For Gas $⇌$ Solid equilibrium; which one is not correct?

Δ G = 0

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

Δ H = T Δ S

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

Δ H = - v e

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

Δ S = + v e

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

Open in App

Solution

The correct option is D $Δ S = + v e$
For Gas $⇌$ Solid equilibrium, $Δ S = + v e$ .
At equilibrium
$Δ G = 0$
$∴ Δ H = T Δ S$ Also, $Δ H = - v e$ and $Δ S = - v e$

Suggest Corrections

Similar questions

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

Δ S

Δ H

Δ S

Δ G

Δ G = Δ H - T Δ S

For $H_{2} O (l) (1 b a r, 373 K) \to H_{2} O (g) (1 b a r, 373 K)$ .The correct set of thermodynamic parameter is

H_{2} O (l) (1 bar, 373 K) ⇌ H_{2} O (g) (1 bar, 373 K)

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

Join BYJU'S Learning Program