Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • Can someone explain - Thermodynamics - NEET

For a given reaction, \Delta H=35.5kJmol^{-1}\ and\ \Delta S=83.6JK^{-1}mol^{-1}. The reaction is spontaneous at :(Assume that \Delta H\and\ \Delta S do not vary with tempearature)

  • Option 1)

    T>425K

  • Option 2)

    All Temperatures

  • Option 3)

    T>298K

  • Option 4)

    T<425K

 

Answers (1)

best_answer

 

Spontanous process -

\Delta G= \Delta H-\Delta T

 

- wherein

For spontanouse process \Delta G must be negative.

 

 For a  Spontaneous reaction 

\Delta G< 0 ie  \Delta H -T\Delta S< 0

T> \frac{\Delta H}{\Delta S}

T> \frac{35.5\times 1000}{83.6}=424.6 =425\:K

T>425K


Option 1)

T>425K

Correct

Option 2)

All Temperatures

Incorrect

Option 3)

T>298K

Incorrect

Option 4)

T<425K

Incorrect

Posted by

Plabita

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

NEET UG 2025 Registration LIVE: NTA NEET form expected today at neet.nta.nic.in; exam date, eligibility
anam-khan

NEET UG registration 2025 form expected today at neet.nta.nic.in; APAAR ID not mandatory
anam-khan

NEET UG application form 2025 expected on January 31 at neet.nta.nic.in: Reports

Latest Question