1. Equilibrium

2. Reactions are reversible
A + B C + D ( forward)
C + D A + B (reverse)
Initially there is only A and B so only the forward reaction is possible
As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
Eventually the rates are equal

3. Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time

4. What is equal at Equilibrium?
Rates are equal
Concentrations are not.
Rates are determined by concentrations and activation energy.
The concentrations do not change at equilibrium.
or if the reaction is very slow.

5. Law of Mass Action
For any reaction
jA + kB lC + mD
K = [C]l[D]m PRODUCTSpower [A]j[B]k REACTANTSpower
K is called the equilibrium constant.
is how we indicate a reversible reaction

6. [C]l[D]m Playing with K If we write the reaction in reverse.
lC + mD jA + kB
Then the new equilibrium constant is
K’ = [A]j[B]k = 1/K
[C]l[D]m

7. K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n
Playing with K
If we multiply the equation by a constant
njA + nkB nlC + nmD
Then the equilibrium constant is
K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n

8. The units for K
Are determined by the various powers and units of concentrations.
They depend on the reaction.
Are usually dropped, unlike rate constant units which are specifically asked for.

9. K is CONSTANT At any temperature. Temperature affects rate.
The equilibrium concentrations don’t have to be the same, only K.
Equilibrium position is a set of concentrations at equilibrium.
There are an unlimited number.

10. UNIQUE FOR EACH TEMPERATURE
Equilibrium Constant
UNIQUE FOR EACH TEMPERATURE

11. Calculate K N2 + 3H2 2NH3 Initial At Equilibrium
[N2]0 =1.000 M [N2] = 0.921M
[H2]0 =1.000 M [H2] = 0.763M
[NH3]0 =0 M [NH3] = 0.157M

12. Calculate K N2 + 3H2 2NH3 Initial At Equilibrium
[N2]0 = 0 M [N2] = M
[H2]0 = 0 M [H2] = M
[NH3]0 = M [NH3] = 0.157M
K is the same no matter what the amount of starting materials

13. Equilibrium and Pressure
Some reactions are gaseous
PV = nRT
P = (n/V)RT
P = CRT
C is a concentration in moles/Liter
C = P/RT

14. Equilibrium and Pressure
2SO2(g) + O2(g) SO3(g)
Kp = (PSO3) (PSO2)2 (PO2)
K = [SO3] [SO2]2 [O2]

15. Equilibrium and Pressure
K = (PSO3/RT) (PSO2/RT)2(PO2/RT)
K = (PSO3)2 (1/RT) (PSO2)2(PO2) (1/RT)3
K = Kp (1/RT)2 = Kp RT (1/RT)3

16. General Equation jA + kB lC + mD
Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m (PA)j (PB)k (CAxRT)j(CBxRT)k
Kp= (CC)l (CD)mx(RT)l+m (CA)j(CB)kx(RT)j+k
Kp = K (RT)(l+m)-(j+k) = Kc (RT)Dn
Dn=(l+m)-(j+k)=change in moles of gas

17. Homogeneous Equilibria
So far every example dealt with reactants and products where all were in the same phase.
We can use K in terms of either concentration or pressure.
Units depend on reaction.

18. Heterogeneous Equilibria
If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.
As long as they are not used up we can leave them out of the equilibrium expression.

19. But the concentration of I2 does not change.
For Example
H2(g) + I2(s) HI(g)
K = [HI] [H2][I2]
But the concentration of I2 does not change.
K[I2]= [HI]2 = K’ [H2]

20. Solving Equilibrium Problems
Type 1

21. Compare Q value to the equilibrium constant
The Reaction Quotient
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
Q = [Products]coefficient
[Reactants] coefficient
Compare Q value to the equilibrium constant

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

23. Example
For the reaction 2NOCl(g) 2NO(g) + Cl2(g) K = 1.55 x 10-5 M at 35ºC In an experiment 10.0 mol NOCl, 1.0 mol NO(g) and .10 mol Cl2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?

24. Solving Equilibrium Problems
Given the starting concentrations and one equilibrium concentration.
Use stoichiometry to figure out other concentrations and K.
Learn to create a table of initial and final conditions.

25. 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 present. Calculate the equilibrium concentrations of O2 and SO2, K and KP
Initial
Change
Equilibrium

27. LARGE K
K >100

28. What if you’re not given equilibrium concentration?
The size of K will determine what approach to take.
First let’s look at the case of a LARGE value of K ( >100).
Allows us to make simplifying assumptions.

29. Example
H2(g) + I2(g) HI(g)
K = 7.1 x 102 at 25ºC
Calculate the equilibrium concentrations if a L
container initially contains 15.9 g of H2 294 g I2 .
[H2]0 = (15.7g/2.02)/5.00 L = 1.56 M
[I2]0 = (294g/253.8)/5.00L = M
[HI]0 = 0

30. Q= 0<K so more product will be formed.
Assumption since K is large reaction will go to completion.
Stoichiometry tells us I2 is LR, it will be smallest at equilibrium let it be x
Set up table of initial, change in and equilibrium concentrations.

