Equilibrium
Chemical equilibrium - a dynamic state in which no net change occurs
-indicated in a chemical reaction by a double arrow
Examples of equilibrium processes: vapor above a liquid and saturated solution of NaCl
Chemical reactions are reversible
A + B C + D (forward)
C + D A + B (reverse)
Initially-only A+B present –only forward reaction possible
As C+D build-up- reverse reaction speeds up while the forward reaction slows.
Eventually rate of forward reaction = rate of reverse reaction
Dynamic Equilibrium
What is equal at equilibrium? RATES, NOT
CONCENTRATION
- Concentrations do not change at equilibrium
Law of Mass Action
- Equilibrium expression shows the relationship between reactant concentration [reactants] and product concentration [products] at equilibrium
jA + kB lC + mD
Example: H2(g)+ I2(g) 2HI(g) K= [HI]2
[H2] [I2]
Expressing the Law of Mass Action
- Homogeneous equilibria- substances are all in the same phase
- Heterogeneous equilibria- substances are in different phases
- Concentration of a pure solid or liquid does not change( density is constant) SO they are not included in an equilibrium expression
S(s)+O2(g) SO2(g) K= [SO2]
[O2]
NH3(aq)+H2O(l) NH4+(aq)+OH-(aq) K= [ NH4+][ OH-]
[NH3]
Example
Write the equilibrium expression for K for the following reactions:
(a) 2 O3(g) 3 O2(g) (b) 2NO(g) + Cl2(g) 2 NOCl(g)
Example
Write the equilibrium expression for each:
(a) CO2(g) + H2(g) CO(g) + H2O(l)(b) SnO2(s) + 2CO(g) Sn (s) + 2CO2(g)
Kc (K)= concentration of solution (M) – dealing with aqueous solutions
Kp= Partial pressure of gases – dealing with gases
The values of Kc and Kp are different from each other
- but you can calculate one from the other: Here is the proof!
- PV=nRT P = n RT P = CRT C= P/RT
V
Partial pressure of a gas is directly proportional to its concentration
- when same number of moles of gas appear on both sides Kp = Kc
When coefficients within an equation are not equal Kp = K (RT)Δn- Where Δ n equals sum of the moles of products – sum of the moles of reactants
N2(g) + 3 H2(g) 2 NH3(g)
Example
According to the reaction for the synthesis of ammonia, K = 9.60 at 300C, calculate the Kp for this reaction at this temperature.
N2(g) + 3H2(g) 2NH3(g)
Playing with K
Using different coefficients in equation- changed by a factor “n”
C(s)+ ½ O2(g) CO(g) 2C(s) + O2(g) 2CO(g)
K1= [CO] = 4.6 x 1023 K2= [CO] =2.1 x 1047
[O2]1/2 [O2]
SO!
K2 = [CO]2 = ( [CO] )2 =K1
[O2] ( [O2]1/2)
Knew= (Kold)n-multiplication factor
Reversing the equation
HCO2H(aq) + H2O(l) HCO2-(aq) + H3O+(aq)
Forward Reverse
Kf= [HCO2-] [H3O+]_ = 1.8 X 10-4 Kr= [HCO2H] = 5.6 X 10+3
[HCO2H] [HCO2-] [H3O+]
Kf =1/Kr knew = 1/Kold
Example
For the formation of NH3 from N2 and H2, Kp = 4.34 x 10-3 at 300C. What is the value of Kp in the reverse direction?
Adding individuals Reactions to attain a net reaction
Multiply all k values together.
knet = k1 x k2 x k3 …
AgCl(s) Ag+(aq) + Cl-(aq) k1 = 1.8 x 10-10=[Ag+] [Cl-]
Ag+(aq)+ 2NH3(aq) Ag (NH3)21+(aq) k2 =1.6 x 107 =[Ag (NH3)2+1]
[Ag+] [NH3+] 2
AgCl(s) +2NH3(aq) Ag (NH3)2+(aq) +Cl-(aq)
[Ag+] [Cl-] times ([Ag (NH3)2+])
([Ag+] [NH3+]2)
[Ag (NH3)2+] [Cl -] *** this is the equilibrium expression of the overall reaction
[NH3]2
k1 x k2 =2.9 x 10-3
Example
Given the following information,
HF(aq) H+(aq) + F-(aq) Keq = 6.8 x 10 –4
H2C2O4(aq) 2H+(aq) + C2O4-2(aq)Keq = 3.8 x 10 –6
Determine the value of the equilibrium constant for the following reaction:
2 HF(aq) + C2O4-2(aq) 2 F-(aq) + H2C2O4(aq)
Units of k – dependent on powers and units of concentration; (it variesso no units of k are present.)
