Chapter 7.8 - Relation Between Equilibrium Constant and Gibbs Energy

In the previous section, we saw the applications of equilibrium constant. In this section, we will see how equilibrium constant is related to Gibbs energy

• In the previous chapter, we saw that a reaction will occur spontaneously in the forward direction if ΔG is negative (Details here)
• In the previous section of the present chapter, we saw that, if Q_c is less than K_c, the reaction will occur spontaneously in the forward direction
◼ So there must be a relation between the three quantities:
(i) ΔG (ii) K_c (or K_p) (iii) Q_c (or Q_p)

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Let us recall the three points that we have learnt about ΔG in the previous chapter:
(i) If ΔG is negative, the reaction is spontaneous and proceeds in the forward direction
(ii) If ΔG is positive, the reaction is non-spontaneous
    ♦ But then, the backward reaction will have a negative ΔG value
    ♦ So the backward reaction will proceed spontaneously
(iii) If ΔG is zero, the reaction is at equilibrium. There is no free energy available to drive the reaction in any direction
◼ Scientists have derived the following equation which gives the relation between ΔG, K and Q
Eq.7.4: $\mathbf\small{\rm{\Delta G= \Delta G^\circleddash + RT\,lnQ}}$
    ♦ Where: $\mathbf\small{\rm{\Delta G^\circleddash}}$ is the standard Gibbs energy

• For convenience of our present discussion,     ♦ We write Kc (or Kp) simply as K     ♦ We write Qc (or Qp) simply as Q

We can write an analysis of Eq.7.4 in 4 steps:
1. At equilibrium, we have: ΔG = 0
    ♦ At equilibrium, we also have Q = K
• So Eq.7.4 becomes: $\mathbf\small{\rm{0= \Delta G^\circleddash + RT\,lnK}}$
2. From this, we get:
Eq.7.5: $\mathbf\small{\rm{\Delta G^\circleddash = -RT\,lnK}}$
• Thus we get:
Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
3. Taking antilog of both sides, we get:
Eq.7.7: $\mathbf\small{\rm{K=e^{\frac{-\Delta G^\circleddash}{RT}}}}$
4. Eq.7.7 gives us a relation between K and $\mathbf\small{\rm{\Delta G^\circleddash}}$
• That means, we can predict the value of K using the value of $\mathbf\small{\rm{\Delta G^\circleddash}}$
• This can be further explained in two steps:
(i) If $\mathbf\small{\rm{\Delta G^\circleddash}}$ is negative, then $\mathbf\small{\rm{\frac{-\Delta G^\circleddash}{RT}}}$ will be positive
    ♦ Consequently, $\mathbf\small{\rm{e^{\frac{-\Delta G^\circleddash}{RT}}}}$ will be greater than 1
    ♦ Consequently, K will be greater than 1
• When K is greater than 1, it means that:
    ♦ the numerator in $\mathbf\small{\rm{\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
    ♦ is greater than
    ♦ the denominator
• That means:
    ♦ concentrations of products
    ♦ will be greater than 
    ♦ concentrations of reactants
• That means:
The reaction will proceed spontaneously in the forward direction

(ii) If $\mathbf\small{\rm{\Delta G^\circleddash}}$ is positive, then $\mathbf\small{\rm{\frac{-\Delta G^\circleddash}{RT}}}$ will be negative
    ♦ Consequently, $\mathbf\small{\rm{e^{\frac{-\Delta G^\circleddash}{RT}}}}$ will be less than 1
    ♦ Consequently, K will be less than 1
• When K is less than, it means that:
    ♦ the numerator in $\mathbf\small{\rm{\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
    ♦ is less than
    ♦ the denominator
• That means:
    ♦ concentrations of products
    ♦ will be less than 
    ♦ concentrations of reactants
• That means:
The reaction will proceed spontaneously in the reverse direction

Let us see some solved examples:
Solved example 7.31
The value of ∆G^⊝ for the phosphorylation of glucose in glycolysis is 13.8 kJ mol^-1.
Find the value of K_c at 298 K.
Solution:
1. We have Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
• Substituting the known values, we get: $\mathbf\small{\rm{lnK=\frac{-13.8\times 10^3}{8.314 \times 298}}}$ = -5.569
⇒ lnK = -5.569
2. Taking antilog of both sides, we get: K = e^-5.569 = 3.81 × 10^-3

Solved example 7.32
Calculate K for the reaction of O₂ with N₂ to give NO at 423 K:
N₂ (g) + O₂ (g) ⇌ 2NO (g)
∆G^⊝ for this reaction is +22.7 kJ/mol
Solution:
1. We have Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
• Substituting the known values, we get: $\mathbf\small{\rm{lnK=\frac{-22.7\times 10^3}{8.314 \times 423}}}$ = -6.45
⇒ lnK = -6.45
2. Taking antilog of both sides, we get: K = e^-6.45 = 0.00158

