Natural gas is very abundant in many Middle Eastern oil fields. However, the cos

Question

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 C. One possible strategy is to oxidize the methane to methanol, CH3OH, which has a boiling point of 65 C and can therefore be shipped more readily. Suppose that 1.06×1010 ft3 of methane at atmospheric pressure and 25 C are oxidized to methanol.

(Part A.) What volume of methanol is formed if the density of CH3OH is 0.791 g/mL?

( Part B) Calculate the total enthalpy change for complete combustion of the 1.06×1010 ft3 of methane described above.

(Part C) Calculate the total enthalpy change for complete combustion of the equivalent amount of methanol, as calculated in part A

Explanation / Answer

PART (A)

CH4 (g) + 1/2 O2 ---------> CH3OH (l)

1 ft3 = 28.32 L

then, 1.06 * 1010 ft3 = 30.02 L

assume ideal behaviour of methane,

P V = n R T
1* 30.02 = n * 0.0821 * 298

n = 1.227 mol

According above stoichiometric equation,

1 mol methane = 1 mol methanol

then, 1.227 mole methane = 1.227 mol methanol

Mass of methanol obtained = 1.227 * 32 = 39.26 g.

It is given that,

0.791 g. of methanol occupies 1 mL of volume

then, 39.26 g. of methanol occupies 1 * 39.26 / 0.791 = 49.63 mL

PART (B)

Standard enthalpy of combustion of methane = -890 kJ/mol

then, enthalpy of combustion of 1.227 mol methane = - 890 * 1.227 = - 1092.03 kJ

PART (C)
Standar enthalpy of combustion of methanol = - 715.0 kJ/mol

the, enthalpy of combustion of 1.227 mol of methanol = 1.227 * (- 715.0) = 877.31 kJ

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