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).

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

Well, propylene sure likes to heat things up! Let's calculate the heat evolved per mole on combustion.

From the balanced combustion reaction of propylene, we see that 3 moles of carbon dioxide and 3 moles of water are produced. The standard enthalpy change of this reaction can be calculated by summing up the standard enthalpies of formation of the products and subtracting the sum of the standard enthalpies of formation of the reactants.

For the products:
3 moles of CO2 = 3 * (0 kJ/mol) = 0 kJ/mol
3 moles of H2O = 3 * (0 kJ/mol) = 0 kJ/mol

For the reactant:
1 mole of C3H6 = +20.4 kJ/mol

Now, let's calculate the change in enthalpy:
ΔH = (ΣΔH_products) - (ΣΔH_reactants)
= (0 kJ/mol + 0 kJ/mol) - (+20.4 kJ/mol)
= -20.4 kJ/mol

Oh, propylene sure knows how to bring the heat! On combustion, it releases 20.4 kJ of energy per mole. So make sure to handle it with care, or it might ignite your funny bone!

To calculate the heat evolved per mole on combustion of propylene, we need to determine the difference in the standard enthalpies of formation of the products and reactants.

The balanced equation for the combustion reaction of propylene is:

C3H6(g) + 9/2 O2(g) → 3CO2(g) + 3H2O(g)

From the equation, we can see that the stoichiometric coefficients of propylene and carbon dioxide are the same.

The heat evolved per mole on combustion of propylene can be calculated by subtracting the sum of the standard enthalpies of formation of the products (3CO2 and 3H2O) from the standard enthalpy of formation of propylene (C3H6).

ΔH = [Σ(n * ΔHf(products))] - [Σ(n * ΔHf(reactants))]

where ΔHf is the standard enthalpy of formation and n is the stoichiometric coefficient.

Using the given standard enthalpies of formation:
ΔHf(C3H6) = +20.4 kJ/mol (given)
ΔHf(CO2) = 0 kJ/mol (by definition)
ΔHf(H2O) = 0 kJ/mol (by definition)

Substituting the values into the equation:

ΔH = [(3 * 0 kJ/mol) + (3 * 0 kJ/mol)] - (1 * +20.4 kJ/mol)
= -20.4 kJ/mol

Therefore, the heat evolved per mole on combustion of propylene is -20.4 kJ/mol.

To calculate the heat evolved per mole on combustion of propylene, we need to use the enthalpy of formation values and the balanced combustion equation for propylene.

The enthalpy change (ΔH) for a reaction can be calculated using the following equation:

ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))

In this case, we need to calculate the enthalpy change for the combustion of propylene.
The enthalpy change can be obtained by summing the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of the reactants.

The balanced combustion equation for propylene is:
C3H6(g) + 9/2 O2 (g) = 3CO2 (g) + 3H2O(g)

Using the enthalpy change equation, we can calculate the heat evolved per mole on combustion of propylene as follows:

ΔH = (3 moles of CO2) * ΔHf (CO2) + (3 moles of H2O) * ΔHf (H2O) - (1 mole of C3H6) * ΔHf (C3H6) - (9/2 moles of O2) * ΔHf (O2)

Now, we substitute the values of the enthalpies of formation:

ΔH = (3 * (-393.5 kJ/mol)) + (3 * (-285.8 kJ/mol)) - (1 * (+20.4 kJ/mol)) - ((9/2) * 0 kJ/mol)

Simplifying the equation:

ΔH = -5394.6 kJ/mol - 857.4 kJ/mol - 20.4 kJ/mol

ΔH = -6269.4 kJ/mol

Therefore, the heat evolved per mole on combustion of propylene is -6269.4 kJ/mol.