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Next we will consider the case when the reaction is not at equilibrium. This can be analyzed in 2 steps:
1. When the reaction is not at equilibrium, we cannot put ΔG = 0 in Eq.7.4
    ♦ Also, we cannot put Q = K
• That means, we keep Eq.7.4 as such
2. Consider Eq.7.4: $\mathbf\small{\rm{\Delta G= \Delta G^\circleddash + RT\,lnQ}}$
• The right side has two terms
   ♦ The first term among them is ∆G^⊝
         ✰ It will be given in the question
        ✰ Or it can be calculated using Eq.7.5 
   ♦ The second term among them can be calculated using concentrations
• So we can easily calculate ΔG

Solved example 7.33
∆G^⊝ for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ is -32.7 kJ mol^-1. Calculate ΔG when the concentrations are as follows:
[N₂] =2.00 M, [H₂] = 7.00 M, [NH₃] = 0.021 M
Temperature = 373 K
Solution:
1. We have Eq.7.4: $\mathbf\small{\rm{\Delta G= \Delta G^\circleddash + RT\,lnQ}}$
The first term on the right = ∆G^⊝ = -32.7 kJ mol^-1
2. The second term on the right is RT lnQ
• So we have to first find Q
• We have: Q = $\mathbf\small{\rm{\frac{[NH_3]^2}{[N_2][H_2]^3}=\frac{[0.021]^2}{[2.00][7.00]^3}}}$ = 6.428 × 10^-7
• Thus we get: RT lnQ = (8.314 × 373 × ln6.428 × 10^-7) = -44214.04 J mol^-1 = -44.21 kJ mol^-1
3.  Adding the two terms, we get: ΔG = (-32.7 -44.21) = -76.91 kJ mol^-1
4. We see that ΔG < 0
So the reaction must proceed spontaneously in the forward direction
5. To confirm this, we have to check the value of K also
• To find K, we can use Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
• Substituting the known values, we get: $\mathbf\small{\rm{lnK=\frac{32.7\times 10^3}{8.314 \times 373}}}$ = 10.545
⇒ lnK = 10.545
• Taking antilog of both sides, we get: K = e^10.545 = 37987.03
• This value of K is greater than the value of Q that we obtained in (2)
• So the reaction will indeed proceed in the forward direction. Result in (4) is confirmed

Solved example 7.34
Calculate ΔG for the reaction N₂(g)+O₂(g) ⇌ 2NO(g) under the conditions: T = 423 K, [NO] = 0.0100 M, [O₂] = 0.200 M, and [N₂] = 1.00 × 10^-4 M. The value of ∆G^⊝ for this reaction is +22.7 kJ. In which direction will the reaction proceed to reach equilibrium?
Solution:
1. We have Eq.7.4: $\mathbf\small{\rm{\Delta G= \Delta G^\circleddash + RT\,lnQ}}$
The first term on the right = ∆G^⊝ = +22.7 kJ mol^-1
2. The second term on the right is RT lnQ
• So we have to first find Q
• We have: Q = $\mathbf\small{\rm{K_c=\frac{[NO]^2}{[N_2][O_2]}=\frac{[0.0100]^2}{[1.00 \times 10^{-4}][0.200]}}}$ = 5
• Thus we get: RT lnQ = (8.314 × 423 × ln5) = 5659.97 J mol^-1 = 5.66 kJ mol^-1
3.  Adding the two terms, we get: ΔG = (22.7 + 5.66) = +28.36 kJ mol^-1
4. We see that ΔG > 0
So the reaction must proceed spontaneously in the backward direction
5. To confirm this, we have to check the value of K also
• To find K, we can use Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
• Substituting the known values, we                 get: $\mathbf\small{\rm{lnK=\frac{-22.7\times 10^3}{8.314 \times 423}}}$ = -6.455
⇒ lnK = -6.455
• Taking antilog of both sides, we get: K = e^-6.455 = 0.00157
• This value of K is less than the value of Q that we obtained in (2)
• So the reaction will indeed proceed in the backward direction. Result in (4) is confirmed

Solved example 7.35
Calculate a) ∆G^⊝ and b) the equilibrium constant for the formation of NO₂ from
NO and O₂ at 298K
NO (g) + 1⁄2 O₂ (g) ⇌ NO₂ (g)
where
∆G^⊝[NO₂] = 52.0 kJ mol^-1
∆G^⊝[NO] = 87.0 kJ mol^-1
∆G^⊝[O₂] = 0 kJ mol^-1
Solution:
Part (a):
1. ∆G^⊝ can be obtained by subtracting item (ii) from item (i) below:
(i) Sum of the ∆G^⊝ values of all products                                                                   
(ii) Sum of the ∆G^⊝ values of all reactants
2. Thus we get:
∆G^⊝ = (52.0 - 87) = -35 kJ
Part (b):
1. We have Eq.7.6: $\mathbf\small{\rm{lnK=\frac{-\Delta G^\circleddash}{RT}}}$
• Substituting the known values, we get: $\mathbf\small{\rm{lnK=\frac{35\times 10^3}{8.314 \times 298}}}$ = 14.13
⇒ lnK = 14.13
2. Taking antilog of both sides, we get: K = e^14.13 = 1.365 × 10⁶

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• In the next section, we will see the factors affecting equilibrium

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