The standard enthalpy of formation of sustance denoted by ΔHf°, is the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference form and in their standard states.
The values of standard enthalpies of formation are determined by direct measurement that has been given in the tabluated form.
You can use the standard enthalpy of formation to find the standard enthalpy change for a reaction.Though you can calculate this from the point of view of Hess's law , but we have noticed a pattern in the result and then developed a simple formula for solving these types of problems.
In general ,you can calculate the ΔH° for the reaction by the equation;
ΔH° = S n ΔHf°(product) - S mHf°(reactant)
where, S is the mathemetical symbol meaning 'the sum of ' and m and n are the coeffcients of the substances in the chemical equation.
Lets see an example :
Find the ΔH° of the following equation;
CH4(g) + 4Cl2 (g) --------------> CCl2 (l) + 4 HCl (g) ΔH°=?
Step 1 : From the table of standdard enthalpies of formation you can find the enthalpies of formation for CH4(g), CCl4(l), and HCl (g);
Step 2: Using the equation :ΔH° = S n ΔHf°(product) - S mHf°(reactant)
note that the ΔHf° for each componenet is multiplies by its coeffcient in the chemical equation whose ΔH you are calculating.You can symoblize the enthalpy of the formation of a substance by writing the formula in parantheses following the ΔHf°,then your calculation can be written as:
ΔH° = [ ΔHf° (CCl4) + 4 ΔHf°(HCl)] - [ΔHf°(CH4) + 4ΔHf°(Cl2)]
= [ (-139) + 4 (-92.3)] kj - [-74.9 + 4(0)] kj
= - 433 kJ