Chapter 6.11 - Solution Process - Exothermic or Endothermic

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

Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com

Chapter 6.10 - Enthalpy of Solution

In the previous section, we saw Lattice enthalpy. In this section, we will see enthalpy of solution

• A solution is formed when a substance (solute) dissolves in another substance (solvent)
• When the solute dissolves in a solvent at constant pressure, two things can happen:
(i) Energy may be absorbed
(ii) Energy may be released
• This absorbed/released energy is called enthalpy of solution
    ♦ It is denoted by the symbol: ΔH^⊖_sol
    ♦ If energy is absorbed, ΔH^⊖_sol will be positive
    ♦ If energy is released, ΔH^⊖_sol will be negative
[Note: For the absorbed/released energy to be designated as ΔH^⊖_sol, one condition must be satisfied:
◼ The solution must be an infinitely dilute solution
• In an infinitely dilute solution, the quantity of solvent will be so large that, addition of more solvent will not result in any further energy absorption /release. We will see more details about this condition, later in this section]
• Our aim is to find the ΔH^⊖_sol for various solutions like:
    ♦ Aqueous solution of NaCl
    ♦ Aqueous solution of MgCl₂ etc.,
• We can devise a general method which can be applied to most of the solutions. It can be written in 26 steps by using the aqueous solution of NaCl as an example

1. The cyan horizontal lines in fig.6.14 below indicates various states in an experiment
• The arrows indicate various processes which transform the system from one state to another

Fig.6.14

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 NaCl molecules
         ✰ These molecules are in the solid state
3. From the datum line, we begin our first process
• Our first process is to convert the NaCl molecules into Na⁺(g) ions and Cl^-(g) ions
The equation is: NaCl(s) → Na⁺(g) + Cl^-(g)
4. We have seen this process in the previous section
• The energy required for this process is the ΔH^⊖_lattice(d) of NaCl
   ♦ It's value is 788 kJ mol^-1
• So the thermochemical equation of this process will be:
NaCl(s) → Na⁺(g) + Cl^-(g); ΔH^⊖_lattice(d) = 788 kJ mol^-1
• This process is numbered as I in the fig.6.14 above
5. So when the process I is complete, we will have:
   ♦ One mol Na⁺ ions in the gaseous state
   ♦ One mol Cl^- ions in the gaseous state

---

6. Our second process is to convert each of the gaseous Na⁺ ions into Na⁺(aq) ions
• Let us see how this is achieved:
7. Consider the H₂O molecules in the liquid state
• We know that, H₂O molecules are polar molecules (Fig.4.225 of section 4.40)
• In each H₂O molecule:
    ♦ The H atoms have a partial positive charge $\mathbf\small{\rm{\delta^+}}$
    ♦ The O atom has a partial negative charge $\mathbf\small{\rm{\delta^-}}$
8. When the NaCl dissolves in water, each of the Na⁺ ions will get surrounded by H₂O molecules as shown in fig.6.15(a) below
    ♦ The red spheres are the O atoms
    ♦ The white spheres are the H atoms
• The partially negative O atoms are attracted towards the Na⁺ ion

Fig.6.15

9. The situation shown in fig.6.15(a) is stable
• Bonds are formed between the H₂O molecules and Na⁺ ions
• These bonds are due to the attraction between $\mathbf\small{\rm{O^{\delta-}}}$ and Na⁺
10. We know that, when bonds are formed, energy is released
◼ The energy released when bonds are formed between ions and H₂O molecules is known as hydration enthalpy
• It's symbol is: ΔH^⊖_hyd
11. The Na⁺ ions, when bonded with water molecules is represented as Na⁺(aq)
• We can represent the process of hydration as:
Na⁺(g) → Na⁺(aq)
12. From the data book, we have:
One mol Na⁺(g) ions release 424 kJ energy when all those ions are converted into Na⁺(aq)
• In other words, ΔH^⊖_hyd for Na⁺(g) = -424 kJ mol^-1
• So the thermochemical equation for this process will be:
Na⁺(g) → Na⁺(aq); ΔH^⊖_hyd = -424 kJ mol^-1
• This process is numbered as II in the fig.6.14 above
13. So when the process II is complete, we will have:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the gaseous state

---

14. Our third process is to convert each of the gaseous Cl^- ions into Cl^-(aq) ions
• Let us see how this is achieved:
15. As before, each of Cl^- will get surrounded by H₂O molecules
• This time, the H atoms with the partial positive charges, get attracted towards the Cl^- ions
• This is shown in fig.6.15(b) above
16. The situation shown in fig.6.15(b) is stable
• Bonds are formed between the H₂O molecules and Cl^- ions
• These bonds are due to the attraction between $\mathbf\small{\rm{H^{\delta+}}}$ and Cl^-
17. As before, hydration enthalpy is released in this case also
18. The Cl^- ions, when bonded with water molecules is represented as Cl^-(aq)
• We can represent this process of this hydration as:
Cl^-(g) → Cl^-(aq)
19. From the data book, we have:
One mol Cl^-(g) ions release 359 kJ energy when all those ions are converted into Cl^-(aq)
• In other words, ΔH^⊖_hyd for Cl^-(g) = -359 kJ mol^-1
• So the thermochemical equation for this process will be:
Cl^-(g) → Cl^-(aq); ΔH^⊖_hyd = -359 kJ mol^-1
• This process is numbered as III in the fig.6.14 above
20. So when the process III is complete, we will have:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the aqueous state

