Question

Which of the following statements is correct for the spontaneous adsorption of a gas?
(A) $\Delta S$ is negative and therefore $\Delta H$ should be highly positive
(B) $\Delta S$ is negative and therefore $\Delta H$ should be highly negative
(C) $\Delta S$ is positive and therefore $\Delta H$ should be highly negative
(D) $\Delta S$ is negative and therefore $\Delta H$ should be highly positive

Answer 1

For a spontaneous adsorption of a gas on a solid or liquid surface, we can say that the change in the free energy $(\Delta G)$ is negative. The relation between free energy ,enthalpy and entropy can be given as
$\Delta G = \Delta H - T\Delta S$

Complete answer:
The accumulation of the molecules at the surface rather than the bulk of the solid or liquid is called adsorption. We will see the effect on thermodynamic parameters for the adsorption of a gas.
- We know that during the process of adsorption, there is always a decrease in the residual forces of the surface. So, there is a decrease in the surface energy which appears as heat. Therefore, we can say that adsorption is an exothermic process.
- Now, we know that in the case of spontaneous adsorption of a gas, the free energy ($\Delta G$) of the reaction can be given as
$\Delta G = \Delta H - T\Delta S$
Now, for a spontaneous reaction, the change in free energy ($\Delta G$) needs to be negative.
- Here, free energy change can be negative only if $\Delta H$ is sufficiently negative which we can see in the equation.
- Also, when a gas is adsorbed, its entropy decreases as the freedom of movement becomes restricted. So, for the adsorption of gas, $\Delta S$ is negative.
- Thus, we can conclude that for a spontaneous adsorption of gas, $\Delta S$ is negative and therefore $\Delta H$ should be highly negative.

So, the correct answer is (B).

Note:
Remember that for all the spontaneous processes, the change is free energy $(\Delta G)$ is negative. For all the non-spontaneous processes, the change in free energy $(\Delta G)$ is positive.