Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask question.(5)

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

ITtutoria

ITtutoria Logo ITtutoria Logo

ITtutoria Navigation

  • Python
  • Java
  • Reactjs
  • JavaScript
  • R
  • PySpark
  • MYSQL
  • Pandas
  • QA
  • C++
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Python
  • Science
  • Java
  • JavaScript
  • Reactjs
  • Nodejs
  • Tools
  • QA
Home/ Questions/When the oxide of generic ametl M is heated at 25.0 ∘C, a negligible amount of M is produced. MO2(s) ⇌ M(s)+O2(g)
Next
Answered
qhtutoria
  • 22
qhtutoria
Asked: April 10, 20222022-04-10T08:10:52+00:00 2022-04-10T08:10:52+00:00In: Chemistry

When the oxide of generic ametl M is heated at 25.0 ∘C, a negligible amount of M is produced. MO2(s) ⇌ M(s)+O2(g)

  • 22

. Advertisement .

..3..

. Advertisement .

..4..

......... ADVERTISEMENT .........

..8..

When the oxide of generic metal M is heated at 25.0 ∘C , a negligible amount of M is produced.

MO2(s) ⇌ M(s) + O2(g)      ΔG∘=290.5 kJmol

When this reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous.

What is the chemical equation of this coupled process? Show that the reaction is in equilibrium. Include physical states and represent graphite as C(s) .

♦ Relevant knowledge
Spontaneous Reaction

The Gibbs free energy variation for spontaneous reactions is always less than zero . This can be measured using the Gibbs-free energy the formation of the species within the reactions.

......... ADVERTISEMENT .........

..8..

negligibleoxide of generic ametlproduced
  • 1 1 Answer
  • 42 Views
  • 0 Followers
  • 0
Answer
Share
  • Facebook
  • Report

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Best Answer
    tonytutoria
    2022-04-10T08:10:54+00:00Added an answer on April 10, 2022 at 8:10 am

    We are given the following reaction:

    MO2(s) ⇌ M(s) + O2(g)  ΔG = 290.5 kJ/mol

    The reaction mentioned above is connected with the transformation of graphite into carbon dioxide. The reaction then turns spontaneous.

    C(s) + O2(g)→ CO2(g)   ΔG = −394.5 kJ/mol

    Chemical reaction:

    MO2(s) + C(s)→M(s) + CO2(g) 

    ΔG = 290.5 kJ/mol − 394.5 kJ/mol = −104.0 kJ/mol

    The Gibbs free energy shift of the reaction is negative once the oxide of the generic metal (M) is joined with graphite. Thus, the reaction is spontaneous.

    The equilibrium constant of the coupled reaction may be calculated by:

    ΔG = −RTln(K)

    ln(K) = ΔG/RT

    ln(K) = (−104.0 kJ/mol) / (−8.314 J/K⋅mol × 298 K)

    ln(K) = 42.0

    K = 1.77×1018

    • 24
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

Sidebar

Ask A Question
  • How to Split String by space in C++
  • How To Convert A Pandas DataFrame Column To A List
  • How to Replace Multiple Characters in A String in Python?
  • How To Remove Special Characters From String Python

Explore

  • Home
  • Tutorial

Footer

ITtutoria

ITtutoria

This website is user friendly and will facilitate transferring knowledge. It would be useful for a self-initiated learning process.

@ ITTutoria Co Ltd.

Tutorial

  • Home
  • Python
  • Science
  • Java
  • JavaScript
  • Reactjs
  • Nodejs
  • Tools
  • QA

Legal Stuff

  • About Us
  • Terms of Use
  • Privacy Policy
  • Contact Us

DMCA.com Protection Status

Help

  • Knowledge Base
  • Support

Follow

© 2022 Ittutoria. All Rights Reserved.

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.