---

Now we can write why it is important to specify 'infinitely dilute solution'. It can be written in 2 steps:
(i) Imagine that, there is not enough water molecules
• Then:
   ♦ All the Na⁺ ions cannot be converted into Na⁺(aq)
         ✰ Some Na⁺ ions will remain as such
         ✰ So all the available ΔH^⊖_hyd for Na⁺ will not be released
   ♦ All the Cl^- ions cannot be converted into Cl^-(aq)
         ✰ Some Cl^- ions will remain as such
         ✰ So all the available ΔH^⊖_hyd for Cl^- will not be released
(ii) If we measure the enthalpies in such a situation:
   ♦ We will be recording a 'lower ΔH^⊖_hyd' than 'actual ΔH^⊖_hyd' for Na⁺
   ♦ We will be recording a 'lower ΔH^⊖_hyd' than 'actual ΔH^⊖_hyd' for Cl^-

---

21. The completion of process III was our goal
• Consider the products obtained at the end of this process:
   ♦ One mol Na⁺ ions in the aqueous state
   ♦ One mol Cl^- ions in the aqueous state
22. These products indicate that, one mol NaCl is completely dissolved in water
• We have achieved our goal
23. This goal is indicated by the third cyan line from top
• So from the datum line, we took the path: (I + II + III) to reach the third cyan line
24. From the datum line, we can take another path also
• It is along the red arrow
25. By Hess's law, we can write:
   ♦ Energy along (I + II + III)
   ♦ is equal to
   ♦ Energy along the red arrow
• Thus we get: (788 - 424 - 359) = X
⇒ X = 5
• This is a positive value
• That means, energy should be supplied
• It is an endothermic process
26. Note that, X is related to the process: Na⁺(g) + Cl^-(g) → Na⁺(aq) + Cl^-(aq)
• This indicates the solution of one mol NaCl in water
• So the red arrow represents the same process that we are investigating
• We can write:
To dissolve one mol NaCl in water, we need to supply 5 kJ energy
◼ In other words, ΔH^⊖_sol of NaCl is 5 kJ mol^-1

---

Next we will find the ΔH^⊖_sol of CaCl₂. The procedure is same as that for NaCl. So we will write only the minimum required 10 steps:

1. The thick cyan horizontal line in fig.6.16 below, is the datum line. It is the state from which we begin our calculations
• At this state, the system consists of:
   ♦ One mol CaCl₂ molecules
         ✰ These molecules are in the solid state

Fig.6.16

 

2. From the datum line, we begin our first process
• Our process I is to convert the CaCl₂ molecules into Ca²⁺(g) ions and Cl^-(g) ions
The equation is: CaCl₂(s) → Ca²⁺(g) + 2Cl^-(g)
3. From the data book, we have: ΔH^⊖_lattice(d) of CaCl₂ = 2258 kJ mol^-1
• So the thermochemical equation of this process will be:
CaCl₂(s) → Ca²⁺(g) + 2Cl^-(g); ΔH^⊖_lattice(d) = 1158 kJ mol^-1
4. So when the process I is complete, we will have:
   ♦ One mol Ca²⁺ ions in the gaseous state
   ♦ Two mol Cl^- ions in the gaseous state
5. Process II is the conversion of Ca²⁺(g) ions into Ca²⁺(aq)
• Using the data book, we write:
Ca⁺(g) → Ca²⁺(aq); ΔH^⊖_hyd = -1650 kJ mol^-1
• This energy is released when bonds are formed between the water molecules and the Ca²⁺ ions
   ♦ This is similar to the case shown in fig.6.15(a) above
6. Process III is the conversion of Cl^-(g) ions into Cl^-(aq)
• Using the data book, we write:
2Cl^-(g) → 2Cl^-(aq); ΔH^⊖_hyd = 2(-359) kJ mol^-1
• This energy is released when bonds are formed between the water molecules and the Cl^- ions
   ♦ This is the same case shown in fig.6.15(b) above
• Note that in the cas of NaCl, there is only one mol of Cl^- ions
• But in the case of CaCl₂, there are two mol Cl^- ions
   ♦ So we multiply -359 by 2
7. The completion of process III was our goal
• This goal is indicated by the cyan line below the datum line
• So from the datum line, we took the path: (I + II + III) to reach the goal
8. From the datum line, we can take another path also to reach the goal
• It is along the red arrow
9. By Hess's law, we can write:
   ♦ Energy along (I + II + III)
   ♦ is equal to
   ♦ Energy along the red arrow
• Thus we get: (1158 - 1650 - 2(359)) = X
⇒ X = -110
• This is a negative value
• That means, energy will be released
• It is an exothermic process
10. Note that, X is related to the process: Ca²⁺(g) + 2Cl^-(g) → Ca²⁺(aq) + 2Cl^-(aq)
• This indicates the solution of one mol CaCl₂ in water
• So the red arrow represents the same process that we are investigating
• We can write:
When one mol CaCl₂ dissolve in water, 110 kJ energy will be released
◼ In other words, ΔH^⊖_sol of CaCl₂ is -110 kJ mol^-1

---

• We have seen the ΔH^⊖_sol of two salts: NaCl and CaCl₂
• We see that:
    ♦ the first is an endothermic process
    ♦ the second is an exothermic process
• We also see that:
    ♦ in the first case, the goal is above the datum line
    ♦ in the second case, the goal is below the datum line
• In the next section, we will see the reason for such differences


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com