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

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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

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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

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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^-

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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

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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

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• 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


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