1. Topic 17: Equilibrium Law
17.1: The position of equilibrium can be quantified by the equilibrium law. The equilibrium constant for a particular reaction only depends on the temperature.

2. Important terms for this section
Homogeneous equilibria
Free energy

3. Calculating Kc from initial and equilibrium concentrations
Homogeneous equilibria:
The reactants and products are all in the same state (gas, liquid, solid)
Ex: No solid and gas phases in the reaction together
Steps for solving I.C.E. problems
Write a balanced equation
Write three rows: initial, change, equilibrium
Initial: original concentration
Amount reacts to reach equilibrium
Equilibrium: [equilibrium] = [initial] +/- change in concentration
Write the expression for Kc from the balanced equation and substitute values for equilibrium concentration and calculate Kc

4. I.C.E. problem 1
A student placed 0.20mol PCl3(g) and 0.01 mol Cl2(g) into a 1.0 dm3 flask at 350℃. The reaction, which produced PCl5, was allowed to come to equilibrium, at which time it was found that the flask contained 0.12 mol PCl3.
What is the value of Kc for this reaction?
Step 1: write a balanced reaction PCl3(g) + Cl2(g) ⇌ PCl5(g)
Step 2:
Step 3:

5. I.C.E. problem 2
Oxidation of NO to form NO2 occurs during smog formation. When 0.60 mol NO was reacted with 0.60 mol O2 in a 2.0 dm3 container at 500℃, the equilibrium mixture was found to contain mol NO2.
What is the value of Kc for this reaction?
Step 1: write a balanced reaction 2NO(g) + O2(g) ⇌ NO2(g)
Step 2:
Step 3:

6. Calc. equilibrium concentrations from Kc
We gotta use Algebra, m’kay!
Follow the same kinds of steps, but now the change in concentrations are x
So this is still an I.C.E. problem, but cooler… 
Example: Kc = 6.78 at a certain temp for the reaction SO3(g) + NO(g) ⇌ NO2(g) + SO2(g)
Initial [NO] and [SO3] are each mol dm-3
Change in NO = -x change in SO3 = -x
Change in NO2 = +x change in SO2 = +x

7. Calc. equilibrium concentrations from Kc

8. Calc. equilibrium conc. from itty bitty Kc
If Kc is less than 10-3, this is considered pretty small
This means the initial reaction concentrations are close to the equilibrium concntrations

9. Calc. equilibrium conc. from itty bitty Kc

10. Calc. equilibrium conc. from itty bitty Kc

11. Free Energy and Equilibrium
Remember that if ∆G is negative, then the reaction is spontaneous in that direction
Thus,
∆Gθ = Gθproducts – Gθreactants
Reaction begins and moves in the forward direction (lots of free energy)
As it proceeds, the forward reaction decreases and reverse reaction increases (loss of free energy)
At equilibrium, there is not more change in reaction rate, (no more free energy)
At equilibrium, since there is a minimum of free energy and entropy is highest

12. Free Energy and Equilibrium
∆G large and negative ∆G large and positive
Equilibrium lies to the right Equilibrium lies to the left

13. Kc can be calculated from thermodynamic data
How are Kc and ∆Gθ related?
∆Gθ = -RT ln K
∆Gθ = standard free energy change of the reaction
R = the gas constant 8.31 J K-1 mol-1
T = the absolute temperature in Kelvin
ln K = the natural logarithm of the equilibrium constant, Kc
This explains why Kc can have different values: ∆Gθ
And ∆Gθ depends on ∆H and ∆S

14. Kinetics and Equilibrium
Remember that Kc tells you where equilibrium lies, but says nothing about the rate!
But, we can look at an example of forward and backward rate constants to tell us some equilibrium info
Example: if we have A + B ⇌ C + D AND
If the rate laws were: forward rate = k[A][B] AND backward rate = k’[C][D]
Then we also have an equilibrium constant expression that looks like this:

15. Kinetics and Equilibrium
So, this means that:
AND for
Concentration:
Increasing [reactant] increases forward rate so shifts equilibrium to the right
Increasing [products] increases backward rate so shifts equilibrium to the left
Kc stays the same
Catalyst:
Increases values of k and k’ by the same factor
Temperature: with Arrhenius equation, temp ⇧ = rate ⇧
But if activation energies of forward and backward RXNs are different, then k and k’ are differently affected
k/k’ is dependent on temperature: ⇧ temp = ⇧ rate of endothermic reaction (specfifically)
Kc will increase if endothermic RXN is the forward RXN