The **standard enthalpy of formation **of sustance denoted by **Δ H_{f}°**, 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 **Δ H° **for the reaction by the equation;

**Δ H° = S n**

**Δ**(product) -

*H*_{f}°**S**

**m**

*H*_{f}°(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 **Δ H°** of the following equation;

CH4(g) + 4Cl2 (g) --------------> CCl2 (l) + 4 HCl (g) **Δ H°**=?

**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 :**Δ H° = S n**

**Δ**(product) -

*H*_{f}°**S**

**m**

*H*_{f}°(reactant)note that the **Δ H_{f}°** for each componenet is multiplies by its coeffcient in the chemical equation whose

**Δ**you are calculating.

*H*__You can symoblize the enthalpy of the formation of a substance by writing the formula in parantheses following the__:

**Δ**,then your calculation can be written as*H*_{f}° **Δ H°** = [

**Δ**(CCl4) + 4

*H*_{f}°**Δ**(HCl)] - [

*H*_{f}°**Δ**(CH4) + 4

*H*_{f}°**Δ**(Cl2)]

*H*_{f}°= [ (-139) + 4 (-92.3)] kj - [-74.9 + 4(0)] kj

= **- 433 kJ**