Question

M.Vammed दि me: Aolullah : Section: Roster #_to Homework 10: Due Sunday, November 17th Show your work. Answers only accepted on printed handout (no photocopies allowed). 1. Use the following data to calculate the AH for magnesium chloride. Lattice energy +2527 kJ/mol First ionization energy of Mg +738 kJ/mol Second Ionization energy of Mg +1451 kJ/mol Electron affinity of Cl +349 kJ/mol Bond enthalpy of Cl2 +242.7 kJ/mol +148 kJ/mol 3 Enthalpy of sublimation of Mg 4 2 32 +195-(3 49

Answer 1

Sol.

Sublimation Reaction of Mg :

Mg(s) -----> Mg(g) , deltaH1° = 148 KJ / mol

First ionization reaction of Mg :

Mg(g) -----> Mg+(g) + e- , deltaH2° = 738 KJ / mol

Second ionization reaction of Mg :

Mg+(g) ----> Mg2+(g) + e- , deltaH3° = 1451 KJ / mol

Bond enthalpy reaction of Cl2 :

Cl2(g) ----> 2Cl(g) , deltaH4° = 242.7 KJ / mol

Electron affinity reaction of Cl :

2Cl(g) + 2e- ----> 2Cl-(g) ,

deltaH5° = - 2 × 349 = - 698 KJ/ mol ( Here energy is released , so , sign is negative and there are 2 Cl atoms , so, energy is multiplied by 2 )

Lattice energy reaction :

Mg2+(g) + 2Cl-(g) -----> MgCl2(s) ,

deltaH6° = - 2527 KJ / mol ( Here energy is released , so, sign is negative )

Now , Enthalpy of formation reaction of MgCl2 :

Mg(s) + Cl2(g) -----> MgCl2(s)

So , deltaHf°

= deltaH1° + deltaH2° + deltaH3° + deltaH4° + deltaH5° + deltaH6°

= 148 + 738 + 1451 + 242.7 - 698 - 2527

= - 645.3 KJ / mol