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

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: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$

02

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.

03

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.

04

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.

05

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: $$ \mathrm{CH}_{4}(g) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(g) $$

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