Get Answers to all your Questions

header-bg qa

For vaporization of water at 1 atmospheric pressure, the values of \Delta H \ and \ \Delta S are 40.63kJmol^{-1} and 108.8JK^{-1}mol^{-1}, respectivily. The temperature when Gibbs energy change \left ( \Delta G \right ) for this transformation will be zero, is:

  • Option 1)

    293.4K

  • Option 2)

    273.4K

  • Option 3)

    393.4K

  • Option 4)

    373.4K

 

Answers (1)

best_answer

 

Gibb's free energy (? G) -

\Delta G= \Delta H-T \Delta S
 

- wherein

\Delta G= Gibb's free energy

\Delta H= enthalpy of reaction

\Delta S= entropy

T= temperature

 

 According to Gibbs equation \Delta G= \Delta H-T\Delta S

when \Delta u= 0

\Delta H= T\Delta S  or T=\frac{\Delta H}{\Delta S}

\Delta H= 40.63 KJ\: mol^{-1}, \Delta S=108.8 JK^{-1}\:mol^{-1}

\therefore =\frac{\Delta H}{\Delta S}= \frac{40.63\times 10^{3}}{108.8}

= 373.43K


Option 1)

293.4K

Incorrect Option

Option 2)

273.4K

Incorrect Option

Option 3)

393.4K

Incorrect Option

Option 4)

373.4K

Correct Option

Posted by

divya.saini

View full answer

NEET 2024 Most scoring concepts

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