Problem 65 - Chemistry The Central Science In Si Units

Question

A particular constant-pressure reaction is barely spontaneous at \(320 \mathrm{~K}\). The enthalpy change for the reaction is + \(15.2 \mathrm{~kJ}\). Estimate \(\Delta S\) for the reaction.

Step-by-Step Solution

Answer

\(\Delta S = 47.5 \, \text{J/K}\).

Step 1: Understanding Spontaneity

For a reaction to be spontaneous at constant pressure and temperature, Gibbs free energy change (\(\Delta G\)) must be zero since the reaction is barely spontaneous. The equation to relate enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and Gibbs free energy is: \[ \Delta G = \Delta H - T \Delta S \] which implies that \(\Delta G = 0\) when the reaction is barely spontaneous. Therefore, \(\Delta H = T \Delta S\).

Step 2: Reorganizing the Formula

Since \(\Delta G = 0\), we need to solve for \(\Delta S\) in the equation \(\Delta H = T \Delta S\). This gives us: \[ \Delta S = \frac{\Delta H}{T} \]

Step 3: Plugging in the Values

Substitute \(\Delta H = 15.2 \text{ kJ}\) and \(T = 320 \text{ K}\) into the equation for \(\Delta S\). Convert \(\Delta H\) into joules because \(1 \text{ kJ} = 1000 \text{ J}\): \[ \Delta H = 15200 \text{ J} \] Therefore, \[ \Delta S = \frac{15200 \text{ J}}{320 \text{ K}} \]

Step 4: Calculating \(\Delta S\)

Perform the division to find \(\Delta S\): \[ \Delta S = \frac{15200}{320} = 47.5 \text{ J/K} \]

Key Concepts

Gibbs Free Energy

Gibbs free energy is a fundamental concept in thermodynamics. It helps us determine if a chemical reaction can occur spontaneously. The symbol used to represent Gibbs free energy is \( \Delta G \). In simple terms, think of it as a kind of energy that can predict the direction a reaction will tend to go.
At a constant temperature and pressure, if \( \Delta G \) is negative, the reaction is spontaneous. That means it can proceed without any external help.
If \( \Delta G \) is zero, like in the example problem, the reaction is barely spontaneous. This means the two states of the reaction are in equilibrium.

• The main equation used is: \[ \Delta G = \Delta H - T \Delta S \]
• Where \( \Delta H \) is enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is entropy change.

When a reaction is barely spontaneous, the balance between enthalpy change and entropy change makes \( \Delta G \) equal to zero.

Enthalpy Change

The enthalpy change, represented as \( \Delta H \), is a crucial aspect of understanding how energy is transferred during a chemical reaction. Enthalpy itself refers to the total heat content of a system.
When we talk about \( \Delta H \), we are looking at how much energy is absorbed or released when a reaction happens.

• If \( \Delta H \) is positive, like in the problem we are considering, the reaction absorbs energy. It is endothermic.
• Conversely, a negative \( \Delta H \) indicates an exothermic reaction, where energy is released.

In the example given, \( \Delta H = +15.2 \) kJ indicates that the reaction requires heat to proceed.

Entropy Change

Entropy is a concept that describes disorder or randomness in a system. When we talk about entropy change, noted as \( \Delta S \), we're interested in how this disorder changes as the reaction takes place.
A positive \( \Delta S \) means the system becomes more disordered, while a negative \( \Delta S \) indicates an increase in order.
In our problem, we use the relationship \[ \Delta S = \frac{\Delta H}{T} \]to calculate the entropy change at the given temperature.

• An increased entropy (\( \Delta S > 0 \)) often helps a reaction be spontaneous, especially at higher temperatures, as it contributes to making \( \Delta G \) more negative.
• In our scenario, the calculated \( \Delta S \) of 47.5 J/K suggests an increase in disorder, aligning with the nearly spontaneous nature of the reaction.

Spontaneity of Reactions

The spontaneity of a reaction is crucial to understanding whether it will occur without outside influence. Generally, a reaction is spontaneous if \( \Delta G \) is negative.
However, if \( \Delta G \) is zero, the reaction is at the brink of spontaneity, which means it can go forward or backward equally.

• For spontaneity, both entropy and enthalpy changes may play a role.
  A favorable entropy change can help overcome a less favorable enthalpy change to make \( \Delta G \) zero or negative.
• The temperature is also key, as seen in the equation \( \Delta G = \Delta H - T \Delta S \).For instance, an endothermic reaction (\( \Delta H > 0 \)) can be made spontaneous at high temperatures if the entropy increases significantly.

In essence, spontaneity depends on the delicate balance between energy input/output and the increase in entropy.