You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibbs Free Energy & Spontaneity

For the proce...

Question

For the process
$H_{2} O (l) ⇌ H_{2} O (g) (1 b a r, 373 K),$ the correct set of thermodynamic parameters is

△ G = 0, △ S = + v e

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

△ G = 0, △ S = - v e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

△ G = + v e, △ S = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

△ G = - v e, △ S = + v e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $△ G = 0, △ S = + v e$
At $373 K, H_{2} O (l)$ is in equilibrium with $H_{2} O (g)$ , and for an equilibrium process we know change in gibbs free energy is zero,
therefore $△ G = 0$ .

As liquid phase is changes to gaseous phase sp molecules entropy increases therefore $△ S = + i v e$ .

Suggest Corrections

Similar questions

H_{2} O (l) ⇌ H_{2} O (g) (1 b a r, 373 K),

H_{2} O (l) ⇌ H_{2} O (g) (1 a t m, 373 K),

H_{2} O (l) ⇌ H_{2} O (g) (1 b a r, 373 K),

For $H_{2} O (l) (1 b a r, 373 K)$ being transformed to $H_{2} O (g) (1 b a r, 373 K)$ , the correct set of thermodynamic parameters is:

H_{2} O (l)

1

373

K

\to H_{2} O (g)

1

373 K

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program