Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A reaction chamber contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{N}\) in equilibrium at a specified temperature and pressure. How will ( \(a\) ) increasing the temperature at constant pressure and (b) increasing the pressure at constant temperature affect the number of moles of \(\mathrm{N}_{2} ?\)

Short Answer

Expert verified
Answer: (a) When the temperature is increased at constant pressure, the number of moles of N₂ will decrease. (b) When the pressure is increased at constant temperature, the number of moles of N₂ will increase.

Step by step solution

01

The equilibrium reaction

First, we need to write the equation representing the equilibrium between N₂ and N. The dissociation of N₂ can be represented as: N₂ (g) ⇌ 2N (g)
02

Applying Le Chatelier's principle

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust itself to counteract the effect of the change and establish a new equilibrium. We will use this principle to analyze the effects of the changes in temperature and pressure.
03

Effect of increasing temperature at constant pressure (a)

For the given reaction, we know that the dissociation of N₂ into N is an endothermic process. When the temperature is increased, the equilibrium will shift in the direction that absorbs the excess heat. In this case, that means the equilibrium will shift towards the right (towards the production of more N). As a result, the number of moles of N₂ will decrease.
04

Effect of increasing pressure at constant temperature (b)

For analyzing the effect of pressure, we need to see the number of moles of gas on both sides of the equilibrium equation. In this case: N₂ (g) ⇌ 2N (g) Since there are 2 moles of gas on the right side and 1 mole on the left side, the equilibrium will shift towards the side with fewer moles of gas to counteract the increased pressure. Thus, the equilibrium will shift towards the left (towards the production of more N₂). As a result, the number of moles of N₂ will increase with an increase in pressure at constant temperature.

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.

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

Consider a mixture of \(\mathrm{CO}_{2}, \mathrm{CO},\) and \(\mathrm{O}_{2}\) in equilibrium at a specified temperature and pressure. Now the pressure is doubled. (a) Will the equilibrium constant \(K_{P}\) change? (b) Will the number of moles of \(\mathrm{CO}_{2}\), \(\mathrm{CO}\), and \(\mathrm{O}_{2}\) change? How?

Determine the equilibrium constant for the reaction \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) when the reaction occurs at \(100 \mathrm{kPa}\) and \(2000 \mathrm{K} .\) The natural logarithms of the equilibrium constant for the reaction \(\mathrm{C}+2 \mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{4}\) and \(\mathrm{C}+\mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) at \(2000 \mathrm{K}\) are 7.847 and 23.839, respectively.

The equilibrium constant for the \(\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\) \(\mathrm{H}_{2} \mathrm{O}\) reaction at 1 atm and \(1200 \mathrm{K}\) is \(K_{P \cdot}\) Use this information to determine the equilibrium constant for the following reactions: \((a)\) at 1 atm \(\quad \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) \((b)\) at \(7 \mathrm{atm} \quad \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) \((c)\) at \(1 \mathrm{atm} \quad 3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\) \((d)\) at 12 atm \(\quad 3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\)

Methane gas is burned with 30 percent excess air. This fuel enters a steady flow combustor at \(101 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\), and is mixed with the air. The products of combustion leave this reactor at \(101 \mathrm{kPa}\) and \(1600 \mathrm{K}\). Determine the equilibrium composition of the products of combustion, and the amount of heat released by this combustion, in \(\mathrm{kJ} / \mathrm{kmol}\) methane.

Air (21 percent O \(_{2}, 79\) percent \(\mathrm{N}_{2}\) ) is heated to \(5400 \mathrm{R}\) at a pressure of 1 atm. Determine the equilibrium composition, assuming that only \(\mathrm{O}_{2}, \mathrm{N}_{2}, \mathrm{O},\) and NO are present. Is it realistic to assume that no N will be present in the final equilibrium mixture?

See all solutions

Recommended explanations on Physics Textbooks

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