Question 12.EP.15: The conversion of methane gas (CH4) to methanol (CH3OH), whi...
The conversion of methane gas (CH_{4}) to methanol (CH_{3}OH), which is a liquid at room temperature, is an area in which considerable research is being done. Still, this process is not yet economically viable. Using tabulated thermodynamic data, calculate the equilibrium constant for the following reaction at 25°C:
CH_{4}(g) +\frac{1}{2} O_{2}(g)\rightleftarrows CH_{3}OH(l)
Comment on whether or not the equilibrium position of this reaction is the likely source of the problem for commercialization of the process.
Strategy
To solve this problem, we must use the relationship between ΔG° and the equilibrium constant given in Equation 12.7. We can use tabulated values of the free energy of formation for the reactants and products to calculate ΔG° and then obtain K.
ΔG° = −RT ln K (12.7)
= −166.27 kJ mol^{−1} − (−50.75 kJ mol^{−1}) − (0 kJ mol^{−1})
= −115.52 kJ mol^{−1}
When using Equation 12.7 we must make sure that the energy units of ΔG° and R are consistent. Here, it is easy to write ΔG° in joules rather than kilojoules.
−115.52 × 10³ J mol^{−1} = −(8.3145 J mol^{−1}K^{−1})(298.15 K) ln K
so
ln K = 46.60
and
K = e^{46.60} = 1.73 × 10^{20}
Discussion
This reaction can be thought of as the partial oxidation of methane, and equilibrium clearly favors the products. The problem that scientists and engineers face with this process is that methane also has a strong tendency to oxidize completely to carbon dioxide and water. This issue is explored further in the “Check Your Understanding” problem.