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
To calculate the standard molar enthalpy of formation of propylene (C3H6), we need to use the given standard heats of combustion and the standard heat of formation of carbon dioxide.
The balanced chemical equation for the combustion of propylene is as follows:
C3H6 + 4.5O2 -> 3CO2 + 3H2O
From the equation, we can see that for every 3 moles of carbon dioxide formed, 1 mole of propylene is burned.
Therefore, the standard heat of combustion of propylene can be expressed as:
2058.44 kJ/mol = (3 mol CO2 / 1 mol C3H6) * 393.5 kJ/mol
So, the standard heat of combustion of propylene is 3 * 393.5 = 1180.5 kJ/mol.
Now, to calculate the standard molar enthalpy of formation of propylene, we use the equation:
ΔHf = Σ(nΔHf products) - Σ(nΔHf reactants)
Since propylene is a product in this case, its coefficient is 1. Therefore:
ΔHf = (1 * ΔHf of propylene) - (3 * ΔHf of CO2) - (3 * ΔHf of H2O)
ΔHf of CO2 is given as 393.5 kJ/mol.
ΔHf of H2O is the standard heat of combustion of hydrogen, which is 141.78 kJ/g. To convert it to kJ/mol, we need to know the molar mass of hydrogen. The molar mass of hydrogen is 2 g/mol. So:
ΔHf of H2O = (141.78 kJ/g) * (18 g H2O/mol) = 2552.04 kJ/mol
Plugging in the values:
ΔHf = (1 * ΔHf of propylene) - (3 * 393.5 kJ/mol) - (3 * 2552.04 kJ/mol)
ΔHf = ΔHf of propylene - 1180.5 kJ/mol - 7656.12 kJ/mol
ΔHf = ΔHf of propylene - 8836.62 kJ/mol
Therefore, the standard molar enthalpy of formation of propylene is 8836.62 kJ/mol.