Problem 100 - Chemistry: The Central Science in SI Units, Global Edition

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?

Step-by-Step Solution

Answer

The equilibrium constants for both reactions at 25°C and 500°C can be calculated using the standard Gibbs free energy change (ΔG°) and the equation: K = \( e^{-\frac{\Delta \mathrm{G}^{\circ}}{\mathrm{RT}}} \). Comparing the ΔH° and -TΔS° terms helps determine if the difference in ΔG° is mainly from the enthalpy or entropy term. These reactions represent driving a nonspontaneous reaction since the spontaneous reaction (Reaction 2) helps drive the overall conversion of methane into ethane and hydrogen. The most likely competing reaction when CH4 and O2 react to form C2H6 and H2O is the combustion of methane, which produces carbon dioxide (CO2) and water (H2O).

Step 1: Calculate the standard Gibbs free energy change of reactions

First, we need to calculate the standard Gibbs free energy change (ΔG°) for both reactions using the equation: ΔG° = ΔH° - TΔS° using the data in Appendix C. We will then use these values to calculate the equilibrium constant (K) at the given temperatures. Reaction 1: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ Reaction 2: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$

Step 2: Calculate the Equilibrium Constant (K) for both reactions at 25°C and 500°C

We can now use the calculated ΔG° values to find the equilibrium constant (K) for both reactions at the specified temperatures, using the equation: K = \( e^{-\frac{\Delta \mathrm{G}^{\circ}}{\mathrm{RT}}} \) Reaction 1: Calculate K at 25°C and 500°C for the first reaction. Reaction 2: Calculate K at 25°C and 500°C for the second reaction.

Step 3: Determine the main cause of ΔG° difference

Comparing the ΔH° and -TΔS° terms for both reactions, we can determine which term is the primary cause of the difference in Gibbs free energy (ΔG°). If the ΔH° terms are significantly different, then enthalpy is the main cause. If the -TΔS° terms are significantly different, then entropy is the main cause.

Step 4: Explain how these reactions are an example of driving a nonspontaneous reaction

To understand how these reactions drive a nonspontaneous reaction, we need to compare the ΔG° values for the two reactions at the provided temperatures. The interaction between the first nonspontaneous reaction (Reaction 1) and the second spontaneous reaction (Reaction 2) can help drive the overall conversion of methane into ethane and hydrogen.

Step 5: Identify the competing reaction

We are asked to find the most likely competing reaction when CH4 and O2 react to form C2H6 and H2O. Combustion of methane is a possible competing reaction, as it can also consume methane and oxygen to produce carbon dioxide (CO2) and water (H2O). This competing reaction can be written as: $$ \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) $$

Key Concepts

Equilibrium Constant

Calculating the equilibrium constant, denoted as \( K \), is a fundamental step in understanding chemical reactions and their behaviors at different temperatures. In this exercise, \( K \) is calculated using the standard Gibbs free energy change \( \Delta G^{\circ} \). The equation used is \( K = e^{-\frac{\Delta G^{\circ}}{RT}} \), where \( R \) is the universal gas constant and \( T \) is the absolute temperature in Kelvin. This relation implies that the equilibrium constant is directly influenced by the change in free energy; the smaller the \( \Delta G^{\circ} \), the larger the \( K \), indicating the position of equilibrium shifts towards the products.

For students, it's essential to understand that a large equilibrium constant (\( K > 1 \)) suggests that a reaction favors product formation at equilibrium, whereas a small equilibrium constant (\( K < 1 \)) implies that reactants are favored. In practical applications, knowing \( K \) helps in designing chemical processes that maximize product yields under specific conditions.

Nonspontaneous Reactions

Nonspontaneous reactions are those which do not occur naturally without external intervention because they have a positive \( \Delta G^{\circ} \). To drive a nonspontaneous reaction, like the conversion of methane to ethane without additional oxygen, an additional spontaneous reaction is often coupled to provide the necessary energy. In this context, the spontaneous reaction involving methane, oxygen, and the formation of water aids in driving the overall process.

This coupling of reactions is a common strategy in industrial chemistry. The energy released from the spontaneous reaction essentially provides the necessary thermodynamic push to drive the nonspontaneous one. Students should note that the overall \( \Delta G^{\circ} \) of the coupled reactions can become negative, making the overall reaction process spontaneous even if one of the steps isn't initially.

Enthalpy vs Entropy

To truly understand what drives a change in Gibbs free energy \( \Delta G^{\circ} \) in a reaction, it's crucial to consider the roles of enthalpy \( \Delta H \) and entropy \( \Delta S \). The relation \( \Delta G^{\circ} = \Delta H - T\Delta S \) shows how both factors influence the spontaneity of a reaction.

When comparing reactions, if \( \Delta H \) differences are substantial, this implies that the heat content or energy change is the main driver. Conversely, if \( -T\Delta S \) differences are more pronounced, it suggests that disorder or randomness, represented by entropy, plays a bigger role.

• For exothermic reactions, a negative \( \Delta H \) can drive the reaction towards spontaneity.
• For reactions increasing in disorder, a positive \( \Delta S \) also assists in making \( \Delta G^{\circ} \) negative, helping spontaneity.

Understanding these concepts is important for predicting how a reaction may be influenced by changing temperatures or pressures, which can adjust either enthalpy or entropy importance in \( \Delta G^{\circ} \).