NTA Abhyas NEET2020ChemistryChemical EquilibriumPractice
For the process H 2 O (l) (1 bar, 373 K) → H 2 O (g) (1 bar, 373 K), the correct set of thermodynamic parameters are
Options
- AΔ G = 0 , Δ S = + v e
- BΔ G = 0 , Δ S = - v e
- CΔ G = + v e , Δ S = 0
- DΔ G = - v e , Δ S = + v e
Correct answer
A. Δ G = 0 , Δ S = + v e
Step-by-step solution
H 2 O ( l ) ( 1 b a r , 373 K ) ⇌ H 2 O ( g ) ( 1 b a r , 373 K ) At 100oC, H 2 O ( l ) has equilibrium with H 2 O ( g ) therefore Δ G = 0 . As liquid molecules are converting into gases molecules therefore Δ S = positive.