The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed as adsorption. There are two types of adsorption. In physisorption, attractive forces are mainly van der Waals while in chemical adsorption adsorbate is held with chemical bonds adsorbent. Adsorption increases with increase in pressure and decrease as temperature is increased.

Adsorption is spontaneous because:

\(\Delta H\) is \(- ve\) and is more than \(T\Delta S\)

The correct answer is (3) \(\Delta H\) is \(- ve\) and is more than \(T\Delta S\).
In chemistry and thermodynamics, the spontaneity of a process is determined by the change in Gibbs free energy (\(\Delta G\)), which is given by the equation:
\[\Delta G = \Delta H - T\Delta S\]
Where:
- \(\Delta G\) is the change in Gibbs free energy.
- \(\Delta H\) is the change in enthalpy (heat energy).
- \(T\) is the temperature in Kelvin.
- \(\Delta S\) is the change in entropy (a measure of disorder).
Now, let's consider option (3): \(\Delta H\) is negative (\(-\Delta H\)) and is more than \(T\Delta S\).
Given that \(\Delta H\) is negative, it means the process is exothermic. Exothermic processes release heat to the surroundings. When \(\Delta H\) is negative, it contributes a negative value to \(\Delta G\). In other words, \(\Delta H\) promotes spontaneity because it tends to make \(\Delta G\) more negative.
Now, let's consider \(T\Delta S\). The second term in the equation, \(T\Delta S\), represents the entropy contribution. When a process increases the entropy of the system (\(\Delta S\) is positive), it also promotes spontaneity. However, this contribution is scaled by the temperature (\(T\)). At higher temperatures, \(T\Delta S\) becomes more significant.
So, when \(\Delta H\) is negative (exothermic), it tends to make \(\Delta G\) more negative and favors spontaneity. However, for spontaneity to occur, the negative contribution from \(\Delta H\) must be greater (in absolute value) than the positive contribution from \(T\Delta S\). In other words, \(\left|\Delta H\right|\) must be greater than \(T\Delta S\) for the process to be spontaneous.
In summary, option (3) is correct because it correctly states that if \(\Delta H\) is negative (exothermic) and its absolute value (\(\left|\Delta H\right|\)) is greater than \(T\Delta S\), then \(\Delta G\) will be negative, indicating a spontaneous process. This is a fundamental concept in thermodynamics, and it applies to many physical and chemical processes, including adsorption reactions.