Many portable gas heaters and grills use propane, C3H8(g).
Using enthalpies of formation, calculate the quantity of heat produced when 17.0g of propane is completely combusted in air under standard conditions. Assume that liquid water is forming.
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C3H8 + 5O2 ==> 3CO2 + 4H2O
dHrxn = (n*Hfproducts)-(n*Hf reactants)
Look up the delta Hf values, substitut and solvef or dHrxn.
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To calculate the quantity of heat produced when propane is completely combusted, we need to use the enthalpies of formation.
Enthalpy of formation (∆Hf) is the heat change when one mole of a compound is formed from its constituent elements in their standard states.
First, we need to write the balanced equation for the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
The coefficients show that one mole of propane (C3H8) reacts with five moles of oxygen (O2) to produce three moles of carbon dioxide (CO2) and four moles of water (H2O).
Next, we'll calculate the enthalpy change (∆H) using the enthalpies of formation (∆Hf) for each compound involved in the reaction. The values are given below:
∆Hf(C3H8) = -103.9 kJ/mol
∆Hf(CO2) = -393.5 kJ/mol
∆Hf(H2O) = -285.8 kJ/mol
Applying Hess's Law (which states that the enthalpy change of a reaction is independent of the pathway), we can calculate the ∆H for the combustion of propane.
∆H = (∆Hf(CO2) + ∆Hf(H2O)) - (∆Hf(C3H8) + ∆Hf(O2))
Substituting the values:
∆H = [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] - [-103.9 kJ/mol + 5(0 kJ/mol)]
Simplifying the equation:
∆H = -1206.3 kJ/mol + 103.9 kJ/mol
∆H = -1102.4 kJ/mol
Now, we'll calculate the quantity of heat produced when 17.0g of propane (C3H8) is completely combusted.
First, we need to convert 17.0g of propane to moles:
Molar mass of C3H8 = (3*12.01 g/mol) + (8*1.01 g/mol) = 44.11 g/mol
Number of moles = Mass / Molar mass = 17.0 g / 44.11 g/mol ≈ 0.3854 mol
Finally, we'll calculate the quantity of heat produced:
Heat produced = ∆H x Number of moles
Heat produced = -1102.4 kJ/mol x 0.3854 mol ≈ -425.0 kJ (rounded to three significant figures)
Therefore, when 17.0g of propane is completely combusted in air under standard conditions, approximately -425.0 kJ of heat is produced. The negative sign indicates that it is an exothermic reaction, meaning heat is released.