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 \(-164^{\circ} \mathrm{C}\). 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 \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) are 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 \(\mathrm{CO}_{2}(\mathrm{~g})\) 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 described above 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 from the given volume of methane is approximately \(5.01 \times 10^8 \; \mathrm{L}\). The enthalpy change for complete combustion of the given volume of methane is approximately \(-1.102 \times 10^{13} \; \mathrm{kJ}\), and for the equivalent amount of methanol is approximately \(-8.998 \times 10^{12} \; \mathrm{kJ}\). Comparing the enthalpy change upon combustion per unit volume of liquid methane and liquid methanol, liquified methane has a higher enthalpy of combustion per unit volume at approximately \(-12.4 \frac{\mathrm{kJ}}{\mathrm{mL}}\), while liquid methanol is approximately \(-8.5 \frac{\mathrm{kJ}}{\mathrm{mL}}\).

Step-by-step solution

** Part (a): Calculate the volume of methanol

** To calculate the volume of methanol, we first need to calculate the moles of methane, then its moles of methanol will be equal and using the density of methanol, find the volume of methanol. Step 1: Convert the given volume of methane to moles. We are given the volume of methane as \(10.7 \times 10^9 \; \mathrm{ft^3}\) at atmospheric pressure and 25℃. Using the ideal gas law, we can calculate the moles of methane. PV = nRT, where P is pressure, V is volume, n is number of moles, R is ideal gas constant, and T is temperature in Kelvin. First, we need to convert the volume to liters, and temperature to Kelvin. \(10.7 \times 10^9 \; \mathrm{ft^3} \times \frac{28.3168 \; \mathrm{L}}{1 \; \mathrm{ft^3}} = 3.03 \times 10^{11} \; \mathrm{L}\) \(25^{\circ} \mathrm{C} = 298 \; \mathrm{K}\) Now, we use the ideal gas law, knowing R = 0.0821 \(\frac{\mathrm{L \; atm}}{\mathrm{mol \; K}}\), and atmospheric pressure (P) = 1 atm, we get: 1 atm × \((3.03 \times 10^{11} \; \mathrm{L}) = n \times 0.0821 (\frac{\mathrm{L \; atm}}{\mathrm{mol \; K}}) \times 298 \; \mathrm{K}\) \(n = \frac{3.03 \times 10^{11} \; \mathrm{L}}{0.0821 \times 298 \; (\frac{\mathrm{L \; atm}}{\mathrm{mol \; K}})} = 1.236 \times 10^{10} \; \mathrm{mol}\) Step 2: Convert moles of methane to moles of methanol. Since the reaction has a 1:1 mole ratio between methane and methanol, the moles of methanol will be equal to moles of methane we calculated earlier: \(n_{methanol} = 1.236 \times 10^{10} \; \mathrm{mol}\) Step 3: Calculate the volume of methanol based on its density. To calculate the volume of methanol, we need to convert the moles to grams, and then use the density to convert to mL. GIVEN: Density of methanol = \(0.791 \frac{\mathrm{g}}{\mathrm{mL}}\), molar mass of methanol = 32.042 g/mol The mass of methanol = moles × molar mass = \((1.236 \times 10^{10} \; \mathrm{mol}) \times (32.042 \frac{\mathrm{g}}{\mathrm{mol}}) = 3.96 \times 10^{11} \; \mathrm{g}\) Now, using the density, we can calculate the volume of methanol in mL and convert it to L (1 L = 1000 mL): Volume of methanol = \(mass \times \frac {1}{density} = (3.96 \times 10^{11} \mathrm{g}) \times \frac {1 \mathrm{mL}}{0.791 \mathrm{g}} = 5.01 \times 10^{11} \; \mathrm{mL} \implies \approx 5.01 \times 10^8 \; \mathrm{L}\) Answer (a): The volume of methanol formed is approximately \(5.01 \times 10^8 \; \mathrm{L}\). **

** Part (b): Write balanced chemical equations for methane and methanol combustion

** Methane combustion: Balanced equation: \(\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\) Methanol combustion: Balanced equation: \(\mathrm{CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\) Now, we will calculate the enthalpy change for complete combustion of given methane and its equivalent methanol using their enthalpies of combustion. Enthalpy of combustion for 1 mol of methane = \(-890.4 \; \mathrm{kJ}\); For 1 mol of methanol = \(-727.3 \; \mathrm{kJ}\) For \(1.236 \times 10^{10} \; \mathrm{mol}\) of methane: Enthalpy change (methane) = \(moles \times enthalpy\_of\_combustion = (1.236 \times 10^{10} \mathrm{mol}) \times (-890.4 \; \mathrm{kJ/mol}) = -1.102 \times 10^{13} \; \mathrm{kJ}\) For \(1.236 \times 10^{10} \; \mathrm{mol}\) of methanol: Enthalpy change (methanol) = \(moles \times enthalpy\_of\_combustion = (1.236 \times 10^{10} \mathrm{mol}) \times (-727.3 \mathrm{kJ/mol}) = -8.998 \times 10^{12} \; \mathrm{kJ}\) Answer (b): Enthalpy change for complete combustion of the given volume of methane is approximately \(-1.102 \times 10^{13} \; \mathrm{kJ}\), and for the equivalent amount of methanol is approximately \(-8.998 \times 10^{12} \; \mathrm{kJ}\). **

