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

The equilibrium constants, \(K\), for the reactions at 25°C and 500°C are as follows: \(K_{1}(298\,\text{K}) = 1.06\times10^{-3}\), \(K_{1}(773\,\text{K}) = 5.68\times10^{-6}\), \(K_{2}(298\,\text{K}) = 2.9\times10^2\), and \(K_{2}(773\,\text{K}) = 2.7\times10^{4}\). The main contributor to the difference in Gibbs free energy is the enthalpy term, \(\Delta H^\circ\). These reactions are an example of driving a nonspontaneous reaction by coupling it with a spontaneous reaction. The most likely competing reaction when reacting CH4 and O2 is the combustion of methane to form CO2 and water (CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(g)).

Step-by-step solution

Extracting \(\Delta H^\circ_\text{f}\) and \(\Delta S^\circ_\text{f}\) values from Appendix C

Using Appendix C, we can find these values: For CH4(g): \(\Delta H^\circ_\text{f}=-74.9\,\text{kJ/mol}\) and \(\Delta S^\circ_\text{f}=186.3\,\text{J/mol}\text{K}\) For C2H6(g): \(\Delta H^\circ_\text{f}=-84.0\,\text{kJ/mol}\) and \(\Delta S^\circ_\text{f}=229.6\,\text{J/mol}\text{K}\) For H2(g): \(\Delta H^\circ_\text{f}=0.0\,\text{kJ/mol}\) and \(\Delta S^\circ_\text{f}=130.7\,\text{J/mol}\text{K}\) For O2(g): \(\Delta H^\circ_\text{f}=0.0\,\text{kJ/mol}\) and \(\Delta S^\circ_\text{f}=205.2\,\text{J/mol}\text{K}\) For H2O(g): \(\Delta H^\circ_\text{f}=-241.8\,\text{kJ/mol}\) and \(\Delta S^\circ_\text{f}=188.8\,\text{J/mol}\text{K}\)

Calculate \(\Delta G^\circ\) for each reaction at 25°C and 500°C

For Reaction 1: At 25°C (298K): \(\Delta H^{\circ}_{\text{Reaction 1}} = 2(-84.0)-2(-74.9) = -18.2\,\text{kJ/mol}\) \(\Delta S^{\circ}_{\text{Reaction 1}} = 229.6+130.7-2(186.3) = -12.3\,\text{J/mol}\text{K}\) \(\Delta G^{\circ}_{\text{Reaction 1}}(298\,\text{K}) = -18.2 - (298\times\frac{-12.3}{1000}) = 17.5\,\text{kJ/mol}\) At 500°C (773K): \(\Delta G^{\circ}_{\text{Reaction 1}}(773\,\text{K}) = -18.2 - (773\times\frac{-12.3}{1000}) = 57.1\,\text{kJ/mol}\) For Reaction 2: At 25°C (298K): \(\Delta H^{\circ}_{\text{Reaction 2}} = -84.0-(-74.9)-\frac{1}{2}(0) = -9.1\,\text{kJ/mol}\) \(\Delta S^{\circ}_{\text{Reaction 2}} = 229.6+188.8-2(186.3)-\frac{1}{2}(205.2) = 36.3\,\text{J/mol}\text{K}\) \(\Delta G^{\circ}_{\text{Reaction 2}}(298\,\text{K}) = -9.1 - (298\times\frac{36.3}{1000}) = -20.0\,\text{kJ/mol}\) At 500°C (773K): \(\Delta G^{\circ}_{\text{Reaction 2}}(773\,\text{K}) = -9.1 - (773\times\frac{36.3}{1000}) = -37.9\,\text{kJ/mol}\)

Calculate \(K\) for each reaction at 25°C and 500°C

We can use the equation \(K = e^{-\frac{\Delta G^\circ}{RT}}\) to calculate the equilibrium constant at each temperature. For Reaction 1: At 25°C: \(K_{1}(298\,\text{K}) = e^{-\frac{17.5\times10^3\,\text{J/mol}}{(8.314\,\text{J/mol}\text{K})(298\,\text{K})}} = 1.06\times10^{-3}\) At 500°C: \(K_{1}(773\,\text{K}) = e^{-\frac{57.1\times10^3\,\text{J/mol}}{(8.314\,\text{J/mol}\text{K})(773\,\text{K})}} = 5.68\times10^{-6}\) For Reaction 2: At 25°C: \(K_{2}(298\,\text{K}) = e^{-\frac{-20.0\times10^3\,\text{J/mol}}{(8.314\,\text{J/mol}\text{K})(298\,\text{K})}} = 2.9\times10^2\) At 500°C: \(K_{2}(773\,\text{K}) = e^{-\frac{-37.9\times10^3\,\text{J/mol}}{(8.314\,\text{J/mol}\text{K})(773\,\text{K})}} = 2.7\times10^{4} \)

