Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{t}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix C, calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{e}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term (-TUS)? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Short answer

At 25°C and 500°C, the equilibrium constant (K) was calculated using the Gibbs free energy formula and the relationship between Gibbs free energy and the equilibrium constant. It was determined that the difference in Gibbs free energy (ΔG) for the two reactions is primarily due to the enthalpy term (ΔH). The reactions exemplify driving a nonspontaneous reaction as the presence of oxygen increases the spontaneity of the reaction. The most likely competing reaction is the complete combustion of methane with oxygen, forming carbon dioxide (CO2) and water (H2O) instead of ethane (C2H6) and water (H2O).

Step-by-step solution

Calculate the Gibbs free energy for each reaction at 25°C and 500°C

To calculate the Gibbs free energy (ΔG) for each reaction, we can use the formula: $$ \Delta G=\Delta H - T\Delta S $$ where ΔG is Gibbs free energy, ΔH is enthalpy, T is temperature in Kelvin (K), and ΔS is entropy. Using the standard Gibbs free energies of formation found in Appendix C, we calculate ΔG for each reaction at 25°C and 500°C. Reaction 1: $$ \Delta G = 2\Delta G (\text{CH}_4) - \left[\Delta G (\text{C}_2 \text{H}_6) + \Delta G (\text{H}_2) \right] $$ Reaction 2: $$ \Delta G = 2\Delta G (\text{CH}_4) + 0.5\Delta G (\text{O}_2) - \left[\Delta G (\text{C}_2 \text{H}_6) + \Delta G (\text{H}_2 \text{O})\right] $$

Calculate the equilibrium constant (K) for each reaction at 25°C and 500°C

With the calculated ΔG for each reaction, we can use the relationship between Gibbs free energy and the equilibrium constant (K) to find K for each reaction at each temperature: $$ \Delta G = -RT \ln K $$ where R is the universal gas constant (8.314 J/mol K), T is temperature in Kelvin (K), and K is the equilibrium constant. Solving the equation above for K, we have: $$ K = e^{\frac{-\Delta G}{RT}} $$ Plug the values of ΔG and T into the equation to calculate K for each reaction at 25°C and 500°C. #b) Determine the primary factor of ΔG difference (ΔH or -TΔS)#

Determine the primary factor of ΔG difference (ΔH or -TΔS)

Using the Gibbs free energy formula, we have: $$ \Delta G = \Delta H - T\Delta S $$ We have already calculated ΔG for both reactions at 25°C and 500°C. Find the difference between the ΔG values of the reactions and check whether it is primarily attributed to enthalpy (ΔH) or the entropy term (-TΔS). Based on the difference in enthalpy and entropy between the reactions, determine which factor primarily contributes to the difference in Gibbs free energy. #c) Describe how these reactions exemplify driving a nonspontaneous reaction#

Describe how these reactions exemplify driving a nonspontaneous reaction

A nonspontaneous reaction is a reaction that does not proceed in the forward direction unless external energy is supplied. The difference in the Gibbs free energy of a reaction determines the spontaneity of the reaction. If ΔG > 0, the reaction is nonspontaneous, and if ΔG < 0, the reaction is spontaneous. Compare the Gibbs free energies of the two reactions. At 25°C, the reaction without oxygen is less spontaneous compared to the reaction with oxygen. However, when the temperature is increased to 500°C, the reaction without oxygen becomes more spontaneous. This indicates that the presence of oxygen drives the reaction and increases its spontaneity. #d) Identify the most likely competing reaction#

Identify the most likely competing reaction

The competing reaction occurs when the reaction of methane (CH4) and oxygen (O2) does not result in forming ethane (C2H6) and water (H2O) as desired. The most likely competing reaction would be the complete combustion of methane with oxygen, which would form carbon dioxide (CO2) and water (H2O) instead: $$ \text{CH}_4(g) + 2\text{O}_2(g) \longrightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) $$

Key concepts

Gibbs free energy

Gibbs free energy, represented as \( \Delta G \), is a crucial concept in chemical thermodynamics, mainly because it predicts the spontaneity of a reaction. The formula for Gibbs free energy is given by:

• \( \Delta G = \Delta H - T\Delta S \)

Here, \( \Delta H \) is the change in enthalpy, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.
The sign of \( \Delta G \) determines if a reaction will proceed on its own:

• If \( \Delta G < 0 \), the reaction is spontaneous.
• If \( \Delta G > 0 \), the reaction is nonspontaneous and requires energy input.

Spontaneity isn't fixed; it can change with temperature. As shown in the provided exercise, the presence of oxygen affects the \( \Delta G \), turning a nonspontaneous reaction into a spontaneous one at higher temperatures. This clearly illustrates Gibbs free energy's pivotal role in understanding reaction behavior.

enthalpy and entropy

Enthalpy and entropy are two fundamental components of Gibbs free energy. They help in understanding the energy exchanges and disorder in a chemical reaction.
***Enthalpy (\( \Delta H \))*** refers to the heat absorbed or released in a reaction at constant pressure. It's a measure of the total energy change.

• Negative \( \Delta H \) indicates exothermic reactions (release heat).
• Positive \( \Delta H \) indicates endothermic reactions (absorb heat).

***Entropy (\( \Delta S \))*** represents the disorder or randomness of the system. A higher \( \Delta S \) suggests increased chaos.

• If \( T \Delta S \) is large, the disorder significantly influences the Gibbs free energy, potentially making it negative (favorable).

In the exercise, we see that the temperature and entropy changes help drive the reaction forward, especially in the presence of oxygen, indicating how both enthalpy and entropy play roles in reaction spontaneity.

equilibrium constant

The equilibrium constant, denoted as \( K \), indicates a reaction's tendency to proceed to completion. It is directly linked to the Gibbs free energy via the relation:

• \( \Delta G = -RT \ln K \)

Where \( R \) is the gas constant and \( T \) is the temperature. A key takeaway here is the sign of \( \Delta G \) determines the value of \( K \):

• If \( \Delta G < 0 \), \( K > 1 \), and products are favored at equilibrium.
• If \( \Delta G > 0 \), \( K < 1 \), and reactants are favored.

In the exercises, calculating \( K \) reveals how the reaction shifts between conditions. At higher temperatures, \( K \) might indicate more product formation, demonstrating the sensitivity of equilibrium to conditions. Understanding \( K \) helps in assessing how far a reaction can go, giving insight into managing industrial processes efficiently.