Q50 / 180 · KCET 2012 · Chemistry

During the adsorption of a gas on the surface of a solid, which of the following is true?

  • A
    $\Delta \mathrm{G}>0, \Delta \mathrm{H} < 0, \Delta \mathrm{S} < 0$
  • Correct answer
    B
    $\Delta \mathrm{G} < 0, \Delta \mathrm{H} < 0, \Delta \mathrm{S} < 0$
  • C
    $\Delta \mathrm{G} < 0, \Delta \mathrm{H} < 0, \Delta \mathrm{S}>0$
  • D
    $\Delta \mathrm{G} < 0, \Delta \mathrm{H}>0, \Delta \mathrm{S} < 0$

Solution

During the adsorption of a gas on the surface of a solid there is decrease in surface energy, i.e., it is an exothermic process. Therefore, $\Delta \mathrm{H} < 0$. Again when a gas is adsorbed, the freedom of movement of its molecules becomes restricted. This leads to decrease in the entropy of the gas after adsorption. Therefore, $\Delta \mathrm{S} < 0$. From equation, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}, \Delta \mathrm{G}$ can be negative if process is spontaneous. Since, adsorption process is a spontaneous process, therefore, $\Delta \mathrm{G} < 0$.

Hence, the correct answer is option: 'B'.

Relevant Questions