In the previous section, we saw the second law of thermodynamics. In this section, we will see Gibbs energy
◼ Gibbs energy G is defined as: Eq.6.16: G = H – TS
♦ H is the enthalpy of the system
♦ T is the temperature of the system
♦ S is the entropy of the system
We know that, H, T and S are state functions. So G is also a state function
Now we will see some interesting calculations based on the above equation. It can be written in 11 steps:
1. Initial and final states of the system:
♦ At the initial state A, we have: Gsys(A) = Hsys(A) - TSsys(A)
♦ At the final state B, we have: Gsys(B) = Hsys(B) - TSsys(B)
2. Now we can find ΔGsys:
ΔGsys = (Gsys(B) - Gsys(A)) = (Hsys(B) - TSsys(B)) - (Hsys(A) - TSsys(A))
⇒ ΔGsys = (Hsys(B) - Hsys(A)) -T(Ssys(B) - Ssys(A))
Thus we get: Eq.6.17: ΔGsys = ΔHsys - T ΔSsys
3. Note that:
♦ Eq.6.16 is related to Gibbs energy
♦ Eq.6.17 is related to Gibbs energy change
4. Let us do a dimensional analysis of Eq.6.16
• On the right side, we have:
$\mathbf\small{\rm{[Energy]-[Temperature]\times \frac{[Energy]}{[Temperature]}}}$
= [Energy] - [Energy]
= [Energy]
• So we can write: G has the dimensions of energy. In other words, G is a quantity of energy
• ΔG in Eq.6.17 is the difference between two G values. So ΔG is also a quantity of energy
◼ Eq.6.17 is known as the Gibbs equation. It is one of the most important equations in chemistry
5. In earlier sections, we saw that:
♦ A decrease in enthalpy (negative ΔH) could be an indication for spontaneity
♦ An increase in entropy of the universe is essential for spontaneity
• Now, Eq.6.17 combines both 'change in enthalpy' and 'change in entropy'
6. Let us see how Gibbs energy is related to spontaneity:
• We have the basic equation: Eq.6.14: ΔSuniverse = [ΔSsys + ΔSsurr]
• We calculated the last term Ssurr as follows:
$\mathbf\small{\rm{\Delta S_{surr}=\frac{Heat\,absorbed/released\,by\,surroundings}{T}}}$
7. We saw that:
$\mathbf\small{\rm{Heat\,absorbed/released\,by\,surroundings}}$ is related to the ΔHsys
• The relation can be written in two ways:
(i) If heat ΔHsys is released by the system, that same heat will be absorbed by the surroundings
♦ We can write: If ΔHsys is negative, $\mathbf\small{\rm{Heat\,absorbed/released\,by\,surroundings}}$ will be positive but with the same magnitude as ΔHsys
(ii) If heat ΔHsys is absorbed by the system, that same heat will be lost by the surroundings
♦ We can write: If ΔHsys is positive, $\mathbf\small{\rm{Heat\,absorbed/released\,by\,surroundings}}$ will be negative but with the same magnitude as ΔHsys
◼ Based on the two points, we can write:
$\mathbf\small{\rm{Heat\,absorbed/released\,by\,surroundings}}$ is the negative of ΔHsys
8. So Eq.6.14 becomes:
ΔSuniverse = [ΔSsys + $\mathbf\small{\rm{\frac{-\Delta H_{sys}}{T}}}$]
• Rearranging this, we get:
TΔSuniverse = TΔSsys - ΔHsys
9. For a spontaneous process, ΔSuniverse will be greater than zero
• T will be always positive because, there are no negative values in the kelvin scale
• Thus we can write:
If the process is spontaneous, (TΔSsys - ΔHsys) will be greater than zero
• (TΔSsys - ΔHsys) can be rearranged as: -(ΔHsys - TΔSsys)
• So we can write:
If the process is spontaneous, -(ΔHsys - TΔSsys) will be greater than zero
• This can be rearranged as:
If the process is spontaneous, (ΔHsys - TΔSsys) will be less than zero
◼ Thus we get an important result:
A process will be spontaneous if: (ΔHsys - TΔSsys) < 0
10. Note that, we are considering the system only. The universe and the surroundings do not come in the equation. So we will drop the subscript 'sys'. We can write:
A process will be spontaneous if: (ΔH - TΔS) < 0
• Also note that, from Eq.6.17, we have: (ΔH - TΔS) = ΔG
◼ So we can write:
A process will be spontaneous if ΔG < 0
11. Consider the equation 6.17: ΔG = ΔH - TΔS
• All three terms are energies
• ΔH is the heat liberated from the system
• But all that ΔH is not available to do work. Because, a quantity of 'TΔS' is being deducted
• The energy remaining after the deduction is the ΔG
• So we can say that, ΔG is the free energy available to do work
◼ For this reason, ΔG is also known as the free energy
Next we will consider the 'possible cases' where a process can be spontaneous or non-spontaneous. It can be written in 6 steps
1. We have Eq.6.17: ΔG = ΔH - TΔS
