standard enthalpy of formation

This time I will post about the subject chemical associated with enthalpy. This lesson I learned in class 3. Laneways more details bca article below.

Enthalpy change for the formation of one mole of a substance directly from its elements called molar enthalpy of formation or the enthalpy of formation. If the measurement is made at the standard state (298 K, 1 atm) and all of its elements in the form of the standard, the enthalpy change is called the standard enthalpy of formation (ΔHf 0). Enthalpy of formation is expressed in kJ per mole (kJ mol -1).
So that there is uniformity, then it should set the standard state, the temperature 25 0 C and a pressure of 1 atm. Thus thermochemical calculations based on the state standards.
In general, in thermochemical equation stated:
AB + CD ----> AC + BD Δ H0 = x kJ / mol
Δ H0 is the epitome of the enthalpy change in circumstances. What is meant by the standard form of an element is the most stable form of the element at standard conditions (298 K, 1 atm).
For elements that have allotropes form, standard form specified by that sense. For example, carbon can form diamond and graphite, a form of default is graphite, because graphite is the most stable form of carbon at 298 K, 1 atm. Two things to note is related to the enthalpy of formation of the substance formed is 1 mole and formed from its elements in standard form.
Example: enthalpy of formation of ethanol (C2H5OH) (l) is -277.7 kJ per mole. This means: the formation of 1 mole (46 grams) of ethanol from elements in standard form, the carbon (graphite), hydrogen gas and oxygen gas, measured at 298 K, 1 atm 277.7 kJ released by termokimianya equation is :
2 C (s, graphite) + 3H2 (g) + ½ O2 (g) -> C2 H5 OH (l) ΔH = -277.7 kJ
Value of enthalpy of formation of various substances and the formation reaction thermochemical equations given in Table 2 below.
Table 2. Value of the enthalpy of formation of various substances and thermochemical reaction equation formation