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\(2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s)\) The reaction above is not thermodynamically favored under standard conditions, but it becomes thermodynamically favored as the temperature decreases toward absolute zero. Which of the following is true at standard conditions? (A) \(\Delta S\) and \(\Delta H\) are both negative. (B) \(\Delta S\) and \(\Delta H\) are both positive. (C) \(\Delta S\) is negative, and \(\Delta H\) is positive. (D) \(\Delta S\) is positive, and \(\Delta H\) is negative.

Short Answer

Expert verified
The correct answer is (A): under standard conditions, both \(\Delta S\) (entropy change) and \(\Delta H\) (enthalpy change) are negative.

Step by step solution

01

Understanding entropy and enthalpy

Entropy, represented as \(\Delta S\), refers to the disorder or randomness of a system. A positive \(\Delta S\) means the reaction leads to a more disordered system, a negative \(\Delta S\) means the reaction leads to a less disordered system. Enthalpy, represented as \(\Delta H\), is related to the heat energy in a reaction. A positive \(\Delta H\) indicates an endothermic reaction (heat is absorbed) while a negative \(\Delta H\) indicates an exothermic reaction (heat is released).
02

Analyzing the given reaction

The given reaction becomes more thermodynamically favored as temperature goes towards absolute zero i.e., it becomes more spontaneous at lower temperatures.
03

Relating reaction spontaneity with entropy and enthalpy

By the Second Law of Thermodynamics, spontaneous reactions tend to increase the total entropy of the universe (or a system and its surroundings). However, the reaction is becoming more spontaneous at lower temperatures, implying that the entropy of the reaction \(\Delta S\), is decreasing, hence, negative. As for enthalpy, the reaction is more favored at lower temperatures, this is usually the case for exothermic reactions which release heat, hence, \(\Delta H\) is negative.
04

Picking the correct answer

From the analysis in the previous steps, it can be inferred that under standard conditions for this reaction, the entropy \(\Delta S\) is negative (as it's becoming more spontaneous as temperature decreases) and enthalpy \(\Delta H\) is also negative (as it is more favorable at lower temperatures, typical of exothermic reactions). The correct answer is therefore, (A) \(\Delta S\) and \(\Delta H\) are both negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy
Entropy, symbolized as \(\Delta S\), is a fundamental concept in thermodynamics representing the level of disorder or randomness within a system. In chemical reactions, entropy helps determine how the arrangement of particles changes. When a reaction results in a more disordered system, \(\Delta S\) is positive, which usually suggests that the reaction products have more freedom of motion or a higher number of different configurations. Conversely, a negative \(\Delta S\) suggests the system becomes more ordered. This often happens when gases transform into solids or liquids, as there is less movement and fewer possible arrangements for the molecules. In the given reaction, the decrease in entropy (negative \(\Delta S\)) indicates a likely change from gases to solid, which is typical as the system loses chaos and gains order.
Enthalpy
Enthalpy, symbolized by \(\Delta H\), measures the heat content in a reaction and indicates whether a reaction absorbs or releases heat. When a reaction has a negative \(\Delta H\), it is exothermic, meaning it releases energy to the surroundings in the form of heat. This is often seen in combustion reactions or reactions forming stronger bonds. A positive \(\Delta H\) corresponds to endothermic reactions, which absorb energy, often seen during bond breaking or phase transitions requiring energy input, like melting or evaporation. In our exercise, the negative enthalpy of the reaction implies it's exothermic: as the temperature approaches absolute zero, this heat-release characteristic enhances the reaction's favorability.
Thermodynamic Favorability
Thermodynamic favorability describes how likely a reaction is to proceed under certain conditions and can be predicted using the Gibbs free energy equation: \[ \Delta G = \Delta H - T\Delta S \]Where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the enthalpy change, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the entropy change. A negative \(\Delta G\) signifies a spontaneous or thermodynamically favorable reaction.In the reaction provided, both \(\Delta H\) and \(\Delta S\) are negative. The reaction becomes more thermodynamically favorable at lower temperatures because the effect of \(T\Delta S\) diminishes as temperature decreases, making \(\Delta G\) more negative. This highlights the crucial balance between enthalpy and entropy in determining reaction spontaneity.
AP Chemistry
In AP Chemistry, understanding the fundamental concepts of entropy, enthalpy, and thermodynamic favorability is important. These concepts are essential for predicting the spontaneity of reactions and understanding how energy changes drive reactions. AP Chemistry students are expected to develop skills to analyze reactions under various conditions and to use equations like the Gibbs free energy equation competently. Moreover, standard conditions often are part of problems presented, where the knowledge of entropy and enthalpy allows students to infer the spontaneity of reactions, helping them to choose correct pathways in solving thermodynamic problems. Overall, these foundational ideas are critical within the AP Chemistry curriculum as they lay the groundwork for more advanced topics, ensuring a solid understanding of how reactions occur and are manipulated in real-world applications.

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