Respuesta :
The entropy of the gaseous substance is more in comparison to the liquid or solid form of the same substance at same temperature and pressure conditions.
More is the number of particles more will be the entropy of the system.
In the first chemical reaction, CH₄ (g) → CH₄ (l)
Since, the reactants are in gaseous phase and the products are in liquid phase, the entropy is decreasing during the reaction.
In the second chemical reaction,
N₂H₄ (g) → N₂ (g) + 2H₂ (g)
Since, one mole of the reactant in gaseous phase is decomposing to produce three moles of product in gaseous phase, the entropy is increasing during the reaction.
In the third chemical reaction, H₂O (s) → H₂O (g)
Since, the reactant is in solid phase and product are in gaseous phase. The entropy is increasing during the reaction.
Thus, the entropy is increasing in the 2nd and 3rd reactions.
Entropy is the unavailable energy of the system that cannot convert the mechanical work. Entropy is increasing in the 2) [tex]\rm N_{2}H_{4} (g) \rightarrow N_{2} (g) + 2H_{2} (g)[/tex] and 3) [tex]\rm H_{2}O (s) \rightarrow H_{2}O (g).[/tex]
What is entropy?
Entropy is the thermal energy per unit of temperature and is not available for the system to use. The entropy of the gaseous substance is more compared to the solid or the liquid substance.
For the first reaction, [tex]\rm CH_{4} (g) \rightarrow CH_{4} (l)[/tex], the entropy will not be positive and instead will be decreasing as the reactant is gaseous and the product is in a liquid state.
In the second reaction, [tex]\rm N_{2}H_{4} (g) \rightarrow N_{2} (g) + 2H_{2} (g)[/tex], the reactants and the products all are in the gaseous state and hence the entropy will increase.
In the third reaction, [tex]\rm H_{2}O (s) \rightarrow H_{2}O (g)[/tex], the entropy will increase as the reactant is solid but the product is in the gaseous state.
Therefore, second and third reactions will have increased entropy.
Learn more about entropy here:
https://brainly.com/question/14038375
