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 STP, must be burned? \(\left(\Delta H_{\text {combustion }}^{\circ}\right.\) for \(\mathrm{CH}_{4}=-891 \mathrm{~kJ} / \mathrm{mol}\) )

Short answer

The volume of methane required to provide the given amount of energy to heat the home is approximately 105,484.56 liters at STP.

Step-by-step solution

Calculate moles of methane needed to produce the given energy

We are given the energy needed to heat the home as 4.19 x 10^6 kJ and the enthalpy of combustion of methane as -891 kJ/mol. We will use these values to find the moles of methane needed to produce the given energy. Energy_needed = Enthalpy_of_combustion x Moles_of_methane Moles_of_methane = Energy_needed / Enthalpy_of_combustion = \( \frac{4.19 \times 10^6 \,\text{kJ}}{-891 \,\text{kJ/mol}} \)

Solve for moles of methane

Now we divide the energy needed by the enthalpy of combustion to find the moles of methane needed: Moles_of_methane = \( \frac{4.19 \times 10^6}{-891} \) Moles_of_methane ≈ 4702.47 mol

Use the ideal gas law to find the volume of methane

Now, we have the moles of methane needed. We will use the ideal gas law, PV=nRT, to find the volume of methane at STP (Standard Temperature and Pressure). At STP, T = 273.15 K and P = 1 atm (1 atm = 101325 Pa). R (universal gas constant) = 8.314 J/(mol*K) Since pressure is given in atm, we will use R in L*atm/mol*K which is equal to 0.0821 L*atm/mol*K. Now we can solve for the volume of methane using the ideal gas law, PV=nRT: V = \( \frac{nRT}{P} \) V = \( \frac{4702.47 \, \text{mol} \times 0.0821\, \text{L*atm/mol*K} \times 273.15\, \text{K}}{1 \, \text{atm}} \)

Solve for the volume of methane

Now we can calculate the volume of methane required: V ≈ \( \frac{4702.47 \times 0.0821 \times 273.15}{1} \) V ≈ 105484.56 L The volume of methane required to provide the given amount of energy to heat the home is approximately 105,484.56 liters at STP.