The lattice energy of lithium chloride

114 downloads 0 Views 6MB Size Report
+122 kJ mol. Electron affinity of chlorine atoms = -349 kJ mol a) Using the usual chemical symbols, the state symbols (s), (l), (g) and the symbol for an electron, e- ...
Dr. Pedro Julio Villegas Aguilar [email protected]

OBJECTIVES • To become familiar with problems about thermo-chemistry (Born-Haber cycle). TEXTBOOK Brown, Lemay & Bursten, Chemistry: The Central Science, 10th Ed. (Chapter 5)

QUESTION 1: What Born-Haber cycle is?

QUESTION 1: What Born-Haber cycle is?

• The Born-Haber cycle consists of several steps of measurable enthalpy values e.g. enthalpy of ionization, enthalpy of atomization, etc. from which the lattice enthalpy Hlatt can be calculated.

BORN-HABER CYCLE Example: Sodium Bromide

BORN-HABER CYCLE Example: Sodium Bromide: Breaking down each of the stages.

Example: Lithium Fluoride: step by step.

Reactant coordinates

QUESTION 2: The Born-Haber cycle for sodium chloride is represented below:

Which one of the following statements is correct? A The electron affinity of chlorine is z. B The first ionization energy of sodium is w. C The lattice energy of sodium chloride is v.

QUESTION 2: The Born-Haber cycle for sodium chloride is represented below:

Which one of the following statements is correct? A The electron affinity of chlorine is z. B The first ionization energy of sodium is w. C The lattice energy of sodium chloride is v. A The electron affinity of chlorine is z. (We are going from Cl(g) to Cl-(g))

PROBLEM 1: LiCl is formed from Li (Nu. 3) and Cl (Nu. 17). Its lattice energy may be calculated from a Born-Haber cycle using the following experimental data: Ionization energy of Li = +520 kJ mol Heat to convert Li(s) to Li(g) = +159 kJ mol Heat of formation of LiCl = -409 kJ mol Heat to convert Cl2(g) to Cl(g) = +122 kJ mol Electron affinity of chlorine atoms = -349 kJ mol a) Using the usual chemical symbols, the state symbols (s), (l), (g) and the symbol for an electron, e-, write an equation in each case to define the following terms: (i) The first ionization energy of lithium. (ii) The heat of formation of lithium chloride. (iii) The electron affinity of chlorine. (iv) The lattice energy of lithium chloride. b) Construct a labeled Born-Haber cycle for the formation of LiCl. c) Calculate the lattice energy of LiCl.

PROBLEM 1: LiCl is formed from Li (Nu. 3) and Cl (Nu. 17). Its lattice energy may be calculated from a Born-Haber cycle using the following experimental data: Ionization energy of Li = +520 kJ mol Heat to convert Li(s) to Li(g) = +159 kJ mol Heat of formation of LiCl = -409 kJ mol Heat to convert Cl2(g) to Cl(g) = +122 kJ mol Electron affinity of chlorine atoms = -349 kJ mol

a) Using the usual chemical symbols, the state symbols (s),

(l), (g) and the symbol for an electron, e-, write an equation in each case to define the following terms: (i) The first ionization energy of lithium: Li(g) → Li+ + 1eDH = +520 kJ mol (ii) The heat of formation of lithium chloride: Li(s) + ½Cl2(g) → LiCl(s) DHf = -409 kJ/mol (iii) The electron affinity of chlorine: Cl(g) + 1e- → Cl-(g) DH = -349 kJ/mol (iv) The lattice energy of lithium chloride: LiCl(s) → Li+(g) + Cl-(g) DH = ???

PROBLEM 1: LiCl is formed from Li (Nu. 3) and Cl (Nu. 17). Its lattice energy may be calculated from a Born-Haber cycle using the following data: Ionization energy of Li = +520 kJ mol Heat to convert Li(s) to Li(g) = +159 kJ mol Heat of formation of LiCl = -409 kJ mol Heat to convert Cl2(g) to Cl(g) = +122 kJ mol Electron affinity of chlorine atoms = -349 kJ mol

b) Construct a labeled Born-Haber cycle for the formation of LiCl.

