    Question

# For adsorption of a gas:

A

$\Delta H<0,\Delta S<0$

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B

$\Delta H<0,\Delta S>0$

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C

$\Delta H>0,\Delta S>0$

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D

$\Delta H=0,\Delta S>0$

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Solution

## The correct option is A $\Delta H<0,\Delta S<0$Explanation for the correct options:A) $\mathbit{\Delta }\mathbit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbit{\Delta }\mathbit{S}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}$According to Thermodynamics, $\Delta G=\Delta H-T\Delta S$$\Delta H<0$, As the residual forces on the surface decrease, this indicates that there is a decrease in surface energy which will appear in form of heat. Thus, adsorption is an exothermic process. $\Delta S<0$ , when the gas particles are adsorbed on the surface, the movement of the particles is very difficult as the particles stick with the surface of the adsorbent. This leads to a decrease in the entropy of the gas particles. So, $\Delta S<0$ is negative.Explanation for the incorrect options: B) $\mathbf{}\mathbit{\Delta }\mathbit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbit{\Delta }\mathbit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$As the entropy of adsorbed gas is positive $\Delta S>0$ ,the gas will not be adsorbed on the surface.C) $\mathbit{\Delta }\mathbit{H}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbit{\Delta }\mathbit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$Adsorption is an exothermic reaction as $\Delta H>0$ the reaction becomes endothermic. As the entropy of adsorbed gas is positive $\Delta S>0$ , the gas will not be adsorbed to the surface. Thus, the adsorption process will not take place.D) $\mathbit{\Delta }\mathbit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbit{\Delta }\mathbit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$In this $\Delta H=0$, so heat will be absorbed on the surface. As the entropy of adsorbed gas is positive $\Delta S>0$,the gas will not be adsorbed on the surface.Hence option (A) is correct, $\mathbit{\Delta }\mathbit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathbit{\Delta }\mathbit{S}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}$  Suggest Corrections  0      Similar questions  Explore more