Chapter 7.1 - Equilibrium in Solutions

In the previous section, we saw solid-liquid and liquid-vapour equilibrium. In this section, we will see solid-vapour equilibrium and equilibrium in solutions

Solid-Vapour Equilibrium

Solid-vapour equilibrium can be studied by taking solid iodine as an example. It can be written in 6 steps
1. Take solid iodine in a closed vessel. Heat it gently
• Since the vessel is closed, it gets filled with violet vapor
2. Also, the intensity of the violet color goes on increasing
• Video of a similar experiment can be seen here:

 

• This ‘increase in intensity’ indicates that, more and more solid iodine is being converted into gaseous iodine
• After some time, the ‘increase in intensity’ stops. At that point, equilibrium is reached
3. Let us analyze the processes taking place inside the vessel from the moment when the iodine is heated. It can be written in 5 steps:
(i) When the iodine is heated, some solid molecules from the surface, escape from the mass of iodine
• Once those molecules escape, they are in the gaseous state
• This process is called sublimation. It can be represented using symbols as:
I₂(s)  → I₂(vap)
(ii) This process continues and so, the quantity of gas inside the vessel increases
• At the same time, the violet color also increases
(iii) All the while when this process is taking place, another process is also taking place simultaneously
• It is the deposition of some of the gaseous iodine molecules back into the solid state
• This process is called deposition. It can be represented using symbols as:
I₂(vap) → I₂(s)
(iv) Since the two processes are taking place simultaneously, we can write:
I₂(s) ⇌ I₂(vap)
• Initially, rate of the forward reaction is greater than rate of backward reaction
• That is.,
   ♦ Number of molecules entering the gaseous phase
   ♦ is greater than
   ♦ Number of molecules leaving the gaseous phase
• So quantity of gas increases, causing the increase in violet color
(v) But after some time, an equilibrium will be reached
• That is.,
   ♦ Number of molecules entering the gaseous phase
   ♦ become equal to
   ♦ Number of molecules leaving the gaseous phase
• So quantity of gas becomes steady, causing no further increase in color
4. So now we know the reason for the following observation:
    ♦ Intensity of violet color becomes steady after some time
• Next, slightly increase the temperature of the water
• Now more iodine molecules will have the required energy to break away from the solid mass
• So the number of gaseous iodine molecules inside the vessel will increase
• This increases the color intensity
• If the new temperature is kept constant, a new equilibrium will be reached
5. At equilibrium,
   ♦ rate of forward reaction
   ♦ is equal to
   ♦ rate of backward reaction
6. The above five steps help us to get a basic understanding about the solid-vapour equilibrium

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Equilibrium Involving Dissolution of Solid or Gases in Liquids

• Next we have to learn about equilibrium in solutions. We consider two types of solutions:
    ♦ Solids in liquids
    ♦ Gases in liquids

