Question

Assume that \(4.19 \times 10^{6} \mathrm{kJ}\) of energy is needed to heat a home. If this energy is derived from the combustion of methane \(\left(\mathrm{CH}_{4}\right),\) what volume of methane, measured at 1.00 atm and \(0^{\circ} \mathrm{C},\) must be burned? \(\left(\Delta H_{\text {combustion }}^{\circ} \text { for } \mathrm{CH}_{4}=-891 \mathrm{kJ} / \mathrm{mol}\right)\).

Short answer

The short answer is: To find the volume of methane that must be burned to provide the given energy, first calculate the moles of methane needed: Moles of methane = \(\frac{4.19 \times 10^{6} \mathrm{kJ}}{-891 \mathrm{kJ/mol}}\) Then, use the ideal gas law to calculate the volume at the given temperature and pressure: V = \(\frac{\frac{4.19 \times 10^{6} \mathrm{kJ}}{-891 \mathrm{kJ/mol}}(0.0821)(273.15)}{1.00}\) Solve for V to find the required volume of methane.

Step-by-step solution

1. Calculate the moles of methane needed for the given energy requirement

We are given that the energy needed to heat the home is \(4.19 \times 10^{6} \mathrm{kJ}\), and the combustion of methane releases \(\Delta H_{\text {combustion}}^{\circ} = -891 \mathrm{kJ/mol}\). To find the moles of methane needed, we can write: Moles of methane = \(\frac{\text{Energy needed}}{\text{Energy released per mole}}\) Plugging in the values, we get: Moles of methane = \(\frac{4.19 \times 10^{6} \mathrm{kJ}}{-891 \mathrm{kJ/mol}}\)

2. Calculate the volume of methane at the given conditions

Now, we can use the ideal gas law to find the volume of methane at the given temperature and pressure. The ideal gas law is given by: \(PV = nRT\) Where, P = Pressure V = Volume n = moles of gas R = Gas constant T = Temperature We have been given the pressure, P = 1.00 atm, and the temperature, T = \(0^{\circ}\mathrm{C}\) (which is 273.15 K in Kelvins). We can also note that the gas constant, R, for this problem is \(0.0821 \frac{L \cdot atm}{K \cdot mol}\). Rearranging the ideal gas law formula to find the volume, we get: \(V = \frac{nRT}{P}\) Now, we can plug in the values for n (moles of methane calculated in step 1), R, T, and P: V = \(\frac{n(0.0821)(273.15)}{1.00}\) Substitute the value of Moles of methane from step 1 and solve for V: V = \(\frac{\frac{4.19 \times 10^{6} \mathrm{kJ}}{-891 \mathrm{kJ/mol}}(0.0821)(273.15)}{1.00}\) Calculate the volume, V, to find the required volume of methane that must be burned to provide the given energy.

Key concepts

Energy Combustion

When considering how energy is derived from combustion, it is important to focus on the concept of enthalpy change, especially with a process like methane combustion. Enthalpy, commonly represented as \(\Delta H\), helps us understand the heat exchange in reactions at constant pressure. During combustion, which is an exothermic reaction, energy is released when chemical bonds are broken and new ones are formed. This release of energy is why combustion is such an effective way to produce heat.
Combustion usually involves a substance (fuel) and oxygen, producing heat, light, and products like carbon dioxide and water. In our example, methane (CH₄), a common fuel, requires an understanding of its specific enthalpy change. The given \(-891 \, \text{kJ/mol}\)\ indicates this energy release per mole of methane combusted. Knowing this value is crucial as it helps us determine how many moles of methane are needed to meet any specific energy requirement, such as heating a home in colder climates.

Methane Combustion

Methane, or CH₄, is a simple hydrocarbon and a principal component of natural gas. It provides a highly efficient combustion process, resulting in the release of significant amounts of energy. Methane combustion is described by the chemical equation:\[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\]The equation shows that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water, releasing energy into the environment. Due to its high energy yield and clean combustion (producing only water and carbon dioxide), methane is widely used for residential and industrial heating. Understanding the complete chemical reaction is vital in quantifying the effects and efficiency of methane as an energy source. Additionally, the efficiency of methane combustion can often outperform other fossil fuels, producing a relatively lower carbon footprint when utilized properly.

Moles Calculation

In any chemical reaction, understanding the amount of reactants and products is vital. This is done by converting given quantities to moles, the unit used to express amounts of a chemical substance. Moles help us measure substances on a molecular level, allowing precise calculations with Avogadro’s number (6.022 x 10²³).
Here, we need to calculate the moles of methane required to produce a certain amount of energy. Using the energy provided per mole of combustion \(\Delta H = -891 \, \text{kJ/mol}\)\ and the total energy required (4.19 x 10⁶ kJ), we find the number of moles by dividing the total energy by the energy per mole release:\[\text{Moles of Methane} = \frac{4.19 \times 10^6 \, \text{kJ}}{-891 \, \text{kJ/mol}}\]
This calculation is a crucial step to solving the problem, showing how the demand for energy translates to a physical quantity of methane.

Gas Constant

The ideal gas law, \(PV = nRT\), is a fundamental equation in chemistry describing the behavior of gases under various temperatures and pressures. The gas constant \(R\) is a crucial component, representing how pressure, volume, temperature, and moles interact. In this context, \(R\) has a value of 0.0821 \(\frac{L \cdot atm}{K \cdot mol}\), commonly used for calculations involving gases at standard conditions.
This constant facilitates our understanding and computation of gas properties, ensuring we correctly calculate the volume (V) when given the amount of substance (n), pressure (P), and temperature (T). It's a standardized value, crucial in calculations where precise gas behavior is expected and required.

Temperature Conversion

Temperature plays a significant role in gas calculations, especially when using the Ideal Gas Law. For accurate results, it's critical to convert temperatures from degrees Celsius to Kelvin, as the Kelvin scale starts at absolute zero, ensuring all calculations remain physically meaningful.
To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature:

• 0°C to Kelvin: \(0 + 273.15 = 273.15 \, K\)

This conversion is non-negotiable when dealing with gas laws. Using Kelvin avoids situations where gas volumes could be calculated as negative or zero, which are not physically possible, thus ensuring our calculations align with the real behavior of gases.