31. H2(g) I2(g) 2HI(g) initial 1.56 M 0.232 M 0 M change eq X
Choose X so it is small.
For I2 the change in X must be X-.232 M
Final must = initial + change

32. H2(g) I2(g) 2HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 eq X
Initial + change = final,
Change = final – initial
Using to stoichiometry we can find:
Change in H2 = = x M
Change in HI = (2(.232-x) = (.464-2x)

33. H2(g). I2(g) HI(g). initial 1. 56 M. 232 M 0 M. change X-0. 232. X-0
H2(g) I2(g) HI(g) initial M M M change X X X eq X
Now we can determine the final concentrations by adding.

34. H2(g). I2(g) HI(g). initial 1. 56 M. 232 M 0 M. change X-0. 232. X-0
H2(g) I2(g) HI(g) initial M M M change X X X
eq X X X
Now plug these values into the equilibrium expression
K = ( X)2 = 7.1 x 102
(1.328+X)(X)
Now you can see why we made sure X was small.

35. Makes the algebra easy Why we chose X K = (0.464-2X)2 = 7.1 x 102
Since X is going to be small, we can ignore it in relation to and 1.328
So we can rewrite the equation
7.1 x 102 = (0.464) (1.328)(X)
Makes the algebra easy

36. H2(g). I2(g) HI(g). initial 1. 56 M. 232 M 0 M. change X-0. 232. X-0
H2(g) I2(g) HI(g) initial M M M change X X X
eq X X X
When we solve for X we get 2.8 x 10-4
So we can find the other concentrations
I2 = 2.8 x 10-4 M
H2 = M
HI = M

37. Checking the assumption
The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid.
If not we would have had to use the quadratic equation.
More on this later………..anyone have it programmed already??????
Our assumption was valid.

38. If the reaction is not at equilibrium, which way will it shift?
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 of all species.

39. Cl2 O2
2 ClO
Initial
Change
equilibrium

40. Problems with small K
K< .01

41. Process is the same
Set up table of initial, change, and final concentrations.
For a small K the product concentration is small.

42. For example
nitrogen + oxygen yields nitrogen monoxide
K= 4.10 x 10-4
If .50 moles of nitrogen and .86 moles of
oxygen are mixed in a 2.0 L container, what
are the equilibrium concentrations of all 3
substance?
Q =

43. K = Because K is small, assume x is negligible compared to .25 and .43
N O 2NO
Initial
Change eq
K = Because K is small, assume x is negligible compared to .25 and .43

44. Simplified K expression: K =
Always check the assumption.
X must be less than 5% of the original concentration.

45. Mid-range K’s
.01<K<100

46. QUADRATIC EQUATION TI-82 PROGRAM
Repeat for the following:
(-B-√(B^2-4*A*C))/(2*A) and hit ENTER
To execute the program:
Hit 2nd and Quit to exit the editing of the
program
Hit PRGM key
Highlight EXEC and use down arrow to highlight
QUAD and hit ENTER
PRGMQUAD should show up at the top of the screen –hit ENTER
ENTER first coefficient after ? (i.e. 1)
ENTER second coefficient after ? (i.e. -5)
ENTER constant term after ? (i.e. 6)
The answers of 3 and 2 DONE should
show up in the lower right corner of the screen.
Hit PRGM key Using right arrow, highlight NEW and hit ENTER Give the program a name (i.e. QUAD) and hit Using right arrow, highlight I/O Highlight Input and hit ENTER Hit ALPHA key and then hit the letter A, then hit ENTER Repeat for B and C Using down arrow, highlight Disp and hit Type in the following formula: (-B+√(B^2-4*A*C))/(2*A) and hit ENTER

47. No Simplification Choose X to be small.
Can’t simplify so we will have to solve the quadratic
H2(g) + I2 (g) HI(g) K=38.6
What are the equilibrium concentrations if mol H2, mol I2 and mol HI are mixed in a L container?

48. H2 I2
2 HI
Initial
Change
equilibrium

49. Problems Involving Pressure
Solved exactly the same, with same rules for choosing X depending on KP
For the reaction
N2O4(g) NO2(g) KP = atm
What are the equilibrium pressures if a flask initially contains atm N2O4?

50. N2O4
2 NO2
Initial
Change
equilibrium

51. Le Chatelier’s Principle
If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.
3 Types of stress
1. Changing amounts of reactants or products
2. Changing pressure
3. Changing temperature

52. 1. Change amounts of reactants and/or products
Adding product makes Q>K
Removing reactant makes Q>K
Adding reactant makes Q<K
Removing product makes Q<K
Determining the effect on Q will tell you the direction of shift

53. 2. Change Pressure By changing volume
System will move in the direction that has the least moles of gas.
Because partial pressures (and concentrations) change a new equilibrium must be reached.
System tries to minimize the moles of gas.

54. Partial pressures of reactants and product are not changed
Change in Pressure
By adding an inert gas
Partial pressures of reactants and product are not changed
No effect on equilibrium position

55. 3. Change in Temperature
Affects the rates of both the forward and reverse reactions.
Doesn’t just change the equilibrium position, changes the equilibrium constant.
The direction of the shift depends on whether it is exo- or endothermic

56. Think of heat as a product
Exothermic
DH<0
Releases heat
Think of heat as a product
Raising temperature pushes toward reactants.
Shifts to left.

57. Think of heat as a reactant
Endothermic
DH>0
requires heat
Think of heat as a reactant
Raising temperature pushes toward products.
Shifts to right.