** Part (c): Compare enthalpy change upon combustion per unit volume of liquid methane and liquid methanol

** First, we need to calculate the enthalpy change per unit volume for both liquid methane and liquid methanol. Enthalpy change per unit volume = \(\frac{Enthalpy change per mol}{Density \times Molar mass}\) Density of liquified methane (GIVEN) = \(0.466 \frac{\mathrm{g}}{\mathrm{mL}}\), molar mass of methane = 16.042 g/mol Enthalpy change per unit volume for liquified Methane = \(\frac{-890.4 \mathrm{kJ/mol}}{0.466 \frac{\mathrm{g}}{\mathrm{mL}} \times 16.042 \mathrm{g/mol}} = -12.4 \frac{\mathrm{kJ}}{\mathrm{mL}}\) Density of methanol at 25℃ (GIVEN) = \(0.791 \frac{\mathrm{g}}{\mathrm{mL}}\) Enthalpy change per unit volume for Methanol = \(\frac{-727.3 \; \mathrm{kJ/mol}}{0.791 \frac{\mathrm{g}}{\mathrm{mL}} \times 32.042 \mathrm{g/mol}} = -8.5 \frac{\mathrm{kJ}}{\mathrm{mL}}\) Answer (c): The enthalpy change upon combustion per unit volume of liquid methane is approximately \(-12.4 \frac{\mathrm{kJ}}{\mathrm{mL}}\), and for liquid methanol is approximately \(-8.5 \frac{\mathrm{kJ}}{\mathrm{mL}}\). From the standpoint of energy production, liquified Methane has a higher enthalpy of combustion per unit volume.

Key concepts

Methane Combustion

Methane combustion is a critical chemical reaction used in energy production and industry. Methane (\(\text{CH}_4\)) is the simplest alkane and is the primary component of natural gas. When methane undergoes combustion, it reacts with oxygen to produce carbon dioxide and water vapor, releasing energy in the form of heat. The balanced chemical equation for methane combustion is:\[\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\]This reaction is exothermic, meaning it releases heat, which is harnessed for various applications like heating and electricity generation.- **Exothermic Reaction**: Releases energy.- **Products**: Carbon dioxide and water.- **Efficiency**: High energy output.Methane is favored in many industries due to its clean-burning nature, meaning it produces less pollution compared to other fossil fuels. Understanding the intricacies of methane combustion helps in optimizing its use as a clean energy source.

Methanol Combustion

Methanol (\(\text{CH}_3\text{OH}\)), also known as wood alcohol, is another compound used in combustion reactions. It can be manufactured from methane through oxidation. Methanol is easier to transport in liquid form due to its higher boiling point compared to methane. During combustion, methanol reacts with oxygen to produce carbon dioxide and water, similar to methane combustion. The balanced equation for methanol combustion is:\[\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)\]Key points about methanol combustion:- **Exothermic Reaction**: Like methane, it releases energy.- **Products**: Carbon dioxide and water.- **Transportability**: Easier to ship because of liquid form.Methanol is used as a fuel, solvent, and antifreeze. Its lower energy content per mol compared to methane makes it less efficient, but its liquid form offers logistic advantages, especially in regions where natural gas transport isn't feasible.

Enthalpy Change

The enthalpy change (\(\Delta H\)) of a reaction is the heat absorbed or released during a chemical reaction at constant pressure. For combustion reactions, this value is typically negative since they are exothermic.- **Methane**: The enthalpy of combustion for methane is \(-890.4 \, \text{kJ/mol}\).- **Methanol**: The enthalpy of combustion is \(-727.3 \, \text{kJ/mol}\).To find the total energy released, the enthalpy change is calculated by multiplying the moles of the substance by their respective enthalpy of combustion. For the combustion of \(1.236 \times 10^{10}\) moles of methane, the calculated enthalpy change is \(-1.102 \times 10^{13} \, \text{kJ}\). Likewise, for methanol, it is \(-8.998 \times 10^{12} \, \text{kJ}\).Large negative values indicate significant energy release, emphasizing the efficiency of these fuels in energy production. Understanding enthalpy change is crucial for effectively harnessing chemical reactions in practical applications.

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry and physics that relates the pressure, volume, temperature, and moles of an ideal gas. It is given by the formula:\[PV = nRT\]where:- \(P\) is the pressure,- \(V\) is the volume,- \(n\) is the number of moles,- \(R\) is the ideal gas constant (\(0.0821 \, \text{L} \, \text{atm/mol} \, \text{K}\)),- \(T\) is the temperature in Kelvin.By using the ideal gas law, we can convert the known volume of methane in cubic feet to moles when given atmospheric pressure and temperature. This relationship is essential for calculating the amount of gas reacting in processes like combustion. The flexibility offered by the ideal gas law makes it a powerful tool for chemists when dealing with gaseous reactants and products, ensuring precise calculations in reactions and energy outputs.