In the previous section, we saw details about enthalpy of solution
• We saw that
♦ ΔH⊖sol of NaCl is positive
♦ ΔH⊖sol of CaCl2 is negative
• In this section, we will see the reason for why some values are positive, while some are negative
• First we will see a side by side comparison between NaCl and LiCl
• The comparison can be written in 7 steps:
1. Drawing the cycles side by side
♦ Fig.6.17(a) below shows the Born-Haber cycle for NaCl
♦ Fig.6.17(b) below shows the Born-Haber cycle for LiCl
 |
| Fig.6.17 |
2. Comparing lattice enthalpies
♦ In fig.a, the yellow arrow shows the ΔH⊖lattice of NaCl
♦ In fig.b, the yellow arrow shows the ΔH⊖lattice of LiCl
• We see that, in fig.b, the yellow arrow is longer
♦ This is because
✰ ΔH⊖lattice of NaCl is 788
✰ ΔH⊖lattice of LiCl is 829
3. Comparing hydration enthalpy of Cl-
• In the previous section , we drew:
♦ ΔH⊖hyd of the positive ion on top
♦ ΔH⊖lattice of negative ion at bottom
• This time, since we have to make a comparison, we will draw the negative ion (Cl-) on top
♦ In fig.a, the top pink arrow shows the ΔH⊖hyd of Cl- ion in NaCl
♦ In fig.b, the top pink arrow shows the ΔH⊖hyd of Cl- ion in LiCl
• The green dashed line indicates that, both the pink arrows are of the same size
4. Comparing hydration enthalpies of Na+ and Li+
♦ In fig.a, the bottom pink arrow shows the ΔH⊖hyd of Na+ ion in NaCl
♦ In fig.b, the bottom pink arrow shows the ΔH⊖hyd of Li+ ion in LiCl
• We see that, in fig.b, the bottom pink arrow is longer
♦ This is because
✰ ΔH⊖hyd of Na+ is -424
✰ ΔH⊖hyd of Li+ is -538
5. The competing items:
• There is always a competition between the following two items:
(i) The yellow arrow
(ii) Sum of the two pink arrows
• This can be explained in two steps:
(i) The yellow arrow indicates the amount of energy that is to be supplied to dissociate the salt into it's ions in gaseous state
• So we can write:
♦ Longer yellow indicates that, we need to supply greater energy
♦ Shorter yellow indicates that, we need to supply only lesser energy
(ii) The pink arrows indicate the amount of energy that will be released during hydration processes
• So we can write:
♦ Longer pink arrow indicates that, we will receive greater energy
♦ Shorter pink arrow indicates that, we will receive only a lesser energy
(iii) So the net energy supplied/received will depend on the relative sizes of arrows
6. In NaCl, we see that, the yellow arrow is longer than the sum of two pink arrows
♦ So we need to supply a net energy
♦ The ΔH⊖sol is thus positive
7. In LiCl, we see that, the yellow arrow is shorter than the sum of two pink arrows
♦ So we will receive a net energy
♦ The ΔH⊖sol is thus negative
The reason for such a difference can be written in 8 steps:
1. In the periodic table, both Li and Na belongs to group I
2. But Li is above Na
• So Li will have a lesser number of main shells
• Indeed we know that:
♦ Li has two main shells: K and L
♦ Na has three main shells: K, L and M
3. So the ionic radius of Li+ will be smaller than the ionic radius of Na+
(We saw those details when we learnt about periodic trends. Details here)
• When the ionic radius decreases, the attractive force on the opposite ion increases
♦ Then it will be more difficult to separate the two ions
4. In our present case, this increased attraction has the following two effects:
(i) In the lattice structures:
♦ Li+ exerts a greater attraction on Cl-
♦ Na+ exerts a lesser attraction on Cl-
• So more energy is required for dissociation in the case of LiCl
(ii) In the hydrated structures:
♦ Li+ exerts a greater attraction on H2O molecules
♦ Na+ exerts a lesser attraction on H2O molecules
• So more energy will be released during hydration of Li+
5. As a result of 4(i):
♦ the yellow arrow of LiCl is longer
♦ the yellow arrow of NaCl is shorter
6. As a result of 4(ii):
♦ the bottom pink arrow of LiCl is longer
♦ the bottom pink arrow of NaCl is shorter
7. How ΔH⊖sol of LiCl becomes negative:
• We see that (when compared to NaCl), both yellow and bottom pink arrow of LiCl become longer
• We would expect the effects to cancel each other
♦ But the increase in the bottom pink arrow
♦ is much larger than
♦ the increase in yellow arrow
• So the 'increase in pink arrow' outweighs the 'increase in yellow arrow'
♦ Thus we receive a net energy
♦ In other words, ΔH⊖sol of LiCl is negative
8. Thus we see that, the ionic radius plays an important role in determining the sign of ΔH⊖sol
In addition to the 'ionic radius', the 'ionic charge' also plays an important role. This can be explained by comparing NaCl and CaCl2. It can be written in steps:
1. Drawing the cycles side by side
♦ Fig.6.18(a) below shows the Born-Haber cycle for NaCl
