What is Haber Cycle???Let Us It Find Out!!

The Definition Of Haber Cycle:
Haber Cycle involves the formation of an ionic compound from the reaction of a metal with a non-metal. Born–Haber cycles are used primarily as a means of calculating lattice energies which cannot otherwise be measured directly.


The lattice enthalpy is the enthalpy change involved in formation of the ionic compound from gaseous ions. Some chemists define it as the energy to break the ionic compound into gaseous ions. The former definition is invariably exothermic and the latter is endothermic.
Enthalpy of formation of LiF:

The sum of the energies for each step of the process must equal the enthalpy of formation of the metal and non-metal, ΔH_f.

• V is the enthalpy of sublimation for metal atoms (lithium)
• B is the bond energy (of F₂). The coefficient 1/2 is used because the formation reaction is Li + 1/2 F₂ → LiF.
• IE_M is the ionization energy of the metal atom: 
• EA_X is the electron affinity of non-metal atom X (fluorine)
• U_L is the lattice energy (defined as exothermic here)

Example of lattice formation enthalpy of NaCl:

Hess Law

Hess Law state that  the enthalpy change for a reaction that is carried out in a series of steps is equal to the sum of the enthalpy changes for the individual steps.

Hess's law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on thechemical equation of reactions using previously determined values for the enthalpies of formation.

How To Calculate Using Hess Law??Let us finding out!!!

1) Calculate the heat released by the burning of sulfur in oxygen given the following steps

2S(s)

+

3O2 (g)

2SO3(g)

H = ?

Given Steps

Step 1	S(s)	+	O2 (g)		SO2 (g)	H=	-297 kJ
Step 2	2SO2 (g)	+	O2 (g)		2SO3 (g)	H=	- 198 kJ

Answer:
1) The overall reaction is written above.

Note that sulfur (S) is a reactant and sulfur trioxide (SO₃) is the product. Therefore neither reaction in steps 1 or 2 needs to be reversed.
Note that the overall equation has 2 moles of S and 2 moles of SO₃. Therefore the first reaction must be multiplied by two , but the second can be left alone.

2) Manipulate equations

Step 1	2 S(s)	+	2O2 (g)		2SO2 (g)	H = - 594 kJ
Step 2	2 SO2(g)	+	O2 (g)		2 SO3 (g)	H = - 198 kJ

Substances in bold above will be products

3) Add equations and heats

Addition:

2S (s)

+

3O2 (g)

+

2SO2 (g)

2SO2 (g)

+

2SO3 (g)

H = - 792 kJ

Substances in bold above are common and are therefore canceled

Net

2S (s)

+

3O2 (g)

2 SO3 (g)

H = - 792 kJ

Example 2:

The oxidation of nitrogen to produce nitrogen dioxide. The enthalpy change ΔH of the reaction is 68 kJ

N₂(g) + 2O₂(g) = 2NO₂(g)                ΔH = 68 kJ

This reaction also can be carried out in two distinct steps:

N₂(g) + O₂(g) = 2NO(g)                   ΔH₁ = 180 kJ

2NO(g) + O₂(g) = 2NO₂(g)              ΔH₂ = -112 kJ

_____________________________________________

Net reaction:    N₂(g) + 2O₂(g) = 2NO₂(g)     

                      ΔH = ΔH₁ + ΔH₂ = 68 kJ

From that example,we can know how to calculate enthalpy change..we should try it!! 

enthalpy

lets warm up with a simple introduction about enthalpy!!! (one of  the most basic sub chapter in thermochemistry ) 

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