Chapter 18: Problem 74
A certain reaction is spontaneous at \(72^{\circ} \mathrm{C}\). If the enthalpy change for the reaction is \(19 \mathrm{~kJ} / \mathrm{mol}\), what is the minimum value of \(\Delta S\) (in \(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) ) for the reaction?
Short Answer
Expert verified
The minimum \( \Delta S \) is approximately \( 55.04 \, \text{J/K} \cdot \text{mol} \).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
Enthalpy change, often represented as \( \Delta H \), measures the heat absorbed or released during a chemical reaction at constant pressure. It indicates whether a reaction is endothermic or exothermic.
- **Endothermic reactions** absorb heat from their surroundings, leading to a positive \( \Delta H \).
- **Exothermic reactions** release heat, resulting in a negative \( \Delta H \).
In the exercise, the given enthalpy change is 19 kJ/mol, indicating the amount of heat absorbed to make the reaction potentially spontaneous at 72°C. When considering spontaneity, the enthalpy change of 19 kJ/mol implies the reaction might need a specific entropy change value to occur spontaneously under these conditions. Converting enthalpy from kJ to J is crucial for consistency: 19 kJ/mol converts to 19000 J/mol. This step ensures that all quantities have compatible units when substituted into the Gibbs Free Energy equation.
- **Endothermic reactions** absorb heat from their surroundings, leading to a positive \( \Delta H \).
- **Exothermic reactions** release heat, resulting in a negative \( \Delta H \).
In the exercise, the given enthalpy change is 19 kJ/mol, indicating the amount of heat absorbed to make the reaction potentially spontaneous at 72°C. When considering spontaneity, the enthalpy change of 19 kJ/mol implies the reaction might need a specific entropy change value to occur spontaneously under these conditions. Converting enthalpy from kJ to J is crucial for consistency: 19 kJ/mol converts to 19000 J/mol. This step ensures that all quantities have compatible units when substituted into the Gibbs Free Energy equation.
Entropy Change
Entropy, indicated as \( \Delta S \), describes the disorder or randomness of a system. Greater entropy implies more disorderly states. For a reaction:
- An increase in entropy \((\Delta S > 0)\) typically indicates more disorder post-reaction.
- A decrease \((\Delta S < 0)\) signals a more ordered state.
Entropy plays a vital role in determining reaction spontaneity, alongside enthalpy. In the exercise, we calculate the minimum equilibrium value of \( \Delta S \) to ensure spontaneity at the given conditions. Using the relationship expressed in the Gibbs Free Energy equation \( \Delta G = \Delta H - T \Delta S \) and knowing that \( \Delta G \leq 0 \), we can rearrange to find \( T \Delta S \geq \Delta H \). Thus, for a reaction at 345.15 K, \( \Delta S \) must satisfy \( \Delta S \geq \frac{19000}{345.15} \approx 55.04\, \text{J/K} \cdot \text{mol} \). This ensures sufficient disorder for the reaction to proceed spontaneously.
- An increase in entropy \((\Delta S > 0)\) typically indicates more disorder post-reaction.
- A decrease \((\Delta S < 0)\) signals a more ordered state.
Entropy plays a vital role in determining reaction spontaneity, alongside enthalpy. In the exercise, we calculate the minimum equilibrium value of \( \Delta S \) to ensure spontaneity at the given conditions. Using the relationship expressed in the Gibbs Free Energy equation \( \Delta G = \Delta H - T \Delta S \) and knowing that \( \Delta G \leq 0 \), we can rearrange to find \( T \Delta S \geq \Delta H \). Thus, for a reaction at 345.15 K, \( \Delta S \) must satisfy \( \Delta S \geq \frac{19000}{345.15} \approx 55.04\, \text{J/K} \cdot \text{mol} \). This ensures sufficient disorder for the reaction to proceed spontaneously.
Spontaneity of Reactions
A spontaneous reaction is one that occurs naturally under specific conditions without external intervention. The Gibbs Free Energy change \( \Delta G \) is used to gauge this spontaneity:
- If \( \Delta G < 0 \), the reaction is spontaneous.
- If \( \Delta G = 0 \), the system is at equilibrium.
- If \( \Delta G > 0 \), the reaction is non-spontaneous.
In the context of the exercise, understanding spontaneity involves using the Gibbs equation: \( \Delta G = \Delta H - T \Delta S \). The given conditions make the reaction spontaneous at 72°C. Thus, it is essential that the sum of the enthalpy and entropy changes results in a negative \( \Delta G \). This is why calculating the correct \( \Delta S \) is necessary: it ensures that \( \Delta G \) remains non-positive, confirming that the reaction can spontaneously proceed at the specified temperature.
- If \( \Delta G < 0 \), the reaction is spontaneous.
- If \( \Delta G = 0 \), the system is at equilibrium.
- If \( \Delta G > 0 \), the reaction is non-spontaneous.
In the context of the exercise, understanding spontaneity involves using the Gibbs equation: \( \Delta G = \Delta H - T \Delta S \). The given conditions make the reaction spontaneous at 72°C. Thus, it is essential that the sum of the enthalpy and entropy changes results in a negative \( \Delta G \). This is why calculating the correct \( \Delta S \) is necessary: it ensures that \( \Delta G \) remains non-positive, confirming that the reaction can spontaneously proceed at the specified temperature.
