# Use the enthalpy of combustion of propene CH2= CHCH3(g), DeltaHCo =-2058 kJ mol-1, and the fact that DeltaHfo[H2O(l)] = -285.83 kJ mol-1, and DeltaHfo[CO2(g)] -393.51 kJ mol-1, to find its enthalpy of formation?

Feb 19, 2018

∆H_"f"˚(C_3H_6)=19.98" kJ"*"mol"^-1

#### Explanation:

First, start by writing the balanced chemical equation

${C}_{3} {H}_{6} \text{(g)"+9/2O_2"(g)"rarr3CO_2"(g)"+3H_2O_"(l)}$

The following equation comes about due to enthalpy being a state function and is important for these types of calculations.

∆H_c˚=sumn∆H_"f"˚("products")-sumn∆H_"f"˚("reactants")=-2058" kJ"

The "n" stands for the stoichiometric coefficients of each component and has units of $\text{mol}$.

The heat of formation of an element in its standard state is zero, so we don't need to worry about oxygen gas

∆H_"f"˚(O_2)=0" kJ"*"mol"^-1

You have to weight each heat of formation by the stoichiometric coefficients, so for sum of the products we get

sumn∆H_"f"˚("products")=3"mol"*∆H_"f"˚(CO_2)+3"mol"*∆H_"f"˚(H_2O)=3*-393.51+3*-285.83=-2038.02" kJ"

And, for the sum of the reactants

sumn∆H_"f"˚("reactants")=1"mol"*∆H_"f"˚(C_3H_6)+9/2∆H_"f"˚(O_2)=1"mol"*∆H_"f"˚(C_3H_6)

Substituting these into our original equation, we get

∆H_c˚=-2058=-2038.02-1"mol"*∆H_"f"˚(C_3H_6)

$\Rightarrow 1 \text{mol"*∆H_"f"˚(C_3H_6)=-2038.02+2058=19.98" kJ}$

Divide both sides of the equation by $1 \text{ mol}$ to get the right units

$\Rightarrow \cancel{1 \text{mol")*(∆H_"f"˚(C_3H_6))/cancel(1"mol")=19.98" kJ"/(1"mol}}$

So from this data, we have calculated the enthalpy of formation of propene as endothermic. You should do some googling (not wikipedia) to look up a data table to verify that this is close to the right value range.

∆H_"f"˚(C_3H_6)=19.98" kJ"*"mol"^-1

This is for propene in its standard state. I.e. gaseous propene at constant pressure of 1 bar and 298.15 K.