Determine which term contributes more to the difference in \(\Delta G^\circ\)

We can evaluate the enthalpy and entropy terms for each reaction by comparing \(\Delta H^\circ\) and \((-T\Delta S^\circ)\). From the \(\Delta G^\circ\) calculations, we can see that the main contributor to the difference in \(\Delta G^\circ\) is the enthalpy term, \(\Delta H^\circ\).

Explaining the connection to driving nonspontaneous reactions

Nonspontaneous reactions can be driven by coupling them with a spontaneous reaction that has a negative Gibbs free energy change, making the overall process favorable. In this case, Reaction 1 is nonspontaneous, while Reaction 2 is spontaneous. Carrying out Reaction 1 in the presence of oxygen (Reaction 2) makes the overall process favorable, as seen in the more negative \(\Delta G^\circ\) values for Reaction 2 compared to Reaction 1.

Identify the competing reaction when reacting CH4 and O2

The most likely competing reaction when reacting CH4 and O2 is the combustion of methane to form carbon dioxide (CO2) and water: $$ \mathrm{CH}_{4}(g) + 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) $$ This reaction is highly exothermic and, if not controlled carefully, could compete with and even dominate the desired reaction to form ethane and water.

Key concepts

Industrial Chemical Processes

The conversion of methane to ethane represents a key example of an industrial chemical process that showcases the ingenuity of human engineering to transform simple molecules into more complex ones. Methane, a primary component of natural gas, is abundant but has limited applications due to its simplicity. Through a controlled chemical reaction, it can be converted into ethane, a substance with numerous industrial uses, such as the production of ethylene for plastic manufacture. This conversion process is not trivial; it requires understanding and control over reaction conditions, catalysts, and reaction kinetics to optimize yields while minimizing unwanted byproducts.

Industrial processes like these rely on meticulous engineering to operate safely and efficiently. When using oxygen in the reaction, there is a delicate balance that must be achieved to prevent the more violent combustion of methane, which produces carbon dioxide and water instead. The art of chemistry in an industrial setting thus lies in directing reactions along desired pathways, often by adjusting temperatures, pressures, or introducing specific catalysts to steer the reaction's progress.

Chemical Thermodynamics

Chemical thermodynamics, which deals with the energy changes in chemical reactions, plays a pivotal role in understanding and predicting the feasibility and extent of chemical processes. At its heart is the concept that every chemical reaction is associated with an energy change, principally involving heat (enthalpy, \( \Delta H \) ) and disorder (entropy, \( \Delta S \) ). These changes dictate whether a reaction is favored to occur (spontaneous) or not.

The calculations involved in determining these energy changes require a deep dive into the likes of enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)), which are garnered from standard tables. For example, at elevated temperatures, reactions that absorb heat (endothermic) or those that increase disorder may be favored, which explains why certain industrial processes operate at high temperatures. A mastery of thermodynamics is not just academic but practical, aiding chemists in predicting whether a reaction is likely to proceed without actually conducting it.

Gibbs Free Energy

Understanding Gibbs free energy (\( \Delta G \)) is crucial for chemists when it comes to predicting the spontaneity of a reaction. It’s a thermodynamic function that combines enthalpy and entropy to provide a single value that can forecast whether a reaction will occur naturally. The equation \( \Delta G = \Delta H - T\Delta S \) succinctly embodies this relationship. If \( \Delta G \) is negative, the reaction is exergonic and will happen spontaneously; if positive, the reaction is endergonic and will not occur without external input.

In the example of methane conversion, we calculate \( \Delta G \) at different temperatures to understand reaction feasibility. At 25 °C and 500 °C, the sought methane conversion is nonspontaneous at both temperatures, but the addition of oxygen makes it spontaneous, highlighting the role of thermodynamic principles in practical applications. These concepts underscore the importance of considering both energy and disorder to gauge the full thermodynamic picture of a chemical process.