Q10 / 180 · KCET 2010 · Chemistry

During the adsorption of krypton on activated charcoal at low temperature

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

Solution

Adsorption is an exothermic process, thus $\Delta H$ is negative (ie, $\Delta \mathrm{H} < 0$ ). Moreover, adsorption results in more ordered arrangements of molecules, thus entropy decreases (ie, $\Delta \mathrm{S} < 0$ ). $$ \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} $$ Hence, low temperature favours the reaction.

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

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