1. Solving Equilibrium Problems
Type 1
Determining if Equilibrium has been reached (find Q and Compare to K)

2. The Reaction Quotient Compare value to equilibrium constant
Q = [Products]coefficient [Reactants] coefficient
Tells you the directing the reaction will go to reach equilibrium
Calculated the same as the equilibrium constant, but for a system not at equilibrium
Compare value to equilibrium constant

3. What Q tells us If Q<K If Q>K If Q=K system is at equilibrium
Not enough products
Shift to right
If Q>K
Too many products
Shift to left
If Q=K system is at equilibrium

4. Example for the reaction 2NOCl(g) 2NO(g) + Cl2(g)
K = 1.55 x 10-5 M at 35ºC
In an experiment mol NOCl, mol NO(g) and mol Cl2 are mixed in 2.0 L flask.
Which direction will the reaction proceed to reach equilibrium?

5. Q= Q= 5x10-9 1.55 x 10-5 M = [Cl2][NO]2 [NOCl ]2
[NOCl] = 0.05 mol/L
[NO] = mol/L
[Cl2] = mol/L
[ mol/L ][ mol/L]2
[0.05 mol/L ]2 Q=
Q= 5x10-9
Q<K reaction proceeds to the right

6. Solving Equilibrium Problems
Type 2
Using equilibrium concentrations to solve for K

7. Solving for K Using equilibrium concentrations to solve for K
A mixture of H2 and I2 is allowed to react at 448 ºC. When equilibrium is established, the concentrations of the participants are found to be;
[H2] = M, [I2] = 0.39 M and [HI] = 3.0 M
Calculate K.

8. Solution Write the balanced equation
Write the equilibrium law expression for this reaction
Substitute the equilibrium concentrations in the K expression and solve.

9. H2(g) + I2(g) HI(g)
K = [HI] [H2][I2]
K = [3.0 M]2
[0.46M][0.39 M ]
= 50.

10. Solving Equilibrium Problems
Type 3
Given the starting concentrations and one equilibrium concentration.
Calculate K

11. Solving Equilibrium Problems
Given the starting concentrations and one equilibrium concentration.
Use stoichiometry to figure out other concentrations and K.
Create a table of initial and final conditions. You may include changes, too I C E

12. Consider the following reaction at 600ºC
2SO2(g) + O2(g) SO3(g)
In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol SO3 were found to be present. Calculate the equilibrium concentrations of O2 and SO2 and K

13. 2SO2(g) + O2(g) 2SO3 (g) I 2.00 mol/L 1.50 mol/L 3.00mol/L
C -2x x x
E x x x
[SO3] = x
3.50= x
X=0.25 mol/L
[SO2]=2.00-2x
= 1.50 mol/L
[O2] = 1.5-x
= 1.25 mol/L

14. K = [3.5mol/L]2 [1.50mol/L]2 [1.25mol/L]
= 4.36
= 4.4

15. Consider the same reaction at 600ºC
In a different experiment mol SO2 and mol SO3 were placed in a L container. When the system reaches equilibrium mol of O2 are present.
Calculate the final concentrations of SO2 and SO3 and K

16. 2SO2(g) + O2(g) 2SO3 (g) I 0.500 mol/L 0 mol/L 0.350 mol/L
C +2x x x
E x x x
[O2] =0.045 mol/L = x
[SO3] = x
=0.260 mol/L
[SO2] = x
=0.590 mol/L

17. K = [0. 260 mol/L]2 [0.590mol/L ]2 [0.045mol/L ]
= 4.32

18. Solving Equilibrium Problems
Type 4
Calculating equilibrium concentration using K and initial concentrations
A- PERFECT SQUARE
B- Simplification of Quadratic
C- solve a quadratic

19. Calculating equilibrium concentration using K and initial concentrations
Write the balanced equation
Write the equilibrium law expression for this reaction
Let x represent the amount of reactant used up or created.
Set up an ICE chart
Solve…possibly
A) mathematically simplify…perfect square situation B)assumption made…if the initial concentration divided by K is greater than 100 and 5% rule met
C) solve the quadratic

20. Example [H2]i = 0.10M [I2]i = 0.10 M [HI]i = 0 H2(g) + I2(g) 2HI(g)
K = 7.1 x 102 at 25ºC
Calculate the equilibrium concentrations if a 5.00 L container initially contains 0.10M of H2 0.10M I2 .
[H2]i = 0.10M
[I2]i = 0.10 M
[HI]i = 0

