To solve this problem, we need to use the standard enthalpy of formation equation:
ΔH°f = Σ(nΔH°f(products)) - Σ(nΔH°f(reactants))
Where ΔH°f is the standard enthalpy of formation, n is the stoichiometric coefficient, and (products) and (reactants) represent the products and reactants, respectively.
In this case, we need to find the standard molar enthalpy of formation of propylene (C3H6). The balanced combustion reaction for propylene is:
C3H6 + 9/2 O2 → 3CO2 + 3H2O
Using the given data, we can calculate the standard enthalpy of formation of propylene:
ΔH°f(C3H6) = [3(ΔH°f(CO2)) + 3(ΔH°f(H2O))] - [ΔH°f(C3H6) + 9/2(ΔH°f(O2))]
Substituting the given values:
ΔH°f(C3H6) = [3(393.5 kJ/mol) + 3(0 kJ/mol)] - [2058.44 kJ/mol + 9/2(141.78 kJ/mol)]
Simplifying the equation:
ΔH°f(C3H6) = 1180.5 kJ/mol - 3052.82 kJ/mol
ΔH°f(C3H6) = -1872.32 kJ/mol
Therefore, the standard molar enthalpy of formation of propylene (C3H6) is -1872.32 kJ/mol.
4. The standard heats of combustion at 25°C (to liquid water) of
propylene (C3H6) and hydrogen are 2058.44 kJ/mol and 141.78 kJ/g, respectively. The standard heat of formation of carbon
dioxide at the same temperature is 393.5 kJ/mol. Calculate the
standard molar enthalpy of formation of propylene.
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1 answer