1. Agenda: Sign off/Discuss Equilibrium constant ws
Collect Beanium Equilibrium Activity
In-Class ICE table practice
HW: Ice Tables ws

2. I can…
Determine the direction of the reaction by the magnitude of the Kc.
Use the ICE tables to predict changes concentration and its effect on the Kc value.

3. Flip Notes
Equilibrium concentrations of each species in 0.2M iodic acid. HIO3, has a Ka = 0.17 HIO3 (aq)  H+ (aq) + IO3- (aq)

4. Flip Lesson

6. 15.4- Calculating Equilibrium Constant
When a reaction has reached equilibrium, we often don’t know HOW the initial concentrations of the species have changed from the equilibrium concentrations.
Or we were given JUST the initial concentrations, how can we determine the equilibrium concentrations?

8. I is initial concentration,
To look at all these variables at the same time we need to create an ICE table.
I is initial concentration,
C is the change in concentration
E is the concentration at equilibrium.

9. There are two types of ICE tables
The initial or equilibrium concentration of some substances must be determined.
B. Initial or equilibrium concentrations of some
substances are given, but not both. Change is
therefore treated as an unknown (x)
C. The equilibrium constant is given.

10. There are two types of ICE tables
A. The equilibrium constant or concentration must be determined.
B. Initial and equilibrium concentrations of at least one substance are given so that change can be calculated directly.
C. All other initial and equilibrium concentrations of substances are determined directly from the table.

11. Walkthrough-(See pg. 571-2 Sample Exercise)
A closed system initially containing
1.000 x 10−3 M H2 and x 10−3 M I2
At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is
H2 (g) + I2 (g)
2 HI (g)

12. STEPS TO ICE ① Make Sure that all concentrations are in
M- molarity!! (Done for you)
② Set up table- ICE (as you see it) and
then species at top.
③Place the known concentrations provided in the question into table.
④Put in the CHANGE for HI (subtract
equilibrium from initial)

13. ② Set up table- ICE (as you see it) and then species at top.
[H2], M
[I2], M
[HI], M
Initially
Change
Equilibrium

14. ③Place the known concentrations provided in the question into table.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
Equilibrium
1.87 x 10-3

15. ④Put in the CHANGE for HI (subtract equilibrium from initial)
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
+1.87 x 10-3
At equilibrium
1.87 x 10-3

16. ⑤NOW we have to use the stoichiometry of the reaction to get the change of H2 and I2. Put a negative sign in front because they are reactants.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
+1.87 x 10-3
At equilibrium
1.87 x 10-3
H2 (g) + I2 (g)
2 HI (g)
1.87 x 10-3mol L
1 mol H2
2 mol of HI
= x 10-3
Same goes for iodine

17. ⑤ Stoichiometry tells us [H2] and [I2] decrease by half as much
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At equilibrium
1.87 x 10-3
The change MUST be in the negative because they are reactants!!

18. ⑥ Subtract the initial concentrations from the change which will provide the equilibrium value.
[H2], M
[I2], M
[HI], M
Initially
1.000 x 10-3
2.000 x 10-3
Change
-9.35 x 10-4
+1.87 x 10-3
At equilibrium
1.87 x 10-3
6.5 x 10-5
1.065 x 10-3
1.000 x10-3 – 9.35 x10-4 =6.5 x 10-5
2.000 x10-3 – 9.35 x10-4 =1.065 x 10-3

19. [HI]2 [H2] [I2] Kc = = (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3)
⑦ Finally, provide the equilibrium expression for this reaction. Substitute the equilibrium values from chart into the expression to solve for Kc.
[HI]2
[H2] [I2]
Kc = =
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Kc = 51

20. Let’s Try another
Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g)  2SO2(g) + O2(g). Initially, the vessel is charged at 1000 K with SO3(g) at a concentration of 6.09 x 10-3 M. At equilibrium the SO3 concentration is 2.44 x10-3M. Calculate the value of Kp at 1000 K.
INITIAL
CHANGE
EQUILIBRIUM
2SO3 
2SO2 O2
6.09 x 10-3 M
0 M
0 M
2.44 x10-3M

