Final answer:
The maximum work obtainable from oxidizing 0.50 mol of methanol under standard conditions is -686 kJ. This calculation is done by multiplying the given Gibbs free energy (ΔG) for the reaction by the number of moles of methanol.
Step-by-step explanation:
The student's question pertains to the calculation of the maximum work that can be obtained by oxidizing methanol under standard conditions, given the Gibbs free energy change (ΔG) of the reaction. To find this, you use the equation ΔG = -nFE, where 'n' is the number of moles of electrons transferred, 'F' is the Faraday constant (96,485 C/mol e-), and 'E' is the electromotive force (emf). However, since ΔG is already provided and is the amount of energy available to do work at constant temperature and pressure, the calculation in this case is simpler and does not require the use of Faraday constantor emf.
The reaction, as given, is 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g) with ΔG = -1372 kJ/mol for the oxidation of 1 mole of methanol.
To calculate the maximum work for 0.50 mol of methanol:
Multiply ΔG by the number of moles of methanol being oxidized (0.50 mol).
Work = -1372 kJ/mol * 0.50 mol = -686 kJ.
Therefore, the maximum work that can be obtained by oxidizing 0.50 mol of methanol under standard conditions is -686 kJ. The negative sign indicates that the reaction is exergonic, releasing energy that can be used to do work.