Electron affinity of Cl Atomization energy of Cl

Ionization energy of Li

Atomization energy of Li

Heat of formation of LiCl

Lattice energy of LiCl

PROBLEM 1: LiCl is formed from Li (Nu. 3) and Cl (Nu. 17). Its lattice energy may be calculated from a Born-Haber cycle using the following data: Ionization energy of Li = +520 kJ mol Heat to convert Li(s) to Li(g) = +159 kJ mol Heat of formation of LiCl = -409 kJ mol Heat to convert Cl2(g) to Cl(g) = +122 kJ mol Electron affinity of chlorine atoms = -349 kJ mol

c) Calculate the lattice energy of LiCl.

Li(s) → Li(g) ΔH° = 159 kJ Li(g) → Li+(g) + 1eΔH° = 520 kJ ½Cl2(g) → Cl(g) ΔH°=122kJ/mol*0.5mol=61.0kJ Cl(g) + 1e- → Cl-(g) ΔH° = -349 kJ Li+(g) + Cl-(g) → LiCl(s) ΔH° = ? (-LE) Li(s) + ½Cl2(g) → LiCl(s)

ΔH°f = -409 kJ

-409kJ = 159kJ + 520kJ + 61.0kJ + -349 kJ + (-LE) LE = 800 kJ

QUESTION 3: The lattice energies for sodium chloride, potassium chloride and potassium bromide are + 780, +710 and +680 kJ/mol, respectively. Comment on these results and that for lithium chloride.

QUESTION 3: The lattice energies for sodium chloride, potassium chloride and potassium bromide are + 780, +710 and +680 kJ/mol, respectively. Comment on these results and that for lithium chloride.

LiCl = 800

Lattice energies (kJ/mol): NaCl = +780 KCl = +710

KBr = 680

•The lattice energy is proportional to the charge/distance. •The charges are the same in all cases. •The sizes of the ions change and hence the distance between the ions change. •Li+ and Cl- are the smallest ions, so the distance is the smallest and LE the biggest. •K+ and Br- are the biggest ions and so the distance is biggest and LE smallest. •The same reasoning tells us that NaCl has a larger LE than KCl (Na+ is smaller than K+).

PROBLEM 2: Answer the questions below regarding the following Born-Haber Cycle.

a) Which step corresponds to ionization energy? A B C D E F b) Which step corresponds to electron affinity? A B C D E F c) What is the value of the lattice energy for LiF(s)? d) Would you expect MgO(s) to have a larger or smaller lattice energy? Explain.

PROBLEM 2: Answer the questions regarding the following Born-Haber Cycle.

a) Which step corresponds to ionization energy? A B C D E F b) Which step corresponds to electron affinity? A B C D E F

PROBLEM 2: Answer the questions regarding the following Born-Haber Cycle.

c) What is the value of the lattice energy for LiF(s)? v + -328 + LE v -618 = 161 + 79.5 + 520 LE = + 1050 kJ

v

v

v

v

v

v

PROBLEM 2: Answer the questions regarding the following Born-Haber Cycle. d) Would you expect MgO(s) to have a larger or smaller lattice energy? Explain.

I would expect MgO to have a larger lattice energy. Lattice energy is proportional to the charge/size of the ions. The charges on the ions are larger for MgO and the sizes are comparable.

DH = -1050 kJ

PROBLEM 3: Calculate the enthalpy change of the reaction: Na(s) + ½Br2(g) → NaBr(s) DΗatomization(Na) = +107 kJ/mol 1 DΗatomization(Br) = +97 kJ/mol 2 DΗionization energy(Na) = +496 kJ/mol DΗelectron affinity(Br) = -325 kJ/mol DΗlattice energy(NaBr) = -742 kJ/mol

3 4 5

DΗ = 107 + 496 +97 -325 -742 = -367 kJ/mol 1 2 3 4 5 Lets do it using a graph a Born-Haber Cycle:

DΗ = 107 + 496 +97 -325 -742 = -367 kJ/mol

CONCLUSIONS • We practice different way to solve problems about Born-Haber cycle associated with first Law of Thermodynamic. • Is very important be careful with the units that we use.