Solids in liquids

First we will consider the dissolution of solids in liquids. It can be explained by taking sugar solution as an example. It can be written in 10 steps:
1. Take some water in a vessel
• Gradually add solid sugar crystals and stir
• The sugar crystals will dissolve in water
2. If we continue adding sugar, the sweetness of the solution will go on increasing
• As we continue adding sugar, a stage will be reached when no more sugar dissolves
    ♦ The sweetness will not increase further
• The extra sugar will be seen as solid crystals at the bottom of the vessel
3. A definite quantity of solvent can dissolve only a definite quantity of a solute
• Any extra solute added, will not dissolve. It will form into a precipitate at the bottom of the vessel
◼ When the precipitation just begins, we say that, the solution has become saturated
4. If we heat the saturated sugar solution, we will be able to add more sugar with out causing precipitation
• So at a higher temperature, the solution will not contain any solid particles
    ♦ The solution will have a greater sweetness than before
5. But if we cool that solution back to the original temperature, the extra sugar will precipitate and form solid sugar at the bottom of the vessel
• While preparing some desserts, we need to dissolve sugar in hot water
• At room temperature, only a lesser quantity of sugar will dissolve. This will not give the required sweetness
• During the cooking process, the sugar in that concentrated solution will be absorbed into the flour. Since it is absorbed, the problem of precipitation on cooling does not arise
6. In our present case, we do not have flour. So the extra sugar will precipitate when it is brought down to room temperature
• This cooled solution will contain two items:
    ♦ Sugar in solution form
    ♦ Sugar in solid form
• We need to study this cooled solution
7. An analysis of the cooled solution can be written in 4 steps:
(i) In the cooled solution, we observe no activity
• But in reality, activities are taking place at the interface between solid sugar and sugar solution
   ♦ Some sugar molecules in the solution form, crystallizes into solid sugar
        ✰ This can be represented as: Sugar(sol) → Sugar(s)
   ♦ Some solid sugar molecules dissolve into the solution
         ✰ This can be represented as: Sugar(s) → Sugar(sol)
   ♦ Both the processes are taking place simultaneously
         ✰ So we can combine them as: Sugar(s) ⇌ Sugar(sol)
(ii) If the temperature remains constant,
   ♦ Number of molecules entering into the solution
   ♦ will be equal to
   ♦ Number of molecules crystallizing into solid form
(iii) That means, if the temperature of the cooled solution remains constant,
   ♦ rate of forward process
   ♦ will be equal to
   ♦ rate of backward process
(iv) That means, if the temperature of the cooled solution remains constant, it will be in a state of equilibrium
8. This equilibrium is not a static equilibrium
   ♦ It is a dynamic equilibrium
   ♦ Because, two processes are still going on
9. It is hard to believe that two processes are still going on at equilibrium
• It is hard to believe because, we see no activity
• But we can prove the two processes using precision instruments
• We can prove it in 8 steps:
(i) To the cooled solution, add some solid radioactive sugar
• We would expect the radioactive sugar to settle to the bottom of the vessel
   ♦ Because, there is no place left for any solid sugar
   ♦ It is a saturated solution
(ii) But if we measure radioactivity using precision instruments, we will see that:
   ♦ (a) The solid precipitate shows radioactivity
   ♦ (b) The solution also shows radioactivity
• Observation (a) is expected
• Observation (b) is not expected
(iii) The only explanation for (b) is that, even at equilibrium, exchange of molecules is taking place:
   ♦ (a) The solid molecules are going into solution form
   ♦ (b) The molecules in solution are going into solid form
(iv) Consider iii(a) above
• During the process mentioned in iii(a), some solid radioactive molecules also go into solution form
• That is how, the solution also attain radioactivity
◼  So it is clear that the forward process in 'Sugar(s) ⇌ Sugar(sol)' takes place even at equilibrium
(v) Next we have to prove that, the reverse process also takes place at equilibrium. The following steps from (vi) to (viii) will help us to prove it
(vi) We know that, at equilibrium, the total number of 'molecules in solution form' will be a constant
• Since we added radioactive solid molecules also, this total number will be the sum of two items:
   ♦ (a) The number of radioactive molecules in solution form
   ♦ (b) The number of non-radioactive molecules in solution form
(vii) If we continuously check the radioactivity of the solution, we will see that:
   ♦ Item vi(a) increases gradually
   ♦ Item vi(b) decreases gradually
• After some time, vi(a) and (vi(b) becomes equal
(viii) Decrease in vi(b) is clear evidence that, the non-radioactive molecules are crystallizing back into solid form
◼  That means the backward process in 'Sugar(s) ⇌ Sugar(sol)' is also taking place at equilibrium
10. The above nine steps help us to get a basic understanding about the equilibrium in dissolution of solids in liquids

Gases in liquids

Now we will consider the dissolution of gases in liquids. It can be explained by taking soda water as an example. It can be written in 8 steps:
1. Consider the making of soda water
• We first take some water in a bottle
• Then we pump CO₂ gas into the bottle
   ♦ The CO₂ begins to dissolve in water
• This process can be represented using symbols as:
CO₂(g) → CO₂(sol)
2. But while this process is going on, some dissolved CO₂, manages to escape back into gaseous form
• This process can be represented using symbols as:
CO₂(sol) → CO₂(g)
3. Since the two processes are taking place simultaneously, we can write:
CO₂(g) ⇌ CO₂(sol)
4. If we increase the pressure, the rate of the forward reaction in 'CO₂(g) ⇌ CO₂(sol)' increases
• That means, at high pressure, more and more CO₂ gets dissolved in water
• This phenomenon is explained by Henry's law

Henry's Law states that: Mass of a dissolved gas in a solvent is proportional to the pressure of the gas above the solvent. This mass decreases with increase in temperature

• So the manufacturer increases the pressure (and reduces the temperature) so that, the required quantity of CO₂ is dissolved in water
• Then the bottle is sealed. So the pressure remains the same
5. Inside the sealed bottle, an equilibrium will be established
• It will be a dynamic equilibrium
   ♦ The quantity of CO₂ escaping into gaseous form
   ♦ will be equal to
   ♦ The quantity of CO₂ dissolving back into solution
• But since the pressure is maintained at a high value, only a small amount of CO₂ will leave the solution and the same quantity dissolves back
6. If we open the lid slightly, let some pressure off and seal the lid again, we will have a lower pressure inside
• This lower pressure favors the reverse process in ‘CO₂(g) ⇌ CO₂(sol)’
• So more CO₂ escapes into the gaseous form
• The taste of soda is lost
• But since the lid is sealed back, a new equilibrium is reached
7. Finally, if we remove the lid and place the bottle open to atmosphere, all the CO₂ will escape. The soda will become flat
8. The above seven steps help us to get a basic understanding about the equilibrium in dissolution of gases in liquids

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• In the next section, we will see the general features of various equilibriums that we have seen so far. We will also see a simple activity to demonstrate the dynamic nature of equilibrium

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