During adsorption which of the following is positive?
A. $\Delta G$
B. $\Delta G + T\Delta S$
C. $T\Delta S$
D. None of these
Answer
597.6k+ views
Hint:Adsorption is a spontaneous process. In Thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
Complete step by step answer:
To understand this let's briefly look at entropy. Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work because work is obtained from ordered molecular motions the amount of entropy is also a major of the molecular disorder or randomness of a system.
If the randomness of the system increases then the value of Entropy increases $(\Delta S > 0)$ and if the randomness of the system decreases then the value of Entropy decreases$(\Delta S < 0)$.
Let's say that a gas is being adsorbed on the solid. Now, randomness is high in the gas phase. When the gas molecules are adsorbed on the surface of the solid, movement of molecules becomes restricted. Thus there is a decrease in randomness. Hence, a decrease in entropy$(\Delta S < 0)$.
Now let's have a look at the impact this has on thermodynamics.
We have a popular thermodynamics relation
$\Delta G = \Delta H - T\Delta S$
Adsorption is accompanied by a decrease in the free energy of the system as it is a spontaneous process. When$\Delta G$ becomes zero, the adsorption equilibrium is established. Since it is an exothermic process$\Delta H$ is negative. Since the adhering of gas molecules to the surface lowers the randomness,$\Delta S$ is negative.
Hence, none of these is the correct option.
Note:If a reaction is favourable for enthalpy $(\Delta {H^0} < 0)$and entropy$(\Delta {S^0} > 0)$, then reaction will be spontaneous $(\Delta {G^0} < 0)$ at any temperature. If a reaction is unfavourable for enthalpy $(\Delta {H^0} > 0)$ and entropy $(\Delta {S^0} < 0)$ then the reaction will be non-spontaneous$(\Delta {G^0} > 0)$ at any temperature. If a reaction is favourable for only one of either entropy or enthalpy, the standard state free energy equation must be used to determine whether the reaction is spontaneous or not.
Complete step by step answer:
To understand this let's briefly look at entropy. Entropy, the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work because work is obtained from ordered molecular motions the amount of entropy is also a major of the molecular disorder or randomness of a system.
If the randomness of the system increases then the value of Entropy increases $(\Delta S > 0)$ and if the randomness of the system decreases then the value of Entropy decreases$(\Delta S < 0)$.
Let's say that a gas is being adsorbed on the solid. Now, randomness is high in the gas phase. When the gas molecules are adsorbed on the surface of the solid, movement of molecules becomes restricted. Thus there is a decrease in randomness. Hence, a decrease in entropy$(\Delta S < 0)$.
Now let's have a look at the impact this has on thermodynamics.
We have a popular thermodynamics relation
$\Delta G = \Delta H - T\Delta S$
Adsorption is accompanied by a decrease in the free energy of the system as it is a spontaneous process. When$\Delta G$ becomes zero, the adsorption equilibrium is established. Since it is an exothermic process$\Delta H$ is negative. Since the adhering of gas molecules to the surface lowers the randomness,$\Delta S$ is negative.
Hence, none of these is the correct option.
Note:If a reaction is favourable for enthalpy $(\Delta {H^0} < 0)$and entropy$(\Delta {S^0} > 0)$, then reaction will be spontaneous $(\Delta {G^0} < 0)$ at any temperature. If a reaction is unfavourable for enthalpy $(\Delta {H^0} > 0)$ and entropy $(\Delta {S^0} < 0)$ then the reaction will be non-spontaneous$(\Delta {G^0} > 0)$ at any temperature. If a reaction is favourable for only one of either entropy or enthalpy, the standard state free energy equation must be used to determine whether the reaction is spontaneous or not.
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