Natural gas (mostly methane, CH4) is very abundant in many Middle Eastern oil fi

Question

Natural gas (mostly methane, CH4) 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 has a boiling point of ?164 ?C. One possible strategy is to oxidize the methane (in a 1:1 mole ratio) to methanol, CH3OH, which has a boiling point of 65 ?C and can therefore be shipped more readily.

Calculate the standard enthalpy changes for the combustion of a mole of methane and the combustion of a mole of methanol.

Explanation / Answer

Main reactions with methane are: combustion, steam reforming to syngas, and halogenation. In general, methane reactions are hard to control. Partial oxidation to methanol, for example, is difficult to achieve; the reaction typically progresses all the way to carbon dioxide and water."
" . . . (methane) . is considered to have an energy content of 39 megajoules per cubic meter, or 1,000 BTU per standard cubic foot."
The heat of combustion of methanol is 5,420 cal/g.
The ideal conversion of methane to methanol is
CH4 + H20 ? CH3OH + H2
but the process follows a very circuitous route.
The ideal yield would be 32 g CH3OH / 16g CH4
or 2:1
(1.07*10^10 ft^3)((1728 in^3/ft^3)(1 in/2.54 cm)^3 = 1.12830 * 10^12 ml
(1.12830 * 10^12 ml)(0.717 kg/10^6 ml) = 8.0899*10^5 g CH4 ==> 1.6180*10^6 g CH3OH
(1.6180*10^6 g) / 0.791 g/10^-3 L) = 2,045.5 L

Note that the conversion is basically endothermic.

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