A) \[\Delta G<O;\,\,\Delta S>O;\,\Delta H<O\]
B) \[\Delta G<O;\,\,\Delta S<O;\,\Delta H<O\]
C) \[\Delta G>O;\,\,\Delta S>O;\,\Delta H<O\]
D) \[\Delta G<O;\,\,\Delta S<O;\,\Delta H>O\]
Correct Answer: B
Solution :
Adsorption is an exothermic process and hence \[\Delta H\] is \[-ve\] for adsorption. On the other hand the molecules of the adsorbate are held on the surface of the adsorbant and therefore they have lesser tendency to move freely. In other words entropy decreases i.e. \[\Delta S\] is \[-ve\] According to Gibbs Helmholtz equation \[\Delta G=\Delta H-T.\Delta S\] Thus for the process of adsorption to occur \[\Delta G\]must be negative. Hence, for adsorption \[\Delta G<O;\] \[\Delta S<O;\] s\[\Delta H<O\]
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