Question

For adsorption of a gas:

• A. $\Delta H<0,\Delta S<0$
• B. $\Delta H<0,\Delta S>0$
• C. $\Delta H>0,\Delta S>0$
• D. $\Delta H=0,\Delta S>0$

Answer 1

The correct option is A

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

Explanation for the correct options:

A) $\mathit{\Delta }\mathit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{\Delta }\mathit{S}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}$

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) $\mathbf{}\mathit{\Delta }\mathit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{\Delta }\mathit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$

As the entropy of adsorbed gas is positive $\Delta S>0$ ,the gas will not be adsorbed on the surface.

C) $\mathit{\Delta }\mathit{H}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{\Delta }\mathit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$

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) $\mathit{\Delta }\mathit{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{\Delta }\mathit{S}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$

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, $\mathit{\Delta }\mathit{H}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}\mathbf{,}\mathbf{}\mathit{\Delta }\mathit{S}\mathbf{}\mathbf{<}\mathbf{}\mathbf{0}$