Chapter 6.14 - Second Law of Thermodynamics

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

◼ The second law of thermodynamics states that:
When a spontaneous process takes place, the entropy of the universe always increases
• This can be explained in 6 steps:
1. Universe is equivalent to (system + surroundings)
• So entropy of universe, S_universe = S_sys + S_surr
2. Initial and final states:
    ♦ At the initial state A, we have: S_universe(A) = S_sys(A) + S_surr(A)
    ♦ At the final state B, we have: S_universe(B) = S_sys(B) + S_surr(B)
3. The second law states that, the entropy of the universe always increases
    ♦ That means: S_universe(B) must be larger than S_universe(A)
    ♦ That means: [S_universe(B) - S_universe(A)] must be greater than zero
    ♦ That means: ΔS_universe must be greater than zero
4. We have:
ΔS_universe = [S_universe(B) - S_universe(A)]
⇒ ΔS_universe = [(S_sys(B) + S_surr(B)) - (S_sys(A) + S_surr(A))]
⇒ ΔS_universe = [(S_sys(B) - S_sys(A)) + (S_surr(B) - S_surr(A))]
• Thus we get Eq.6.14: ΔS_universe = [ΔS_sys + ΔS_surr]
5. So based on the second law, we can write:
For a process to be spontaneous, ΔS_universe > 0
6. From (4), it is clear that:
To calculate ΔS_universe, we must know ΔS_sys and ΔS_sys
    ♦ We know how to find ΔS_sys
        ✰ see solved examples 6.23 and 6.24 of the previous section
    ♦ Later in this section, we will see the method to find ΔS_surr

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• Using the second law, we can predict the direction of a process
• For that, we look at the entropy in two directions
1. First we consider the direction from left to right
• When the process proceeds in this direction, if we get ΔS_universe > 0, then the  process will occur spontaneously in this direction
• That means, no external energy is required for the reaction to proceed from left to right
• The opposite is also true:
When the process proceeds in this direction, if we get ΔS_universe < 0, then the  process will not occur spontaneously in this direction
• That means, external energy is required for the reaction to proceed from left to right
2. Next we consider the direction from right to left
• When the process proceeds in this direction, if we get ΔS_universe > 0, then the  process will occur spontaneously in this direction
• That means, no external energy is required for the reaction to proceed from right to left 
• The opposite is also true:
When the process proceeds in this direction, if we get ΔS_universe < 0, then the process will not occur spontaneously in this direction
• That means, external energy is required for the reaction to proceed from right to left

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• As mentioned in (6) above, our next aim is to find ΔS_surr. It can be written in 5 steps:
1. Initial and final entropies:
• We have:
    ♦ Entropy of the surroundings in the initial state = S_surr(A)
    ♦ Entropy of the surroundings in the final state = S_surr(B)
2. Then change in entropy ΔS_surr = S_surr(B) – S_surr(A)
3. This change in entropy is due to the absorption of a heat Q
• This Q can be large or small:
• If Q is large, S_surr(B) will be large
    ♦ (Because, large Q causes large disorder)
    ♦ Then ΔS_surr will be large
• If Q is small, S_surr(B) will be small
    ♦ Then ΔS_surr will be small
◼ We can write:
ΔS_surr is directly proportional to Q
4. Based on experiments, scientists obtained another information. It can be written in 6 steps:
(i) Let the surroundings be at a lower temperature T₁
    ♦ Let a heat Q be added to the surroundings
    ♦ Let the resulting change in entropy be ΔS_surr(1)
(ii) Let the surroundings be at a higher temperature T₂
    ♦ Let the same heat Q be added to the surroundings
    ♦ Let the resulting change in entropy be ΔS_surr(2)
(iii) Here we see an interesting fact:
    ♦ ΔS_surr(2) will be smaller than ΔS_surr(1)
• Let us analyze the reason:
(iv) At the lower temperature T₁, the surroundings is somewhat calm
    ♦ To the calm surroundings, we are adding Q
• At the higher temperature T₂, the surroundings will already have some disorder
    ♦ To that disordered surroundings, we are adding the same Q
(v) Observing the 'change in disorder' at two temperatures:
    ♦ The 'change in disorder' created by Q at the lower temperature T₁
    ♦ Will be more observable than
    ♦ The 'change in disorder' created by the same Q at the higher temperature T₂
• This is because, at T₂, the surroundings already have some disorder
(vi) So we can write:
    ♦ ΔS_surr(1) at lower temperature is high
    ♦ ΔS_surr(2) at higher temperature is low
• That means:
ΔS_surr is inversely proportional to the temperature
5. So we have two information:
    ♦ From (3), we have: ΔS_surr is directly proportional to Q
    ♦ From (4), we have: ΔS_surr is inversely proportional to T
◼ Combining the two, we get:
Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
• From this equation, it is clear that, units of entropy is J K^-1

