeds up both forward and reverse reactions equally without affecting the equilibrium, minimizing the time required to achieve a certain amount of ammonia. d. High pressure shifts the equilibrium towards the side with fewer gas moles, resulting in a higher yield of ammonia according to Le Chatelier's principle.

Step by step solution

01

Understanding the chemical reaction

The synthesis of ammonia gas from nitrogen gas and hydrogen gas can be represented by the following chemical equation: \[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\] This is a reversible reaction, meaning that the reaction can proceed in both directions, where the reactants form products and vice versa. It achieves a state of equilibrium where the rates of both the forward (product formation) and reverse (reactant formation) reactions are equal.
02

Effect of elevated temperature

a. Running the reaction at an elevated temperature increases the rate of reaction due to the increased average kinetic energy of particles, which leads to more frequent and effective collisions. However, according to Le Chatelier's principle, increasing the temperature for an exothermic reaction like the synthesis of ammonia shifts the equilibrium in the reverse direction, leading to lower yields. Therefore, running the reaction at an elevated temperature might increase the reaction rate, but the yield of ammonia would not necessarily be maximized.
03

Effect of removing ammonia from the reaction mixture

b. Removing the ammonia from the reaction mixture as it forms helps to maximize the yield of ammonia because it prevents the reverse reaction (ammonia breaking back down into nitrogen and hydrogen) from occurring. According to Le Chatelier's principle, if we remove a product, the equilibrium will shift in the direction of the product (forward direction) to re-establish the equilibrium. This leads to more reactants being converted into ammonia and an increased yield.
04

Effect of using a catalyst

c. Using a catalyst helps to maximize the yield of ammonia, because a catalyst speeds up both the forward and reverse reactions without getting consumed in the process. It works by providing an alternative pathway with a lower activation energy, increasing the rate of reaction. Since the catalyst increases the rate of both reactions equally, the position of the equilibrium remains unaffected. However, the reaction achieves equilibrium more quickly, so the time required to make a certain amount of product (in this case, ammonia) is significantly reduced.
05

Effect of running the reaction at high pressure

d. Running the reaction at high pressure helps to maximize the yield of ammonia. According to Le Chatelier's principle, when the pressure is increased, the equilibrium will shift in the direction that has fewer moles of gas. In the synthesis of ammonia, there are 4 moles of gas molecules on the reactants side (1 mole of nitrogen gas and 3 moles of hydrogen gas) and 2 moles of gas molecules on the product side (2 moles of ammonia gas). Therefore, increasing the pressure causes the equilibrium to shift toward the product side (forward direction) resulting in a higher yield of ammonia.

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

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

Consider the reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\) in a 1.0-L rigid flask. Answer the following questions for each situation \((\mathrm{a}-\mathrm{d})\) : i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between \(95 M\) and \(100 M\) ii. Explain how you decided on the limits for the estimated range. iii. Indicate what other information would enable you to narrow your estimated range. iv. Compare the estimated concentrations for a through d, and explain any differences. a. If at equilibrium \([\mathrm{A}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for \([\mathrm{A}]\) once equilibrium is reestablished. b. If at equilibrium \([\mathrm{B}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for [B] once equilibrium is reestablished. c. If at equilibrium \([\mathrm{C}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for \([\mathrm{C}]\) once equilibrium is reestablished. d. If at equilibrium \([\mathrm{D}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for [D] once equilibrium is reestablished.

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

The gas arsine, \(\mathrm{AsH}_{3}\), decomposes as follows: $$ 2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g) $$ In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of \(392.0\) torr. After 48 hours the pressure in the flask was observed to be constant at \(488.0\) torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(\mathrm{~g})\). b. Calculate \(K_{\mathrm{p}}\) for this reaction.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0 \mathrm{~mol} \mathrm{NO}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed into a \(1.0\) - \(\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Organic 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