Calculate the heat of formationof benzene if the heat of combustion of benzene is 3260 kj/mol, given the following enthalpies of formation of the product.
enthalpies of formation for O2= 0k (element)
enthalpies of formation CO2 = -394kj/mol
enthalpies of formation H2O= -285 kJ/mol
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.
To calculate the heat of formation of benzene, we need to consider the change in enthalpy during the combustion reaction of benzene. The reaction equation for the combustion of benzene is:
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.
To calculate the heat of formation of benzene (C6H6), we need to use the enthalpies of formation of the products and reactants involved in the combustion reaction.
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.