Question

Which of the following statements is correct for the spontaneous adsorption of a gas?

• A. △S is negative and therefore, △H should be highly negative
• B. △S is positive and, therefore, △H should be negative
• C. △S is positive and, therefore, △H should also be highly positive
• D. △S is negative and, therefore, △H should be highly positive

Answer 1

The Correct Option is A

To determine the correct statement for the spontaneous adsorption of a gas, we must analyze the thermodynamic aspects of adsorption.

Adsorption is a process in which gas molecules accumulate on the surface of a solid. The spontaneity of a process is determined by the Gibbs free energy change, \(\Delta G\). For a spontaneous process, \(\Delta G\) must be negative.

The Gibbs free energy change is related to enthalpy change \(\Delta H\) and entropy change \(\Delta S\) by the equation:

\(\Delta G = \Delta H - T\Delta S\),

where \(T\) is the temperature in Kelvin.

In the case of adsorption:

• The entropy change \(\Delta S\) is generally negative because the disorder decreases when gas molecules adhere to a solid surface.
• To ensure that \(\Delta G\) is negative (for the process to be spontaneous), \(\Delta H\) must be significantly negative. This compensates for the negative (- T\Delta S) term.

Therefore, the correct statement is: △S is negative and therefore, △H should be highly negative.

Let's briefly discuss why other options are incorrect:

• △S is positive and, therefore, △H should be negative: This is not typical for adsorption processes, as the entropy usually decreases.
• △S is positive and, therefore, △H should also be highly positive: This contradicts the common requirement of a negative \(\Delta H\) for the spontaneity of adsorption.
• △S is negative and, therefore, △H should be highly positive: A positive \(\Delta H\) does not support spontaneity when \(\Delta S\) is negative.