The standard change in internal energy for the combustion of buckyball is -25,968 kJ/mol. What is the enthalpy of combustion? What is the enthalpy of formation of buckyball? What is the molar enthalpy associated with converting buckyball to graphite? How does this compare to the enthalpies associated with converting diamond to graphite?
C60 + 60 O2 ==> 60CO2
dE = q + w
Combustions are done at constant and W = -pdV. 60 mols CO2 come from 60 mols O2 so there no change in volume and no work. Thus dE = q = dH.
dH formation would be the reverse; i.e., 60C --> C60 and
dHrxn = dHformation C60. We don't know that but can calculate that from the above.
C60 + 60O2 ==> 60CO2
dHrxn = (n*dHf CO2)-(dHfC60)-(dHf O2)
-26988 kJ/mol = 60*-393.5) - 0
Calculate dHf C60 and use that to calculate dH formation. Post your work if you get stuck.
To answer these questions, we need to understand the concepts of enthalpy, enthalpy of combustion, enthalpy of formation, and molar enthalpy.
1. Enthalpy of Combustion:
The enthalpy of combustion (ΔHcomb) is the heat released or absorbed during the complete combustion of one mole of a substance. It can be determined by using the equation:
ΔHcomb = -ΔHsys
Therefore, the enthalpy of combustion is equal to the negative change in internal energy (-ΔU).
Given that the standard change in internal energy for the combustion of buckyball is -25,968 kJ/mol, the enthalpy of combustion for buckyball is also -25,968 kJ/mol.
2. Enthalpy of Formation:
The enthalpy of formation (ΔHf) is the standard enthalpy change when one mole of a compound is formed from its elements in their standard states. It can be determined by using the equation:
ΔHf = ΣnΔHf(products) - ΣmΔHf(reactants)
Unfortunately, the information provided does not directly state the enthalpy of formation for buckyball. This value is usually given in relation to a reference compound. However, if you have the enthalpies of formation for all the reactants and products involved in the formation of buckyball, you can calculate the enthalpy of formation using the equation above.
3. Molar Enthalpy associated with converting buckyball to graphite:
To determine the molar enthalpy associated with converting buckyball to graphite, we need to know the enthalpy change for this conversion, which is not provided. Therefore, we cannot calculate this value given the information provided.
4. Enthalpies associated with converting diamond to graphite:
The enthalpies associated with converting diamond to graphite can vary depending on the conditions, such as temperature and pressure. At standard conditions, the enthalpy change for converting diamond to graphite is approximately 1 kJ/mol. This value indicates that the conversion is slightly endothermic, meaning it requires a small amount of energy.
In summary:
- The enthalpy of combustion for buckyball is -25,968 kJ/mol.
- The enthalpy of formation of buckyball cannot be determined without additional information.
- The molar enthalpy associated with converting buckyball to graphite cannot be calculated with the information provided.
- The enthalpy associated with converting diamond to graphite is approximately 1 kJ/mol, indicating a slightly endothermic process.