For a particular isomer of C8H18, The following reaction produces 5093.7 KJ of heat per mole of C8H18(g) consumed under standered conditions. What is the standard enthalpy of formation of this isomer of C8H18(g)?
Answer:- The enthalpy change for the combustion reaction of octane is -5093.7 kJ/mol. However, this is not the standard enthalpy of formation of octane. The standard enthalpy of formation is the enthalpy change for the formation of one mole of a substance from its constituent elements in their standard states. In the case of octane, the standard states of the elements are carbon (graphite) and hydrogen (gas). The combustion reaction does not involve the formation of octane from its constituent elements in their standard states. Therefore, the enthalpy change for the combustion reaction cannot be used to calculate the standard enthalpy of formation of octane.
To calculate the standard enthalpy of formation of octane, we can use the following equation:
ΔH°f = ΣΔH°f(products) – ΣΔH°f(reactants)
The standard enthalpies of formation of the products are:
- CO2(g): -393.5 kJ/mol
- H2O(l): -285.8 kJ/mol
The standard enthalpies of formation of the reactants are:
- C(graphite): 0 kJ/mol
- H2(g): 0 kJ/mol
Substituting these values into the equation, we get:
ΔH°f(C8H18(g)) = (8 * -393.5 kJ/mol) + (9 * -285.8 kJ/mol) – (0 kJ/mol) – (0 kJ/mol)
= -230 kJ/mol
Therefore, the standard enthalpy of formation of octane is -230 kJ/mol.
Leave a Reply