Calculate ∆Hºf /mol for propane, C3H8(g), given the following reaction: (10 points)
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ ); ∆Hºrxn = -2219.9 kJ
dH = delta H.
dHrxn = (3*dHf CO2 + 4*dHf H2O) - (dHf C3H8)
Solve for dHf propane.
To calculate ΔHºf (standard molar enthalpy of formation) for propane, C3H8(g), we need to use the given reaction and the enthalpy of formation values for the other substances involved.
The standard enthalpy of formation (ΔHºf) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states under standard conditions. The standard state means that the substance is in its most stable form at a specified temperature and pressure (usually 298 K and 1 atm).
We can use the given reaction as follows:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ); ΔHºrxn = -2219.9 kJ
The reaction equation provides the stoichiometric coefficients, which correspond to the number of moles needed to balance the equation. In this case, 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O.
To calculate the standard enthalpy of formation of propane (ΔHºf(C3H8)), we need to consider the balanced reaction equation and the enthalpy of formation values of CO2 and H2O.
Using the given reaction, we can write the equation for the formation of 1 mole of propane:
1/3 * [C3H8(g) + 5O2(g) - 3CO2(g) - 4H2O(ℓ)] → ΔHºf(C3H8)
The coefficient of CO2 and H2O are adjusted to make 1 mole of propane.
Using this equation, we can now calculate ΔHºf(C3H8):
ΔHºrxn = 1/3 * [ΔHºf(C3H8) + 5ΔHºf(O2) - 3ΔHºf(CO2) - 4ΔHºf(H2O)]
-2219.9 kJ = 1/3 * (ΔHºf(C3H8) + 5 * 0 kJ/mol - 3 * ΔHºf(CO2) - 4 * ΔHºf(H2O))
We can look up the enthalpy of formation values for CO2 and H2O from a reliable source, such as a thermochemical table or database.
Assuming the standard enthalpy of formation values are:
ΔHºf(CO2) = -393.5 kJ/mol
ΔHºf(H2O) = -285.8 kJ/mol
Plugging in the values:
-2219.9 kJ = 1/3 * (ΔHºf(C3H8) + 0 - 3 * (-393.5) - 4 * (-285.8))
Now, solve for ΔHºf(C3H8):
-2219.9 kJ = ΔHºf(C3H8) + 1180.5 kJ + 1143.2 kJ
ΔHºf(C3H8) = -2219.9 kJ - 1180.5 kJ - 1143.2 kJ
ΔHºf(C3H8) = -4543.6 kJ/mol
Therefore, the standard enthalpy of formation of propane, C3H8(g), is -4543.6 kJ/mol.