♦ Fig.6.18(b) below shows the Born-Haber cycle for CaCl2
 |
| Fig.6.18 |
2. Comparing lattice enthalpies
♦ In fig.a, the yellow arrow shows the ΔH⊖lattice of NaCl
♦ In fig.b, the yellow arrow shows the ΔH⊖lattice of CaCl2
• We see that, in fig.b, the yellow arrow is longer
♦ This is because
✰ ΔH⊖lattice of NaCl is 788
✰ ΔH⊖lattice of LiCl is 2258
3. Comparing hydration enthalpies of Cl-
♦ In fig.a, the top pink arrow shows the ΔH⊖hyd of Cl- ion in NaCl
♦ In fig.b, the top pink arrow shows the ΔH⊖hyd of Cl- ion in CaCl2
• We see that, in fig.b, the top pink arrow is longer
♦ This is because
✰ In CaCl2, two moles of Cl- ions are available
✰ Naturally, greater energy will be released during hydration of Cl-
4. Comparing hydration enthalpies of Na+ and Ca2+:
♦ In fig.a, the bottom pink arrow shows the ΔH⊖hyd of Na+ ion in NaCl
♦ In fig.b, the bottom pink arrow shows the ΔH⊖hyd of Ca+ ion in CaCl2
• We see that, in fig.b, the bottom pink arrow is longer
♦ This is because
✰ ΔH⊖hyd of Na+ is -424
✰ ΔH⊖hyd of Ca2+ is -1650
5. The competing items:
• There is always a competition between the following two items:
(i) The yellow arrow
(ii) Sum of the two pink arrows
• This can be explained in two steps:
(i)
The yellow arrow indicates the amount of energy that is to be supplied
to dissociate the salt into it's ions in gaseous state
• So we can write:
♦ Longer yellow indicates that, we need to supply greater energy
♦ Shorter yellow indicates that, we need to supply only lesser energy
(ii) The pink arrows indicate the amount of energy that will be released during hydration processes
• So we can write:
♦ Longer pink arrow indicates that, we will receive greater energy
♦ Shorter pink arrow indicates that, we will receive only a lesser energy
(iii) So the net energy supplied/received will depend on the relative sizes of arrows
6. In NaCl, we see that, the yellow arrow is longer than the sum of two pink arrows
♦ So we need to supply a net energy
♦ The ΔH⊖sol is thus positive
7. In CaCl2, we see that, the yellow arrow is shorter than the sum of two pink arrows
♦ So we will receive a net energy
♦ The ΔH⊖sol is thus negative
The reason for such a difference can be written in 8 steps:
1. Comparison of charges:
♦ In Na+, the charge is +1
♦ In Ca2+, the charge is +2
• So Ca2+ has a greater charge
• When the ionic charge increases, the attractive force on the opposite ion increases
♦ Then it will be more difficult to separate the two ions
2. In our present case, this increased attraction has the following two effects:
(i) In the lattice structures:
♦ Ca2+ exerts a greater attraction on Cl-
♦ Na+ exerts a lesser attraction on Cl-
• So more energy is required for dissociation in the case of CaCl2
(ii) In the hydrated structures:
♦ Ca2+ exerts a greater attraction on H2O molecules
♦ Na+ exerts a lesser attraction on H2O molecules
• So more energy will be released during hydration of Ca2+
3. As a result of 2(i):
♦ the yellow arrow of CaCl2 is longer
♦ the yellow arrow of NaCl is shorter
4. As a result of 2(ii):
♦ the bottom pink arrow of CaCl2 is longer
♦ the bottom pink arrow of NaCl is shorter
7. How ΔH⊖sol of CaCl2 becomes negative:
• We see that (when compared to NaCl), both yellow and pink arrows of CaCl2 become longer
• We would expect the effects to cancel each other
♦ But the increase in the pink arrows
♦ is much larger than
♦ the increase in yellow arrow
• So the 'increase in pink arrow' outweighs the 'increase in yellow arrow'
♦ Thus we receive a net energy
♦ In other words, ΔH⊖sol of CaCl2 is negative
8. Thus we see that, the 'ionic charge' also plays an important role in determining the sign of ΔH⊖sol
• In the above discussions we compared compounds having the same negative ion but different positive ions:
♦ NaCl and LiCl
♦ NaCl and CaCl2
• Next we will compare two compounds having the same positive ion but different negative ions:
♦ NaCl and NaF
• First we will see a side by side comparison between NaCl and NaF
• The comparison can be written in 7 steps:
1. Drawing the cycles side by side
♦ Fig.6.19(a) below shows the Born-Haber cycle for NaCl
♦ Fig.6.19(b) below shows the Born-Haber cycle for NaF
 |
| Fig.6.19 |
2. Comparing lattice enthalpies
♦ In fig.a, the yellow arrow shows the ΔH⊖lattice of NaCl
♦ In fig.b, the yellow arrow shows the ΔH⊖lattice of NaF
• We see that, in fig.b, the yellow arrow is longer
♦ This is because
✰ ΔH⊖lattice of NaCl is 788
✰ ΔH⊖lattice of NaF is 904
3. Comparing hydration enthalpy of Na+
• Since we have to make a comparison, we will draw the positive ion (Na+) on top
♦ In fig.a, the top pink arrow shows the ΔH⊖hyd of Na+ ion in NaCl
♦ In fig.b, the top pink arrow shows the ΔH⊖hyd of Na+ ion in NaF
