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. Substance Hf (kJ/mol) CH4(g) 74.8 CH3OH(l) 238.6 CO2(g) 393.5 H2O(l) 285.83 O2(g) 0 Part B Calculate the standard enthalpy changes for the combustion of a mole of methane and the combustion of a mole of methanol. Enter the enthalpy for methane followed by the enthalpy for methanol in kilojoules separated by a comma using four significant figures.

Explanation / Answer

For methane, the equation is

CH4(g) + O2(g) --> 2H2O (l) + CO2 (g)

For methanol:

2CH3OH(g) + 3O2(g) --> 2CO2(g) + 4H2O (l)

Methane standard enthalpy change

2(-285.83) - 393.5 + 74.8 = -890.36 kJ

Methanol standard enthalpy change

for 2 moles

2(-238.6) + 3(0) - (2(-393.5) + 4(-285.83)) = 1453.12 kJ

for 1mole :

= 726.56 kJ

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