Many portable gas heaters and grills use propane, C3H8(g).
Using enthalpies of formation, calculate the quantity of heat produced when 16.0g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming.
C3H8 + 5O2 ==> 3CO2 + 4H2O
dHrxn = (n*dHf products) - (n*dHf reactants). Let's say dHrxn = x and that is the heat of combustion. You can get the dHf values from your text or notes.
So you can get x kJ of heat from 44 g (1 mole from the equation) C3H8. How much heat can you get from 16.0g.
That's x kJ x (16.0/44.0) = ? kJ for 16.0 g propane.
-806 kJ
To calculate the quantity of heat produced when propane is completely combusted, we need to use the enthalpies of formation. The enthalpy of formation (ΔH°f) is the heat change accompanying the formation of one mole of a compound from its constituent elements, with all substances in their standard states.
First, let's write the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
Now, we need to calculate the enthalpy change (ΔH°) for this reaction by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. The value of ΔH° will represent the quantity of heat produced.
The enthalpies of formation for the compounds involved are as follows:
ΔH°f(C3H8) = -103.85 kJ/mol
ΔH°f(CO2) = -393.5 kJ/mol
ΔH°f(H2O) = -285.8 kJ/mol
Now, we can calculate ΔH° for the reaction:
ΔH° = (3 × ΔH°f(CO2) + 4 × ΔH°f(H2O)) - (ΔH°f(C3H8) + 5 × ΔH°f(O2))
ΔH° = (3 × -393.5 kJ/mol + 4 × -285.8 kJ/mol) - (-103.85 kJ/mol + 5 × 0 kJ/mol)
ΔH° = -4094.7 kJ/mol - (-103.85 kJ/mol)
ΔH° = -4094.7 kJ/mol + 103.85 kJ/mol
ΔH° = -3990.85 kJ/mol
Now, we can calculate the quantity of heat produced when 16.0g of propane is combusted under standard conditions. We'll use the molar mass of propane (C3H8) to convert grams to moles:
Molar mass of C3H8 = 12.01 g/mol (C) + (1.01 g/mol (H)) × 8 = 44.10 g/mol
Moles of propane = Mass of propane / Molar mass of propane
Moles of propane = 16.0 g / 44.10 g/mol
Finally, we can calculate the quantity of heat produced using the stoichiometric coefficients in the balanced equation and the molar enthalpy change:
Heat produced = Moles of propane × ΔH°
Heat produced = (16.0 g / 44.10 g/mol) × -3990.85 kJ/mol
Now, calculate the result to determine the quantity of heat produced.