1. Equilibrium
Chapter 12

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, no significant change will occur.

5. Law of Mass Action For any reaction aA + bB cC + dD
K = [C]c[D]d PRODUCTSpower [A]a[B]b REACTANTSpower
K is called the equilibrium constant.
is how we indicate a reversible reaction

6. calculating K If we write the reaction in reverse. cC + dD aA + bB
Then the new equilibrium constant is
K’ = [A]a[B]b = 1/K [C]c[D]d

7. K’ = [A]na[B]nb = ([A] a[B]b)n = Kn [C]nc[D]nd ([C]c[D]d)n
Calculating with K
If we multiply the equation by a constant
naA + nbB ncC + ndD
Then the equilibrium constant is
K’ = [A]na[B]nb = ([A] a[B]b)n = Kn [C]nc[D]nd ([C]c[D]d)n

8. Adding K
When we have two equations which we are adding to get a resulting equation
We will need to multiply the K from each equation.
K1 X K2 = K3 (sum of equations 1 +2)

9. The units for K
Are determined by the various powers and units of concentrations.
They depend on the reaction.
Will usually cancel out to leave no units.

10. K is a CONSTANT
At each temp, the reaction has a unique K that won’t change for that temp..
Temperature affects rate and K.
The equilibrium concentrations won’t be the same, only K.
Equilibrium position is a set of concentrations at equilibrium.
There are an unlimited number equilibrium positions.

11. Calculate K N2 + 3H2 2NH3 Initially 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] = 0.203M
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 aA + bB cC + dD
Kp= (PC)c (PD)d= (CCxRT)c (CDxRT)d (PA)a (PB)b (CAxRT)a (CBxRT)b
Kp= (CC)c (CD)d x(RT)c+d (CA)a (CB)b x(RT)a+b
Kp = K (RT)(c+d)-(a+b) = K (RT)Dn
Dn=(c+d)-(a+b)=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.

18. Heterogeneous Equilibria
If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.
So, we can leave them out of the equilibrium expression.

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

20. The Reaction Quotient (Q)
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 value to equilibrium constant

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

22. 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?

23. Solving Equilibrium Problems
Given the starting concentrations and one equilibrium concentration.
Use stoichiometry to figure out other concentrations and K.

24. 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 of SO3 were found to be present. Calculate
The equilibrium concentrations of O2 and SO2, K

25. Consider the same reaction at 600ºC
In a different experiment .500 mol SO2 and .350 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

26. What if you’re not given an 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.

27. Example 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 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

28. Q= 0, which is <K so more product will be formed.
Assume that since K is large, then the reaction will go to completion.
Set up table of initial, final and equilibrium in concentrations.

29. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2X equil.
Using to stoichiometry we can find
Change in H2 = 1.56M -X
Change in HI = twice change in H2
Change in HI = 0 +2X

30. H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2X
equil X X X
Now plug these values into the equilibrium expression
K = (2X) = 7.1 x 102 (1.56-X)(0.232-X)

31. When we solve for X we get 0.232
So we can find the other concentrations
I2 = 0 M
H2 = M
HI = M

32. Practice For the reaction Cl2 + O2 2ClO(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.

33. For example For the reaction 2NOCl 2NO +Cl2 K= 1.6 x 10-5
If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl2 are mixed in a 1 L container
What are the equilibrium concentrations
Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 M [NOCl]2 (1.20)2

34. 2NOCl 2NO Cl2
Initial
Change
Equil

35. 2NOCl 2NO Cl2
Initial
Change -2X X +X
Equil X X X

36. Now we can write equilibrium constant
2NOCl 2NO Cl2
Initial
Change -2X X +X
Equil X X X
Now we can write equilibrium constant
K = ( X)2 ( X) (1.20 – 2X)2 = 1.6E-5

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

38. Example H2(g) + I2 (g) 2 HI(g) K=38.6
What is the equilibrium concentrations if mol H2, mol I2 and mol HI are mixed in a L container?

39. Problems Involving Pressure
Solved exactly the same, with same rules for choosing X depending on KP
For the reaction N2O4(g) NO2(g) KP = .131 atm. What are the equilibrium pressures if a flask initially contains atm N2O4?

40. 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

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

42. 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.

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

44. 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

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

46. Endothermic DH>0 Produces heat Think of heat as a reactant
Raising temperature push toward products.
Shifts to right.