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

If \((\mathrm{x} / \mathrm{m})\) is the mass of adsorbate adsorbed per unit mass of adsorbent, \(p\) is the pressure of the adsorbate gas and a and \(\mathrm{b}\) are constants, which of the following represents Langmuir adsorption isotherm? (a) \(\log (\mathrm{x} / \mathrm{m})=\log (\mathrm{a} / \mathrm{b})+(1 / \mathrm{a}) \log \mathrm{p}\) (b) \(\mathrm{x} / \mathrm{m}=\mathrm{b} / \mathrm{a}+1 / \mathrm{ap}\) (c) \(\mathrm{x} / \mathrm{m}=1+\mathrm{bp} / \mathrm{ap}\) (d) \(1 /(x / m)=b / a+1 / a p\)

Short Answer

Expert verified
The answer is (d) \(1/(x/m) = b/a + 1/ap\).

Step by step solution

01

Understand Langmuir Adsorption Isotherm

Langmuir adsorption isotherm is typically given by the equation \[\frac{1}{x/m} = \frac{1}{ab} + \frac{1}{a} \cdot \frac{1}{p}\]. This equation represents the relationship between the mass of adsorbate per unit mass of adsorbent and the pressure of adsorbate gas.
02

Analyze Provided Options

We must compare the standard Langmuir isotherm form \( \frac{1}{x/m} = \frac{1}{ab} + \frac{1}{a} \cdot \frac{1}{p} \) with each given option: (a) \( \log (x/m) = \log (a/b) + (1/a) \log p \) (b) \( x/m = b/a + 1/ap \) (c) \( x/m = 1 + bp/ap \) (d) \( 1/(x/m) = b/a + 1/ap \).
03

Match Equations

Compare each option to the Langmuir adsorption isotherm form:- (a) uses logarithms, not matching the Langmuir form.- (b) and (c) represent equations in the form of \(x/m\), not \(1/(x/m)\), hence don't match.- (d) is \(1/(x/m) = b/a + 1/ap\), which fits the standard Langmuir form after recognizing that \(\frac{1}{x/m}\) is equal to \(\frac{1}{ab} + \frac{1}{a} \cdot \frac{1}{p}\).
04

Conclusion

Given the analysis, option (d) \(1/(x/m) = b/a + 1/ap\) correctly matches the Langmuir adsorption isotherm.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Adsorption
Adsorption is a fundamental concept in surface chemistry, where atoms, ions, or molecules from a gas, liquid, or dissolved solid stick to a surface. It differs from absorption, where a substance integrates into another medium or liquid.
Adsorption involves creating a film of the adsorbate on the surface of the adsorbent, which forms due to attractive forces like van der Waals forces or chemical bonds.
  • The process is usually reversible, meaning the adsorbate can be detached, which is called desorption.
  • Common examples of adsorption are seen in air filters, where pollutants stick to the filter surface, and in dehumidifiers, where moisture adheres to absorbent materials.
  • Key factors influencing adsorption include surface area, temperature, and pressure.
Adsorption generally decreases with increasing temperature because the adsorbate has higher kinetic energy, leading to desorption.
Adsorbate
The adsorbate is the substance that adheres or bonds to the surface of another material, known as the adsorbent. Understanding the nature of the adsorbate is essential to evaluate how effective an adsorption process will be.
Adsorbates can be gases, liquids, or dissolved solids.
  • The interaction between adsorbate and adsorbent is critical, as stronger interactions result in better adsorption capacity.
  • Common adsorbates include gases like oxygen, nitrogen, and various organic compounds, or contaminants in water treatment processes.
  • Properties, such as polarity and molecular size, play significant roles in determining how well a substance will act as an adsorbate. Stronger attractions usually occur with smaller, polar molecules.
Understanding the properties of both the adsorbate and adsorbent can optimize industrial applications, such as using activated charcoal for removing toxins.
Pressure Equilibrium
Pressure equilibrium is a key concept when studying adsorption processes, particularly in gases. It refers to the state where the rate of adsorption equals the rate of desorption, leading to a stable concentration of adsorbate on the adsorbent's surface.
This balance is crucial in the Langmuir Adsorption Isotherm, where pressure is a driving force for adsorption.
  • Equilibrium is reached quicker at higher pressures because increased pressure generally results in more particles striking the adsorbent surface.
  • Modeling the relationship between pressure and adsorption helps predict how substances will behave under different conditions.
  • It is crucial in applications like catalyst design, where maintaining a certain adsorption level is necessary for reaction efficiency.
Pressure equilibrium helps determine the optimal pressure conditions for maximizing adsorption while minimizing energy expenditure.
Surface Chemistry
Surface chemistry is the study of chemical processes at interfaces, primarily involving solid surfaces. It forms the basis for understanding adsorption and other surface-related phenomena.
In surface chemistry, the properties of the surfaces and their interactions with adsorbates are vital.
  • This area of chemistry helps in developing materials like catalysts and adsorbents by tailoring surface properties to enhance performance.
  • It involves techniques like spectroscopy to study the chemical composition and dynamics at surfaces.
  • Concepts such as surface energy, surface area, and surface tension are central to this field and affect how substances are adsorbed.
Surface chemistry is critically important in industries, from catalysis for efficient chemical reactions to designing better adsorptive materials for environmental cleanup.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free