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Now that we know the two terms of Eq.6.14, we will see a practical application of the second law. The following five solved example demonstrates the application

Solved example 6.25
Prove that:
   ♦ Ice will not melt spontaneously at -5 ^०C
   ♦ Ice will begin to melt spontaneously at 0 ^०C
   ♦ Ice will melt spontaneously at 5 ^०C
Given that: For the process H₂O(s)  → H₂O(l), ΔS^⊖ = 22.0 J K^-1 mol^-1
Solution:
1. We have to analyze the process: H₂O(s)  → H₂O(l)
• We have seen this process in an earlier section. (see 'Enthalpy changes during phase transformations' in section 6.6)
• The thermochemical equation is: H₂O(s)  → H₂O(l) ΔH^⊖_fusion = 6.00 kJ mol^-1
2. To determine whether a process is spontaneous, we have to apply Eq.6.14:
ΔS_universe = [ΔS_sys + ΔS_surr]
3. First we will find ΔS_sys
• But it is already given in the question. ΔS_sys = ΔS^⊖ = 22.0 J K^-1 mol^-1
• Let us consider one mol ice. Then we can ignore the 'mol^-1'
• We can write:
For our present system, ΔS_sys = 22.0 J K^-1
4. Next we will find ΔS_surr
• We have Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
5. From the thermochemical equation, we see that, when 1 mol ice melts, the surroundings will lose 6.00 kJ energy
   ♦ Let Q_A be the initial heat of the surroundings
   ♦ Let Q_B be the final heat of the surrounding
• Q_B will be less than Q_A because, heat is lost by the surroundings
   ♦ This lost heat is used up for melting the ice
• The change of heat = (Q_B -Q_A) = -6 kJ
   ♦ The -ve sign is to be given for 6 because, Q_B will be less than Q_A
6. Substituting the known values in (4), we get:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{-6000(J)}{T(K)}}}$
7. Now the result in (2) becomes:
ΔS_universe = [22.0 J K^-1 + $\mathbf\small{\rm{\frac{-6000(J)}{T(K)}}}$]
• Now we can take up each case
8. Case 1: When temperature is -5 ^०C
• -5 ^०C is 268 K
• So the result in (7) becomes:
ΔS_universe = [22.0 J K^-1 + $\mathbf\small{\rm{\frac{-6000(J)}{268(K)}}}$] = -0.3881
◼  This is a negative value. So we can write:
If this process takes place at -5 ^०C, entropy of the universe decreases
• Any process which causes a decrease in entropy of the universe will not be spontaneous
   ♦ So the process H₂O(s)  → H₂O(l) is not spontaneous at -5 ^०C
   ♦ The reverse process H₂O(l)  → H₂O(s) will be spontaneous at -5 ^०C
9. Case 2: When temperature is 0 ^०C
• 0 ^०C is 273 K
• So the result in (7) becomes:
ΔS_universe = [22.0 J K^-1 + $\mathbf\small{\rm{\frac{-6000(J)}{273(K)}}}$] = 0.022
◼  This value is close to zero. So we can write:
If this process takes place at 0 ^०C, entropy of the universe neither increases nor decreases
• Any process which neither increases nor decreases the entropy of the universe will be at equilibrium
   ♦ So the process H₂O(s)  → H₂O(l) is at equilibrium at 0 ^०C
   ♦ If the temperature rises just above 0 ^०C, melting will begin
10. Case 3: When temperature is +5 ^०C
• +5 ^०C is 278 K
• So the result in (7) becomes:
ΔS_universe = [22.0 J K^-1 + $\mathbf\small{\rm{\frac{-6000(J)}{278(K)}}}$] = 0.4172 J K^-1
◼  This is a positive value. So we can write:
If this process takes place at +5 ^०C, entropy of the universe increases
• Any process which causes an increase in entropy of the universe will be spontaneous
   ♦ So the process H₂O(s)  → H₂O(l) is spontaneous at +5 ^०C
   ♦ The reverse process H₂O(l)  → H₂O(s) will not be spontaneous at +5 ^०C

