Thermodynamics may be defined as the branch of science which deals with the quantitative relationship between heat and other forms of energies. Chemical thermodynamics deals with the change in heat during chemical processes. The energy change during the chemical reaction at constant temperature and volume is given by internal energy change.
However chemical reactions are generally involved some change in volume, therefore in order to study the heat changes of chemical reaction at constant temperature and pressure, a new function enthalpy is introduced.
Enthalpy is the total energy associated with any system which includes its internal energy and also energy due to environmental factor such as pressure-volume conditions.
Mathematically enthalpy can be written as -
H = U + PV
Where U stands for internal energy and PV is additional energy possesses by the substance. Enthalpy is a state function and an extensive property. It is also known as heat content of the system whose value depends upon amount and chemical nature of substance as well as on pressure and temperature.
Born Haber cycle is a simplified method developed by Max Born and Fritz Haber in 1919 to correlate the lattice energies of ionic solids to other thermodynamic data. Lattice enthalpy or lattice energy is defined as enthalpy change which occurs when one mole of ionic solid is formed by close packing of constituent ions in gaseous state. Or it may be defined as energy released when one mole of ionic solid dissociates into its gaseous state.
For example, the formation of sodium chloride from sodium and chloride ion involves 788 kj/mol of energy and dissociation of sodium chloride into it constituent ions requires same amount of energy.
NaCl(s) $\to$ Na+(g) + Cl- (g) $\Delta_L$H°= 788 kj/molNa+(g) + Cl-(g) $\to$ NaCl(s) $\Delta_L$H° = -788 kj/mol
When oppositely charged ions interact to form ionic solid, a large amount of energy is released and dissociation of crystal also require a lot of energy which makes melting point and boiling points of crystal very high. Born Haber cycle is used to determine this lattice energy by using other energy values like ionization energy, electron affinity, dissociation energy, sublimation energy and heat of formation.
- Ionization Energy: It is the energy required to remove an electron from a neutral gaseous atom or an ion.
- Electron Affinity: It is the energy released when an electron is added to an isolated neutral gaseous atom or an ion.
- Dissociation energy: The energy required to dissociate a compound is called as dissociation energy. Dissociation of a compound is always an endothermic process and requires an input of energy.
- Sublimation energy: The energy required to change the phase from solid to gas, by passing the liquid phase is called as sublimation energy.
- Heat of formation: The energy change during the formation of a compound from its elements is known as heat of formation.
Born Haber cycle is based on Hess law which states that the standard enthalpy change of reactions is the sum of algebraic sum of the standard enthalpies of reactions into which the overall reaction may be split or divided.
For example, the formation of C from A and B can take place in two ways. Both ways involve same initial and final states, but one is single step and other one is multi-step reaction which involves intermediates, D and E. Both ways has an enthalpy change of 160 kJ.
The formation of ionic compounds from alkali metals like sodium, potassium with halogen such as chlorine, bromine to form alkali halide can be represented by using Born Haber cycle. Remember the heat of formation of ionic halide is equal to the sum of other energies involves in various steps.
The energy change during the formation of sodium chloride crystal from metallic sodium and chlorine gas can be used to calculate lattice enthalpy of NaCl(s). The net enthalpy change for the formation of NaCl (
ΔfH° ) is 411.2 kj/mol.
Νa(s) +1/2Cl2(g) → NaCl(s) ΔfH° = -411.2 kj/mol
The overall process can be explained in following steps.
- Formation of sodium chloride: The formation of one mole of sodium chloride from its elements under standard conditions releases 411 kj/mol of heat. This is known as standard enthalpy of formation and represented by ∆fH°.
Na(s) + ½ Cl2 (g) → NaCl(s) ∆Hf°(NaCl) = - 411 kJ/mol
- Standard enthalpy of atomization of chlorine (∆Ha°): The energy required to atomize one mole of gaseous chlorine molecules into one mole of gaseous chlorine atoms is known as standard enthalpy of atomization of chlorine. It’s an endothermic step as the bonds between Cl-Cl atoms are being broken.
½ Cl2 (g) → Cl (g) ∆Ha° (Cl2) = +121 kJ/mol
- Standard enthalpy of electron affinity of chlorine (∆Hea°): The energy released when one mole of gaseous chlorine atoms gains one mole of electrons to form chloride ion. The electron affinity of chlorine atom is 349 kj/mol.
Cl (g) + e- → Cl-(g) ∆Hea° (Cl) = -349 kJ/mol
- Sublimation energy of sodium (∆Hs°): This is the energy required to change one mole of solid sodium atoms into one mole of gaseous atoms. Sublimation of sodium is an endothermic process and requires energy to change the state from a solid to a gas.
Na (s) → Na (g) ∆Hs° (Na) = +107 kJ/mol
- Ionization energy of sodium (∆HIE° (Na): Removal of one mole of electron from a gaseous metal atom involve some energy change, known as ionization energy. Endothermic, energy needs to be absorbed to remove the electron.
Na (g) → Na+(g)+ e- (∆HIE° (Na) = +496 kJ/mol)
- Lattice enthalpy of sodium chloride (U): The enthalpy change during the formation of one mole of sodium chloride from its constituent ions is called as lattice energy of lattice enthalpy. It is an exothermic step and releases about 786 kj energy.
Na+(g) + Cl-(g) → NaCl (s) U = -786 kj/molApplying Hess’s law we get.
Or Lattice enthalpy
∆ HL° =107 + 496 + 121 - 349 - 786 = -411 kj/mol
Born Haber cycle for the formation of sodium chloride can be represented as given.
The Born Haber cycle for the formation of calcium chloride form its constituent elements involves following steps.
- Atomization enthalpy of calcium: This step involves the conversion of solid calcium to gaseous state.
