Chapter 6.13 - Entropy

In the previous section, we saw some basics about spontaneity. In this section, we will see entropy and the second law of thermodynamics

• We saw that, being exothermic/endothermic is not the only criterion for spontaneity
• So to investigate further, we consider a system which cannot be exothermic or endothermic
    ♦ For that, we consider a system in which there is no heat transfer
    ♦ Such a system is called an isolated system
• It can be written in 18 steps

1. Consider the closed container in fig.6.21(a) below. It is an isolated container
• It is ‘isolated’ because, there is no flow of energy into or out of the container

Fig.6.21

2. On the left side of the container, there are some cold molecules of an ideal gas
    ♦ They are represented in blue color
• The fact that ‘they are cold’ is emphasized by the 'short arrows' indicating 'small velocities'
3. On the right side, there some hot molecules of the same ideal gas
    ♦ They are represented in red color
• The fact that ‘they are hot’ is emphasized by the 'long arrows' indicating 'large velocities'
4. The two sides are separated by a temporary partition shown in green color
• So the left side of the container will be cold and the right side will be hot
5. Let us remove the temporary partition at the middle
• The cold and hot molecules will begin to collide with each other
    ♦ The red molecules will lose some energy
    ♦ The blue molecules will gain some energy
• As a result, all the molecules will attain an intermediate velocity
• Thus, all the molecules will attain an intermediate temperature
• The fact that ‘they are at intermediate temperature’ is emphasized by the 'intermediate arrows' in fig.b, indicating 'intermediate velocities'
6. Since there is no net loss or gain, we can say: the energy is conserved
• But the energy is now distributed evenly
7. Now, two questions arise in our minds:
• After removal of the partition and intermixing,
    ♦ Why don't the blue molecules become cold and return to the left side?
    ♦ Why don't the red molecules become hot and return to the right side?
• If they return to their original sides, it would look like as in fig.c
◼ So the question is this:
Why don't we see the arrangement in fig.c?
• We can find the answer with the help of simple statistics
• The following steps from (8) to (11) will give the answer
8. Once the partition is removed, there are many possible arrangements
• The arrangement in fig.c is only one among the many possible arrangements
9. In fact, the intermixed arrangement in fig.b is also only one among the many possible arrangements
    ♦ A blue molecule at top left in fig.b, may move to bottom right
    ♦ A red molecule at bottom left in fig.b, may move to top right
    ♦ so on
• Such movements will give different intermixed arrangements
10. But the ordered arrangement in fig.c do not have many options
• This is because:
    ♦ The blue molecules have to stay on the left side
    ♦ The red molecules have to stay on the right side
11. So it is clear that:
    ♦ The number of possible intermixed arrangements as in fig.b
    ♦ is far greater than
    ♦ The number of ordered arrangements as in fig.c
12. We know that, even one mole of a gas will contain 6.023  × 10²³ molecules
• Remember that one million is only 1 × 10⁵
• So in a practical situation,
    ♦ The probability of finding intermixed arrangements as in fig.b is nearly 1 (100%)
        ✰ Like the decimal: 0.9999
    ♦ The probability of finding an ordered arrangement as in fig c is very close to zero
        ✰ Like a fraction with one in the numerator and 10²³ in the denominator
        ✰ An arrangement with such a low probability will not occur
• So we never see cold molecules on one side and hot molecules on the other side
13. If heat is to flow from cold to hot, it means that:
• The cold portions get colder and colder and settle down at a portion of the container
• The hot portion get warmer and warmer and occupy the remaining portion of the container
◼ Such an orderly arrangement have only a very low probability
• That is why we never see spontaneous flow of heat from a cold body to a hot body
14. Now consider a cold body and a hot body of same material and size
• The molecules in the cold body will be more or less stationary
    ♦ They do not move much
• The molecules in the hot body will be moving greater distances
15. We can compare the ‘number of possible arrangements’
• The ‘number of possible arrangements’ for the molecules in the hot body is larger
    ♦ Because, the hot molecules have greater mobility
• The ‘number of possible arrangements’ for the molecules in the cold body is smaller
    ♦ Because, the cold molecules have lesser mobility
16. ‘Number of possible arrangements’ is called entropy
• It’s symbol is ‘S’
◼ We can say:
    ♦ A hot body
    ♦ Has greater entropy
    ♦ Than a cold body
17. Besides temperature, the 'natural arrangement' of molecules also has a major role in deciding entropy of a body
• We know that:
    ♦ Molecules in a gas, naturally have very high mobility
    ♦ Molecules in a liquid, naturally have intermediate mobility
    ♦ Molecules in a solid, naturally have very low mobility
• So it is clear that
    ♦ Gases have the highest entropy
    ♦ Liquids have intermediate entropy
    ♦ Solids have the least entropy
◼ We can write: S_Gas > S_Liquid >S_Solid
18. Comparison between a neat room and a messy room is often used to explain entropy. It can be written in 6 steps:
(i) Consider a neat room
• We call it neat because, every object in that room is in it’s proper place
   ♦ Books are stacked neatly in the book shelf
   ♦ The table-lamp is at the left side of the table
   ♦ Monitor, keyboard and mouse are kept together as a system
   ♦ so on . . .
(ii) Consider the messy room
• We call it messy because, objects are not in their proper places
   ♦ Some books are on the shelf, some on the table, some on the floor etc.,
   ♦ The table-lamp is placed at a corner of the room
   ♦ Monitor is on the table
   ♦ But keyboard is on the floor
   ♦ Mouse is below the bed
    ♦ so on . . .
(iii) Now let us consider the possible arrangements in the neat room
• The neat room can have many different arrangements and still be neat
   ♦ In the book shelf
         ✰ The physics text book can be stacked below the chemistry text book
         ✰ Or the chemistry text book can be stacked below the physics text book
         ✰ In either arrangement, the room will be neat
   ♦ The table-lamp can be either on left or right side of the table according to convenience
         ✰ In either arrangement, the room will be neat
   ♦ The mouse can be either on the left or right side of the monitor according to convenience
         ✰ In either arrangement, the room will be neat
• Thus we see that, there are many possible arrangements in a neat room. The room will still be neat
(iv) Next let us consider the possible arrangements in the messy room
• The messy room can also have many different arrangements and still be messy
   ♦ The books can be any where in the room
         ✰ Above the shelf, below the shelf, north east corner, south west corner etc.,
         ✰ The room will still be messy
   ♦ Like wise, the table-lamp, monitor, keyboard, mouse etc., can also be any where
         ✰ The room will still be messy
(v) Now we can write about the number of possible arrangements
◼  The neat room has many number of possible arrangements by which it can be neat
◼  The messy room has infinite number of possible arrangements by which it can be messy
(vi) Applying entropy:
   ♦ The number of possible arrangements for the messy room
   ♦ is greater than
   ♦ The number of possible arrangements for the neat room
• So we say that:
   ♦ The messy room
   ♦ has greater entropy
   ♦ Than the neat room