K (equilibrium constant)
K is constant for each set of conditions for a given reaction in equilibrium.
At any temperature
Equilibrium positions sets of equilibrium concentrations (unlimited numbers); with ONLY one equilibrium constant at constant temperature.
For example
N2 + 3H2 2NH3
I. [Initial][Equilibrium] K
N2 1.00 M.9216.02 x 10-2
H2 1.00 M.763
NH3 0 M.157
II
N2 0 M.3996.02 x 10-2
H2 0 M1.197
NH3 1.00 M.203
III.
N2 2.00 M2.596.02 x 10-2
H2 1.00 M2.77
NH3 3.00 M1.82
From K we can predict :
the tendency of a reaction to occur (but not speed)
whether or not the given concentrations represent an equilibrium condition
the equilibrium position from a given set of initial concentrations
The magnitude of K indicates whether reaction is product or reactant favored( how far the reaction proceeds to form products)
- K > 1 Reaction is product favored ;[equilibrium] of products is larger than the reactants; equilibrium lies to the right
- K < 1 (MUCH LESS THAN ONE) Reaction is reactant favored ;[equilibrium] of reactants is larger than the products; equilibrium lies to the left
Example
The equilibrium constant for the reaction H2(g) + I2(g) 2HI(g) varies with
temperature in the following ways: K = 54 at 700 K and K = 794 at 298 K. Is the
formation of HI favored more at higher or lower temperatures?
Calculating an Equilibrium Constant
- Substitute equilibrium concentrations into the equilibrium expression
Example
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472C. The equilibrium mixture of gases was analyzed and found to contain .1207 M H2 and .0402 M N2 and .00272 M NH3. From these data, calculate the equilibrium constant K for N2(g) + 3 H2(g) 2NH3(g)
Reaction Quotient
- Tells you if the reaction is at equilibrium or not; ( if not: will show the direction the reaction will shift to in order to reach equilibrium
- Calculated the same way as the equilibrium constant using equilibrium concentrations EXCEPT you use initial concentrations
Q = [products]coefficient
[reactants]coefficient
- Compare Q to the equilibrium constant (K)
If Q less than Kfewer products than at equilibrium; reaction shifts to the right
If Q greater than Kmore products than at equilibrium; reaction shifts to the left
Q = Kthe reaction is at equilibrium
Example
At 448C the equilibrium constant Keq is 51 for the following reaction
H2 (g)+ I2(g) 2HI (g)
Predict how the reaction will proceed to reach equilibrium at 448C if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mol H2, and 3.0 x 10-2 mol I2 in a 2.00 L container.
Solving Equilibrium Problems
- If we know the equilibrium concentrations of at least one species, we can use the stoichiometry of the reaction to figure out the others
- Use ICE tables as the problem solving approach
- Must know which direction the reaction will go: it will tell you or you will need to use the reaction quotient Q
- Tabulate the initial and equilibrium concentrations of all species in the expression.
- Calculate the change in concentration that occurs as the system reaches equilibrium of all species that have an initial and equilibrium concentration known
- Use the coefficients of the balanced chemical equation to calculate the changes in concentration for all other species
- From the initial concentrations, calculate the equilibrium concentrations.
Procedure:
write balanced chemical equation
write the equilibrium expression using the law of mass action
list initial concentrations
calculate Q and determine which way equilibrium lies
ICE TABLE: define change needed to reach equilibrium and define the equilibrium concentrations
substitute in the equilibrium concentrations into the equilibrium expression and solve for the unknown
check your answers by using them to solve for K to see if it agrees with the accepted value of K (5% rule - if an answer is within 5% it is considered valid)
Example
Enough ammonia is dissolved in 5.00 L of water at 25 C to produce a solution that is .0124 M in ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows the concentration of OH- is 4.64 x 10-4 M. Calculate Keq at 25C for the reaction NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
1.Set up ICE table
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Initial.0124 M 0 0
Change
Equilibrium 4.64 x 10-4 M
2.Calculate the change in hydroxide
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Initial.0124 M 0 0
Change+4.64 x 10-4 M
Equilibrium 4.64 x 10-4 M
3.Use stoichiometry to calculate changes of other species
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Initial.0124 M 0 0
Change-4.64 x 10-4 M +4.64 x 10-4 M +4.64 x 10-4 M
Equilibrium 4.64 x 10-4 M
4. Finish table- calculate the equilibrium concentrations and solve for Keq
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Initial.0124 M 0 0
Change-4.64 x 10-4 M +4.64 x 10-4 M +4.64 x 10-4 M
Equilibrium .0119 M 4.64 x 10-4 M 4.64 x 10-4 M
K = [NH4+] [OH-] = (4.64 x 10-4)2 = 1.81 x 10-5
[NH3] .0119
Equilibrium problems range from simple to complex. Read the problems carefully to identify the given information.
