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 has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, $\mathrm{CH}_{3} \mathrm{OH},\( which has a boiling point of \)65^{\circ} \mathrm{C}$ and can therefore be shipped more readily. Suppose that $3.03 \times 10^{8} \mathrm{~m}^{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 $0.791 \mathrm{~g} / \mathrm{mL} ?(\mathbf{b})$ 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 \(3.03 \times 10^{8} \mathrm{~m}^{3}\) 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?

Step by step solution

01

Step 1:Calculate the volume of methanol formed by oxidizing the given volume of methane

First, let's use the ideal gas law to find the amount of methane in moles: PV = nRT Here, P= 1 atm, V = \(3.03 \times 10^8\) m^3, R = 0.0821 L atm /(K mol), T= 25 + 273 = 298 K n = PV / RT = (1 * \(3.03 \times 10^8\) * 1000) / (0.0821 * 298) = 1.2 * 10^7 moles Now, let's use the stoichiometry of methane oxidizing into methanol to find the amount of methanol formed: CH\(_4\) + \(\frac{1}{2}\) O\(_2\) → CH\(_3\)OH Since the reaction has a 1:1 stoichiometry, 1.2 * 10^7 moles of methanol are formed. Next, let's find the volume of formed methanol. The density of CH\(_3\)OH is \(0.791 \frac{g}{mL}\). Hence, we can calculate the volume as follows: Volume of CH\(_3\)OH = \(\frac{1.2 \times 10^7\text{mol}\times 32 \frac{g}{\text{mol}}}{0.791 \frac{g}{\text{mL}}}\) = \(4.87 \times 10^8 \)mL.

02

Write balanced chemical equations for the oxidation of methane and methanol to CO\(_2\) and H\(_2\)O

For the oxidation of methane: CH\(_4\) + 2O\(_2\) → CO\(_2\) + 2H\(_2\)O For the oxidation of methanol: CH\(_3\)OH + \(\frac{3}{2}\) O\(_2\) → CO\(_2\) + 2H\(_2\)O

03

Calculate the total enthalpy change for complete combustion of the given volume of methane and the equivalent amount of methanol

The given enthalpies of combustion are as follows: ΔH\(_c\) (CH\(_4\)) = -890 kJ/mol ΔH\(_c\) (CH\(_3\)OH) = -726 kJ/mol For methane combustion: ΔH = n * ΔH\(_c\) (CH\(_4\)) = \(1.2\times 10^7\) mol * -890 kJ/mol = -\(1.07\times 10^{10}\) kJ For methanol combustion (the equivalent amount from part a): ΔH = n * ΔH\(_c\) (CH\(_3\)OH) = \(1.2\times 10^7\) mol * -726 kJ/mol = -\(8.71\times 10^9\) kJ

04

Compare the enthalpy change of a unit volume of liquid methane and liquid methanol

First, we need to calculate the number of moles of each substance in each unit volume (1 mL) since densities are in g/mL. Molarity of liquid methane = \(\frac{0.466 \frac{g}{\text{mL}}}{16 \frac{g}{\text{mol}}}\)= 0.0291 mol/mL Molarity of liquid methanol = \(\frac{0.791 \frac{g}{\text{mL}}}{32 \frac{g}{\text{mol}}}\)= 0.0247 mol/mL The enthalpy of combustion per unit volume would be ΔH/volume for each substance: Enthalpy of combustion per unit volume for methane = ΔH\(_c\) (CH\(_4\)) * Molarity(CH\(_4\)) = -890 kJ/mol * 0.0291 mol/mL = -25.87 kJ/mL Enthalpy of combustion per unit volume for methanol = ΔH\(_c\) (CH\(_3\)OH) * Molarity(CH\(_3\)OH) = -726 kJ/mol * 0.0247 mol/mL = -17.92 kJ/mL Methane has a higher enthalpy of combustion per unit volume (25.87 kJ/mL) compared to methanol (17.92 kJ/mL). Thus, from the standpoint of energy production per unit volume, methane is the preferable choice.

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