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 are high because it is necessary to point at atmospheric pressure of \(-164^{\circ} \mathrm{C} .\) One possible strategy is to oxidize the methane to methanol, CH \(_{3} \mathrm{OH}\) , which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is 0.791 \(\mathrm{g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of 0.466 \(\mathrm{g} / \mathrm{mL}\) ; the density of methanol at \(25^{\circ} \mathrm{C}\) is 0.791 \(\mathrm{g} / \mathrm{mL} .\) Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

First, we'll need to find the number of moles of methane in the given volume. The molar volume of an ideal gas under standard temperature and pressure (STP: 273.15K and 1 atm) is 22.41 L/mol. We can convert the volume given from ft³ to L and then find the number of moles using the ideal gas law. Given, volume of methane = \(10.7 × 10^9\) ft³. To convert ft³ to liters (L), we use the conversion factor: \(1 \: ft^3 = 28.3168466 \: L\) Volume of methane in liters = \(10.7 × 10^9 \: ft^3 * 28.3168466 \: L/ft^3 = 3.0272 × 10^{11} \: L\) Now, use the ideal gas law to find the moles of methane: \(PV = nRT\), where P = pressure (atm), V = volume (L), n = moles, R = gas constant 0.08206 (\(L \cdot atm / (mol \cdot K)\)), and T = temperature in Kelvin (K). Temperature in Kelvin = \(273.15 + 25 = 298.15 \: K。\) We know that it is at atmospheric pressure, i.e., P = 1 atm. So, we can calculate n as follows: \(n = \frac{PV}{RT} = \frac{(1 \: atm)(3.0272 \times 10^{11} \: L)}{(0.08206 \: L \cdot atm / (mol \cdot K))(298.15 \: K)} = 1.236 \times 10^{10} \: mol\) The mole ratio of methane to methanol in the reaction is 1:1; thus, the same number of moles will be formed. Next, we will use the density formula \(mass = (density)(volume)\) to find the volume of methanol formed. First, find the mass of methanol: \(mass\ of\ methanol = (1.236 × 10^{10} \: mol)(32.04 \: g/mol) = 3.960 \times 10^{11} \: g\) Now, calculate the volume using the given density of methanol: \(\frac{mass}{density} = \frac{3.960 \times 10^{11} \: g}{0.791 \: g/mL} = 5.007 \times 10^{11} \: mL\)

For the combustion of methane: \(CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\) For the combustion of methanol: \(CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\)

Use the balanced chemical equation and available enthalpy change data for the reactions to calculate the total enthalpy change for both the complete combustion of methane and methanol. Enthalpy change for complete combustion of methane: ΔH°(CH₄) = -890 kJ/mol (enthalpy change for the combustion of one mole of methane) Total enthalpy change for methane = ΔH°(CH₄) × moles of methane = \((-890 \: kJ/mol)(1.236 \times 10^{10} \: mol) = -1.101 \times 10^{13} \: kJ\) ΔH°(CH₃OH) = -726 kJ/mol (enthalpy change for the combustion of one mole of methanol) Total enthalpy change for complete combustion of methanol = ΔH°(CH₃OH) × moles of methanol = \((-726 \: kJ/mol)(1.236 \times 10^{10} \: mol) = -8.983 \times 10^{12} \: kJ\)

Given the density of liquid methane, 0.466 g/mL, and liquid methanol, 0.791 g/mL, calculate enthalpy changes per unit volume for both. Enthalpy change per unit volume of methane = ΔH°(CH₄) × (\(\frac{g}{mL}\)) \(-890 \: kJ/mol \cdot \frac{0.466 \: g/mL}{16.04 \: g/mol} = -25.73 \: kJ/mL\) Enthalpy change per unit volume of methanol = ΔH°(CH₃OH) × (\(\frac{g}{mL}\)) \(-726 \: kJ/mol \cdot \frac{0.791 \: g/mL}{32.04 \: g/mol} = -17.88 \: kJ/mL\) Comparing the enthalpy changes per unit volume, liquid methane (-25.73 kJ/mL) has a higher enthalpy of combustion compared to liquid methanol (-17.88 kJ/mL). From the standpoint of energy production, methane is a more efficient substance to combust per unit volume.