Problem 8 - Basic Principles of Organic Chemistry

Question

Formation of the addition product of \(\mathrm{SO}_{2}\) and 1,3 -butadiene has \(\Delta H^{0}=-16.5 \mathrm{kcal} \mathrm{mol}^{-1}\) for the vapor phase. Assuming the equilibrium constant \(K\) is unity at \(00^{\circ} \mathrm{C}\), calculate \(\Delta S^{0}\) for the reaction. Compare this value with the \(\Delta S^{0}\) that you can calculate for addition of ethene to 1,3 -butadiene, which has \(\Delta G^{0}=-27: \mathrm{kcal}\) and \(\Delta H^{0}=-47 \mathrm{kcal} .\) Estimate the temperature in \({ }^{\circ} \mathrm{C}\) that would be required for the equilibrium between ethene and 1,3 -butadiene to have \(K=1\). (You may be interested to know that an early route for preparation of 1,3-butadiene involved passing cyclohexane through a tube containing a red-hot wire spiral, \(\sim 900^{\circ} \mathrm{C}\).)

Step-by-Step Solution

Answer

1) \( \Delta S^0_{\text{SO}_2} = -60.44 \text{ cal/mol K} \); 2) \( \Delta S^0_{\text{ethene}} = 67.11 \text{ cal/mol K} \). 3) Equilibrium at \( 427^{\circ} \text{C} \).

Step 1: Understanding the Problem

We need to find \( \Delta S^0 \) for the addition of \( \text{SO}_2 \) to 1,3-butadiene and compare it with the same reaction involving ethene. We will also estimate the temperature where the equilibrium constant \( K \) is 1 for the ethene reaction.

Step 2: Finding \( \Delta S^0 \) for \( \text{SO}_2 \) Reaction

For the reaction of \( \text{SO}_2 \) and 1,3-butadiene \( \Delta H^0 = -16.5 \text{ kcal/mol} \) and \( K = 1 \) at \( 0^{\circ} \text{C} \). Use \( \Delta G^0 = -RT \ln K \) to find \( \Delta G^0 = 0 \text{ kcal/mol} \). Use the Gibbs free energy equation: \( \Delta G^0 = \Delta H^0 - T \Delta S^0 \) to solve for \( \Delta S^0 \).

Step 3: Calculate \( \Delta S^0 \) for \( \text{SO}_2 \) Reaction

Set \( \Delta G^0 = 0 \):\[ 0 = -16.5 - (273)(\Delta S^0/1000) \]This leads to: \[ \Delta S^0 = \frac{-16.5}{273/1000} = -60.44 \text{ cal/mol K} \].

Step 4: Find \( \Delta S^0 \) for Ethene Reaction

Given \( \Delta G^0 = -27 \text{ kcal/mol} \) and \( \Delta H^0 = -47 \text{ kcal/mol} \), use the Gibbs energy equation:\[ \Delta G^0 = \Delta H^0 - T \Delta S^0 \]\[ -27 = -47 - (298)(\Delta S^0/1000) \]Solve for \( \Delta S^0 \): \( \Delta S^0 = \frac{-27 + 47}{298/1000} \approx 67.11 \text{ cal/mol K} \).

Step 5: Estimate Temperature for \( K = 1 \) in Ethene Reaction

Set \( \Delta G^0 = 0 \):\[ 0 = -47 - T(67.11/1000) \]Solve for \( T \):\[ T = \frac{47}{67.11/1000} = 700 \text{ K} \]. Convert to Celsius: \( 700 - 273 = 427^{\circ} \text{C} \).

Step 6: Compare and Conclude

\( \Delta S^0 \) for \( \text{SO}_2 \) reaction is \(-60.44 \text{ cal/mol K}\), while for ethene, it's \(67.11 \text{ cal/mol K}\). The equilibrium requires \(427^{\circ} \text{C}\) for the ethene reaction to have \(K=1\).

Key Concepts

Gibbs Free Energy

In thermodynamics, Gibbs Free Energy is a very important concept. It helps predict how feasible a chemical reaction is. The formula is given by: \( \Delta G^0 = \Delta H^0 - T \Delta S^0 \). Here, \( \Delta G^0 \) is the Gibbs Free Energy change, \( \Delta H^0 \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S^0 \) is the entropy change.

When \( \Delta G^0 \) is negative, the reaction will be spontaneous, meaning it can occur by itself. If \( \Delta G^0 \) is positive, the reaction is non-spontaneous, and it won’t happen without external energy. It's key to note that at \( \Delta G^0 = 0 \), the reaction is at equilibrium.

In the example given, for the vapor phase reaction between SO\(_2\) and 1,3-butadiene, the equilibrium constant is 1 at **0°C**, meaning \( \Delta G^0 = 0 \). This allows us to link the temperature inside the energy change equation to find entropy. By setting \( \Delta G^0 = 0 \), we easily calculate \( \Delta S^0 \) using the other known values.

Equilibrium Constant (K)

The equilibrium constant, denoted as \( K \), is a number that expresses the ratio of products to reactants at equilibrium. It tells us which side of the reaction is favored under given conditions. If \( K \) is greater than one, products are favored; if less than one, reactants are favored.

To find the equilibrium constant at a certain temperature, we can use the relation: \( \Delta G^0 = -RT \ln K \). At standard conditions where \( K = 1 \), the reaction is perfectly balanced, and \( \Delta G^0 = 0 \). This provides a way to manipulate the Gibbs Free Energy equation to derive necessary thermodynamic properties.

In our example, the fact that SO\(_2\) and 1,3-butadiene form an equilibrium (where \( K = 1 \) at **0°C**) simplifies the calculations. It implies an exact balance between reactants and products, representing a state where energy is optimized.

Entropy (S)

Entropy is a measure of disorder or randomness in a system. In chemical reactions, changes in entropy (\( \Delta S^0 \)) provide insight into the degree of disorder introduced by a reaction. The more positive \( \Delta S^0 \) is, the more randomness or disorder is introduced during the reaction.

For the reaction between SO\(_2\) and 1,3-butadiene, entropy change is critical to understanding the nature of the reaction at equilibrium, as expressed in the Gibbs Free Energy formula. With \( \Delta G^0 \) set to zero at \( K = 1 \), entropy can be calculated when both the temperature and enthalpy change \( \Delta H^0 \) are known. Negative entropy in this case suggests a net decrease in system disorder.

Contrast this with the ethene addition reaction to 1,3-butadiene, which displays a positive \( \Delta S^0 \) indicating an increase in disorder. This difference in entropy values gives clues about the structural and energetic shifts that each reaction undergoes.

Temperature and Equilibrium

Temperature plays a pivotal role in chemical equilibrium. It influences the energy balance and the value of the equilibrium constant \( K \). In the context of our problem, the equilibrium temperature for the ethene reaction can be calculated using the entropy and enthalpy values.

For instance, by setting the standard Gibbs Free Energy \( \Delta G^0 = 0 \), we solved for the temperature needed for \( K = 1 \). This occurred at approximately **427°C** for the reaction between ethene and 1,3-butadiene. Such calculations help us estimate reaction conditions for equilibrium under different scenarios.

The interplay between energy, entropy, and equilibrium at various temperatures enables chemists to predict and control reactions, aligning them with practical and industrial needs. Understanding how to manipulate these variables is vital for optimizing reaction conditions.