We are asked which substance has the higher enthalpy of combustion per unit volume.
The standard heats of formation for selected substances are shown here.
Calculate molar enthalpy change:
CH4 + 2 O2 → CO2 + 2 H2O
ΔH∘combustion = -890.36 kJ/mol CH4
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
ΔH∘combustion = -1453.12 kJ/2 mol CH3OH
Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of -164 oC. One possible strategy is to oxidize the methane to methanol, CH3OH, which has a boiling point of 65 oC and can therefore be shipped more readily. Suppose that 1.07×1010 ft3 of methane at atmospheric pressure and 25 oC are oxidized to methanol.
Methane, when liquefied, has a density of 0.466 g/mL; the density of methanol at 25 oC is 0.791 g/mL. Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?
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