In the previous section, we saw Hess's law and bond enthalpy. In this section, we will see lattice enthalpy
Lattice enthalpy can be explained in 7 steps:
1. Consider the ionic compound NaCl
♦ We know that, NaCl has a crystal structure
♦ The crystal is made up of Na+ and Cl- ions
♦ Those ions are arranged in a symmetric 3D form
♦ This 3D form is called crystal lattice
♦ (some images can be seen here)
2. The crystal lattice of NaCl is very stable because of the strong electrostatic attractions between the Na+ and Cl- ions
• We want to know how much energy will be required for the following process:
♦ Separating all the Na+ and Cl- ions in one mol NaCl
◼ The
resulting individual ions should be scattered far away from each other
so that, there will be no attractive or repulsive forces between the
resulting ions. In other words, the resulting ions must be in the
gaseous state
3. It is obvious that:
• When the lattice is turned into gaseous ions, we will need to supply energy. In other words, it will be an endothermic process
• For the reverse process, that is., when
the gaseous ions come together and form the lattice, we will receive
energy. In other words, it will be an exothermic process
◼ This energy that we supply/receive is called lattice enthalpy
4. Lattice enthalpy can be defined in two ways:
◼ Lattice dissociation enthalpy
The
lattice dissociation enthalpy is the enthalpy change needed to convert 1
mole of solid crystal lattice into its scattered gaseous ions
♦ We will denote it by the symbol: ΔH⊖lattice(d)
◼ Lattice formation enthalpy
The
lattice formation enthalpy is the enthalpy change when 1 mole of solid
crystal lattice is formed from its separated gaseous ions
♦ We will denote it by the symbol: ΔH⊖lattice(f)
5. It is obvious that:
♦ Both ΔH⊖lattice(d) and ΔH⊖lattice(f) will have the same magnitude
♦ The sign of ΔH⊖lattice(d) will be positive
♦ The sign of ΔH⊖lattice(f) will be negative
6. To find ΔH⊖lattice(d) of NaCl means, to find X in the following thermochemical equation:
♦ NaCl(s) → Na+(g) + Cl-(g); ΔH⊖lattice(d) = X
7. To find ΔH⊖lattice(f) of NaCl means, to find X in the following thermochemical equation:
♦ Na+(g) + Cl-(g) → NaCl(s); ΔH⊖lattice(f) = X
• So we will see how ΔH⊖lattice(d) or ΔH⊖lattice(f) is determined
• It is not possible to carry out the experiment in (6) above. Neither is it possible to carry out the experiment in (7)
◼ That means, we cannot measure lattice enthalpy from any experiments
• So we use an indirect method. It can be written in 22 steps:
1. The cyan horizontal lines in fig.6.13 below indicates various states in an experiment
• The arrows indicate various processes which transform the system from one state to another
 |
| Fig.6.13 |
2. Consider the thick cyan horizontal line. It is the datum line. It is the state from which we begin our calculations
• At this state, the system consists of:
♦ One mol Na atoms
✰ These atoms are in the solid state
♦ Half mol Cl2 molecules
✰ These molecules are in the gaseous state
◼ The above solid and gaseous states are obvious because:
♦ At standard temperature and pressure
✰ a sample of sodium consists of Na atoms in the solid state
✰ a sample of chlorine consists of Cl2 molecules in the gaseous state
3. From the datum line, we begin our first process
• Our first process is to convert the sodium into gaseous state
• So it will be a sublimation process
4. From the data book, we have:
One mol Na(s) requires 108.4 kJ energy, for complete sublimation
• In other words, ΔH⊖sub for Na(s) = 108.4 kJ mol-1
• So the thermochemical equation for this process will be:
Na(s) + 1/2Cl2(g) → Na(g) + ½ Cl2(g); ΔH⊖sub = 108.4 kJ mol-1
• This process is numbered as I in the fig.12.13 above
5. So when the process I is complete, we will have:
♦ One mol Na atoms in the gaseous state
♦ Half mol Cl2 molecules in the gaseous state
6. Our second process is to convert each of the gaseous Na atoms into gaseous Na+ ions
• So it will be an ionization process
7. From the data book, we have:
One mol Na(g) requires 495.6 kJ energy for complete ionization
• In other words, ΔH⊖i for Na(g) = 495.6 kJ mol-1
• So the thermochemical equation for this process will be:
Na(g) + 1/2Cl2(g) → Na+(g) + ½ Cl2(g); ΔH⊖i = 495.6 kJ mol-1
• This process is numbered as II in the fig.12.13 above
8. So when the process II is complete, we will have:
♦ One mol Na+ ions in the gaseous state
♦ Half mol Cl2 molecules in the gaseous state
9. Our third process is to separate each of the gaseous Cl2 molecules into two gaseous Cl atoms
• So it will be a bond dissociation process
10. From the data book, we have:
One mol Cl2(g) requires 242 kJ energy for complete bond dissociation
• In other words, ΔH⊖Cl-Cl = 242 kJ mol-1
♦ In our present case, we have half mol Cl2 molecules
♦ It will give one mol Cl atoms
♦ It will require (242/2) = 121 kJ
• So the thermochemical equation for this process will be:
