Question

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

Answer 1

`∆H_"f"˚(C_3H_6)=19.98" kJ"*"mol"^-1`

Explanation:

First, start by writing the balanced chemical equation

`C_3H_6"(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 `"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)`

`rArr1"mol"*∆H_"f"˚(C_3H_6)=-2038.02+2058=19.98" kJ"`

Divide both sides of the equation by `1" mol"` to get the right units

`rArrcancel(1"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.