• The green dashed line indicates that, both the pink arrows are of the same size
4. Comparing hydration enthalpies of Cl- and F-:
♦ In fig.a, the bottom pink arrow shows the ΔH⊖hyd of Cl- ion in NaCl
♦ In fig.b, the bottom pink arrow shows the ΔH⊖hyd of F- ion in NaF
• We see that, in fig.b, the bottom pink arrow is longer
♦ This is because
✰ ΔH⊖hyd of Cl- is -359
✰ ΔH⊖hyd of F- is -504
5. The competing items:
There is always a competition between the following two items:
(i) The yellow arrow
(ii) Sum of the two pink arrows
• This can be explained in two steps:
(i)
The yellow arrow indicates the amount of energy that is to be supplied
to dissociate the salt into it's ions in gaseous state
• So we can write:
♦ Longer yellow indicates that, we need to supply greater energy
♦ Shorter yellow indicates that, we need to supply only lesser energy
(ii) The pink arrows indicate the amount of energy that will be released during hydration processes
• So we can write:
♦ Longer pink arrow indicates that, we will receive greater energy
♦ Shorter pink arrow indicates that, we will receive only a lesser energy
(iii) So the net energy supplied/received will depend on the relative sizes of arrows
6. In NaCl, we see that, the yellow arrow is longer than the sum of two pink arrows
♦ So we need to supply a net energy
♦ The ΔH⊖sol is thus positive
7. In NaF, we see that, the yellow arrow is shorter than the sum of two pink arrows
♦ So we will receive a net energy
♦ The ΔH⊖sol is thus negative
The reason for such a difference can be written in 8 steps:
1. In the periodic table, both Cl and F belong to group 17
2. But F is above Cl
• So F will have a lesser number of main shells
• Indeed we know that:
♦ F has two main shells: K and L
♦ Cl has three main shells: K, L and M
3. So the ionic radius of F- will be smaller than the ionic radius of Cl-
(We saw those details when we learnt about periodic trends. Details here)
• When the ionic radius decreases, the attractive force on the opposite ion increases
♦ Then it will be more difficult to separate the two ions
4. In our present case, this increased attraction has the following two effects:
(i) In the lattice structures:
♦ F- exerts a greater attraction on Na+
♦ Cl- exerts a lesser attraction on Na+
• So more energy is required for dissociation in the case of NaF
(ii) In the hydrated structures:
♦ F- exerts a greater attraction on H2O molecules
♦ Cl- exerts a lesser attraction on H2O molecules
• So more energy will be released during hydration of F-
5. As a result of 4(i):
♦ the yellow arrow of NaF is longer
♦ the yellow arrow of NaCl is shorter
6. As a result of 4(ii):
♦ the bottom pink arrow of NaF is longer
♦ the bottom pink arrow of NaCl is shorter
7. How ΔH⊖sol of NaF becomes negative:
• We see that (when compared to NaCl), both yellow and bottom pink arrow of NaF become longer
• We would expect the effects to cancel each other
♦ But the increase in the bottom pink arrow
♦ is much larger than
♦ the increase in yellow arrow
• So the 'increase in pink arrow' outweighs the 'increase in yellow arrow'
♦ Thus we receive a net energy
♦ In other words, ΔH⊖sol of NaF is negative
8. Thus we see that, the 'ionic radius' of 'negative ions' also play an important role in determining the sign of ΔH⊖sol
Solubility of salts
• We have seen that, the ionic radius plays an important role in determining the lattice enthalpy
♦ Smaller the ionic radius, greater will be the bond between the ions
• So in the lattice structure, bonds in the fluorides will be stronger than bonds in the chlorides
♦ That is why it is more difficult to dissolve fluorides when compared to chlorides
General formula for enthalpy of solution
A general formula can be derived in 5 steps:
1. We saw that, there is always a competition between the following two items:
(i) The yellow arrow
(ii) Sum of the two pink arrows
2. Mathematical form:
♦ The yellow arrow is ΔH⊖lattice
♦ Sum of two pink arrows is: ∑ ΔH⊖hyd
3. Result of the competition:
♦ The yellow arrow is in the upward direction
♦ Pink arrows are in the downward direction
• So result will be given by subtraction: ΔH⊖lattice - ∑ ΔH⊖hyd
4. This result is the enthalpy of solution. So we can write:
Eq.6.12: ΔH⊖sol = ΔH⊖lattice - ∑ ΔH⊖hyd
5. This Eq.6.12 can be used for both exothermic and endothermic processes
• Consider the two terms on the right side of the equation
♦ If the first term is larger, we will get a positive result
✰ This indicates an endothermic process
♦ If the first term is smaller, we will get a negative result
✰ This indicates an exothermic process
• In the next section, we will see spontaneity
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