You need the heats of formation of CO2 and H2O. That of O2 is zero by definition since is a naturally occuring state of an element.
At 298 K, Hf of H2O is -57.8 kcal/mol and that of CO2 is -94.1 kcal/mol. Multiply those by 4.18 for kJ/mol. That makes -241.6 for H2O and -393.3 for CO2
The heat evolved per mole is the sum of the heats of formation of the reactants minus the sum of the heats of formation of the products. Each heat of formation must be multiplied by the number of moles appearing in the reaction equation for that compound.
For propylene, I get
+20.4 -[3*(-393.3) +3*(-241.6)]
= 20.4 + 1904.7 = 1925.1 kJ/mol
for the heat of reaction.
This reference gives 1926.1 kJ/mol for the value, so I was pretty close:
http://books.google.com/books?id=bL86AAAAIAAJ&pg=PA27&lpg=PA27&dq=propylene+%22heat+of+combustion%22+kJ/mol&source=bl&ots=v3kzFiWQsY&sig=OOLK8ZuOF8uXkwwonQuZEOB3foE&hl=en&ei=6WXkSeGkBomEtgPjjrytCQ&sa=X&oi=book_result&ct=result&resnum=9
The standard enthalpies of formation of gaseous propylene (C3H6) and propane (C3H8) are +20.4kJ/mol and -103.8 kJ/mol. Their respective combustion reactions may be written as follows, where one mole of either gas is consumed by oxyen to yield carbon dioxide and water.
Propylene:C3H6(g) + 9/2 O2 (g) = 3CO2 (g) + 3H2O(g)
Propane: C3H8(g) + 5O2(g) = 3CO2(g) + 4 H2O(g)
Calculate the heat evolved per mole on combustion of propylene(units of kJ/mol).
1 answer
1 answer