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• Now we will see the role of entropy in a chemical/physical process
• During chemical reactions, there will be rearrangements of atoms and ions
   ♦ If the products have a more disordered state than the reactants, we can say:
         ✰ Entropy increases
   ♦ If the products have a more ordered state than the reactants , we can say:
         ✰ Entropy decreases
• The following solved example demonstrates this aspect

Solved example 6.22
Predict in which of the following, entropy increases/decreases:
(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from 0 K to 115 K
(iii) 2NaHCO₃(s)  → Na₂CO₃(s) + CO₂(g) + H₂O(g)
(iv) H₂(g) → 2H(g)
Solution:
Part (i)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in liquid state
   ♦ At state B, the system is the crystalline (solid) state
(iii) Comparing entropies:
   ♦ Liquid state
   ♦ has a higher entropy than
   ♦ Solid state
• So S_B will be smaller than S_A
• We can predict that, in this process, the entropy decreases
Part (ii)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in solid state but the temperature is 0 K
   ♦ At state B also, the system is the solid state, but the temperature is higher
(iii) Comparing entropies:
   ♦ At 0 K, all molecules become motion less
   ♦ So the solid at 0 K has lesser disorder
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases
Part (iii)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in solid state
   ♦ At state B, the system has one solid and two gases
(iii) Comparing entropies:
   ♦ At state B, there is greater entropy due to the presence of two gases
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases
Part (iv)
(i) Let A be the initial state and B the final state
   ♦ Let S_A be the entropy of  the system at state A
   ♦ Let S_B be the entropy of  the system at state B
(ii) Comparing the states:
   ♦ At state A, the system is in gaseous state
   ♦ At state B also, the system is in gaseous state
(iii) Comparing entropies:
   ♦ At state B, there is greater entropy due to the presence of more particles
         ✰ Each H2 molecule is dissociated into two H atoms
         ✰ Larger number of particles increases the 'possible number of arrangements'
• So S_B will be larger than S_A
• We can predict that, in this process, the entropy increases

