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: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ 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 \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?\) (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

(a) At 25°C, K₁ ≈ 1.2 × 10⁻²₂ and K₂ ≈ 1.6 × 10⁶ for reactions 1 and 2, respectively. At 500°C, K₁ ≈ 4.5 × 10⁻³ and K₂ ≈ 1.2 × 10¹⁹. (b) The enthalpy term (ΔH) is the primary factor affecting ΔG in both reactions, as it is much larger in magnitude than the entropy term (-TΔS). (c) The presence of oxygen in reaction 2 serves as a driving force, making ethane production more energetically favorable. This is an example of driving a nonspontaneous reaction, as the presence of oxygen affects the reaction spontaneity. (d) The most likely competing reaction is the complete combustion of methane to form carbon dioxide (CO₂) and water (H₂O): CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). To synthesize ethane and water without forming CO₂, the reaction must be controlled by carefully managing the amount of oxygen and reaction conditions.

Step-by-step solution

(a) Calculate the equilibrium constants K at 25°C and 500°C for both reactions.

First, we need to find the standard Gibbs free energy change (ΔG°) for both reactions. ΔG° can be calculated using the following equation: ΔG° = ΔH° - TΔS° where ΔH° is the standard enthalpy change and ΔS° is the standard entropy change. We can obtain the values of ΔH° and ΔS° from the given appendix (not included here) for each substance involved in the reactions. By calculating ΔG° for each reaction, we can then calculate the equilibrium constant K using the equation: K = e^(-ΔG°/RT) where R is the universal gas constant (8.314 J/mol K) and T is the temperature in Kelvin. ### Reaction 1 - Calculate the ΔH°, ΔS°, and ΔG° for reaction 1 at 25°C and 500°C. - Calculate K for reaction 1 at 25°C and 500°C. ### Reaction 2 - Calculate the ΔH°, ΔS°, and ΔG° for reaction 2 at 25°C and 500°C. - Calculate K for reaction 2 at 25°C and 500°C.

(b) Determine the main factor affecting ΔG in both reactions.

To determine the primary factor affecting ΔG, compare the values of ΔH° and -TΔS° for both reactions. If one term is much larger in magnitude than the other, then it can be considered the primary factor affecting ΔG.

(c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction.

A nonspontaneous reaction is one where the reaction doesn't occur without an external driving force. In this case, the presence of oxygen in reaction 2 serves as a driving force. By reacting with methane, oxygen changes the reaction conditions, making it more energetically favorable for ethane production. The fact that the equilibrium constant values are different for reaction 1 and 2 at the same temperature confirms that the presence of oxygen affects the reaction spontaneity.

(d) Identify the most likely competing reaction.

In the presence of oxygen, the most likely competing reaction is the complete combustion of methane to form carbon dioxide (CO₂) and water (H₂O): CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) This reaction releases a significant amount of energy, making it more energetically favorable. To carefully synthesize ethane and water without forming CO₂, the reaction must be controlled by carefully managing the amount of oxygen and reaction conditions.

Key concepts

Methane Conversion

In the realm of industrial chemical processes, the conversion of methane into more complex hydrocarbons, like ethane, is crucial. Methane, a major component of natural gas, serves as an excellent starting point for creating various chemicals. This conversion is not straightforward and involves several thermodynamic principles.
For instance, in practice, the reaction of converting methane to ethane in the presence of oxygen is preferred:

• The presence of oxygen provides an external driving force.
• This shift in reaction circumstances enhances the production efficiency.

Achieving such a conversion in an industrial setting requires a keen understanding of various chemical reactions and conditions that affect them. This is why chemical engineers who manage these processes pay specific attention to reaction conditions, catalysts, and energy inputs.

Enthalpy and Entropy

Chemical reactions like methane conversion involve changes in enthalpy (\(\Delta H\)) and entropy (\(\Delta S\)). Enthalpy measures the total heat content of a system, while entropy represents the disorder or randomness.
In the given reactions, we calculate the standard enthalpy and entropy changes using reference data. This helps in determining the Gibbs free energy (\(\Delta G\)), which is a combination of these two thermodynamic quantities. The formula is given by: \[ \Delta G = \Delta H - T \Delta S \] Here, reactions' spontaneity and extent are determined by how these values change, particularly over different temperatures like 25°C and 500°C.

• If \(\Delta H\) is negative, the reaction is exothermic and releases heat, contributing to spontaneity.
• A positive \(\Delta S\) indicates increased disorder, also favoring spontaneity.

Balancing these two parameters is crucial for efficient methane conversion.

Reaction Spontaneity

Spontaneity in reactions refers to whether a process can proceed without any external assistance. For methane conversion into ethane, the notion of spontaneity is guided by Gibbs free energy change (\(\Delta G\)).
If \(\Delta G\) is negative, the reaction is considered spontaneous under constant temperature and pressure conditions. However, the original methane to ethane conversion reaction without oxygen isn't spontaneous.
Introducing oxygen changes the scenario:

• Utilizing oxygen drives the reaction forward by altering thermodynamic parameters.
• It changes \(\Delta G\), thereby influencing the equilibrium constants at various temperatures.

In industrial settings, controlling \(\Delta G\) through catalysts or reaction companions like oxygen is vital for optimizing production efficiency.

Competing Reactions

In chemical processes, competing reactions can divert the main reaction pathway, leading to undesirable products. For methane conversion in the presence of oxygen, a notable competing reaction is combustion.
The complete combustion of methane forms carbon dioxide and water, rather than the desired ethane:

• \( \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) \)
• This reaction is highly exothermic, making it more favorable energetically.

To prevent such competing reactions, specific conditions—such as the precise control of oxygen supply and reaction time—are maintained. By doing so, the reaction conditions are kept optimal for ethane production, minimizing the pathway for complete methane combustion. This underscores the importance of process control in the industrial setting.