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 Qc is less than Kc, the reaction will occur spontaneously in the forward direction
◼ So there must be a relation between the three quantities:
(i) ΔG (ii) Kc (or Kp) (iii) Qc (or Qp)
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: ΔG=ΔG⊝+RTlnQ
♦ Where: ΔG⊝ 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: 0=ΔG⊝+RTlnK
2. From this, we get:
Eq.7.5: ΔG⊝=−RTlnK
• Thus we get:
Eq.7.6: lnK=−ΔG⊝RT
3. Taking antilog of both sides, we get:
Eq.7.7: K=e−ΔG⊝RT
4. Eq.7.7 gives us a relation between K and ΔG⊝
• That means, we can predict the value of K using the value of ΔG⊝
• This can be further explained in two steps:
(i) If ΔG⊝ is negative, then −ΔG⊝RT will be positive
♦ Consequently, e−ΔG⊝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 [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 ΔG⊝ is positive, then
−ΔG⊝RT will be negative
♦ Consequently, e−ΔG⊝RT will be less than 1
♦ Consequently, K will be less than 1
• When K is less than, it means that:
♦ the numerator in [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 Kc at 298 K.
Solution:
1. We have Eq.7.6: lnK=−ΔG⊝RT
• Substituting the known values, we get: lnK=−13.8×1038.314×298 = -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: lnK=−ΔG⊝RT
• Substituting the known values, we get: lnK=−22.7×1038.314×423 = -6.45
⇒ lnK = -6.45
2. Taking antilog of both sides, we get: K = e-6.45 = 0.00158
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: ΔG=ΔG⊝+RTlnQ
• 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: ΔG=ΔG⊝+RTlnQ
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 = [NH3]2[N2][H2]3=[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: lnK=−ΔG⊝RT
• Substituting the known values, we get: lnK=32.7×1038.314×373 = 10.545
⇒ lnK = 10.545
• Taking antilog of both sides, we get: K = e10.545 = 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: ΔG=ΔG⊝+RTlnQ
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 = Kc=[NO]2[N2][O2]=[0.0100]2[1.00×10−4][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: lnK=−ΔG⊝RT
• Substituting the known values, we get: lnK=−22.7×1038.314×423 = -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: lnK=−ΔG⊝RT
• Substituting the known values, we get: lnK=35×1038.314×298 = 14.13
⇒ lnK = 14.13
2. Taking antilog of both sides, we get: K = e14.13 = 1.365 × 106
•
In the next section, we will see the factors affecting equilibrium
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment