Use Table 9.4 to calculate an approximate enthalpy or heat of reaction for the combustion of one mole of methane gas (CH4) to form gaseous H2O and CO2. What Volume (in L) of methane is needed to produce 3287 kJ of heat for methane gas at 11 oC and 3.40 atm pressure?

determine z as in the previous question for the reaction CH4 + O2 --> CO2 + H2O.

z kJ/mol CH4 x # mols CH4 = 3287 kJ. Solve for # mols CH4.
Then use PV = nRT and solve for V at the conditions listed.
# mols CH4

To calculate the volume of methane needed to produce 3287 kJ of heat, we first need to calculate the heat of reaction for the combustion of one mole of methane.

Table 9.4 provides standard enthalpy of formation values for various compounds, including methane (CH4), water vapor (H2O), and carbon dioxide (CO2). Since the question asks for an approximate enthalpy or heat of reaction, we can use these standard enthalpy values as estimates.

The balanced chemical equation for the combustion of methane is:

CH4 + 2O2 -> CO2 + 2H2O

Based on the equation, the enthalpy change for the combustion of one mole of methane is equal to the sum of the enthalpy changes for the formation of one mole of CO2 and two moles of H2O, minus the enthalpy change for the formation of one mole of CH4.

Using Table 9.4, we find the following standard enthalpy of formation values:

ΔHf°(CH4) = -74.80 kJ/mol
ΔHf°(CO2) = -393.51 kJ/mol
ΔHf°(H2O) = -241.82 kJ/mol

The enthalpy change for the combustion of one mole of methane can be calculated as follows:

ΔHrxn = [ΔHf°(CO2) + 2ΔHf°(H2O)] - ΔHf°(CH4)
= [-393.51 kJ/mol + 2(-241.82 kJ/mol)] - (-74.80 kJ/mol)
= -802.95 kJ/mol

Now, to calculate the volume of methane needed to produce 3287 kJ of heat, we can use the ideal gas law equation in combination with the given temperature and pressure:

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.

First, we need to convert the given temperature of 11 oC to Kelvin:

T = 11 oC + 273.15 = 284.15 K

We also need to convert the given pressure of 3.40 atm to SI units (Pascals):

P = 3.40 atm * 101325 Pa/atm = 345.17 × 10^3 Pa

Now, we rearrange the ideal gas law equation to solve for the volume (V):

V = (nRT) / P

We have the heat change (ΔHrxn) in kJ/mol, so we need to convert it to Joules per mole:

ΔHrxn = -802.95 kJ/mol = -802950 J/mol

Finally, we can calculate the volume of methane (V) in liters by dividing the heat change by the ideal gas constant (R = 8.314 J/(mol·K)):

V = (ΔHrxn * n) / (R * T)

V = (802950 J/mol) / (8.314 J/(mol·K) * 284.15 K)
V ≈ 116 L

Therefore, approximately 116 liters of methane gas is needed to produce 3287 kJ of heat at 11 oC and 3.40 atm pressure.