Question

Gas pipelines are used to deliver natural gas (methane, \(\mathrm{CH}_{4}\) ) to the various regions of the United States. The total volume of natural gas that is delivered is on the order of \(2.7 \times 10^{12} \mathrm{L}\) per day, measured at STP. Calculate the total enthalpy change for combustion of this quantity of methane. (Note: Less than this amount of methane is actually combusted daily. Some of the delivered gas is passed through to other regions.)

Short answer

The total enthalpy change for the combustion of \(2.7 \times 10^{12}\text{ L}\) methane at STP is approximately \( -9.63 \times 10^{13} \text{ kJ}\).

Step-by-step solution

Convert the volume of methane to moles

Since the volume of methane is given at STP (standard temperature and pressure), we can use the ideal gas law to convert the given volume to moles: \[PV = nRT\] At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm. The ideal gas constant (R) is 0.0821 L atm K^{-1} mol^{-1}. We are given the volume (V) as \(2.7 \times 10^{12} L\) and need to find the number of moles (n). Rearranging the ideal gas law and plugging in the values, we get: \[n = \frac{PV}{RT} = \frac{(1)(2.7 \times 10^{12} L)}{(0.0821)(273.15)} \approx 1.2 \times 10^{11} \text{ mol}\]

Determine moles of reactants and products

The balanced chemical equation for the combustion of methane is: \[\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] From this equation, we can see that for every mole of methane combusted, one mole of carbon dioxide is produced and two moles of water are produced. Since we have \(1.2 \times 10^{11} \text{ mol}\) of methane, we will have the same amount of carbon dioxide and twice the amount of water produced.

Calculate the enthalpy change for the combustion of methane

We will now calculate the enthalpy change for the combustion of methane using the standard enthalpies of formation (\(\Delta H_f^\circ\)) for the reactants and products. For this, we have: \[\Delta H^\circ = \sum n_\text{products} \Delta H_f^\circ(\text{products}) - \sum n_\text{reactants} \Delta H_f^\circ(\text{reactants})\] The standard enthalpies of formation for the reactants and products are as follows: \[\Delta H_f^\circ(\mathrm{CH}_{4}) = -74.87 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{O}_{2}) = 0 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{CO}_{2}) = -393.51 \frac{\text{kJ}}{\text{mol}}\] \[\Delta H_f^\circ(\mathrm{H}_{2}\mathrm{O}) = -241.83 \frac{\text{kJ}}{\text{mol}}\] Applying these values to the equation above, we have: \[\Delta H^\circ = \left[ (1)(-393.51) + (2)(-241.83) \right] - \left[ (1)(-74.87) + (2)(0) \right] = -802.33 \frac{\text{kJ}}{\text{mol}}\]

Calculate the total enthalpy change

Finally, we will multiply the enthalpy change for the combustion of one mole of methane by the total moles of methane to obtain the total enthalpy change: \[\Delta H_\text{total} = (1.2 \times 10^{11} \text{mol}) \times (-802.33 \text{ kJ/mol}) = -9.63 \times 10^{13} \text{ kJ}\] So, the total enthalpy change for the combustion of this quantity of methane is approximately \( -9.63 \times 10^{13} \text{ kJ}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. This relationship is expressed with the formula: \[ PV = nRT \] where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹).
• \(T\) is the temperature in Kelvin.

The ideal gas law is crucial when calculating the amount of gas in moles if the gas is at standard temperature and pressure (STP). At STP, which is 273.15 K and 1 atm, the ideal gas law simplifies the conversion from volume to moles. For instance, in the exercise scenario, it allowed us to convert the massive volume of methane into a workable number of moles to proceed with further calculations.

Methane Combustion

Methane combustion is a process that occurs when methane (\(\mathrm{CH}_{4}\)) reacts with oxygen (\(\mathrm{O}_{2}\)) to produce carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). This reaction is both exothermic and significant in energy production. The equation representing this combustion reaction is:\[ \mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \]In real-world applications, such as the combustion in gas-powered vehicles or electricity generation, this reaction releases a considerable amount of energy. Understanding methane combustion is essential for improving energy efficiency and reducing greenhouse gas emissions.

Standard Enthalpy of Formation

The standard enthalpy of formation (\(\Delta H_f^\circ\)) is a concept in thermodynamics that refers to the heat change when one mole of a compound is formed from its elements under standard conditions (1 atm and 298 K). The values of standard enthalpies of formation are tabulated for common substances, allowing chemists to calculate the enthalpy change for a reaction, provided they know the \(\Delta H_f^\circ\) for both reactants and products. For methane combustion, the reaction enthalpy change \(\Delta H^\circ\) can be determined with the formula:\[ \Delta H^\circ = \sum n_\text{products} \Delta H_f^\circ(\text{products}) - \sum n_\text{reactants} \Delta H_f^\circ(\text{reactants}) \]By plugging in respective \(\Delta H_f^\circ\) values for carbon dioxide, water, methane, and oxygen, the net enthalpy change of the reaction can be evaluated.

Moles of Gas

Moles of gas refer to the quantity of gas measured in moles, which is a standard unit for expressing amounts of a chemical substance. By understanding the mole, chemists can quantify and relate the amount of gaseous reactants and products in a chemical equation.To calculate moles from a given volume of gas using the ideal gas law, the equation is rearranged to:\[ n = \frac{PV}{RT} \]In our exercise, knowing the volume and conditions at standard temperature and pressure (STP) allows us to calculate the moles of methane. This, in turn, facilitates the determination of the reaction extent and energy changes during combustion.

Balanced Chemical Equation

A balanced chemical equation is vital in representing the stoichiometry of a reaction, showing the exact relationship between reactants and products. Achieving a balanced equation involves ensuring that the number of atoms for each element is the same on both sides of the equation. For methane combustion:\[ \mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O} \]This equation indicates that one mole of methane reacts with two moles of oxygen to form one mole of carbon dioxide and two moles of water. Such relationships are crucial for calculating theoretical yields and energy changes, ensuring that the reaction is completely and efficiently described.