Understanding Lattice Energy and Its Implications in Ionic Compounds
This document explores the concept of lattice energy (L.E.), defined as the energy released when one mole of a solid ionic compound forms from its gaseous constituents. It delves into the necessity of considering ionic compounds as solids at standard temperature and pressure (STP) and explains the significance of gaseous states in bond formation. The interaction of ions in different states, such as gaseous to solid through hydration processes, is also analyzed through calculations of heat energy changes for lithium chloride (LiCl) and sodium chloride (NaCl), demonstrating their exothermic nature and implications on temperature changes.
Understanding Lattice Energy and Its Implications in Ionic Compounds
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Presentation Transcript
Qn. 10 To do or not to do?---------That is a question too…. Liu Jiani Sherry (14)
a) L.E.: It is the energy evolved when 1 mole solid ionic compound formed from its gaseous constituents. WHY?
Standard conditions • Why solid? Recall what is missing from the definition? Ionic compounds are operationally defined as solids at stp, with the exceptions of a few composed of cation and anionic complex….
Why gaseous constituents? L.E. refers to the energy released from a bond formation. In gaseous state, ions have minimal interaction when they form the solid. So they bond energy evolved is then a good indication of the actual strength.
b) LE is proportional to (q+ * q-)/ (r+ + r-) • Charges of Li =Na • Ionic radius of Li > Na • smaller inter-ionic distance between LiCl and bigger electrostatic forces of attraction between its ions • Thus, larger magnitude and more exothermic.
-L.E: Solid ionic compound to gaseous ions • Delta E of solution: dissolution of ionic compound in water to get aqueous ions • Delta E of Hydration: hydration of individual gaseous ions to form Ion-Dipole interactions with water
0 Energy/ KJ per mol Li+ (g)+ Cl- (g) ∆H hyd (Li+) =-499 L.E of LiCl=-848 Li+ (aq)+ Cl- (g) ∆H hyd (Cl-) =-381 LiCl (s) ∆H sol LiCl(s) =-36 Li+ (aq)+ Cl- (aq) ∆H sol LiCl(s) = -L.E.+∆H hyd (Li+) + ∆H hyd (Cl-) =848 + (-499) + (-381) = -36KJ/mol
0 Energy/ KJ per mol Na+ (g)+ Cl- (g) L.E of NaCl=-776 ∆H hyd (Na+) =-390 Na+ (aq)+ Cl- (g) ∆H hyd (Cl-) =-381 Na+ (aq)+ Cl- (aq) ∆H sol NaCl(s) =+5 NaCl (s) ∆H sol NaCl(s) = -L.E.+∆H hyd (Na+) + ∆H hyd (Cl-) =776 + (-390) + (-381) = +5KJ/mol
Last part… Final temp. solution=? Q=mc ∆T -36000J/mol * [1.0/(6.9+35.5)]mol = (100*4.2* ∆T)J ∆T=2.02 degree celsius Final temp.=298K+2.02K =300.02K =300K (3sf)
