1. “ICE” CALCULATIONS

2. EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature, catalysts, and changes in concentration affect equilibria.
The outcome is governed by LE CHATELIER’S PRINCIPLE
“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

3. Le Chatelier’s Principle
Change T
change in K
therefore change in P or concentrations at equilibrium
Use a catalyst: reaction comes more quickly to equilibrium. K not changed.
Add or take away reactant or product:
K does not change
Reaction adjusts to new equilibrium “position”

4. NH3 Production o N2(g) + 3 H2(g) g 2 NH3(g) + heat
K = 3.5 x 108 at 298 K

5. EQUILIBRIUM AND EXTERNAL EFFECTS
Concentration changes
no change in K
only the position of equilibrium changes. (concentration changes to maintain ratio = “shift” )

6. EQUILIBRIUM AND EXTERNAL EFFECTS
Add catalyst ---> no change in K
A catalyst only affects the RATE of approach to equilibrium.
Catalytic exhaust system

7. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Increase P in the system by reducing the volume (at constant T).

8. Increase P in the system by reducing the volume.
N2O4(g) NO2(g) o
Increase P in the system by reducing the volume.
In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P).
Therefore, reaction shifts LEFT and P of NO2 decreases and P of N2O4 increases.

9. Temperature Effects on Equilibrium
Kc (273 K) =
Kc (298 K) =
N2O4 (colorless) + heat 2 NO2(brown) ∆Ho = kJ (endo) o

10. EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature change ---> change in K
Consider the fizz in a soft drink H2CO3(aq) + HEAT CO2(g) + H2O(liq)
K = P (CO2) / [CO2]
Increase T. What happens to equilibrium position? To value of K?
K increases as T goes up because P(CO2) increases and [CO2] decreases.
Decrease T. Now what?
Equilibrium shifts left and K decreases. o

12. “ICE” CALCULATIONS

13. THE EQUILIBRIUM CONSTANT
a A + b B c C + d D
(at a given T) o
Pure substances are not included in equation.

14. K = K = H+ (aq)+ OH- (aq) e H2O(l) CaO (s) + CO2 (g) e CaCO3 (s)
WRITE THE EQUILIBRIUM CONSTANT EXPRESSION
H+ (aq)+ OH- (aq) e H2O(l)
K =
CaO (s) + CO2 (g) e CaCO3 (s)
K =

15. Butane-Isobutane Equilibrium

16. butane
Butane Isobutane
isobutane
At equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. K = 2.5.
Add 1.50 M butane.
When the system comes to equilibrium again, what are [iso] and [butane]?

17. Butane Isobutane o Solution
Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5
Solution
Calculate Q immediately after adding more butane and compare with K.
Q is LESS THAN K. Therefore, the reaction will shift to the ____________.

18. Butane Isobutane o Solution
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
Solution
Q is less than K, so equilibrium shifts right — away from butane and toward isobutane.
Set up ICE table
[butane] [isobutane]
Initial
Change
Equilibrium
1.25
- x
+ x x x

19. Butane Isobutane Equilibrium has shifted toward isobutane.
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
Solution
x = 1.07 M
At the new equilibrium position,
[butane] = 0.93 M and [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.

20. - - + Q = 0/12 = 0 Typical Calculations X X e 2 X
PROBLEM:
Place 1.00 mol each of H2 and I2 in a 1.0 L flask. Calc. equilibrium concentrations.
H2(g) + I2(g) e 2 HI(g) - - +
X X e X
Q = 0/12 = 0

21. H2(g) + I2(g) g 2 HI(g) Kc = 55.3 Step 1. Set up ICE table
[H2] [I2] [HI]
Initial
Change
Equilib

22. Step 2. define “X”
[H2] [I2] [HI]
Initial
Change -x -x +2x
Equilib 1.00-x 1.00-x 2x
where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

23. H2(g) + I2(g) e 2 HI(g) Kc = 55.3
Step 3. Put equilibrium concentrations into Kc expression.

24. H2(g) + I2(g) ---> 2 HI(g) Kc = 55.3
Step 4. Solve Kc expression - take square root of both sides.
x = 0.79
Therefore, at equilibrium
[H2] = [I2] = x = M
[HI] = 2x = M

25. Pg 132 CPSC

26. Nitrogen Dioxide Equilibrium N2O4(g) ---> 2 NO2(g)
X X e

27. Nitrogen Dioxide Equilibrium N2O4(g) ---> 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Step 1. Set up an ICE table
[N2O4] [NO2]
Initial
Change
Equilib

28. Nitrogen Dioxide Equilibrium N2O4(g) ---> 2 NO2(g)
Step 2. Define “X”
[N2O4] [NO2]
Initial
Change -x +2x
Equilib x 2x

29. APPROXIMATE: K is small: x will be small
Step 2. Define “X”
[N2O4] [NO2]
Initial
Change -x +2x
Equilib x

30. X = 0.027 Rearrange: 0.0059 (0.50) = 4x2 0.0029 = 4x2 0.0029/4 = x2
Step 3. Substitute into Kc expression and solve.
Rearrange: (0.50) = 4x2
= 4x2
0.0029/4 = x2
X = 0.027
Step 4. check if assumption is correct
0.50 – = same?

31. This is a QUADRATIC EQUATION ax2 + bx + c = 0
Step 3. Substitute into Kc expression and solve.
Rearrange: ( x) = 4x2
x = 4x2
4x x = 0
This is a QUADRATIC EQUATION
ax bx c = 0
a = 4 b = c =

32. Solve the quadratic equation for x.
ax bx c = 0
a = 4 b = c =
x = ± 1/8(0.046)1/2
= ± 0.027

33. Nitrogen Dioxide Equilibrium N2O4(g) ---> 2 NO2(g)
x = or
negative value is not reasonable.
Conclusion: x = M
[N2O4] = x = M
[NO2] = 2x = M

34. Solving Quadratic Equations
Recommend you solve the equation exactly on a calculator