21. Let’s Try another 6.09 x 10-3 M 0 M 0 M 2.44 x10-3M
INITIAL
CHANGE
EQUILIBRIUM
2SO3 
2SO2 O2
6.09 x 10-3 M
0 M
0 M
-3.65 x 10-3 M
2.44 x10-3M
What information can we fill in with what we are given?
We can fill in the change of SO3.
6.09 x 10-3 M x10-3M = x 10-3 M

22. Let’s Try another 6.09 x 10-3 M 0 M 0 M 2.44 x10-3M
INITIAL
CHANGE
EQUILIBRIUM
2SO3 
2SO2 O2
6.09 x 10-3 M
0 M
0 M
-3.65 x 10-3 M
+3.65 x 10-3 M
+1.83 x 10-3 M
2.44 x10-3M
If SO3 went down by x 10-3 M we have to use stoichiometry to find out the relationship the products have with the reactant.
2 moles SO2
2 mole of SO3
3.65 x 10-3 M of SO3
= x 10-3 M of SO2
1 moles O2
2 mole of SO3
3.65 x 10-3 M of SO3
= x 10-3 M of O2

23. Let’s Try another 6.09 x 10-3 M 0 M 0 M 2.44 x10-3M 1.83 x 10-3 M
INITIAL
CHANGE
EQUILIBRIUM
2SO3 
2SO2 O2
6.09 x 10-3 M
0 M
0 M
-3.65 x 10-3 M
+3.65 x 10-3 M
+1.83 x 10-3 M
2.44 x10-3M
3.65 x 10-3 M
1.83 x 10-3 M
Now subtract the initial from the change to get the equilibrium.
x 10-3 M = 3.65 x 10-3 M
0 – 1.83 x 10-3 M = 1.83 x 10-3 M

24. Let’s Try another 6.09 x 10-3 M 0 M 0 M 2.44 x10-3M 1.83 x 10-3 M
INITIAL
CHANGE
EQUILIBRIUM
2SO3 
2SO2 O2
6.09 x 10-3 M
0 M
0 M
-3.65 x 10-3 M
+3.65 x 10-3 M
+1.83 x 10-3 M
2.44 x10-3M
3.65 x 10-3 M
1.83 x 10-3 M
Use the equilibrium concentrations to find Kc-
plug n chug
( x10-5)(1.83 x10-3) = x10-8
x10-6
Kc= 4.11 x 10-3

25. In many situations we will know the value of the equilibrium constant and the initial concentrations of all species. We must then solve for the equilibrium concentrations. We have to treat them as variables or “x”

26. A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g)  2 AB (g) Kc = 9.0 at 100°C If 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0 L container, what are the equilibrium concentrations of A2, B2, and AB?
① Convert to molarity!!
1 mol
2.0 L
= 0.50 M

27. A chemist has a container of A2 and B2 and they react as given: A2 (g) + B2 (g)  2 AB (g) Kc = 9.0 at 100°C If 1.0 mole A2 and 1.0 mole B2 are placed in a 2.0 L container, what are the equilibrium concentrations of A2, B2, and AB?
[A2], M
[B2], M
[AB], M
Initially
Change
Equilibrium 
0.50 M
0.50 M
0.0 M

28. -x -x + 2x 0.50-x 0.50-x 2x 0.50 M 0.50 M 0.0 M [A2], M [B2], M
[AB], M
Initially
Change
Equilibrium 
0.50 M
0.50 M
0.0 M -x -x
+ 2x
0.50-x
0.50-x 2x
Fill in the chart with variables!!
A2(g) + B2(g)  2AB (g)
You need to (for this ICE table) to factor in the stoich relationship)

29.  [AB]2 [A2] [B2] Kc = 9.0= (2x)2 (0.50-x)(0.50-x) -x -x + 2x 0.50-x
[A2], M
[B2], M
[AB], M
Initially
0.50M
0.50 M
0 M
Change
At equilibrium  -x -x
+ 2x
0.50-x
0.50-x 2x
We have to work backwards!!
[AB]2
[A2] [B2]
Kc =
Provide the Kc expression
9.0=
(2x)2
(0.50-x)(0.50-x)

