Question

Given a reaction and enthalpies of combustion, how does one indirectly calculate enthalpy of reaction? i.e.,...

Given a reaction and enthalpies of combustion, how does one indirectly calculate enthalpy of reaction? i.e., if I take stoichiometric coefficients into account and subtract reactant combustion enthalpies from products, I end up with an overall combustion enthalpy for the reaction. How does this relate to enthalpy of reaction? Is it the same?

Homework Answers

Answer #1

Let us take an example of the combustion of methane.

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

The enthalpy of reaction enthalpy of reaction Hrxn is given as sum of Hformation (products) - sum of Hformation (reactants)

Where Hformation is the enthalpy change of formation of the compound.

So, Hrxn = Hformation (products) - Hformation (reactants)

Hrxn = Hformation(CO2(g) ) + 2*Hformation(H2O(l)) - ( Hformation(CH4(g)) + 2*Hformation(O2(g)) )

As you can see for the overall enthalpy of combustion of the CH4 is same as enthalpy of reaction of combustion of CH4

If Hformation of reactant or products is not given directly we may need to get it by the individual reactions of each of them.

For example for H2O its

H2(g) + (1/2)O2(g) -> H2O(l)

Hformation(H2O(l)) = 2*Bond energy of H-O - (Bond Energy of H2(g) in H-H + (1/2)*Bond energy of O2(g) in O=O)

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