Lattice enthalpy

            Ionic bonds are strong

            Due to strong electrostatic forces of attraction

Definition Lattice enthalpy

 

The lattice enthalpy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions (298K, 100kPa).

 

 

Na+(g) + Cl-(g) g NaCl(s) DHqLE = -781 KjMol-1

Features of lattice enthalpy:

  • Lattice enthalpy is an exothermic process as large amounts of energy is released upon the formation of the lattice from gaseous ions.

  • Lattice enthalpy indicates the strength of ionic bonds in a compound lattice structure.

  • Lattice enthalpy is impossible to measure directly as it is impossible to have gaseous ions.

  • Covalent compounds cannot have lattice enthalpies due to the nature of the bonding.

 

 

Key enthalpy changes:

 

1)  Standard enthalpy change of formation, DH qf :

 

1 mole of compound is formed from its constituent elements in their standard state.

 

Example - usually exothermic

 

                                                              K(s)     +    1/2Cl2(g)           KCl(s)            DH qf  =  - 437 KJ Mol-1

 

2)  The standard enthalpy change of atomisation,DH qat :

1 mole of gaseous atoms are formed from the element in its standard state.

 

Example - Always endothermic as bonds have to be broken:

 

                                                              Mg(s)    Mg(g)            DH qat  =  +148 KJ Mol-1

 

3)  First ionisation energy, DH qI1 :

 

1 mole of gaseous 1+ ions is formed from gaseous atoms

 

Example - Always endothermic as an electron has to overcome the attraction from the nucleus:

 

                                                              Li(g)      Li+(g)    +  e-            DH qI1  =  + 520 KJ Mol-1

 

4)  Second ionisation energy, DH qI2 :

 

1 mole of gaseous 2+ ions is formed from 1 mole gaseous 1+ ions

 

Example - Always endothermic as an electron has to overcome the attraction from the nucleus:

 

                                                              Ca1+(g)      Ca2+(g)    +  e-            DH qI2  =  + 1145 KJ Mol-1

 

5)  First Electron affinity, DH qEA1 :

1 mole of gaseous 1- ions formed from gaseous atoms

 

Example - Is exothermic as an electron is attracted to the outer shell by the nucleus:

 

                                                             Cl(g)  +  e-    Cl-(g)              DH qEA1  =  - 349 KJ Mol-1

 

 6)  Second Electron affinity, DH qEA2 :

1 mole of gaseous 2- ions formed from 1 mole gaseous 1- ions

 

Example - Is endothermic as the electron being put in is repelled by the 1- ion:

 

                                                             O-(g)  +  e-    O2-(g)              DH qEA2  =  - 798 KJ Mol-1

 

Qu 1 - 3  P 167

 

Constructing Born - Haber cycles

 

Calculation:

 

Route 1     =     - Route 2  +  Route 3

 

Route 1     =     Route 3  -  Route 2

 

Lattice enthalpy, DHqLE         =         DHqf      -         Route 2 (the rest)

 

Or better - rearranged:

 

  DHqLE = DHqf - (SUM OF DHq THE REST)

 

Other rules:

 

2 ions of the same formula, MgCl2 You need to count the enthalpies from the atom in its standard state to the ion in its gaseous state TWICE.

 

More than 1 charge on the ion you need to 2 ionisation energies or electron affinities:

 

eg  Mg    Mg+    Mg2+                      or                            O    O-    O2-

 

Qu 1 - 3  P169

 

Born - Haber cycle calculation

1)  Caesium chloride:

 

Calculation:

 

  DHqLE = DHqf - (SUM OF DHq THE REST)
  DHqLE = - 433 - ( +79 + 121 + 376 +  - 346)
  DHqLE = - 433 - ( +79 + 121 + 376 - 346)
  DHqLE = - 663

Kj mol-1

 

2)  Sodium oxide:

 

 

 

Calculation:

 

