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 $3.03\times {10}^8{\mathrm{m}}^3$ 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?

Short answer

The volume of methanol formed is $4.87\times {10}^8$ mL. The balanced chemical equations for the oxidation of methane and methanol to CO${}_2$ and H${}_2$O are: 1. CH${}_4$ + 2O${}_2$ → CO${}_2$ + 2H${}_2$O 2. CH${}_3$OH + $\frac{3}{2}$ O${}_2$ → CO${}_2$ + 2H${}_2$O The total enthalpy change for complete combustion of methane is -$1.07\times {10}^{10}$ kJ and for methanol is -$8.71\times {10}^9$ kJ. The enthalpy of combustion per unit volume for methane and methanol is -25.87 kJ/mL and -17.92 kJ/mL, respectively. Thus, methane has a higher enthalpy of combustion per unit volume and is a preferable choice from an energy production standpoint.

Step-by-step solution

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.

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

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

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.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and amount of gas in moles. It’s expressed with the formula: $PV=nRT$. Here, $P$ stands for pressure, $V$ for volume, $n$ for number of moles, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

In the exercise, the Ideal Gas Law was employed to find out how many moles of methane were present at given conditions, specifically atmospheric pressure and a temperature of ${25}^{\circ }\text{C}$.

• Convert the temperature to Kelvin by adding 273.
• Use the given values: $P=1\text{ atm}$, $V=3.03\times {10}^8{\text{ m}}^3$, and $R=0.0821\text{ L atm/(K mol)}$.

This allows the calculation of $n$, the moles of methane, which is crucial for further chemical calculations in the exercise.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and formation of chemical bonds, resulting in new products. In the given problem, the focus is on the oxidation of methane and later methanol.

• Oxidation of methane: ${\text{CH}}_4+2{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$.
• Oxidation of methanol: ${\text{CH}}_3\text{OH}+\frac{3}{2}{\text{O}}_2\to {\text{CO}}_2+2{\text{H}}_2\text{O}$.

Each equation represents a balanced chemical reaction because the number of atoms for each element is the same on both sides, adhering to the Law of Conservation of Mass. In these reactions, methane and methanol are being oxidized, indicating that they react with oxygen and form carbon dioxide and water.

Enthalpy Change

Enthalpy change, $\mathrm{\Delta }H$, is a measure of the heat energy change in a chemical reaction at constant pressure, often associated with combustion reactions. It’s commonly expressed in kilojoules per mole (kJ/mol).

For combustion reactions in the exercise:

• Enthalpy of combustion for methane, $\mathrm{\Delta }{H}_c({\text{CH}}_4)=-890\text{ kJ/mol}$.
• Enthalpy of combustion for methanol, $\mathrm{\Delta }{H}_c({\text{CH}}_3\text{OH})=-726\text{ kJ/mol}$.

The enthalpy change for a substance indicates how much energy is released or absorbed. Negative values here show that both reactions are exothermic, meaning they release energy. Methane has a higher enthalpy value, requiring more energy release upon complete combustion compared to methanol.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict how much product will form from a given amount of reactants.

In the exercise, the conversion of methane to methanol is represented by the stoichiometric equation:

• ${\text{CH}}_4+\frac{1}{2}{\text{O}}_2\to {\text{CH}}_3\text{OH}$.

This shows a 1:1 mole ratio, meaning each mole of methane produces one mole of methanol. Using stoichiometry, the exercise calculates the theoretical yield of methanol from the initial moles of methane. This principle ensures that reactions are predictable and can be scaled accurately, supporting industrial applications like energy production.