Topic 5: Energetics - Science at Yorkdale with...

Topic 5: Energetics

Born Haber Cycles & Lattice Enthalpy

Thursday, May 16, 2013

Born-Haber Cycle• Why are NF3 and PCl3 stable while NCl3 is

dangerously explosive? • Born-Haber cycle provides a useful way

to account for the relative stabilities of chemical compounds

• Relative stability of the compound is determined by the lattice enthalpy of a compound


Born-Haber Cycle

• How can we determine the lattice enthalpy of an ionic compound?

• We can break up the formation of an ionic compound into a series of steps

• Then take all those steps and add them up so it is really a Hess’ Law problem!

• Recall that the enthalpy of formation for an uncombined element is therefore = 0


Some Definitions• The enthalpy of atomization is the enthalpy change that

occurs when one mole of gaseous atoms is formed from the element in the standard state under standard conditions

Example: ½ Cl2 (g) Cl (g) ΔHoat = 121 kJ mol-1

• The electron affinity is the enthalpy change that occurs when an electron is added to an isolated atom in the gaseous state:

O (g) + e- O- (g) ΔHo = -142 kJ mol-1

O- (g) + e- O2- (g) ΔHo = +844 kJ mol-1

• The lattice enthalpy is the enthalpy change that occurs from the conversion of an ionic compound in the gaseous state into its gaseous ions

LiCl (g) Li+ (g) + Cl- (g) ΔHo = +846 kJ mol-1


Born Haber Cycle for NaCl

The formation of NaCl can be considered as a five step process

Na (s) + 1/2 Cl2 (g) NaCl (s)1. The vaporization of sodium metal to form the gaseous

element. Na(s) Na(g)2. The dissociation of chlorine gas to gaseous chlorine

atoms is equal to one half of the bond energy for a Cl-Cl covalent bond 1/2 Cl2(g) Cl(g)

3. The ionization of gaseous sodium atoms to Na(g) Na+(g) + e-

4. The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.) Cl(g) + e- Cl-(g)

5. The lattice energy on the formation of sodium chloride from the gaseous ions Na+(g) + Cl-(g) NaCl(s)


Born Haber Cycle Diagram

The stepwise energy changes for the formation of NaClThursday, May 16, 2013

Born-Haber Cycle for NaClThe stepwise energy changes for the formation of NaCl:

The vaporization of sodium metal to form the gaseous element.

Na (s) Na (g) ∆H°sublimation = + 109 kJ mol-1

The dissociation of chlorine gas to gaseous chlorine atoms is equal to one half of the bond energy for a Cl-Cl covalent bond

1/2 Cl2 (g) Cl (g) ∆H°diss = + 122 kJ mol-1

The ionization of gaseous sodium atoms to: Na (g) Na+ (g) + e- ∆H°ionization = + 496 kJ mol-1

The ionization of chlorine atoms. (This quantity is the negative electron affinity for the element chlorine.)

Cl (g) + e- Cl- (g) ∆H°elect.affinity = - 368 kJ mol-1


Try it for MgCl2

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Try it! • 1. Calculate the energy released (in kJ mol-1 ) in the reaction K (s) + 1/2 Br2 (l) --------> KBr (s)

The energy of vaporization of potassium is +89 kJ mol-1. The first ionization enthalpy of potassium is +419 kJ mol-1.

The sum of the dissociation energy and vaporization energy for Br2 (l) is 223 kJ mol-1, the negative electron affinity of bromine is -324 kJ mol-1, and the lattice energy of KBr is -688 kJ mol-1.

2. Calculate the energy released (in kJ mol-1 ) in the reaction

Na (s) + 1/2 I2 (l) --------> NaI (s)

The energy of vaporization of sodium is +109 kJ mol-1. The first ionization energy of sodium is +496 kJ mol-1.

The sum of the dissociation energy and vaporization energy for I2 (s) is 214 kJ mol-1, the negative electron affinity of iodine is -295 kJ mol-1, and the lattice energy of NaI is - 704 kJ mol-1.


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Size & Charge • How does size and charge effect the lattice

enthalpy?


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Which has larger lattice enthalpy?

KCl 701 kJ mol-1, NaCl 777 kJ mol -1K is larger

The cation remains the same, the anion gets biggerThe larger the anion, the smaller the enthalpy

MgF2 2957 kJ mol-1NaF 902 kJ mol-1


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Why the difference?


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