Problem 10 - Basic Principles of Organic Chemistry

Question

a. Calculate \(\Delta H^{0}\) from bond energies for the conversion of 1 -hexene to cyclohexane at \(25^{\circ}\) and from this, with \(\Delta S^{0}\) as \(-20.7\) e.u. \(\mathrm{mol}^{-1}\), calculate the equilibrium constant \(K_{\text {eq from Equation } 4-2 . \text { For comparison, calculate the equilibrium }}\) constant that would be expected if the degrees of disorder of the reactants and the products were equal (i.e., \(\Delta S^{0}=0\) ). b. How large can \(\Delta S^{0}\) be at \(25^{\circ}\) for a reaction before our \(\pm 15\) kcal rule starts to give incorrect answers?

Step-by-Step Solution

Answer

Calculate \( \Delta H^0 \) using bond energies, then \( K_{eq} \) using \( \Delta S^0 \). Re-calculate \( K_{eq} \) for \( \Delta S^0 = 0 \). \( \Delta S^0 \) impact begins when \( \pm 15 \) kcal rule is close.

Step 1: Determine the Reaction and Bond Energies

The conversion of 1-hexene to cyclohexane involves the breaking of the double bond in 1-hexene and forming the additional C-C bonds to close the ring and form cyclohexane. We use standard bond energy values: C=C bond energy is approximately 146 kcal/mol, C-C bond is 83 kcal/mol, and for every C-H, it remains unchanged. Here you are forming new C-C bonds to close the ring.

Step 2: Calculate \( \Delta H^0 \) Using Bond Energies

To calculate \( \Delta H^0 \), subtract the bond energies of bonds formed from the bond energies of bonds broken. Assuming we break one C=C bond to form one C-C bond, the change in enthalpy \( \Delta H^0 = (1 \times 146\, \text{kcal/mol}) - (1 \times 83\, \text{kcal/mol}) = 63\, \text{kcal/mol}.\) This indicates the net energy change of forming cyclohexane from 1-hexene.

Step 3: Calculate Equilibrium Constant \( K_{eq} \)

We use the equation \( \Delta G^0 = \Delta H^0 - T\Delta S^0 \) and \( \Delta G^0 = -RT\ln K_{eq} \) to find \( K_{eq} \). First, with \( \Delta S^0 = -20.7 \text{ e.u.} = -20.7 \times 10^{-3} \text{ kcal/mol.K} \), calculate \( \Delta G^0 = 63 - 298 \times (-0.0207) = 69.17\, \text{kcal/mol}. \) Substitute \( \Delta G^0 \) back to find \( K_{eq} = e^{-69.17/(0.001987 \times 298)} \), where \( R = 1.987 \text{ cal/mol.K} = 0.001987\, \text{kcal/mol.K} \).

Step 4: Calculate \( K_{eq} \) with \( \Delta S^0 = 0 \)

Re-calculate \( \Delta G^0 \) with \( \Delta S^0 = 0 \): \( \Delta G^0 = \Delta H^0 = 63\, \text{kcal/mol}. \) Substitute back to find \( K_{eq} = e^{-63/(0.001987 \times 298)} \). This will provide a comparison to the previous \( K_{eq} \) assuming no entropy changes.

Step 5: Address \( \Delta S^0 \) Impact on \( \pm 15 \) kcal Rule

The \( \pm 15 \) kcal rule suggests that any increase or decrease beyond this value in \( \Delta S^0 \) could give misleading \( K_{eq} \) calculations due to significant shifts in equilibrium favorability. Evaluate this by seeing when \( T\Delta S^0 \) significantly affects \( \Delta G^0 \), estimated to be around when \( \Delta S^0 = \pm 15/298 \approx \pm 0.0503 \text{ kcal/mol.K}. \)

Key Concepts

Bond Energies

Bond energies represent the amount of energy required to break a bond between two atoms. In the context of thermodynamics in organic chemistry, calculating bond energies is crucial because these values allow us to determine the enthalpy change, \(\Delta H^0\), of a chemical reaction.

In the conversion of 1-hexene to cyclohexane, we focus on the bonds involved. A C=C double bond in 1-hexene needs to be broken, which requires approximately 146 kcal/mol. On forming cyclohexane, new C-C bonds are established, each requiring about 83 kcal/mol. Because bond energies release or absorb significant energy, understanding how to calculate the net change is key. In our exercise, the enthalpy change is computed as the energy of bonds broken minus bonds formed: \(\Delta H^0 = (1 \times 146 \text{ kcal/mol}) - (1 \times 83 \text{ kcal/mol}) = 63 \text{ kcal/mol}.\)

• C=C bond breaking takes considerable energy.
• New C-C bonds formation recoups some energy.
• Net enthalpy gives insights into reaction heat dynamics.

Bond energies help explain why some reactions absorb heat (endothermic) while others release it (exothermic). This forms the foundation for understanding reaction spontaneity and equilibrium.

Equilibrium Constant

The equilibrium constant, \(K_{eq}\), quantifies the relative concentrations of products and reactants at equilibrium in a chemical reaction. In thermodynamics, the equilibrium constant becomes particularly significant as it links \(\Delta G^0\), the change in free energy, with reaction favorability.

To find \(K_{eq}\), we first calculate the change in Gibbs free energy using the relationship: \(\Delta G^0 = \Delta H^0 - T\Delta S^0\). For the conversion of 1-hexene to cyclohexane, the enthalpy change \(\Delta H^0\) is known, and with known entropy change \(\Delta S^0\), \(\Delta G^0\) is computed as: \(\Delta G^0 = 63 - 298 \times (-0.0207) = 69.17\text{ kcal/mol}.\)

We use this value with the equation \(\Delta G^0 = -RT\ln K_{eq}\) to find \(K_{eq}\): \(K_{eq} = e^{-69.17/(0.001987 \times 298)}\). Exploring scenarios with varying \(\Delta S^0\) values, such as zero entropy change, helps to understand how certain thermodynamic factors may influence chemical equilibria.

• Positive \(\Delta G^0\) indicates non-spontaneity.
• Negative \(\Delta G^0\) suggests reaction spontaneity.
• \(K_{eq}\) is sensitive to both temperature and entropy changes.

Entropy Change

Entropy change, \(\Delta S^0\), describes the change in disorder or randomness within a system during a chemical reaction. It is a critical concept in determining how the energy within a reaction system is distributed, impacting both \(\Delta G^0\) and thereby \(K_{eq}\).

In the exercise, we explore \(\Delta S^0 = -20.7\) e.u. for the reaction, impacting the free energy with \(\Delta G^0 = \Delta H^0 - T\Delta S^0\). The degree of disorder between reactants and products shapes the equilibrium position and determines reaction spontaneity. A negative \(\Delta S^0\), for instance, suggests a decrease in disorder, potentially creating less favorable conditions for a spontaneous reaction.

• Positive \(\Delta S^0\) enhances spontaneity.
• Zero \(\Delta S^0\) suggests no change in disorder.
• Evaluating \(\Delta S^0\) can predict reaction behavior under varying conditions.

The exercise also prompts us to consider if disorder changes were neutral (\(\Delta S^0 = 0\)). Such insights reveal the sensitivity of reactions to entropy changes and highlight scenarios where typical predictive rules might falter, such as with the \(\pm 15 \) kcal rule.