- reaction A: Your answer is incorrect.
- reaction B: Your answer is incorrect.
- reaction C: Your answer Is Incorrect.

Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy [tex]$\Delta H$[/tex] and reaction entropy [tex]$\Delta S$[/tex].
Note: if you have not been given enough information to decide a sign, select the "unknown" option.

| reaction | observations | conclusions |
|---|---|---|
| A | The reverse of this reaction is always spontaneous but proceeds slower at temperatures below [tex]$140 .^{\circ} C$[/tex]. | [tex]$\Delta H$[/tex] is negative [tex]$\Delta S$[/tex] is |
| B | This reaction is spontaneous only below [tex]$45 .^{\circ} C$[/tex]. | [tex]$\Delta H$[/tex] is [tex]$\Delta S$[/tex] is |
| C | This reaction is slower below [tex]$-3 .^{\circ} C$[/tex] than above. | [tex]$\Delta H$[/tex] is [tex]$\Delta S$[/tex] is |

Reaction A: 0"> Δ H > 0 , Δ S < 0 ; Reaction B: Δ H < 0 , Δ S < 0 ; Reaction C: Δ H is unknown, Δ S is unknown.

Explanation

Understanding the Gibbs Free Energy Equation Let's analyze each reaction to determine the signs of Δ H and Δ S . We'll use the Gibbs free energy equation: Δ G = Δ H − T Δ S where:


Δ G is the Gibbs free energy (negative for spontaneous reactions, positive for non-spontaneous reactions, and zero at equilibrium).
Δ H is the enthalpy change (heat absorbed or released during the reaction).
T is the temperature in Kelvin.
Δ S is the entropy change (change in disorder of the system).


Analyzing Reaction A Reaction A: The reverse reaction is always spontaneous. This means Δ G re v erse ​ < 0 , so 0"> Δ G > 0 for the forward reaction at all temperatures. Thus, the forward reaction is non-spontaneous at all temperatures. 0"> Δ G = Δ H − T Δ S > 0 The reaction proceeds slower at temperatures below 140°C. Since the reverse reaction is spontaneous at all temperatures, increasing the temperature makes the forward reaction less unfavorable, suggesting the forward reaction is endothermic. Therefore, 0"> Δ H > 0 .
If 0"> Δ H > 0 , then − T Δ S must be positive to make Δ G positive (since 0"> Δ G = Δ H − T Δ S > 0 ). This means Δ S < 0 .

Analyzing Reaction B Reaction B: This reaction is spontaneous only below 45°C. This means Δ G < 0 when T < 45° C and 0"> Δ G > 0 when 45°C"> T > 45° C . At T = 45° C (318.15 K), Δ G = 0 .
Therefore, 0 = Δ H − T Δ S Δ H = T Δ S Since the reaction is spontaneous at low temperatures, it must be exothermic ( Δ H < 0 ) and have a negative entropy change ( Δ S < 0 ). If both Δ H and Δ S were positive, the reaction would be spontaneous at high temperatures.

Analyzing Reaction C Reaction C: The reaction is slower below -3°C than above. This suggests that increasing the temperature favors the reaction. However, this information alone is not sufficient to determine the signs of Δ H and Δ S . We cannot determine whether the reaction is spontaneous or non-spontaneous at any temperature. Therefore, Δ H is unknown and Δ S is unknown.

Stating the Final Answer Final Conclusions: Reaction A: 0"> Δ H > 0 (positive), Δ S < 0 (negative) Reaction B: Δ H < 0 (negative), Δ S < 0 (negative) Reaction C: Δ H is unknown, Δ S is unknown


Examples
Understanding the spontaneity of chemical reactions is crucial in many real-world applications. For example, in designing industrial processes, chemists need to know whether a reaction will proceed spontaneously under certain conditions. Similarly, in environmental science, understanding the thermodynamics of reactions helps predict the fate of pollutants in the environment. In biology, enzyme-catalyzed reactions must be thermodynamically favorable to occur in living organisms. By controlling temperature and other conditions, we can influence the spontaneity and rate of reactions to achieve desired outcomes.

Answered by GinnyAnswer | 2025-07-06

For Reaction A, 0"> Δ H > 0 and Δ S < 0 ; for Reaction B, Δ H < 0 and Δ S < 0 ; for Reaction C, both signs are unknown.
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Answered by Anonymous | 2025-07-18