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!

  1. 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 300C, 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 300C. 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 472C. 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 448C 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 448C 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
  1. Tabulate the initial and equilibrium concentrations of all species in the expression.
  2. Calculate the change in concentration that occurs as the system reaches equilibrium of all species that have an initial and equilibrium concentration known
  3. Use the coefficients of the balanced chemical equation to calculate the changes in concentration for all other species
  4. 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 25C 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

448C. 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