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Qn. 10

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?

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Qn. 10

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  1. Qn. 10 To do or not to do?---------That is a question too…. Liu Jiani Sherry (14)

  2. a) L.E.: It is the energy evolved when 1 mole solid ionic compound formed from its gaseous constituents. WHY?

  3. 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….

  4. 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.

  5. 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.

  6. When a bond forms, PE is released as KE.

  7. -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

  8. 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

  9. 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

  10. 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)

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