  DHqLE = DHqf - (SUM OF DHq THE REST)
  DHqLE = - 414 - ( +216 + 249 + 992 + - 141 + 790)
  DHqLE = - 414 - ( +216 + 249 + 992 - 141 + 790)
  DHqLE = - 2520

Kj mol-1

 

Qu 1  P 171

 

Further examples of Born - Haber cycles

 

3)  Calcium chloride:

Calculation:

 

  DHqLE = DHqf - (SUM OF DHq THE REST)
  -2258 = - 795 - ( +178 + 242 + 590 + 1145 + (2 x DHqEA))
  -2258 = - 795 - (+ 2155 + (2 x DHqEA))
  -2258 = - 795 - 2155 - (2 x DHqEA)
  -2258 = - 2950   - (2 x DHqEA)
  2 x DHqEA = - 2950   2258
  2 x DHqEA = - 692    
  DHqEA = - 346

Kj mol-1

 

4)  Copper (II) oxide:

 

 

Calculation:

 

  DHqLE = DHqf - (SUM OF DHq THE REST)
  DHqLE = - 155 - ( +339 + 249 + 745 + 1960 + -141 + 790)
  DHqLE = - 155 - ( +339 + 249 + 745 + 1960 - 141 + 790)
  DHqLE = - 4097

Kj mol-1

 

Qu 1  P173

 

Enthalpy change of solution

 

Practical

Video clip

 

  • If the electrostatic forces of attraction are so strong, how is this possible?

  • The clue comes with the temperature changes that occur when they dissolve.

  • Calcium chloride gives us an exothermic reaction.

  • Ammonium nitrate gives us an endothermic reaction.

  • The energy produced when water surrounds the ions must be about the same as the electrostatic forces of attraction between the ions.

  • If there is energy left over it is given out as heat energy.

  • If there is not enough energy it is taken in from the surroundings

  • These can all be calculated in another type of Hess's cycle but some other enthalpy changes are required first:

 

 

1)  Enthalpy change of solution, DHqs

 

1 mole of a compound is dissolved in water under standard conditions

 

Example - This process can be endothermic or exothermic:

 

                                                             KCl(s)  +  aq     K+(aq)   +      Cl-(aq)

 

2)  Enthalpy change of hydration, DHqhyd

 

1 mole of aqueous ions are formed from their gaseous ions under standard conditions

 

Example - This process is an exothermic process:

 

                                                             K+(g)  +  aq     K+(aq)

 

                                                             Cl-(g)  +  aq     Cl-(aq)

 

What happens when a solid dissolves:

     1)  Ionic lattice breaks down into gaseous ions, Ein:

K+(g) + Cl-(g) g KCl(s) DHqLE = - 771 KjMol-1
KCl(s) g K+(g) + Cl-(g) DHqLE = + 771 KjMol-1

 

     2)  Hydration of gaseous ions, Eout:

 

                                                             K+(g)  +  aq     K+(aq)

 

                                                             Cl-(g)  +  aq     Cl-(aq)

Calculating a lattice enthalpy from enthalpy changes of solution and hydration:

 

Route 1    =     Route 2        +        - Route 3

Route 1    =     Route 2        -          Route 3

DHqLE      =      Route 2        -          DHqs

  • Route 2 is made up of 2 steps:

DHqLE     =       (SUM OF DHqhyd ions)       -          DHqs

DHqLE     =       (- 322     +     - 363)       -          +26

DHqLE     =       (- 322     -      363)       -          26

DHqLE     =       - 685            -          26

DHqLE     =       - 711  Kj mol-1

Qu 1  P175

Understanding hydration and lattice enthalpies

Lattice enthalpies

        1.      Ionic size

        2.      Ionic charge

 

Ionic size

 

Ionic charge

            

 

Enthalpy change of hydration:

        1.      Ionic size

        2.      Ionic charge

 

Ionic size

Down the group, size increases, DHqhyd gets less exothermic

Ion DHqhyd
Cl- -363
Br- -336
I- -295

 

Ionic charge

Across a Period, charge increases and size decreases, DHqhyd gets more exothermic

Ion DHqhyd
Na+ -406
Mg2+ -1921
Al3+ -4665

 

Qu 1 - 3  P177

 

Entropy

 

What is entropy:

Entropy:

 

                Melting ice cream

                Smell of cooking spreading

                        Your bedroom

                        Expanding Universe

 

 

 

What can entropy tell us?

