1. Value for K
Large K (>1) = equilibrium will contain mostly products (equilibrium position is to the right)
Small K (<1) = equilibrium will contain mostly reactants (equilibrium position is to the left)

2. More… K is NOT related to time (rate/activation energy)
When either the [reactants/products] = 0, it is easy to determine which way the reaction will shift
If not (some of each is present), Reaction Quotient (Q) must be determined

3. Reaction Quotient (Q)
Use the Law of Mass Action with initial concentrations instead of equilibrium concentrations.
N2(g) + 3H2(g) <-> 2NH3(g)
Q = [NH3]02
[N2]0[H2]03
Compare Q and K to determine which way the reaction will shift…

4. Q vs K Q = K: System is at equilibrium - no shift
Q > K: System shifts to the left, consuming products and forming reactants
Q < K: System shifts to the right, consuming reactants and forming products

5. H2(g) + I2(g) <-> 2HI(g), K = 7.1 X 102 at 25°C
Example
H2(g) + I2(g) <-> 2HI(g), K = 7.1 X 102 at 25°C
Predict the direction that the system will shift in order to reach equilibrium given each of the following initial conditions:
Q = 427
Q = 1522
[H2]0 = 0.81 M, [I2]0 = 0.44 M, [HI]0 = 0.58 M
[H2]0 = M, [I2]0 = M, [HI]0 =1.35 M
[H2]0 = M, [I2]0 = M, [HI]0 =1.50 M

6. Calculating Equilibrium Pressures and [ ]
N2O4(g) <-> 2NO2(g)
Kp = 0.133, pressure of N2O4 = 2.71 atm, determine equilibrium pressure of NO2(g)

7. ICE Tables
In order to determine concentrations at equilibrium, ICE tables can be used
I = INITIAL
C = CHANGE
E = EQUILIBRIUM

8. PCl5(g) <-> PCl3(g) + Cl2(g)
Example #1
At a certain temperature a 1.00L flask initially contained mol PCl3(g) and 8.70 X 10-3 mol PCl5(g). After the system had reached equilibrium, 2.00 X 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the reaction:
PCl5(g) <-> PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the value of K.

9. Example #2
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask.

10. Example #3
Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 X 102 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.

11. Steps… Write the balanced equation
Write the equilibrium expression using law of mass action
List initial concentrations
Calculate Q and determine the shift
Define change for equilibrium, apply the change to the initial concentrations
Substitute equilibrium concentrations into the equilibrium expression and solve for unknown
Check that calculated values make correct K

12. Problem-OH NO!
We are not always given easy numbers to work with and solve for
Sometimes the QUADRATIC EQUATION is necessary!!
When ax2 + bx + c = 0
X = -b ± √(b2 - 4ac) 2a
You will get two values for x. Only ONE will work to give a positive value for concentration once subtracted from the original values

13. Example
For synthesis of HF from its elements, mol H2 and mol F2 are mixed in a L flask. Assume that the equilibrium constant for the synthesis reaction at this temperature is 1.15 X Calculate the equilibrium concentration of each component.

14. Example with Pressures
Assume that gaseous hydrogen iodide is synthesized from its elemental vapors at a temperature where the equilibrium constant is 1.00 X Suppose HI at X 10-1 atm, H2 at X 10-2 atm, and I2 at X 10-3 atm are mixed in a L flask. Calculate the equilibrium pressures of all species.

15. Example with Small Equilibrium Constant
Usually these situations have you solving an x3 equation
Gaseous NOCl decomposes to form gases NO and Cl2. At 35°C the equilibrium constant is 1.6 X In an experiment in which 1.0 mol NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations?
2NOCl(g) <-> 2NO(g) + Cl2(g)
For this assumption to work, the difference between the initial and equilibrium must must be <5%