Write the equation and balance it.
2C6H6 + 15O2 ==> 12CO2 + 6H2O
DHrxn = (n*DHfproducts) - (n*DHfreactants)
Divide by 2 to find the DHf for one mole.
enthalpies of formation for O2= 0k (element)
enthalpies of formation CO2 = -394kj/mol
enthalpies of formation H2O= -285 kJ/mol
2C6H6 + 15O2 ==> 12CO2 + 6H2O
DHrxn = (n*DHfproducts) - (n*DHfreactants)
Divide by 2 to find the DHf for one mole.
C6H6 + 15/2O2 -> 6CO2 + 3H2O
Given that the heat of combustion of benzene is 3260 kJ/mol, we can relate it to the enthalpies of formation using Hess's Law. Hess's Law states that the change in enthalpy for a chemical reaction is the same regardless of the pathway between the initial and final states.
First, we need to determine the change in enthalpy for the formation of 6 moles of CO2 and 3 moles of H2O:
ΔH1 = 6 × enthalpy of formation of CO2 + 3 × enthalpy of formation of H2O
= 6 × (-394 kJ/mol) + 3 × (-285 kJ/mol)
Next, we need to determine the change in enthalpy for the combustion of benzene:
ΔH2 = -3260 kJ/mol
Finally, we can calculate the heat of formation of benzene (ΔHf) using the equation:
ΔHf = ΔH2 - ΔH1
= -3260 kJ/mol - (6 × -394 kJ/mol + 3 × -285 kJ/mol)
Calculating this expression will give us the heat of formation of benzene.
The combustion of benzene (C6H6) can be represented by the following equation:
C6H6 + xO2 → yCO2 + zH2O
From the given data:
Enthalpy of formation of O2 = 0 kJ/mol (since it is in its elemental form)
Enthalpy of formation of CO2 = -394 kJ/mol
Enthalpy of formation of H2O = -285 kJ/mol
We can balance the equation to determine the coefficients:
C6H6 + 15O2 → 6CO2 + 3H2O
Now, we can calculate the heat of combustion (∆Hcomb) of benzene using the enthalpies of formation of the products and reactants:
∆Hcomb = (6 × ∆Hf(CO2)) + (3 × ∆Hf(H2O)) - (15 × ∆Hf(O2))
∆Hcomb = (6 × -394 kJ/mol) + (3 × -285 kJ/mol) - (15 × 0 kJ/mol)
∆Hcomb = -2364 kJ/mol - 855 kJ/mol
∆Hcomb = -3219 kJ/mol
The negative sign indicates that the combustion reaction is exothermic. The heat of combustion of benzene is -3219 kJ/mol.
However, to find the heat of formation of benzene (∆Hf), we'll need to rearrange the equation to isolate the heat of formation term:
∆Hf = ∆Hcomb - (∆Hf(CO2) × 6) - (∆Hf(H2O) × 3) + (∆Hf(O2) × 15)
∆Hf = -3219 kJ/mol - (-394 kJ/mol × 6) - (-285 kJ/mol × 3) + (0 kJ/mol × 15)
∆Hf = -3219 kJ/mol + 2364 kJ/mol + 855 kJ/mol + 0 kJ/mol
∆Hf = -3219 kJ/mol + 3219 kJ/mol
∆Hf = 0 kJ/mol
The heat of formation of benzene is 0 kJ/mol.