Q. During the adsorption of krypton on activated charcoal at low temperature
A) $\mathrm{\Delta}H > 0\ and\mathrm{\Delta}\ S < 0$
B) $\mathrm{\Delta}H < 0\ and\ \mathrm{\Delta}\ S < 0$
C) $\mathrm{\Delta}H > 0\ and\ \mathrm{\Delta}\ S > 0$
D) $\mathrm{\Delta}H < 0\ and\ \mathrm{\Delta}S > 0$
Explanation
(b)
Adsorption is an exothermic process, thus$\mathrm{\Delta}H$ is negative (i.e., $\mathrm{\Delta}H$<0). Moreover, adsorption results in more ordered arrangement of molecules, thus entropy decreases (i.e.,$\mathrm{\Delta}S\ < 0$).
$\mathrm{\Delta}\ G = \mathrm{\Delta}H - T\mathrm{\Delta}S$
Hence, low temperature favours the reaction.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.