# 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?

##### 1 Answer

#### Explanation:

First, start by writing the balanced chemical equation

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

The "n" stands for the stoichiometric coefficients of each component and has units of

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

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

And, for the sum of the reactants

Substituting these into our original equation, we get

Divide both sides of the equation by

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.

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