Problem 117
Question
Which of the following conditions may lead to a nonspontaneous change? (a) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=+\mathrm{ve}\) (b) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=-\mathrm{ve}\) (c) \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are both \(+\mathrm{ve}\) (d) \(\Delta H=+v e ; \Delta S=-v e\)
Step-by-Step Solution
Verified Answer
Condition (d) leads to a nonspontaneous change.
1Step 1: Understanding Spontaneity
A change (or reaction) is considered spontaneous if it occurs naturally without being driven by some external force. The spontaneity of a process is determined by the Gibbs free energy change, represented as \( \Delta G \). Spontaneity is predicted using the equation: \( \Delta G = \Delta H - T\Delta S \), where \( \Delta H \) is the change in enthalpy, \( \Delta S \) is the change in entropy, and \( T \) is the temperature in Kelvin.
2Step 2: Analyzing the Conditions
To determine if a process is spontaneous, \( \Delta G \) must be negative (\( \Delta G < 0 \)). For each set of \( \Delta H \) and \( \Delta S \):(a) \( \Delta H = -\text{ve} \), \( \Delta S = +\text{ve} \), would typically lead to negative \( \Delta G \), hence spontaneous.(b) \( \Delta H = -\text{ve} \), \( \Delta S = -\text{ve} \), depends on temperature; might be spontaneous at low temperatures.(c) both \( \Delta H \) and \( \Delta S \) are positive; spontaneity depends on high temperature.(d) \( \Delta H = +\text{ve} \), \( \Delta S = -\text{ve} \), would lead to positive \( \Delta G \), hence nonspontaneous.
3Step 3: Identifying Nonspontaneous Condition
From step 2, we observe that condition (d) has \( \Delta H = +\text{ve} \) and \( \Delta S = -\text{ve} \), leading to a positive \( \Delta G \). This means that the process continues to have a positive Gibbs free energy change regardless of temperature, making it nonspontaneous.
Key Concepts
SpontaneityEnthalpy ChangeEntropy Change
Spontaneity
Spontaneity in chemical reactions is a key concept that determines whether a reaction can happen without needing additional energy. When we say a reaction is spontaneous, it means it can proceed on its own at given conditions such as temperature and pressure. A spontaneous process tends to result in the release of free energy, which means it occurs naturally without external intervention. The spontaneity of a process is closely linked to Gibbs free energy, represented by the symbol \( \Delta G \).
The equation \( \Delta G = \Delta H - T\Delta S \) helps us calculate spontaneity. Here, \( \Delta H \) stands for change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) denotes change in entropy. By making sense of the signs and values of these variables, we can predict if and when a process will occur spontaneously.
- If \( \Delta G < 0 \), the reaction is spontaneous.
- If \( \Delta G > 0 \), the reaction is nonspontaneous.
- If \( \Delta G = 0 \), the system is at equilibrium.
The equation \( \Delta G = \Delta H - T\Delta S \) helps us calculate spontaneity. Here, \( \Delta H \) stands for change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) denotes change in entropy. By making sense of the signs and values of these variables, we can predict if and when a process will occur spontaneously.
Enthalpy Change
Enthalpy change, denoted by \( \Delta H \), is all about the heat exchange between the system and its surroundings during a chemical reaction or a physical change. It tells us if the reaction absorbs heat (endothermic) or releases heat (exothermic).
Enthalpy is crucial for understanding spontaneity because it impacts the overall energy balance of a reaction. To determine if a reaction is spontaneous, a negative \( \Delta H \) often contributes to a negative Gibbs free energy \( \Delta G \), which favors spontaneity, particularly if the system also has a positive entropy change \( \Delta S \) or is at a low temperature when \( \Delta S \) is negative.
- When \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to the surroundings.
- When \( \Delta H \) is positive, the reaction is endothermic, meaning it absorbs heat from the surroundings.
Enthalpy is crucial for understanding spontaneity because it impacts the overall energy balance of a reaction. To determine if a reaction is spontaneous, a negative \( \Delta H \) often contributes to a negative Gibbs free energy \( \Delta G \), which favors spontaneity, particularly if the system also has a positive entropy change \( \Delta S \) or is at a low temperature when \( \Delta S \) is negative.
Entropy Change
Entropy change, symbolized as \( \Delta S \), is a measure of the disorder or randomness within a system. In simple terms, it indicates how spread out or chaotic the energy in a system is. As a fundamental concept in thermodynamics, entropy provides insights into the thermodynamic favorability and spontaneity of a process.
A positive entropy change suggests that a system is moving towards greater disorder, which often aligns with the natural tendency of processes and contributes favorably to spontaneity at higher temperatures. The change in entropy is critical in the Gibbs free energy equation \( \Delta G = \Delta H - T\Delta S \), where a positive \( \Delta S \) can drive a reaction to spontaneous if coupled with a favorable \( \Delta H \) or high temperature. Understanding these principles helps predict how a reaction or change will behave under different conditions.
- When \( \Delta S \) is positive, the disorder in the system increases.
- When \( \Delta S \) is negative, the system becomes more ordered.
A positive entropy change suggests that a system is moving towards greater disorder, which often aligns with the natural tendency of processes and contributes favorably to spontaneity at higher temperatures. The change in entropy is critical in the Gibbs free energy equation \( \Delta G = \Delta H - T\Delta S \), where a positive \( \Delta S \) can drive a reaction to spontaneous if coupled with a favorable \( \Delta H \) or high temperature. Understanding these principles helps predict how a reaction or change will behave under different conditions.
Other exercises in this chapter
Problem 115
The dissociation energies of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) to convert them into gaseous atoms are 360 and \(620 \mathrm{kcal}\) mol
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Calculate \(\mathrm{Q}\) and \(\mathrm{W}\) for the isothermal reversible expansion of one mole of an ideal gas from an initial pressure of \(1.0\) bar to a fin
View solution Problem 119
Which of the following is /are true about the isothermal expansion of an ideal gas? (a) \(\Delta \mathrm{U}=0\) (b) \(\Delta \mathrm{T}=0\) (c) \(\mathrm{q}=2.3
View solution Problem 120
For which of the following reactions, is \(\Delta \mathrm{H}\) equal to \(\Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \r
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