Order

State

Explanation

Entropy

 

Ordered (O)

Solids

Vibrating in fixed position

Low

 
 

Disordered (DO)

Liquids/ Aqueous

Free moving but closely packed

Medium

 
 

Very Disordered (VDO)

Gases

As far apart as possible, moving lots

High

 
 

Using Entropy:

Reaction NH4NO3(s)  + aq NH4+(aq)  + NO3-(aq) 
Order Ordered   Disordered Disordered   Disordered
Overall Ordered Disordered
  LIKELY TO HAPPEN

 

Reaction MgCO3(s)  MgO(s)  + CO2(g) 
Order Ordered Ordered   Very disordered
Overall Ordered Disordered
  LIKELY TO HAPPEN

 

Calculating entropy:

                         DSqsystem        =         SSqproducts         -           SSqreactants

Worked example:

Reaction N2(g)  + 3H2(g)  2NH3(g) 
Entropy content + 192   + 131   + 193

 

                         DSqsystem        =         SSqproducts         -           SSqreactants

                         DSqsystem        =       (2 x +193)   -       [ (+192)  +  (3 x +131) ]

                         DSqsystem        =       -199 J K-1 mol-1

Qu 1 - 2  P179

 

Free energy

 

Spontaneous process

Exothermic reactions:

 

  • Are usually spontaneous at room temperature.

  • The enthalpy content decreases during the reaction and the excess energy being released to the surroundings.

  • There is an increase in stability.

 

 

Endothermic reactions:

 

  • Some endothermic reactions are also spontaneous at room temperature.
  • The enthalpy content increases during the reaction with energy being taken in from the surroundings.

  • Overall this must also increase in stability.

  Enthalpy Deals with the contribution of energy of the surroundings
  Entropy Deals with the contribution of energy of the system

Energy from entropy

                Energy derived from entropy  =  TDS

Free energy and feasibility:

1 Temperature in k T
2 Entropy change of the system DS
3 Enthalpy change with the surroundings DH
DG = DH - TDS
    Surroundings   System

                                DG < 0    Negative

DH DS DG Feasibility / spontaneous
(-)ve (+)ve (-)ve Feasible
(+)ve (-)ve (+)ve Never feasible
(-)ve (-)ve (-)ve at low temps Feasible at low temps
(+)ve (+)ve (-)ve at high temps Feasible at high temps

How do endothermic reactions take place:

DG = DH - TDS
DG = DH - TDS
    DH < TDS
    (+)ve < TDS

Example:

At what temperature will ZnCO3 decompose?

Reaction ZnCO3(s)  ZnO(s)  + CO2(g)  DH = +71 KJ mol-1
Entropy, DS +82   +44   +214  

1)  Calculate DSqsystem:

                         DSqsystem        =         SSqproducts         -           SSqreactants

                         DSqsystem        =       (+44  +  +214)   -        +82

                         DSqsystem        =       176 J K-1 mol-1

                         DSqsystem        =       0.176 KJ K-1 mol-1  COMMON ERROR - MUST COVERT TO KJ

 

2)  Assume DG = 0 to calculate T:

DG = DH - TDS
0 = + 71 - T  x  0.176
T  x  0.176 = + 71    
T = + 71    
    0.176    
T = 403 k (-273 for oC)  
T = 130 oC    

Qu 1 - 2  P181  /  1 - 4  P195  /  1 - 7  P196 - 198