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For chemical reactions, a more precise method can be used. It can be explained in 8 steps:
1. We have the general form of a balanced equation:
a₁R₁ + a₂R₂ + a₃R₃ + . . . →  b₁P₁ + b₂P₂ + b₃P₃ + . . .   
• R₁, R₂, R₃, . . . are the reactants
    ♦ a₁, a₂, a₃, . . . are the ‘number of mol’ of each of those reactants
• P₁, P₂, P₃, . . . are the products
    ♦ b₁, b₂, b₃, . . . are the ‘number of mol’ of each of those products
2. Let us write the S^⊖ values (standard entropy values). They can be looked up from the data book
• S^⊖ values of the reactants R₁, R₂, R₃, . . .
    ♦ They can be written as: S^⊖[R₁], S^⊖[R₂], S^⊖[R₃], . . .
• S^⊖ values of the products P₁, P₂, P₃, . . .
    ♦ They can be written as: S^⊖[P₁], S^⊖[P₂], S^⊖[P₃], . . .
3. The above S^⊖ values that we obtain from the data book, are related to one mole
• For example, the S^⊖ value of Al₂O₃(s) is 50.92 J K^-1 mol^-1
    ♦ The unit 'J K^-1 mol^-1' indicates that, the values are related to one mol
    ♦ (In the next section, we will see how this unit 'J K^-1 mol^-1' is obtained)
• But in our present case, we have:
    ♦ a₁ mol of reactant R₁
    ♦ a₂ mol of reactant R₂ . . . so on
• So we must multiply the values by the corresponding mol number
• Thus the step (2) can be modified as:
    ♦ S^⊖ values of the reactants: a₁S^⊖[R₁], a₂S^⊖[R₂], a₃S^⊖[R₃], . . .
    ♦ S^⊖ values of the products: b₁S^⊖[P₁], b₂S^⊖[P₂], b₃S^⊖[P₃], . . .
4. Now we find two sums:
(i) Sum for the reactants:
$\mathbf\small{\rm{\sum\limits_{i}{a_i {S^\circleddash}[R_i]}}}$ = a₁S^⊖[R₁] + a₂S^⊖[R₂] + a₃S^⊖[R₃] . . .
(ii) Sum for the products:
$\mathbf\small{\rm{\sum\limits_{i}{b_i {S^\circleddash}[P_i]}}}$ = b₁S^⊖[P₁] + b₂S^⊖[P₂] + b₃S^⊖[P₃] . . .
5. Next we subtract the first sum from the second sum
• The result is: standard entropy change of reaction
    ♦ It is denoted as: ΔS^⊖_r
6. Thus we get:
Eq.6.13: $\mathbf\small{\rm{{\Delta S^\circleddash }_r=\sum\limits_{i}{b_i {S^\circleddash}[P_i]}-\sum\limits_{i}{a_i {S^\circleddash}[R_i]}}}$
7. If 𝚺b_iS^⊖[P_i] is larger than 𝚺a_iS^⊖[R_i], we can say: Products have a greater entropy
• Also, we will get a positive value for ΔS^⊖_r
◼  So we can write:
A positive ΔS^⊖_r indicates an increase in entropy
8. If 𝚺b_iS^⊖[P_i] is smaller than 𝚺a_iS^⊖[R_i], we can say: Products have a lesser entropy
• Also, we will get a negative value for ΔS^⊖_r
◼  So we can write:
A negative ΔS^⊖_r indicates a decrease in entropy

Let us see two solved examples:
Solved example 6.23
Find the entropy change for the following reaction:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
Solution:
1. From the data book, we have:
   ♦ S^⊖ of Fe(s) = 27.28 J K^-1 mol^-1
   ♦ S^⊖ of O₂(g) = 205.14 J K^-1 mol^-1
   ♦ S^⊖ of Fe₂O₃(s) = 87.40 J K^-1 mol^-1
2. Applying the coefficients, we get:
   ♦ S^⊖ of the first reactant = 4 × 27.28
   ♦ S^⊖ of the second reactant = 3 × 205.14
   ♦ S^⊖ of the product = 2 × 87.40
3. Applying Eq.6.13, we get:
ΔS^⊖_r of the reaction = [(2 × 87.40) - (4 × 27.28) - (3 × 205.14)] = -549.74 J K^-1 mol^-1
• This is a negative value. So there is a decrease in entropy

Solved example 6.24
Find the entropy change for the following reaction:
N₂(g) + 3H₂(g) → 2NH₃(g)
Solution:
1. From the data book, we have:
   ♦ S^⊖ of N₂(g) = 191.61 J K^-1 mol^-1
   ♦ S^⊖ of H₂(g) = 130.68 J K^-1 mol^-1
   ♦ S^⊖ of NH₃(s) = 192.45 J K^-1 mol^-1
2. Applying the coefficients, we get:
   ♦ S^⊖ of the first reactant = 1 × 191.61
   ♦ S^⊖ of the second reactant = 3 × 130.68
   ♦ S^⊖ of the product = 2 × 192.45
3. Applying Eq.6.13, we get:
ΔS^⊖_r of the reaction = [(2 × 192.45) - (1 × 191.61) - (3 × 130.68)] = −198.75 J K^-1 mol^-1
• This is a negative value. So there is a decrease in entropy

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• In the next section, we will see the second law of thermodynamics


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