- Simple: given the equilibrium constant and the initial and/or equilibrium concentrations and solve for an unknown equilibrium concentration
- Complex- use the ice tables seen above, and stoichiometry. Not given any equilibrium concentrations, only initial
- Extremely complex- use quadratic formula- Typical AP Problem
Example SIMPLE TYPE
For the Haber process, N2(g) + 3H2(g) 2NH3(g) Keq = 1.45 x 10-5 at 500 C.
in an equilibrium mixture of the three gases at 500 C. the partial pressure of H2
is .928 atm and that of N2 is .432 atm. What is the partial pressure of NH3 in
this equilibrium mixture?
Example : EXTREMELY COMPLEX
A 1.000 L flask is filled with 1.000 mol of hydrogen and 2.000 moles of iodine gas at
448C. The value of the equilibrium constant is 50.5. What are the partial pressures
of hydrogen, iodine, and hydrogen iodide at equilibrium?
Systems that have small equilibrium constants
Simplications can be used to solve complicated calculations
- If K is extremely small, the reaction will not proceed far to the right to reach equilibrium
- X= extremely small number- it becomes inconsequential
Example:
K = 1.6 x 10-5 at 35 C . 1.0 mol NOCl is placed in a 2.0 liter flask, what are the equilibrium concentrations? 2 NOCl 2 NO + Cl2
2 NOCl 2 NO + Cl2
Initial .50 M 0 0
Change-2x +2x +x
Equilibrium .50 – 2x 2x x
So k = (2x2)(x) too complicated
(.50 – 2x)2
make an assumption: .50 – 2x is changed to .50
So k = (2x2)(x) 4x3 = 1 x 10-2 so .50 – 2X = .48
(.50)2 .25
After completion of this simplification, I need you to make sure that the change does not increase larger than 5%. If it does, then you will have to do the problem the hard way, quadratic all the way baby!
5% rule amount of change over initial concentration i.e. from the example above
1 x 10-2 / .50 x 100 = 2% so is o.k.
Le Chatelier’s Principle
- If a change is imposed on a system in equilibrium, the position of the equilibrium will shift to reduce the change
- 3 factors control equilibrium:
1. Change in concentration: system to shift away (toward) from the added reactant or product.
Add more reactants shifts right
Add more products shifts left
Complete removal of a substance will have the opposite effect
2. Change in pressure
To change pressure :
Remove or add a gas reactant or product- see above
Add an inert gas – no effect
e.g. if you have 2.0 moles of a reactant gas in a one liter vessel, the concentration of the reactant will be 2.0 M. Adding helium to the one liter vessel will not change the concentration of the reactant gas - there is still 2.0 moles in a one liter vessel
Change volume of container –volume is proportional to moles – so as decrease in volume is a decrease in moles
Decrease volume: The new equilibrium position will shift towards the side that
has the smallest number of gas molecules
Increase volume largest # of gas molecules
3. Change in temperature ( also changes K and equilibrium positions)
Treat temperature as energy
- Exothermic reaction N2 + H2 2 NH3 + 92 kJ
Add Energy & the equilibrium shifts left to increase temperature
- Endothermic reaction 556 kJ + CaCO3 CaO + CO2
Add energy, equilibrium shifts right to increase temperature
Increase temperature: If reaction is exo, shifts left and if endo, shifts right
Decrease temperature: If reaction is exo, shifts right and if endo, shifts left
Examples 58 kJ + N2O4 2 NO2
Addition of N2O4
Addition of NO2
Removal of N2O4
Removal of NO2
Addition of He
Decrease contaniner volume
Increase container volume
Increase temperature
Decrease temperature
See additional page for more example