Use enthalpy of formation data to calculate the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions (1atm and 25 degrees C)
a. coal C(s, graphite) = _____mole-MJ
b. natural gas, CH4(g)=
c. propane, C3H8(g)=
d. octane, C8H18(l) (H(f)= -250.1kJ)=
a.
C(s) + O2 ==> CO2
dHfrxn = (n*dHf products) - (n*dHf reactants)
Then
dHfrxn in kJ/mol x (#mols) = 1000 kJ
Solve for # mols.
The others are done the same way but the eq2uation is combustion equation isn't quite as simple as CO2.
Could you do the first one for me so I can see please?
To calculate the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel, we will use the enthalpy of formation data.
The balanced chemical equation for the combustion of each fuel is as follows:
a. Coal C(s, graphite) combustion equation: C(s) + O2(g) -> CO2(g)
b. Natural gas, CH4(g) combustion equation: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
c. Propane, C3H8(g) combustion equation: C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
d. Octane, C8H18(l) combustion equation: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(g)
We will use the given enthalpy of formation data to calculate the enthalpy change (ΔH) for each reaction and then divide it by the heat released in megajoules (MJ) to find the number of moles of CO2(g) produced per MJ.
a. Coal C(s, graphite):
The enthalpy change (ΔH) for the reaction is zero since coal is the reference material. Therefore, the number of moles of CO2 produced per MJ of heat released is 0 moles-MJ.
b. Natural gas, CH4(g):
ΔH = [ΔHf(CO2(g))] - [ΔHf(CH4(g))] + 2[ΔHf(H2O(g))] - 2[ΔHf(O2(g))]
= [-393.5 kJ/mol] - [-74.8 kJ/mol] + 2[-241.8 kJ/mol] - 2[0 kJ/mol]
= -393.5 kJ/mol + 74.8 kJ/mol + 2(-241.8 kJ/mol)
= -803.4 kJ/mol
Since 1 MJ = 1000 kJ, the number of moles of CO2 produced per MJ of heat released for natural gas combustion is:
= -803.4 kJ/mol / (1000 kJ/MJ)
= -0.8034 mol-MJ
c. Propane, C3H8(g):
ΔH = 3[ΔHf(CO2(g))] - [ΔHf(C3H8(g))] + 4[ΔHf(H2O(g))] - 5[ΔHf(O2(g))]
= 3[-393.5 kJ/mol] - [-103.9 kJ/mol] + 4[-241.8 kJ/mol] - 5[0 kJ/mol]
= -1180.5 kJ/mol + 103.9 kJ/mol + 4(-241.8 kJ/mol)
= -1180.5 kJ/mol + 103.9 kJ/mol - 967.2 kJ/mol
= -2043.8 kJ/mol
Therefore, the number of moles of CO2 produced per MJ of heat released for propane combustion is:
= -2043.8 kJ/mol / (1000 kJ/MJ)
= -2.0438 mol-MJ
d. Octane, C8H18(l):
ΔH = 8[ΔHf(CO2(g))] - [ΔHf(C8H18(l))] + 9[ΔHf(H2O(g))] - 12.5[ΔHf(O2(g))]
= 8[-393.5 kJ/mol] - [-250.1 kJ/mol] + 9[-241.8 kJ/mol] - 12.5[0 kJ/mol]
= -3148 kJ/mol + 250.1 kJ/mol + 9(-241.8 kJ/mol)
= -3148 kJ/mol + 250.1 kJ/mol - 2176.2 kJ/mol
= -4074.1 kJ/mol
Therefore, the number of moles of CO2 produced per MJ of heat released for octane combustion is:
= -4074.1 kJ/mol / (1000 kJ/MJ)
= -4.0741 mol-MJ
Note: The negative sign indicates the heat released during combustion.
To calculate the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel, we can use the enthalpy of formation data.
Enthalpy of formation is defined as the enthalpy change for the formation of one mole of a substance from its elements in their standard states under standard conditions.
The balanced combustion reactions for each fuel are as follows:
a. Coal (C(s, graphite)):
C(s, graphite) + O2(g) -> CO2(g)
b. Natural gas (CH4(g)):
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)
c. Propane (C3H8(g)):
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
d. Octane (C8H18(l)):
C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(g)
To calculate moles of CO2 produced per megajoule of heat released, we need to find the enthalpy change for each reaction and convert it to moles of CO2.
Step 1: Calculate the enthalpy change for the combustion reaction.
Use the enthalpy of formation data to find the enthalpy change for each reaction.
a. Enthalpy change for coal (C(s, graphite)):
ΔH = -393.5 kJ/mol (enthalpy of formation of CO2(g))
b. Enthalpy change for natural gas (CH4(g)):
ΔH = -74.8 kJ/mol (enthalpy of formation of CH4(g))
ΔH = -393.5 kJ/mol (enthalpy of formation of CO2(g))
c. Enthalpy change for propane (C3H8(g)):
ΔH = -103.8 kJ/mol (enthalpy of formation of C3H8(g))
ΔH = -393.5 kJ/mol (enthalpy of formation of CO2(g))
d. Enthalpy change for octane (C8H18(l)):
ΔH = -250.1 kJ/mol (given enthalpy of formation of octane)
ΔH = -393.5 kJ/mol (enthalpy of formation of CO2(g))
Step 2: Convert enthalpy change to moles of CO2.
Convert the enthalpy change to moles using the given enthalpy of formation of CO2 (ΔH = -393.5 kJ/mol).
a. Moles of CO2 produced from coal:
Moles CO2 = ΔH / enthalpy of formation of CO2
Moles CO2 = -393.5 kJ/mol / -393.5 kJ/mol = 1 mole CO2 per MJ
b. Moles of CO2 produced from natural gas:
Moles CO2 = ΔH / enthalpy of formation of CO2
Moles CO2 = -393.5 kJ/mol / -74.8 kJ/mol = 5.26 moles CO2 per MJ
c. Moles of CO2 produced from propane:
Moles CO2 = ΔH / enthalpy of formation of CO2
Moles CO2 = -393.5 kJ/mol / -103.8 kJ/mol = 3.79 moles CO2 per MJ
d. Moles of CO2 produced from octane:
Moles CO2 = ΔH / enthalpy of formation of CO2
Moles CO2 = -393.5 kJ/mol / -250.1 kJ/mol = 1.57 moles CO2 per MJ
So, the number of moles of CO2(g) produced per megajoule of heat released from the combustion of each fuel under standard conditions are:
a. Coal: 1 mole CO2 per MJ
b. Natural gas: 5.26 moles CO2 per MJ
c. Propane: 3.79 moles CO2 per MJ
d. Octane: 1.57 moles CO2 per MJ