Solved example 6.26
The process White Tin(s) → Gray Tin(s) occurs when the surrounding temperature falls just below 13.2 ^०C. The ΔH for the process is -2.1 kJ mol^-1. What is the ΔS for the process? Which one has greater order? White tin or Gray tin?
Solution:
1. We have to analyze the process: White Tin(s)  → Gray Tin(s); ΔH = -2.1 kJ mol^-1
2. To determine whether a process is spontaneous, we have to apply Eq.6.14:
ΔS_universe = [ΔS_sys + ΔS_surr]
3. First we will write ΔS_sys
• We have: ΔS_sys = S[Product] - S[Reactants]
⇒ ΔS_sys = S[Gray] - S[White]
4. Next we will find ΔS_surr
We have Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
5. From the thermochemical equation, we see that, when one mol white Tin gets converted into one mol gray Tin, 2.1 kJ is released
• So the surroundings will gain 2.1 kJ energy
   ♦ Let Q_A be the initial heat of the surroundings
   ♦ Let Q_B be the final heat of the surrounding
• Q_B will be greater than Q_A because, heat is gained by the surroundings
• The change of heat = (Q_B - Q_A) = +2.1 kJ
   ♦ The +ve sign is to be given for 2.1 because, Q_B will be greater than Q_A
6. Substituting the known values in (4), we get:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{+2100(J)}{T(K)}}}$
7. Now the result in (2) becomes:
ΔS_universe = [S[Gray] - S[White] + $\mathbf\small{\rm{\frac{+2100(J)}{T(K)}}}$]
8. Given that, the conversion just begins at 13.2 ^०C
    ♦ That means, at 13.2 ^०C, the system is in equilibrium
    ♦ That means, at 13.2 ^०C, there is no change in entropy of the universe
    ♦ That means, at 13.2 ^०C, ΔS_universe = 0
• 13.2 ^०C is equal to (273.15 + 13.2) = 286.4 K
9. So the result in (7) becomes:
0 = [S[Gray] - S[White] + $\mathbf\small{\rm{\frac{+2100(J)}{286.4(K)}}}$]
⇒ S[Gray] - S[White] = $\mathbf\small{\rm{\frac{-2100(J)}{286.4(K)}}}$ = -7.33 J K-1
10. So we can write:
ΔS for the process White Tin(s) → Gray Tin(s) is -7.33 J K^-1
11. The -ve value of the ΔS indicates that, the entropy decreases
    ♦ That means, the reactant has greater entropy
    ♦ That means, the reactant has greater disorder
    ♦ That means the product has greater order
• So we can write:
Gray Tin has greater order

Solved example 6.27
The process Orthorhombic Sulfur(s) → Monoclinic Sulfur(s) occurs when the surrounding temperature rises just above 95.3 ^०C. The ΔH for the process is +0.401 kJ mol^-1. What is the ΔS for the process? Which one has greater order? Orthorhombic Sulfur or Monoclinic Sulfur?
Solution:
1. We have to analyze the process:
Orthorhombic Sulfur(s) → Monoclinic Sulfur(s); ΔH = +0.401 kJ mol-1
2. To determine whether a process is spontaneous, we have to apply Eq.6.14:
ΔS_universe = [ΔS_sys + ΔS_surr]
3. First we will write ΔS_sys
• We have: ΔS_sys = S[Product] - S[Reactants]
⇒ ΔS_sys = S[Mono] - S[Ortho]
4. Next we will find ΔS_surr
• We have Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
5. From the thermochemical equation, we see that, when one mol ortho gets converted into one mol mono, 0.401 kJ is absorbed
• So the surroundings will lose 0.401 kJ energy
   ♦ Let Q_A be the initial heat of the surroundings
   ♦ Let Q_B be the final heat of the surrounding
• Q_B will be lesser than Q_A because, heat is lost by the surroundings
• The change of heat = (Q_B - Q_A) = -0.401 kJ
   ♦ The -ve sign is to be given for 0.401 because, Q_B will be lesser than Q_A
6. Substituting the known values in (4), we get:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{-401(J)}{T(K)}}}$
7. Now the result in (2) becomes:
ΔS_universe = [S[Mono] - S[Ortho] + $\mathbf\small{\rm{\frac{-401(J)}{T(K)}}}$]
8. Given that, the conversion just begins at 95.3 ^०C
    ♦ That means, at 95.3 ^०C, the system is in equilibrium
    ♦ That means, at 95.3 ^०C, there is no change in entropy of the universe
    ♦ That means, at 95.3 ^०C, ΔS_universe = 0
• 95.3 ^०C is equal to (273.15 + 95.3) = 368.45 K
9. So the result in (7) becomes:
0 = [S[Mono] - S[Ortho] + $\mathbf\small{\rm{\frac{-401(J)}{368.45(K)}}}$]
⇒ S[Mono] - S[Ortho] = $\mathbf\small{\rm{\frac{401(J)}{368.45(K)}}}$ = 1.09 J K^-1
10. So we can write:
ΔS for the process Orthorhombic Sulfur(s) → Monoclinic Sulfur(s) is 1.09 J K-1
11. The +ve value of the ΔS indicates that, the entropy increases
    ♦ That means, the product has greater entropy
    ♦ That means, the product has greater disorder
    ♦ That means the reactant has greater order
• So we can write:
Orthorhombic sulfur has greater order