Na+(g) + ½ Cl2(g) → Na+(g) + Cl(g); 1/2ΔH⊖Cl-Cl = 121 kJ mol-1
• This process is numbered as III in the fig.12.13 above
11. So when the process III is complete, we will have:
♦ One mol Na+ ions in the gaseous state
♦ One mol Cl atoms in the gaseous state
12. Our fourth process is to convert each of the gaseous Cl atoms into gaseous Cl- ions
• So it will be an electron accepting process
• We have seen electron gain enthalpy in an earlier chapter (details here)
13. From the data book, we have:
One mol Cl(g) releases 348.6 kJ energy when all the atoms accept one electron each
• In other words, ΔH⊖eg for Cl(g) = 348.6 kJ mol-1
• So the thermochemical equation for this process will be:
Na+(g) + Cl(g) → Na+(g) + Cl-(g); ΔH⊖eg = -348.6 kJ mol-1
• This process is numbered as IV in the fig.12.13 above
14. So when the process IV is complete, we will have:
♦ One mol Na+ ions in the gaseous state
♦ One mol Cl- ions in the gaseous state
15. The completion of process IV is a milestone
• Consider the products obtained at the end of this process:
♦ One mol Na+ ions in the gaseous state
♦ One mol Cl- ions in the gaseous state
◼ It is from these products that, we calculate the of ΔH⊖lattice(f) NaCl. This fact is clear from the 'definition of ΔH⊖lattice(f)' that we wrote in (4) at the beginning of this section
16. So, our fifth process is to make the system release an energy equal to ΔH⊖lattice(f)
• When this energy is released, we will get NaCl(s)
• The ΔH⊖lattice(f) value of NaCl is available in the data book
♦ But remember that, our aim itself is to find this ΔH⊖lattice(f)
♦ For the time being, we will ignore the value given in the data book, and find it ourselves
• The thermochemical equation of this process will be:
Na+(g) + Cl-(g) → NaCl(s); ΔH⊖lattice(f) = X kJ mol-1
• This process is numbered as V in the fig.12.13 above
• Our aim is to find X
17. To find X, we split it into X1 and X2
• This is shown in fig.6.13
◼ It is clear that:
After completing process IV, if the system releases X1, the datum line can be reached
• The process in which X1 is released, is marked as V1
◼ When this X1 is released, we can say:
The net energy absorbed/released by the system becomes zero
• Thus we get:
108.4 + 495.6 + 121 - 348.6 - X1 = 0
⇒ X1 = 376.4
18. So when the process V1 is complete, we will have:
♦ One mol Na(s)
♦ Half mol Cl2(g)
19. The completion of process V1 is another milestone
• We have reached back at the datum
• From here, we have to release some more energy
♦ This energy is marked as X2
• The process in which X2 is released, is marked as V2
20. We have to find the magnitude of X2
♦ For this, the red arrow gives us a valuable clue. It can be explained in 4 steps:
(i) Consider the products obtained at the end of process V1:
♦ One mol Na(s)
♦ Half mol Cl2(g)
◼ It is from these products that, we calculate the of ΔH⊖(f) NaCl(s)
◼ This fact is clear from the definition of ΔH⊖(f) that we wrote in an earlier section
♦ It is the enthalpy of formation from constituent elements in the standard states
♦ It is available in the data book: -411.2 kJ mol-1
◼ It is clear that:
From the datum line, if the system releases 411.2 kJ, the bottom most cyan line can be reached
(ii) The process in which this ΔH⊖(f) is released, is marked with the downward red arrow
♦ The down ward red arrow reaches upto the bottom most cyan line
♦ After completing process V1 also, we have to reach the bottom most cyan line
(iii) So we can write:
♦ The process represented by the downward red arrow
♦ is equivalent to
♦ The process V2
(iv) Thus we get: X2 = 411.2 kJ
21. So we have calculated both X1 and X2
♦ Then X = (X1 + X2) = (376.4 + 411.2) = 787.6 kJ
♦ Thus we can write: ΔH⊖lattice(f) of NaCl(s)= -787.6 kJ mol-1
22. The cyclic process shown in fig.6.13 is known as the Born-Haber cycle
◼ It is based on the Hess's law of constant heat summation
◼ Whatever be the path, the energy of the system at a particular state, will be the same
◼ This method can be used to determine those ΔH values which are impossible to find experimentally
• Note:
The above result of -787.6 kJ mol-1 can be obtained using another approach also. It can be explained in 4 steps:
1. From the datum line, we travel along the path: I - II - III - IV - V
• We reach the bottom most cyan line
• The net energy along this path = (108.4 + 495.6 + 121 - 348.6 - X) = (376.4 - X)
2. From the datum line, we travel along the red arrow
• We reach the bottom most cyan line
• The net energy along this path = -411.2
3. In both the paths:
♦ The starting states are the same: The datum line
♦ The ending states are the same: The bottom most cyan line
4. So, according to Hess's law, the energies must be the same
• Thus we can write: (376.4 - X) = -411.2
⇒ X = 787.6 kJ mol-1
• In the next section, we will see enthalpy of solution
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