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 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 ${25}^{\circ }\mathrm{C}$ 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

In summary, the volume of methanol formed is $V_{C{H}_{3}OH}$ (calculated in Step 3) and the enthalpy change upon combustion of 1 mL of liquid methane is ΔH_methane, while for 1 mL of liquid methanol is ΔH_methanol (calculated in Step 5). Comparing the values of ΔH_methane and ΔH_methanol, we can determine which substance has the higher enthalpy of combustion per unit volume and thus the higher energy production.

Step-by-step solution

Convert the volume of methane to moles.

To convert the given volume of methane at atmospheric pressure and 25 °C to the number of moles, we will use the ideal gas law formula, $PV=nRT$ where P is the pressure (we will use 1 atm as atmospheric pressure), V is the volume, n is the number of moles, R is the gas constant in the appropriate units (0.0821 L atm/mol K for this case), and T is the temperature in Kelvins. First, convert the volume of methane from cubic feet to Liters: $1f{t}^3=28.3168L$ $V=(10.7\times {10}^{9}f{t}^3)\times (28.3168L/f{t}^3)$ Now, convert the temperature to Kelvins: $T=25+273.15=298.15K$ Using the ideal gas law, we can calculate the number of moles of methane: $n_{C{H}_4}=\frac{PV}{RT}=\frac{(1atm)(V)}{(0.0821Latm/molK)(298.15K)}$

Write the balanced chemical equation for the oxidation of methane to methanol.

The balanced equation for methane oxidation to methanol is: $C{H}_4(g)+\frac{1}{2}{O}_2(g)\to C{H}_{3}OH(g)$

Calculate the moles and volume of methanol formed.

In the balanced equation for methane oxidation to methanol, the stoichiometry is 1:1, which means one mole of methane produces one mole of methanol. Therefore, the moles of methanol formed are equal to the moles of methane. Now, convert the moles of methanol to volume using the given density of methanol, 0.791 g/mL. First, calculate the mass of methanol using its molar mass (32.04 g/mol): $mass=n_{C{H}_{3}OH}\times (32.04\frac{g}{mol})$ Next, calculate the volume of methanol in mL: $V_{C{H}_{3}OH}=\frac{mass}{density}=\frac{mass}{0.791\frac{g}{mL}}$

Write balanced chemical equations for complete combustion and calculate the total enthalpy change.

Balanced chemical equations for complete combustion of methane and methanol: $C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(l)$ $C{H}_{3}OH(g)+\frac{3}{2}{O}_2(g)\to C{O}_2(g)+2H_{2}O(l)$ We need to find the enthalpy change for complete combustion of the given amount of methane and the equivalent amount of methanol. We can use the standard enthalpy of combustion, ΔH_c for both substances: ΔH_c(CH_4) = -890.4 kJ/mol ΔH_c(CH_3OH) = -726.0 kJ/mol The total enthalpy change for methane can be calculated by: $\Delta H_{total}^{C{H}_4}=n_{C{H}_4}\times \Delta H_c^{C{H}_4}$ The total enthalpy change for the methanol can be calculated by: $\Delta H_{total}^{C{H}_{3}OH}=n_{C{H}_{3}OH}\times \Delta H_c^{C{H}_{3}OH}$

Compare the enthalpy change per unit volume for liquid methane and liquid methanol.

To compare the enthalpy change per unit volume for liquid methane and liquid methanol at 25 °C, we can use their densities and molar masses: Density_methane = 0.466 g/mL Density_methanol = 0.791 g/mL Molar_mass_methane = 16.04 g/mol Molar_mass_methanol = 32.04 g/mol First, find the moles of each substance in 1 mL: n_methane = (Density_methane / Molar_mass_methane) mol/mL n_methanol = (Density_methanol / Molar_mass_methanol) mol/mL Next, calculate the enthalpy change upon combustion of 1 mL of each substance: ΔH_methane = n_methane × ΔH_c(CH_4) kJ/mL ΔH_methanol = n_methanol × ΔH_c(CH_3OH) kJ/mL Finally, compare the values of ΔH_methane and ΔH_methanol to determine which substance has the higher enthalpy of combustion per unit volume.

Key concepts

Methane to Methanol Conversion

Methane to methanol conversion is a process that has gathered interest due to the storage and transportation benefits of methanol. Unlike methane, which requires extremely low temperatures to liquefy, methanol is a liquid at room temperature, making it more convenient to handle and transport.

When methane is oxidized to become methanol, the process typically involves a reaction with oxygen. The conversion can be represented by the equation:
$C{H}_4(g)+\frac{1}{2}{O}_2(g)\to C{H}_{3}OH(g)$
This implies for every one mole of methane, one mole of methanol is produced, maintaining the conservation of mass and moles as indicated by stoichiometry.

Oxidation Reactions

Oxidation reactions are fundamental in chemical processes and energy production. They involve the transfer of electrons from one substance to another, resulting in the production of energy. In the context of the combustion of methane and methanol, the oxidation reactions are those that convert these substances to carbon dioxide and water, releasing energy in the form of heat.

For example:
$C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(l)$
The combustion of methane is a highly exothermic process, and the energy released is the enthalpy change of the reaction. Oxidation reactions are essential for understanding the energy content of different fuels.

Stoichiometry

Stoichiometry plays a critical role in chemical equations and reactions. It is the quantitative relationship between reactants and products in a chemical reaction. Using stoichiometry, we can calculate the amounts of reactants needed and products formed. For the conversion of methane to methanol, stoichiometry tells us that the reaction is a one-to-one conversion. Similarly, the stoichiometry of the combustion reactions allows us to determine the necessary amount of oxygen for complete combustion and the amount of carbon dioxide and water produced.

This principle also helps us convert between the volumes, masses, and moles of substances involved in a reaction, making it indispensable for calculations related to chemical processes.

Ideal Gas Law

The ideal gas law is a critical equation in chemistry and physics, represented as $PV=nRT$, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. This law describes the behavior of ideal gases and establishes a relationship between these variables.

In this problem, we use the ideal gas law to convert the given volume of methane to moles at standard atmospheric pressure and a specific temperature. The number of moles of methane dictates the number of moles of methanol formed after the conversion, due to the stoichiometric relationship in the chemical equation.

Standard Enthalpy of Combustion

The standard enthalpy of combustion is the amount of heat released when one mole of a substance is burned in the presence of oxygen under standard conditions. It is an important parameter for comparing the energy content of different fuels. Methane and methanol have different standard enthalpies of combustion; methane's is significantly higher due to its simpler structure and higher hydrogen to carbon ratio, meaning it releases more energy per mole when combusted.

In comparing the fuels, one must consider the energy density by volume. The enthalpy changes per unit volume are calculated using the densities and molar masses of the substances to determine which fuel is more energy-efficient. Thus, the standard enthalpy of combustion is not just about the energy released, but also about understanding energy content in relation to volume, which is crucial for practical applications like fuel storage and transport.