♦ ΔH can be positive or negative
♦ ΔS can be positive or negative
♦ T can only be positive
✰ This is because, there are no negative values in the kelvin scale
2. So four possible cases arise
• Two of them are when ΔH is positive:
♦ ΔH positive, ΔS positive
♦ ΔH positive, ΔS negative
• Remaining two are when ΔH is negative:
♦ ΔH negative, ΔS positive
♦ ΔH negative, ΔS negative
3. Let us consider the first case: ΔH +ve, ΔS +ve
• Substituting in Eq.6.17, we get:
ΔG = (+ve) - T(+ve)
• On the right side, first term is positive and second term is negative
♦ At small values of T, the first term will be larger
✰ Then the result will be a +ve ΔG
♦ At large values of T, the first term will be smaller
✰ Then the result will be a -ve ΔG
• We know that:
♦ -ve ΔG indicates spontaneous process
♦ +ve ΔG indicates non-spontaneous process
◼ So for case 1, we can write:
♦ ΔH is +ve, ΔS is +ve
✰ The process will be spontaneous at large values of T
✰ The process will be non-spontaneous at small values of T
◼ This 'case 1' is a special case. It is an endothermic process. That is., heat is absorbed by the system. We see that, if T is increased to a high level, even an endothermic process can be made to take place spontaneously
4. Let us consider the second case: ΔH +ve, ΔS -ve
• Substituting in Eq.6.17, we get:
ΔG = (+ve) - T(-ve)
• On the right side, first term is positive and second term is also positive
♦ So whatever be the value of T, ΔG will be always +ve
• We know that:
♦ -ve ΔG indicates spontaneous process
♦ +ve ΔG indicates non-spontaneous process
◼ So for case 2, we can write:
♦ ΔH is +ve, ΔS is +ve
✰ Whatever be the value of T, the process will be always non-spontaneous
5. Let us consider the third case: ΔH -ve, ΔS +ve
• Substituting in Eq.6.17, we get:
ΔG = (-ve) - T(+ve)
• On the right side, first term is negative and second term is also negative
♦ So whatever be the value of T, ΔG will be always -ve
• We know that:
♦ -ve ΔG indicates spontaneous process
♦ +ve ΔG indicates non-spontaneous process
◼ So for case 3, we can write:
♦ ΔH is -ve, ΔS is +ve
✰ Whatever be the value of T, the process will be always spontaneous
6. Let us consider the fourth case: ΔH -ve, ΔS -ve
• Substituting in Eq.6.17, we get:
ΔG = (-ve) - T(-ve)
• On the right side, first term is negative and second term is positive
♦ At small values of T, the first term will be larger
✰ Then the result will be a -ve ΔG
♦ At large values of T, the first term will be smaller
✰ Then the result will be a +ve ΔG
• We know that:
♦ -ve ΔG indicates spontaneous process
♦ +ve ΔG indicates non-spontaneous process
◼ So for case 4, we can write:
♦ ΔH is -ve, ΔS is -ve
✰ The process will be spontaneous at small values of T
✰ The process will be non-spontaneous at large values of T
◼ This 'case 4' is a special case. It is an exothermic process. That
is., heat is released by the system. We see that, if T is increased to a
high level, even an exothermic process can be made to take place
non-spontaneously
Let us see some solved examples:
Solved example 6.30
A reaction, A + B → C + D + q is found to have a positive entropy change. The
reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(v) possible at any temperature
Solution:
1. The given equation is: A + B → C + D + q
• It is not a thermochemical equation because, the energy is written along with the equation
• The products constitute of: C, D and q
♦ That means, q is also produced. So it is an exothermic reaction
• We can write the thermochemical equation as: A + B → C + D; ΔH = -q
2. So ΔH is negative. We are given that, ΔS is positive
• Thus this process falls under case 3: ΔH -ve, ΔS +ve
3. We can write:
Whatever be the temperature, the given process will be always spontaneous
Solved example 6.31
For the reaction at 298 K,
2A + B → C
∆H = 400 kJ mol-1 and ∆S = 0.2 kJ K-1 mol-1
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range
Solution:
1. We have E.6.17: ΔG = ΔH - TΔS
• For the reaction to be spontaneous, ∆G must be less than zero
2. To find the temperature at which the reaction just becomes spontaneous, we put: ∆G = 0
• Thus we get: 0 = ΔH - TΔS
3. Substituting the known values, we get:
0 = 400 (kJ mol-1) - [T (K) × 0.2 (kJ K-1 mol-1)]
⇒ T = 2000 K
Solved example 6.32
For the reaction,
2Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?
