Question

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

The minimum \( \Delta S \) is approximately \( 55.04 \, \text{J/K} \cdot \text{mol} \).

Step-by-step solution

Convert Temperature to Kelvin

To solve this problem, we first need to convert the given temperature from Celsius to Kelvin. The formula for conversion is \( T(\text{K}) = T(^{\circ}C) + 273.15 \). Substituting the given temperature: \( T = 72 + 273.15 = 345.15 \, \text{K} \).

Use the Gibbs Free Energy Equation

Spontaneity of a reaction can be determined using the Gibbs Free Energy change equation: \( \Delta G = \Delta H - T \Delta S \). For a reaction to be spontaneous, \( \Delta G \leq 0 \).

Solve for \( \Delta S \)

Since the reaction is spontaneous at 72°C, we set \( \Delta G \leq 0 \). Rearrange the equation \( \Delta G = \Delta H - T \Delta S \) to \( T \Delta S \geq \Delta H \). Solving for \( \Delta S \), we have \( \Delta S \geq \frac{\Delta H}{T} \).

Substitute Values into the Rearranged Equation

We substitute the given \( \Delta H = 19 \, \text{kJ/mol} = 19000 \, \text{J/mol} \) and \( T = 345.15 \, \text{K} \) into the equation \( \Delta S \geq \frac{19000}{345.15} \).

Calculate \( \Delta S \)

Perform the division: \( \Delta S \geq \frac{19000}{345.15} \approx 55.04 \, \text{J/K} \cdot \text{mol} \). Thus, the minimum value of \( \Delta S \) is approximately \( 55.04 \, \text{J/K} \cdot \text{mol} \).

Key concepts

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.

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.

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.