1. Solving Equilibrium Problems
7.5

2. We then compare the value to the equilibrium constant
How do we know if a system has reached equilibrium? The Reaction Quotient
Q = [Products]coefficient [Reactants] coefficient
Tells you the direction the reaction will go to reach equilibrium
Calculated the same as the equilibrium constant, but for a system not at equilibrium
We then compare the value to the 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
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
I C E Chart

7. 1. Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(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

8. 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

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

10. 2. Consider the same reaction at 600ºC In a different experiment 0
2. 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

11. 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] =0.500+x
=0.590 mol/L

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

13. 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 using the quadratic formula, unless you can simplify if the initial concentration divided by k is greater than 100

14. Example [H2]i = (15.7g/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 15.9 g of H2 294 g I2 .
[H2]i = (15.7g/2.02)/5.00 L = 1.56 mol/L
[I2]i = (294g/253.8)/5.00L = mol/L
[HI]i = 0

15. 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=0.231
[H2] = 1.33mol/L
[I2] almost 0
[HI] = mol/L
[H2] [I2]
= [2x]2
[1.56-x] [0.232 –x ]

16. Practice
For the reaction Cl2 + O ClO(g) K = 156
In an experiment mol ClO, 1.00 mol O2 and mol Cl2 are mixed in a 4.00 L flask.
If the reaction is not at equilibrium, which way will it shift?
Calculate the equilibrium concentrations.

17. Let x(mol/L) represent the amount of H2 used up
Cl2(g) O2(g) ClO(g)
I M M M Q=1 which is < Ke; forward reaction results
C -x -x x
E x –x x

18. 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

19. k = [NO]2[Cl2] [NOCl]2 x =1.14 x 10 -4 .

20. 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.

21. [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

22. 6.4 x 10-3 = [0.025+x ][ x]
[ x ]2
X = M
[Cl2] = M
[O2] = M
[ClO]= M

23. 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]

24. 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?

25. 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

26. 3.0 = [2.0+x ][ 3.0+x]
[1.5 –x ][4.0-x]
x = 0.59 M
[SO2] = 3.6 M
[SO3] = 3.0 M
[NO2] = M
[NO] = M