Calculate the standard entalphy of formation of naphthalene(C10H8) from the following data:
1) standard enthalphy of combustion of naphtalene is -5153kJ/mol
2) standard enthalphy of formation of CO2 is -393.5kJ/mol
3) standard enthalphy of formation of H2O is -285.8kJ/mol
C10H8 +12O2 ==>10CO2 + 4H2O
dHrxn = (n*dHo formation products) - (n*dHo formation reactants)
To calculate the standard enthalpy of formation of naphthalene (C10H8), we will use the Hess's Law, which states that the change in enthalpy for a chemical reaction is the same whether it occurs in one step or through a series of steps.
The standard enthalpy of combustion of naphthalene gives us the enthalpy change for the reaction:
C10H8 + 12.5O2 → 10CO2 + 4H2O (-5153 kJ/mol)
We can break down this reaction into two steps:
1) Combustion of naphthalene to produce carbon dioxide (CO2):
C10H8 + 12O2 → 10CO2 (ΔH1)
2) Combustion of hydrogen gas (H2) to produce water (H2O):
H2 + 0.5O2 → H2O (ΔH2)
The enthalpy change for the first reaction is the negative of the enthalpy change of combustion of naphthalene, which is -5153 kJ/mol.
The enthalpy change for the second reaction is the negative of the enthalpy change of formation of water, which is -285.8 kJ/mol.
We can combine these two equations to get the overall reaction:
C10H8 + 12.5O2 + 2H2 → 10CO2 + 4H2O (Overall equation)
To find the standard enthalpy of formation of naphthalene, we need to subtract the enthalpy changes of the two reactions from the overall reaction:
ΔHf(C10H8) = ΔH1 + ΔH2
ΔHf(C10H8) = (-5153 kJ/mol) + (-4)(-285.8 kJ/mol)
ΔHf(C10H8) = -5153 kJ/mol + 1143.2 kJ/mol
ΔHf(C10H8) = -4010.8 kJ/mol
Therefore, the standard enthalpy of formation of naphthalene (C10H8) is approximately -4010.8 kJ/mol.
To calculate the standard enthalpy of formation of naphthalene (C10H8), we can use the equation:
ΔHf°(C10H8) = ΣnΔHf°(products) - ΣmΔHf°(reactants)
where:
ΔHf°(C10H8) is the standard enthalpy of formation of naphthalene,
ΣnΔHf°(products) is the sum of the standard enthalpies of formation of the products, and
ΣmΔHf°(reactants) is the sum of the standard enthalpies of formation of the reactants.
Given the data:
1) standard enthalpy of combustion of naphthalene (-5153 kJ/mol),
2) standard enthalpy of formation of CO2 (-393.5 kJ/mol), and
3) standard enthalpy of formation of H2O (-285.8 kJ/mol),
we first need to determine the products and reactants in the combustion reaction of naphthalene.
The balanced equation for the combustion of naphthalene can be written as follows:
C10H8 + mO2 → nCO2 + xH2O
From the equation, we can see that the products include CO2 and H2O, and the reactants are naphthalene (C10H8) and oxygen (O2).
Now, let's plug the given values into the equation mentioned earlier:
ΔHf°(C10H8) = ΣnΔHf°(products) - ΣmΔHf°(reactants)
ΔHf°(C10H8) = [nΔHf°(CO2) + xΔHf°(H2O)] - [ΔHc°(naphthalene) + mΔHf°(O2)]
We know the standard enthalpies of formation of CO2 and H2O:
ΔHf°(CO2) = -393.5 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol
Now, based on the stoichiometry of the balanced equation, we can write:
n = 10 (since there are 10 moles of CO2 for every mole of C10H8)
x = 4 (since there are 4 moles of H2O for every mole of C10H8)
We also know the standard enthalpy of combustion of naphthalene:
ΔHc°(naphthalene) = -5153 kJ/mol
Lastly, we need to determine the number of moles of oxygen consumed in the combustion reaction (m). This can be done by balancing the equation or referring to a chemical database. Let's assume that we require 15 moles of oxygen.
m = 15 (number of moles of O2)
Substituting these values into the equation, we get:
ΔHf°(C10H8) = [(10)(-393.5 kJ/mol) + (4)(-285.8 kJ/mol)] - [(-5153 kJ/mol) + (15)(0 kJ/mol)]
Simplifying further:
ΔHf°(C10H8) = [-3935 kJ/mol - 1143.2 kJ/mol] - [-5153 kJ/mol]
ΔHf°(C10H8) = -6943.2 kJ/mol + 5153 kJ/mol
ΔHf°(C10H8) = -1790.2 kJ/mol
Therefore, the standard enthalpy of formation of naphthalene (C10H8) is -1790.2 kJ/mol.