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 1.06x1010 ft of methane at atmospheric pressure and 25 °C are oxidized to methanol.

Answer 1

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

Part A

no of mol of methane(n) = PV/RT

   P = atmospheric pressure = 1 atm

   v = volume of methane = 1.06*10^10 ft3 = 3*10^11 L

   R = 0.0821 l.atm.k-1.mol-1

   T = 25c = 298 k

n = (1*3*10^11)/(0.0821*298)

   = 1.23*10^10 mol

from equation, 2 mol CH4 = 2 mol CH3OH

No of mol of CH3OH formed = 1.23*10^10 mol

mass of CH3OH formed = 1.23*10^10*32 = 3.94*10^11 g

volume of CH3OH formed = 3.94*10^11/0.791 = 4.98*10^11 ml

                       = 4.98*10^8 L

Part B

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

Part C

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

Part D

DH0c,methane = -802.3 kj/mol

enthalpy of combustion of methane = 1.23*10^10*802.3

                                  = 9.87*10^12 kj

part E

DH0c,methanol = -726.56 kj/mol

enthalpy of combustion of methanol = 1.23*10^10*726.56

                                   = 8.94*10^12 kj