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Discuss the different sign combinations of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) that are possible for a process carried out at constant temperature and pressure. For each combination, state whether the process must be spontaneous or not, or whether both situations are possible. Explain.

Short Answer

Expert verified
Spontaneity depends on \( T \), with cases: 1) high \( T \) for both positive, 2) low \( T \) for both negative, 3) non-spontaneous if \( \Delta H > 0, \Delta S < 0 \), 4) always spontaneous if \( \Delta H < 0, \Delta S > 0 \).

Step by step solution

01

Understanding the Parameters

First, understand the terms involved: - \( \Delta H^{\circ} \) is the change in enthalpy, representing the heat absorbed or released during a process at constant pressure. - \( \Delta S^{\circ} \) is the change in entropy, representing the disorder change in the system. - A process is spontaneous if the Gibbs free energy change \( \Delta G < 0 \). At constant temperature and pressure, \( \Delta G = \Delta H - T \Delta S \).
02

Analyzing Different Sign Combinations

We will analyze four possible combinations of \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \): 1) Both are positive, 2) Both are negative, 3) \( \Delta H^{\circ} \) is positive and \( \Delta S^{\circ} \) is negative, 4) \( \Delta H^{\circ} \) is negative and \( \Delta S^{\circ} \) is positive.
03

Case 1: Both Positive

If both \( \Delta H^{\circ} > 0 \) and \( \Delta S^{\circ} > 0 \), the process requires energy input (endothermic) and has increasing disorder. The spontaneity depends on \( T \Delta S^{\circ} \) being greater than \( \Delta H^{\circ} \). The process is spontaneous at high temperatures.
04

Case 2: Both Negative

If \( \Delta H^{\circ} < 0 \) and \( \Delta S^{\circ} < 0 \), the process releases energy (exothermic) but decreases disorder. It is spontaneous at low temperatures because \( |\Delta H^{\circ}| > |T \Delta S^{\circ}| \).
05

Case 3: \( \Delta H^{\circ} \) Positive and \( \Delta S^{\circ} \) Negative

For \( \Delta H^{\circ} > 0 \) and \( \Delta S^{\circ} < 0 \), the process is non-spontaneous at all temperatures because both factors are unfavorable: it requires energy and becomes more ordered, making \( \Delta G \) always positive.
06

Case 4: \( \Delta H^{\circ} \) Negative and \( \Delta S^{\circ} \) Positive

With \( \Delta H^{\circ} < 0 \) and \( \Delta S^{\circ} > 0 \), the process is always spontaneous. Releasing energy and increasing disorder is favorable considering \( \Delta G = \Delta H - T \Delta S < 0 \) for all temperatures.

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \( \Delta H^{\circ} \), represents the heat exchanged in a process occurring at constant pressure. It tells us whether a system absorbs or releases energy. A positive \( \Delta H^{\circ} \) indicates an endothermic process—alluding to energy absorption. Conversely, a negative \( \Delta H^{\circ} \) signifies an exothermic process, which releases energy.
If the process is endothermic, energy is absorbed by the system which can affect the spontaneity depending on other conditions such as temperature and entropy change. If it is exothermic, it spontaneously releases energy, generally making spontaneity more favorable especially when combined with other favorable conditions.
Entropy Change
Entropy change, noted as \( \Delta S^{\circ} \), pertains to the measure of disorder or randomness in a system. When the entropy change is positive, \( \Delta S^{\circ} > 0 \), it means that the disorder is increasing. This is often seen as a favorable condition because nature tends to move towards more disorder.
A negative entropy change, \( \Delta S^{\circ} < 0 \), means the system is becoming more ordered, generally seen as unfavorable for spontaneity. The interplay between enthalpy and entropy greatly determines whether a given process is spontaneous or not at specific temperatures.
Spontaneity Conditions
The spontaneity of a process is linked to the Gibbs Free Energy change, \( \Delta G \). A process is considered spontaneous when \( \Delta G < 0 \). At constant temperature and pressure, Gibbs Free Energy is calculated using the formula: \[ \Delta G = \Delta H - T \Delta S \]
This formula highlights that both the enthalpy and entropy changes significantly impact whether a process will naturally occur. If \( \Delta H^{\circ} \) is negative and \( \Delta S^{\circ} \) is positive, the process is spontaneous because both conditions contribute towards a negative \( \Delta G \). However, if both \( \Delta H^{\circ} \) is positive and \( \Delta S^{\circ} \) is negative, the process is non-spontaneous as both factors drive \( \Delta G \) to be positive.
Thermodynamic Processes
Thermodynamic processes are driven by energy changes and the creation or disintegration of disorder within the system. The direction and extent to which these processes occur depends on the interplay between \( \Delta H^{\circ} \), \( \Delta S^{\circ} \), and temperature.
The formula \( \Delta G = \Delta H - T \Delta S \) underscores how these components come together.
  • If \( \Delta H^{\circ} > 0 \) and \( \Delta S^{\circ} > 0 \), the process may become spontaneous at high temperatures as the term \( T \Delta S^{\circ} \) can dominate.
  • For \( \Delta H^{\circ} < 0 \) and \( \Delta S^{\circ} < 0 \), spontaneity is seen at low temperatures since \( |\Delta H^{\circ}| \) surpasses \( |T \Delta S^{\circ}| \).
  • A positive \( \Delta H^{\circ} \) with negative \( \Delta S^{\circ} \) results in non-spontaneity under all conditions.
  • When \( \Delta H^{\circ} < 0 \) and \( \Delta S^{\circ} > 0 \), the process is spontaneous irrespective of temperature.
The conditions determine the ease with which processes like chemical reactions and phase transitions occur and progress over time.

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Most popular questions from this chapter

The free energy of formation of one mole of compound refers to a particular chemical equation. For each of the following, write that equation. a. \(\mathrm{MgO}(s)\) b. \(\mathrm{COCl}_{2}(g)\) c. \(\mathrm{CF}_{4}(g)\) d. \(\mathrm{PCl}_{5}(g)\)

You run a reaction that has a negative entropy change and is exothermic. Assuming that the entropy and enthalpy do not change with temperature, you could predict that as you increase the temperature: I. the equilibrium shifts to favor the reaction products. II. the reaction becomes more spontaneous. III. \(\Delta G\) for the reaction increases (becomes more positive). a. I only b. II only c. III only d. I and III only e. II and III only

The combustion of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is a spontaneous reaction given by the equation $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ As expected for a combustion, the reaction is exothermic. What is the sign of \(\Delta H^{\circ} ?\) What do you expect for the sign of \(\Delta S^{\circ}\) ? Explain the spontaneity of the reaction in terms of the enthalpy and entropy changes.

How is the concept of coupling of reactions useful in explaining how a nonspontaneous change could be made to occur?

Acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), boils at \(56^{\circ} \mathrm{C}\). The heat of vaporization of acetone at this temperature is \(29.1 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy change when \(1 \mathrm{~mol}\) of liquid acetone vaporizes at \(56^{\circ} \mathrm{C} ?\)

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