Solved example 6.28
Calculate the entropy change for oxidation of iron, 4Fe(s) + 3O₂ (g) → 2Fe₂O₃(s). Inspite of negative entropy change of this reaction, why is the reaction spontaneous at 298 K?
(ΔH^⊖_r of the reaction is –1648 × 10³ J mol^-1)
Solution:
1. We have already calculated the entropy change for this reaction
(see solved example 6.23 of the previous section)
• We got: ΔS^⊖_r = -549.74 J K^-1
2. To determine whether a process is spontaneous, we have to apply Eq.6.14:
ΔS_universe = [ΔS_sys + ΔS_surr]
3. The result in (1) is ΔS_sys
So we have: ΔS_sys = -549.74 J K^-1
4. Next we will find ΔS_surr
• We have Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
5. From the data given, we see that, during the reaction, 1648 × 10³ J is released
• So the surroundings will gain 1648 × 10³ J energy
   ♦ Let Q_A be the initial heat of the surroundings
   ♦ Let Q_B be the final heat of the surrounding
• Q_B will be greater than Q_A because, heat is gained by the surroundings
• The change of heat = (Q_B - Q_A) = +1648 × 10³ J
   ♦ The +ve sign is to be given for 1648 × 10³ because, Q_B will be greater than Q_A
6. Substituting the known values in (4), we get:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{+1648 × 10^3 (J)}{T(K)}}}$
7. Now the result in (2) becomes:
ΔS_universe = [-549.74 J K^-1 + $\mathbf\small{\rm{\frac{+1648 × 10^3(J)}{T(K)}}}$]
8. We have to examine the reaction at 298 K. So T = 298 K
9. So the result in (7) becomes:
ΔS_universe = [-549.74 J K^-1 + $\mathbf\small{\rm{\frac{+1648 × 10^3(J)}{298(K)}}}$]
⇒ ΔS_universe = 4980.5
10. This is a positive value
• That means, the entropy of the universe increases
• So the process will be spontaneous

Solved example 6.29
Calculate the temperature at which the following reaction becomes spontaneous:
CaCO₃(s) → CaO(s) + CO₂(g); ΔH^⊖_r = +179 kJ mol^-1
Solution:
1. We have to analyze the process:
CaCO₃(s) → CaO(s) + CO₂(g); ΔH^⊖_r = +179 kJ mol^-1
2. To determine whether a process is spontaneous, we have to apply Eq.6.14:
ΔS_universe = [ΔS_sys + ΔS_surr]
3. We have: ΔS_sys = S[Product] - S[Reactants]
⇒ ΔS^⊖_r = S^⊖[CaO(s)] + S^⊖[CO₂(g)] - S^⊖[CaCO₃(s)]
⇒ ΔS^⊖_r = 39.75 + 213.74 - 92.9 = 160.59
4. Next we will find ΔS_surr
• We have Eq.6.15: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Q}{T}}}$
5. From the data given, we see that, during the reaction, 179 × 10³ J is absorbed
• So the surroundings will lose 179 × 10³ J energy
   ♦ Let Q_A be the initial heat of the surroundings
   ♦ Let Q_B be the final heat of the surrounding
• Q_B will be lesser than Q_A because, heat is lost by the surroundings
• The change of heat = (Q_B - Q_A) = -179 × 10³ J
   ♦ The -ve sign is to be given for 179 × 10³ because, Q_B will be lesser than Q_A
6. Substituting the known values in (4), we get:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{-179 × 10^3 (J)}{T(K)}}}$
7. Now the result in (2) becomes:
ΔS_universe = [160.59 J K^-1 + $\mathbf\small{\rm{\frac{-179 × 10^3(J)}{T(K)}}}$]
8. We want ΔS_universe to be positive. Only then will the reaction become spontaneous
• The sign of ΔS_universe is decided by two terms:
   ♦ 160.59 and $\mathbf\small{\rm{\frac{-179 × 10^3(J)}{T(K)}}}$
• 179 × 10³ is very large when compared to 160.59
   ♦ So we will want a large 'T' to make the second term small
9. We will first find the equilibrium temperature by equating ΔS_universe to zero. We get:
0 = [160.59 J K^-1 + $\mathbf\small{\rm{\frac{-179 × 10^3(J)}{T(K)}}}$]
⇒ T = 1114.64 K
10. If we raise T a little above 1114.64 K, the reaction will become spontaneous

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• In the next section, we will see Gibbs energy


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