1. Lab Manual & Schedule
Tutorial
Formula sheet
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3. Chapter 7: Chemical Equilibrium Chapter 9 in the 7th edition

4. 7.1 The Gibbs energy minimum
1. Extent of reaction (ξ):
The amount of reactants being converted to products. Its unit is mole.
Consider: A ↔ B
Assume an infinitesimal amount dξ of A turns into B,
dnA = -dξ
On the other hand, dnB = dξ
Why do we need a new quantity?
consider a generic reaction: 2A ↔ 3B
In a very general way, the extent of reaction is calculated
as dξ = dnA/vA (where vA is the stoichiometric number of the reactant A, which is negative for the reactant!!)

5. Example 1: N2(g) + 3H2(g) ↔ 2NH3(g)
when the extent of reaction changes from ξ = 0 to ξ = 1.0 mole, what are the changes of each reagent?
Solution: identify vj:
v(N2) = -1; v(H2) = -3; v(NH3) = 2.
since dξ = 1.0 mole,
dn(N2) = -1x1.0 mole = -1.0 mole,
dn(H2) = -3x1.0 mole = -3.0 moles,
dn(NH3) = 2x1.0 mole = 2.0 moles,
Example 2: CH4(g) Cl2(g) ↔ CHCl3(l) HCl(g), in which the amount of reactant Cl2(g) decreases by 2 moles. What is the extent of the reaction?
Solution:

6. Variation of Gibbs energy during a reaction process

7. The Reaction Gibbs energy: ΔrG
the slope of the Gibbs energy plotted against the extent of reaction:
ΔrG =
here Δr signifies a derivative
A reaction for which ΔrG < 0 is called exergonic.
A reaction for which ΔrG > 0 is called endergonic.
ΔrG < 0, the forward reaction is spontaneous.
ΔrG > 0, the reverse reaction is spontaneous.
ΔrG = 0, the reaction is at equilibrium!!!

8. An useful description of the reaction Gibbs energy (ΔrG)
Consider the reaction A ↔ B
initial amount nA nB0
final amount nAf nBf
Ginitial = uBnB0 + uAnA0
Gfinal = uBnBf + uAnAf
ΔG = Gfinal - Ginitial = (uBnBf + uAnAf ) – (uBnB0 + uAnA0 )
= uB(nBf- nB0) + uA (nAf - nA0) = uBΔξ + uA(-Δξ)
= (uB - uA )Δξ
ΔrG = = uB - uA
When uA > uB, the reaction A → B is spontaneous.
When uB > uA, the reverse reaction (B → A) is spontaneous.
When uB = uA, the reaction is spontaneous in neither direction (equilibrium condition).

9. Molecular interpretation of the minimum in the reaction Gibbs energy

10. 7.2 The description of equilibrium
1. Perfect gas equilibrium:
A(g) ↔ B(g)
ΔrG = uB – uA

11. At equilibrium, ΔrG = 0, therefore
Note: The difference in standard molar Gibbs energies of the products and reactants is equal to the difference in their standard Gibbs energies of formation, thus,
ΔrGθ = ΔfGθ(B) - ΔfGθ(A)

12. Equilibrium for a general reaction
Example: A + B ↔ C D
The reaction Gibbs energy, ΔrG, is defined in the same way as discussed earlier:
where the reaction quotient, Q, has the form:
Q = activities of products/activities of reactants
in a compact expression Q = Π αj νj
vj are the corresponding stoichiometric numbers; positive for products and negative for reactants.
vA = -2; vB = -1; vC = 1; vD = 3

13. Justification of the equation
dG = ∑ujdnj
Assuming that the extent of reaction equals dξ,
one gets dnj = vjdξ
then dG = ∑ujdnj
= ∑ujvjdξ
(dG/dξ) = ∑ujvj  ΔrG = ∑ujvj
because uj = ujθ + RTln(aj)
ΔrG = ∑{vj(ujθ + RTln(aj))}
= ∑(vjujθ) + ∑vj(RTln(aj))
= ΔrGθ RT ∑ln(aj)vj
= ΔrGθ RT ln (∏(aj)vj) = ΔrGθ RT ln (Q)

14. K = Qequilibrium = (∏ajvj)equilibrium
Again, we use K to denote the reaction quotient at the equilibrium point,
K = Qequilibrium = (∏ajvj)equilibrium
K is called thermodynamic equilibrium constant. Note that until now K is expressed in terms of activities)
Example: calculate the quotient for the reaction: A + 2B ↔ 3C D
Solution: first, identify the stoichiometric number of each reactant:
vA = -1, vB = -2, vC = 3, and vD = 4.
At the equilibrium condition:

15. Examples of calculating equilibrium constants
Consider a hypothetical equilibrium reaction
A(g) B(g) ↔ C(g) D(g)
While all gases may be considered ideal. The following data are available for this reaction:
Compound μθ(kJ mol-1)
A(g)
B(g)
C(g)
D(g)
Calculate the value of the equilibrium constant Kp for the reaction at K.
Solution:

16. Example 2: Using the data provided in the data section, calculate the standard Gibbs energy and the equilibrium constant at 25oC for the following reaction
CH4(g) Cl2(g) ↔ CHCl3(g) HCl(g)
Solution: (chalkboard)
ΔfGθ(CHCl3, g) = kJ mol-1
ΔfGθ(HCl, g) = kJ mol-1
ΔfGθ(CH4, g) = kJ mol-1