Many portable gas heaters and grills use propane, C3H8(g).
Using enthalpies of formation, calculate the quantity of heat produced when 15.0g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming.
C3H8 + 5O2 ==> 3CO2 +4H2O
mols Propane = grams/molar mass = estimated 0.34.
Convert mols propane to mols CO2 and mols H2O, then apply the following. You will need to look up the values for dH of CO2 and H2O(liquid) and C3H8. dH for O2 = 0
Post your work if you get stuck.
dHrxn = (n*dH products) - (n*dH reactants)
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To calculate the quantity of heat produced when propane is completely combusted, we will use the enthalpies of formation. The enthalpy of formation (ΔHf) is defined as the heat change when one mole of a compound is formed from its constituent elements in their standard states at a given temperature and pressure.
The balanced equation for the combustion of propane can be written as follows:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)
To use the enthalpies of formation, we need to know the values for the reactants and products involved in the reaction:
ΔHf(C3H8) = -103.9 kJ/mol (enthalpy of formation for propane)
ΔHf(CO2) = -393.5 kJ/mol (enthalpy of formation for carbon dioxide)
ΔHf(H2O) = -286.0 kJ/mol (enthalpy of formation for liquid water)
Now, we can calculate the quantity of heat produced when 1 mole of propane is combusted:
ΔHrxn = (3ΔHf(CO2) + 4ΔHf(H2O)) - (ΔHf(C3H8) + 5ΔHf(O2))
ΔHrxn = (3*(-393.5 kJ/mol) + 4*(-286.0 kJ/mol)) - (-103.9 kJ/mol + 5*0 kJ/mol)
ΔHrxn = -2043.7 kJ/mol
This means that when 1 mole of propane is combusted, 2043.7 kJ of heat is released.
To calculate the quantity of heat produced when 15.0g of propane is combusted, we need to convert grams to moles. The molar mass of propane (C3H8) is 44.1 g/mol (3*12.0 + 8*1.0 = 44.1).
Number of moles of propane = mass of propane / molar mass of propane
Number of moles of propane = 15.0 g / 44.1 g/mol = 0.340 moles
Now, we can calculate the quantity of heat produced:
Quantity of heat produced = ΔHrxn * Number of moles of propane
Quantity of heat produced = -2043.7 kJ/mol * 0.340 mol = -694.5 kJ
Therefore, when 15.0g of propane is completely combusted in air under standard conditions, approximately 694.5 kJ of heat is produced.