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 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?

Short answer

The volume of methanol formed is \(5.007 \times 10^{11} \: mL\). The balanced chemical equations for the combustion of methane and methanol are \(CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\) and \(CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\), respectively. The total enthalpy change for the complete combustion of methane is \(-1.101 \times 10^{13} \: kJ\), and for methanol is \(-8.983 \times 10^{12} \: kJ\). 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). Therefore, from the standpoint of energy production, methane is a more efficient substance to combust per unit volume.

Step-by-step solution

Calculate the volume of methanol formed

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: \(\frac{mass}{density} = \frac{3.960 \times 10^{11} \: g}{0.791 \: g/mL} = 5.007 \times 10^{11} \: mL\)

Write balanced chemical equations for the combustion of methane and methanol

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: \(CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\)

Calculate the enthalpy change for the combustion of methane and methanol

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 = \((-726 \: kJ/mol)(1.236 \times 10^{10} \: mol) = -8.983 \times 10^{12} \: kJ\)

Compare the enthalpy change per unit volume of methane and methanol

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.

Key concepts

Methane Combustion

The combustion of methane is a chemical reaction where methane (CH _4) reacts with oxygen (O_2) to produce carbon dioxide (CO_2) and water (H_2O). This is a classic exothermic reaction, meaning it releases energy in the form of heat. This energy release is quantified by the enthalpy change, a key aspect when considering methane as a fuel source.

The balanced equation for methane combustion is:\[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l)\]To understand the efficiency of methane, it's important to grasp the concept of enthalpy of combustion. This is the energy released when one mole of methane burns completely in oxygen. For methane, the enthalpy of combustion is approximately -890 kJ/mol. The negative sign indicates energy release.

In practical terms, methane combustion is foundational in energy production. Its high energy yield makes it an effective and popular choice for heating and electricity generation across the globe.

Methanol Production

Methanol (CH_3OH) is a versatile chemical compound produced from methane primarily through oxidation processes. This production method transforms methane into methanol, enabling easier transportation and storage compared to liquefied natural gas. Methanol's boiling point is significantly higher at 65°C compared to methane's requirement for liquefaction at -164°C.

The conversion of methane to methanol is often represented as:\[CH_4(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)\]This process involves carefully controlled conditions to ensure conversion efficiency and safety, as the reaction must be prevented from proceeding to complete combustion.

Methanol not only facilitates transportation but also serves as a building block in producing chemicals, solvents, and fuels. This makes methanol production from natural gas a strategic operation in the energy and chemical sectors.

Enthalpy of Combustion

Enthalpy of combustion is a critical concept in understanding the energy efficiency of fuels like methane and methanol. It represents the amount of heat released when a specific amount of substance burns in oxygen.

• Methane: The enthalpy of combustion is about -890 kJ/mol.
• Methanol: This value is approximately -726 kJ/mol.

The difference in these values indicates the higher energy yield of methane per mole compared to methanol.

Calculating total enthalpy change helps compare energy production capabilities. For instance, the total energy released from combusting a large volume of methane will typically exceed that from an equivalent methanol volume, assuming similar conditions.

Energy producers often reference enthalpy changes to determine the most efficient and economically viable fuel options for power generation.

Gas to Liquid Conversion

Converting gas to liquid involves turning natural gas, primarily methane, into liquid forms like methanol. This gas-to-liquid conversion serves multiple industrial purposes, chiefly making transportation and storage more feasible and cost-effective.

• Transport Efficiency: Liquids are generally easier to transport than gases, enabling more efficient distribution of energy resources.
• Economic Benefits: By converting methane to methanol, industries can reduce the massive costs associated with transporting liquefied natural gas at very low temperatures.

These conversions leverage the flexibility of liquid fuels and chemicals in various applications, ranging from energy production to chemical manufacturing.

Understanding the principles behind gas-to-liquid technologies is vital for students and professionals in the energy sector, reflecting ongoing innovations in sustainable fuel processing and global energy logistics.