Solution:
1. Sign of ∆H:
• In this process, individual Cl atoms combine to form Cl2 molecules
♦ Bond enthalpy will be released
♦ So ∆H will be negative
2. Sign of ∆S:
• In the initial state, there are more number of particles
• In the final state, number of particles become half
♦ This reduces the disorder/randomness
♦ So S decreases
♦ Thus ∆S will be negative
Solved example 6.33
For the reaction
2A(g) + B(g) → 2D(g)
∆U = –10.5 kJ and ∆S = –44.1 J K-1
Calculate ∆G for the reaction, and predict whether the reaction may occur
spontaneously
Solution:
• We have seen this type of problems in section 6.3
• We will solve this problem using Method I only
• The reader may try Method II also
Method I:
1. The given balanced equation is:
2A(g) + B(g) → 2D(g)
• Let A be the initial state
• In this state, the system is:
♦ 2 mol of A and 1 mol B at 298 K and P atm pressure
2. Let B be the final state
• In this state, the system is:
♦ 2 mol of D at 298 K and P atm pressure
3. We have seen that:
♦ Enthalpy in the initial state, HA = (UA + PVA)
♦ Enthalpy in the final state, HB = (UB + PVB)
♦ So enthalpy change = (HB - HA) = [(UB - UA) + P(VB - VA)]
4. Here, (UB - UA) is given as -10.5 KJ
• So we can write:
[(-10.5) + P(VB - VA)] = HB - HA
5. In the above equation, there are 3 terms: (-10.5), P(VB - VA) and (HB - HA)
• So, if we can find P(VB - VA), we can calculate (HB - HA)
6. P(VB - VA) can be calculated in steps:
(i) From ideal gas equation, we have:
♦ PAVA = nARTA
♦ PBVB = nBRTB
• In our present case:
♦ PA = PB = P
♦ TA = TB = T = 298 K
(ii) So P(VB - VA) = (nB - nA)RT
♦ nA = number of gaseous moles in the initial state = total number of gaseous moles of A and B = (2+1) = 3
♦ nB = number of gaseous moles in the final state = number of gaseous moles of D = 2
♦ So (nB - nA) = (2 - 3) = -1
(iii) Substituting the known values, the right side of (ii) becomes:
(-1 mol) × (8.3 J mol-1 K-1) × (298 K) = -2473.4 J = -2.473 kJ
7. We can use the value obtained above, in the place of P(VB - VA) in (4)
• We get: [(-10.5) -2.473] = (HB - HA)
• So (HB - HA) = ΔH = -[12.973] kJ
8. Now we use Eq.6.17: ΔG = ΔH - TΔS
• Substituting the known values, we get:
ΔG = -12.973 × 103 - (298 × -44.1) = 168.8 J
• Since this value is positive, the reaction is not spontaneous
Solved example 6.34
Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed
under standard conditions. ΔH⊖f = –286 kJ mol-1
Solution:
1. Given that, ΔH⊖f = –286 kJ mol-1
• So 286 kJ will be absorbed by the surroundings
2. We have: $\mathbf\small{\rm{\Delta S_{surr}=\frac{Heat\,absorbed\,by\,surroundings}{T}}}$
• The 286 should be given a +ve sign because, heat content of the surroundings increases due to the absorption
• We get: $\mathbf\small{\rm{\Delta S_{surr}=\frac{286 \times 10^3}{298}}}$ = 959.73 J K-1
• In the next chapter, we will see equilibrium
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