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.

Part A What volume of methanol is formed if 3.03×1011 L of methane at standard atmospheric pressure and 25 C is oxidized to methanol? The density of CH3OH is 0.791 g/mL. Assume that the oxidation of methane to methanol occurs in a 1:1 stoichiometry.

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

The standard heats of formation for selected substances are shown here.

Substance Hf
(kJ/mol) CH4(g) 74.8 CH3OH(l) 238.6 CO2(g) 393.5 H2O(l) 285.83 O2(g) 0

Explanation / Answer

a)

The oxidation of methanol is written as

CH4 + 0.5O2 = CH3OH

where 3.03×1011 L of methane at standard atmospheric pressure and 25C is oxidized to methanol.

The density of CH3OH is 0.791 g/mL.

And the oxidation of methane to methanol occurs in a 1:1 stoichiometry.

3.03×1011 L of methane at standard atmospheric pressure(101.325 kPa) and 25C(i.e. 298.15K) is oxidized to methanol.

using ideal gas law,

PV = nRT

101.325*3.03×1011 = n*8.314*298.15

n = 1.24×1010 mols.

Hence  1.24×1010 mols of methane oxidizes to methanol. Since stoichiometry methane to methanol is 1:1,

1.24×1010 mols of methanol are formed.

Molecular weight of methanol = 32 g/mol

weight of methanol = 32* 1.24×1010 = 3.968×1011 gms

density of methanol is 0.791 g/ml

volume of methanol formed = 3.968×1011 /0.791 = 5.016×1011 ml = 5.016×108 L

B)

the standard enthalpy changes for the combustion of a mole of methane

CH4 + 0.5O2 = CH3OH

Hcombustion =Hf product - Hf reactants

Hcombustion = Hf CH3OH - Hf CH4 - 0.5Hf O2

Hcombustion = -238.6 - (-74.8) - 0

Hcombustion = -163.8 kJ/mol

the standard enthalpy changes for the combustion of a mole of methanol

CH3OH + 2O2 = CO2 + 2H2O

Hcombustion =Hf product - Hf reactants

Hcombustion =Hf CO2 + 2Hf H2O - Hf CH3OH + 2Hf O2

Hcombustion = -393.5 - 2*285.83 - (-238.6) + 2*0

Hcombustion = -726.56 kJ/mol

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