30. 9.0= (2x)2 (0.50-x)(0.50-x) 3.0= (2x) (0.50-x) (0.50-x)2Or
(0.50-x)2
Root both sides
3.0=
(2x)
(0.50-x)
Get 0.50-x out by multiply on both sides

31. 3.0=
(2x)
(0.50-x)
Get 0.50-x out by multiply on both sides
3.0(0.50-x)= 2x
Multiply 3 through
1.5-3x= 2x
Add 3x to both sides to get rid of -3x

32. 0.3 = x 1.5-3x= 2x 1.5 = 5x Add 3x to both sides to get rid of -3x
Divide both sides by 5 to get 5 out of there
0.3 = x
SO back to the chart with our x value of 0.3

33. -x -x + 2x 0.50 M 0.50 M 0.0 M [A2], M [B2], M [AB], M Initially
Change
At equilibrium 
0.50 M
0.50 M
0.0 M -x -x
+ 2x
= 0.2M
= 0.2M
2(0.3) =
0.6M
Now plug in our “x” value of 0.3

34. Sample Exercise 15.11
A L flask is filled with mol of H2 and mol of I2 at 448°C. The value of the equilibrium constant (Kc) is 50.5.
H2(g) + I2(g)  2HI (g)
What are the concentrations of H2, I2, and HI in the flask at equilibrium?

35.  -x -x + 2x 1.000-x 2.000-x 2x [H2], M [I2], M [HI], M Initially
Change
At equilibrium  -x -x
+ 2x
1.000-x
2.000-x 2x
Fill in the chart with variables!!
H2(g) + I2(g)  2HI (g)
You need to (for this ICE table) to factor in the stoich relationship)

36.  -x -x + 2x 1.000-x 2.000-x 2x [H2], M [I2], M [HI], M Initially
Change
At equilibrium  -x -x
+ 2x
1.000-x
2.000-x 2x
Fill in the chart with variables!!
H2(g) + I2(g)  2HI (g)
You need to (for this ICE table) to factor in the stoich relationship)

37. 50.5= (2x)2 (1.000-x)(2.000-x) 50.5 = 4x2 2-3x+x2 50.5 (2-3x+x2)= 4x2
Factor the denominator
50.5 =
4x2
2-3x+x2
Get the denominator out of there multiply both sides
50.5 (2-3x+x2)=
4x2
Multiply 51 though

38. 50.5 (2-3x+x2)=
4x2
Multiply 51 though
Subtract 4x2 from both sides
x+50.5x2=
4x2
x+46.5x2=
Quadratic equation
C - Bx Ax2 =
You have program your calculator OR do it the LONG WAY

39. x+46.5x2=
C - Bx Ax2 =
X= -(-151.5)± √ – 4(46.5)(101)
2(46.5)

40. x+46.5x2=
C - Bx Ax2 =
X= -(-151.5)± √ –18786 93

41. x+46.5x2=
C - Bx Ax2 =
X= -(-151.5)± √ 93

42. x+46.5x2=
C - Bx Ax2 =
X= -(-151.5)+ 64.5
= 2.32 93
X= -(-151.5)- 64.5
= 0.935 93

43. Which is the correct X= -(-151.5)+ 64.5 = 2.32 93 X= -(-151.5)- 64.5
= 0.935 93
Substitute the value for X. If we get a negative concentration that value would NOT be the correct on.

44.  [H2] 1.000- 2.32= -1.32 -x -x + 2x 1.000-x 2.000-x 2x [H2], M
[I2], M
[HI], M
Initially
1.000M
2.000 M
0 M
Change
At equilibrium  -x -x
+ 2x
1.000-x
2.000-x 2x
NOT correct value!!
[H2] = -1.32
From Quadratic 1

45. [H2], M
[I2], M
[HI], M
Initially
1.000M
2.000 M
0 M
Change
At equilibrium 
-0.935
-0.935
2(0.935)
0.065M
1.065 M
1.87 M
Correct value!!
[H2] = 0.065M
From Quadratic 2

46. [H2], M
[I2], M
[HI], M
Initially
1.000M
2.000 M
0 M
Change
At equilibrium 
-0.935
-0.935
2(0.935)
0.065M
1.065 M
1.87 M
Plug the equilibrium concentrations into Kc expression see if its 50.5
Kc= 50.5