21. H2(g) I2(g) HI(g) I 0.10 M 0.10 M 0 M C -x -x +2x
Let x(mol/L) represent the amount of H2 used up
H2(g) I2(g) HI(g)
I M 0.10 M M C -x -x x
E M-x M –x 2x
Solve for x
2x/o.1-x= square root of K
2x= 0.1-x (square root K)
x= 0.093M
[H2] = M
[I2] = M
[HI] = 0.186M
K = [HI]2
= 7.1 x 102
[H2] [I2]
= [2x]2
[0.1-x] [0.1 –x ]
(Perfect square!...simplify)

22. For example
For the reaction NOCl NO +Cl2
K= 1.6 x 10-5
If 1.20 mol NOCl, and 0.45 mol of NO are mixed in a 1 L container
What are the equilibrium concentrations?

23. x =1.14 x 10 -4M (x/initialx100 less than 5%)
k = [NO]2[Cl2]
[NOCl]2
1.6 x 10-5 = (0.45+2x)2(x) (1.20-2x)2
Initial conc./K is GREATER than 100 …simplify
1.6 x 10-5 = (0.45)2(x) (1.20)2
x =1.14 x 10 -4M (x/initialx100 less than 5%)
….calculate equilibrium concentrations
[NO]=0.45+2x=0.45M, [Cl2]= 1.14 x 10-4M, [NOCL] = x=1.20M

24. Example [H2]i = (15.8g/2.02)/5.00 L = 1.56 mol/L
H2(g) + I2(g) HI(g)
K = 7.1 x 102 at 25ºC
Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.8 g of H2 294 g I2 .
[H2]i = (15.8g/2.02)/5.00 L = 1.56 mol/L
[I2]i = (294g/253.8)/5.00L = mol/L
[HI]i = 0

25. H2(g) I2(g) HI(g) I 1.56 M 0.232 M 0 M C -x -x +2x
Let x(mol/L) represent the amount of H2 used up
H2(g) I2(g) HI(g)
I M M M C -x -x x
E x –x x
K = [HI]2
= 7.1 x 102
Solve for x
x=0.2318
[H2] = 1.33mol/L
[I2] almost 0…0.0002M
[HI] = mol/L
[H2] [I2]
= [2x]2
[1.56-x] [0.232 –x ]
USE quadratic formula…plug in a,b,c, in calculator

26. Solving Equilibrium Problems
Type 5
Calculating equilibrium concentrations when INItial reactant and product concentrations both given…CALCULATE q FIRST BEFORE STARTING

27. What are the equilibrium concentrations?
Practice Problem
For the reaction
2ClO(g) Cl2 (g) + O2 (g)
K = 6.4 x 10-3
In an experiment 1.00 mol ClO(g), mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container.
What are the equilibrium concentrations?

28. [ClO] [O2 ] [Cl2] 6.4 x 10-3 = [O2 ][Cl2] [ClO]2 I 0.250 0.025 0.0025
Q= .001; reaction proceeds in the forward direction.
Let x(M) be the amount of Cl2 made.
C -2x x x
E x x x

29. 6.4 x 10-3 = [0.025+x ][ x]
[ x ]2
X = M
[Cl2] = M
[O2] = M
[ClO]= M

30. Solving Equilibrium Problems
Type 6
Calculating new concentrations when the equilibrium has been disturbed.

31. Calculating new concentrations when the equilibrium has been disturbed.
Consider the effect of introducing more product after this equilibrium has been established.
SO2 (g) + NO2(g) SO3 (g) + NO(g)
K = [NO][SO3]
[NO2 ][SO2]

32. Analysis of the equilibrium mixture show the following:
[SO2] = 4.0 M [SO3] = 3.0 M [NO2] = 0.50 M [NO] = 2.0 M What is the new equilibrium concentration for NO when 1.5 mol of NO2 is added?

33. K = [NO][SO3] [NO2 ][SO2] find K K = [2.0 M][3.0 M] [0.50 M ][4.0 M]
= 3.0
Let x(M) be the amount of NO2 used up in the forward rxn
[NO2 ] [SO2]
[NO ] [SO3] I
C x x x x
E –x x x x

34. 3.0 = [2.0+x ][ 3.0+x]
[2.0–x ][4.0-x]
x = 0.59 M
[SO2] = 3.6 M
[SO3] = 3.0 M
[NO2] = M
[NO] = M