Ca(s) $\to$ Ca(g) $\Delta$Ha°(Ca) =178 kj/mol
- Ionization enthalpy of calcium: Calcium forms di-positive ion (Ca2+), therefore the energy involve in the removal of first electron is called as first ionization energy and valued 590 kj/mol. The removal of second electron requires more energy as it’s difficult to remove electron from a cation. Hence the second ionization energy for calcium ion is 1145 kj/mol.
Ca(g) $\to$ Ca+(g) + e-$\Delta$ HIE°= 590 kj/molCa+(g) $\to$ Ca2+(g) + e- $\Delta$4HIE°= 1145 kj/mol
- Atomization enthalpy of Chlorine: This step involves dissociation of Cl2(g) into Cl(g) atoms. The reaction enthalpy is half of the bond dissociation enthalpy of chlorine.
½ Cl2(g) $\to$ Cl(g) ;$\Delta$ Ha°= ½$\Delta$ HCl-Cl°=121 kj/mol
- For the formation of calcium chloride, two Cl(g) requires, therefore total atomization enthalpy will be double that is 242 kj/mol.
- Electron affinity of chlorine: This is the amount of energy released during the addition of electron in an isolated neutral gaseous chlorine atom.
Cl(g) + e- $\to$ Cl-(g) ;$\Delta$ Hea°= -364 kj/mol
- Lattice enthalpy: The combination of one Ca2+ ion and two chloride ions (Cl-)to form one mole of calcium chloride release lattice energy ($\Delta$HLE).
Ca2+ (g) + 2Cl-(g) $\to$ CaCl2(s)
The lattice energy of calcium chloride can be calculated by using the Born Haber cycle in which the sum of enthalpy in a cycle is zero. By applying the Hess’s Law,
Heat of formation ( $\Delta$ Hf°) = Heat of atomization( $\Delta$ Ha°)+ Dissociation energy( $\Delta$ Hd°)+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy
Or can be rearranging as;
Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities)
= -796 -178-242-(590+1145) –(2 x -364)
= - 796 - 1427 = - 2223 kJ/mol
Below you could see some problems
Calculate the standard free energy for the following reaction by using given data.
(l) + H2
O(l) → COCl2
(g) + 2HCl(g)
- Heat of formation of CCl4= -135.44 kj/mol
- Heat of formation of H2O= -285.83 kj/mol
- Heat of formation of COCl2= -218.8 kj/mol
- Heat of formation of HCl= -92.307 kj/mol
By applying Hess’s law, given data can be written as follow.
CCl4(l) → C(s) + 2Cl2(g) ΔH1 = -(-135.44kJ/mol
H2O(l) → H2(g)+ ½O2(g) ΔH2 = -(-285.83 kJ/mol)
C(s)+ ½O2(g) + Cl2(g) → COCl2(g) ΔH3= (-218.8 kJ/mol)
H2(g) + Cl2(g) → 2HCl(g) ΔH4 = 2(-92.307 kJ/mol)
Overall reaction: CCl4(l) + H2O(l) → COCl2(g) + 2HCl(g)
= 135.44+ 285.83-218.8+ (2x -92.307)
= 17.9 kJ/mol
Write all energy changes involve in Born-Haber cycle for the formation of calcium oxide Solution:
Enthalpy of atomization of Ca(s): Ca(s) → Ca(g) ; Δ Ha°
- First ionization energy of Ca(g): Ca(g) → Ca+(g) + e- ; Δ HIE°
- Second ionization energy of Ca(g): Ca+(g) → Ca2+(g) + e- ; Δ HIE°
- Enthalpy of atomization of O2(g): ½ O2(g) → 2O(g) ; Δ Ha°
- First electron affinity of O(g): O(g)+ e-→ O(g)-; Δ Hea°
- Second electron affinity of O(g): O-(g) + e- → O2-(g) ; Δ Hea°
- Enthalpy of formation of CaO(s): Ca(s) + ½ O2(g) → CaO(s) ; Δ Hf°
- Lattice enthalpy of CaO(s) : Ca2+(g) + O2-(g) → CaO(s); Δ HLE°
Calculate the lattice enthalpy of calcium oxide from the following data.
- Δ Ha°of Ca(s): 178 kJ/mol
- 1st Δ HIE°of Ca(g): 590 kJ/mol
- 2nd Δ HIE°of Ca(g): 1150 kJ/mol
- Δ Ha° of O2(g): 249 kJ/mol
- 1st Δ Hea°of O(g): -141 kJ/mol
- 2nd Δ Hea°of O(g): 844 kJ/mol
- Δ Hf°of CaO(s): -635 kJ/mol
Δ Hf°+ ∆HLE° = Δ Ha°(Ca) + ∑ Δ HIE° + Δ Ha°(O) + ∑ Δ Hea°
∆HLE° = Δ Ha°(Ca)+ ∑Δ HIE°+ Δ Ha°(O) + ∑Δ Hea° - Δ Hf°
= 178 + (590 + 1150) + 249 + (-141) + 844 –(-635)
= 2870 – (-635)
= 3505 kJ/ mol
By using Born Haber cycle for the formation of silver chloride, calculate the lattice enthalpy of silver chloride from the following data.
- Enthalpy of atomization of Ag(s): 284 kJ/mol
- 1st ionization energyof Ag(g): 731 kJ/mol
- Enthalpy of atomization of Cl2(g): 122 kJ/mol
- 1st electron affinity of Cl(g): -349 kJ/mol
- Enthalpy of formation of AgCl(s): -127 kJ/mol
∆HLE° = ∆Ha°(Ag)+ ∑∆HIE°+ ∆Ha°(Cl2)+ ∆Hea° - ∆Hf°
= 284 + (731+ 122) + (-349) – (-127)
= 788– (-127)