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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?

Step by step solution

01

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

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

02

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

To determine the primary factor affecting 螖G, compare the values of 螖H掳 and -T螖S掳 for both reactions. If one term is much larger in magnitude than the other, then it can be considered the primary factor affecting 螖G.

03

(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.

04

(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 (CO鈧�) and water (H鈧侽): CH鈧�(g) + 2O鈧�(g) 鈫� CO鈧�(g) + 2H鈧侽(g) This reaction releases a significant amount of energy, making it more energetically favorable. To carefully synthesize ethane and water without forming CO鈧�, the reaction must be controlled by carefully managing the amount of oxygen and reaction conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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掳C and 500掳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鈥攕uch as the precise control of oxygen supply and reaction time鈥攁re 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.