Many portable gas heaters and grills use propane, C3H8(g).
Using enthalpies of formation, calculate the quantity of heat produced when 19.0g 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*dHf products) - (n*dhf reactants)
Then convert dHrxn from 44g (1 mol in the equation above) to 19.0g.
? = dHrxn x 19.0/44
If you carry the sign over from the first calculation you will get the correct sign for the smaller amount.
-1001.7
To calculate the quantity of heat produced when propane is completely combusted, we need to use the enthalpies of formation of the reaction. The balanced chemical equation for the combustion of propane is:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)
To calculate the heat produced, we need to find the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.
The enthalpies of formation are typically listed in standard conditions, where the temperature is 25°C (298 K) and the pressure is 1 atm.
The enthalpies of formation for the elements in their standard states are defined as zero. Therefore, we only need to consider the enthalpies of formation for the compounds involved in the reaction.
The enthalpy of formation for carbon dioxide (CO2) is -393.5 kJ/mol, and the enthalpy of formation for water (H2O) is -285.8 kJ/mol.
To calculate the heat produced, we need to convert the mass of propane (19.0g) to moles. The molar mass of propane (C3H8) is approximately 44.1 g/mol.
Number of moles of propane = 19.0g / 44.1 g/mol ≈ 0.431 mol
Now, using the balanced equation, we can determine the quantity of heat produced when 19.0g of propane is completely combusted:
Quantity of heat produced = (number of moles of propane) × (ΔHf of CO2) + (number of moles of water formed) × (ΔHf of H2O)
Number of moles of CO2 formed = 3 × 0.431 mol = 1.29 mol
Number of moles of water formed = 4 × 0.431 mol = 1.72 mol
Quantity of heat produced = (0.431 mol) × (-393.5 kJ/mol) + (1.72 mol) × (-285.8 kJ/mol)
Simplifying this expression, we find:
Quantity of heat produced ≈ -169.5 kJ
Therefore, when 19.0g of propane is completely combusted under standard conditions, it produces approximately -169.5 kJ (or 169.5 kJ of heat is released). Note that the negative sign indicates that the reaction is exothermic, meaning that heat is released during the combustion process.