standard enthalpy of formation method

[Nicole Ju 3H]

Can someone explain to me why the standard enthalpy of formation method works? I believe he said something about enthalpy being a state function, but I'm still kind of confused. Thanks!

[Tala Ayoub 1H]

I believe that since the standard enthalpy of a reaction is the enthalpy change that occurs in a system when one mole of matter is transformed by a chemical reaction under standard conditions, and the standard enthalpy of formation is the change in enthalpy that accompanies the formation of one mole of a compound from its elements, with all substances in their standard states and since enthalpy is a state function meaning it is additive, therefore, enthalpy values can be added or subtracted. The enthalpy of components in a reaction is the same as the enthalpy of their formation, thus, since we are forming products and getting rid of the reactants in enthalpy changes for reactions we can simply subtract the enthalpy changes of formation of reactants from the enthalpy changes of formation for the products.

[Elaine Steinberg 3H]

As Dr. Lavelle mentioned in Wednesday's lecture, a state function is one where only the initial and final values matter (not the process in between). Since we know that enthalpy is a state function, we can say that the standard enthalpy of reaction is the difference between the standard enthalpy of formation for the products and the standard enthalpy of formation for the reactants. Another way of saying this is that the sum of the energies it takes to form each products - the sum of the energies it took to form each reactant is equal to change in enthalpy.

[905695298]

State function means the path is not important. So you can basically use Hf(P)-Hf(R) to find H.