Adsorption of a gas on solid metal surface is spontaneous and exothermic, then :
ΔH increases
ΔS increases
ΔG increases
ΔS decreases
A
ΔH increases
B
ΔS increases
C
ΔG increases
D
ΔS decreases
Open in App
Solution
Verified by Toppr
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then ΔS decreases.
Since for spontaneous and exothermic process, ΔG=−ve,ΔH=−ve at all temperatures.
Therefore, from ΔG=ΔH−TΔS,ΔS should be −ve.
Was this answer helpful?
1
Similar Questions
Q1
Assertion :For adsorption ΔG,ΔH,ΔSall have negative values. Reason: Adsorption is a spontaneous exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate.
View Solution
Q2
Assertion :For adsorption, ΔG,ΔH and ΔS all have negative values. Reason: Adsorption is a spontaneous exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate.
View Solution
Q3
Adsorption of a gas on solid metal surface is spontaneous and exothermic, then :
View Solution
Q4
Assertion :For adsorption ΔG,ΔS and ΔH all have negative values. Reason: Adsorption is a spontaneous process accompanied by decrease in randomness.
View Solution
Q5
In a process, temperature of 2 mole of Ar gas is increased by 1∘Cthen.