Lecture Notes 112      Weeks 3 and 4   Equilibrium   Chapter 14    Chemistry by Chang  

January 25- 31, 2005

 

What is equilibrium?  Equilibrium is when there is a balance of changes involving particles so that the change in one direction is equal to the change in the opposite direction.

            Example:   Physical Equilibrium such as the evaporation of water

                                    H2O(l) ó H2O(g)

 

                             Chemical Equilibrium such as this reaction

                                    N2O4 (g) ó 2 NO2 (g)

                                   

 

                       

 

 

 

 

 

What is equal at equilibrium for a system?  

       The rate of the forward reaction is equal to the rate of the reverse reaction

 

What is constant at equilibrium?

      When a system reaches equilibrium, the concentrations of the species

      stay the same value since, as they are formed in the forward reaction, they

      are decomposed in the reverse reaction and vice versa.

 

 

 

 

 

 

 

 

 

 

 

 


When people studied systems that had come to equilibrium, they found that there was a relationship that is quite useful.  It is called the Law of Mass Action.  The ratio of the concentrations of the products raised to a power equal to the coefficients in the balanced equations divided by the concentrations of the reactants raised to a power equal to the coefficients in the balanced equation is a constant for a given reaction at a given temperature.  The ration is called the equilibrium constant, K.  It is a lot easier to illustrate than to put into words:

Consider a system in which this reaction is at equilibrium: 

 aA  +  bB  <=>   cC  +  dD    

The equilibrium constant expression is:

                                                [C]c[D]d

            K=       [A]a[B]b

           

K is the equilibrium constant.  What does K mean?  

If K is a large number?  Equilibrium will lie to the right, i.e. at equilibrium

there will be high concentrations of products compared to the concentrations of the reactants.

                                       K> 1         LOT OF PRODUCT: NOT MUCH REACTANT

 

 If K is a small number?  Equilibrium will lie to the left, i.e. at equilibrium

there will be high concentrations of reactants compared to

the concentrations of the products.

                                       K< 1         LOT OF REACTANTS: NOT MUCH PRODUCT

 

Write equilibrium constant expressions for each of the following reactions:

            a.  2 NOCl(g)  <=>  2 NO(g)  +  Cl2(g)

 

 

 

            b.    2  H2O(l) <=>  2 H2(g)   +    O2(g)

 

 

 

            c.    AgCl(s) <=>  Ag+(aq)   +   Cl –(aq)

 

 

 

EQUILIBRIUM CONSTANTS ARE CONSTANT FOR A REACTION AT A GIVEN TEMPERATURE.  IF YOU CHANGE THE TEMPERATURE, THE VALUE OF K WILL CHANGE.

 

Equilibrium constants can be expressed in either Kc  or   Kp

             For  Kc values:

The  concentrations are in molarities for solutions and gases.

             

For  Kp values, the concentrations are in partial pressures for gases.

           

The relationship between them for a reaction in which either could be used:

                        K p = K c (RT)Dn    (where Dn is the change in the # of moles of gas= #moles of gas in products - #moles of gas in reactants)

 

 

 

Quiz yourself:    Chemistry 112

Calculating K, given equilibrium concentrations

 

1.  Example:  Given            2 SO2(g)  +  O2(g)  <=>  2  SO3(g)

A mixture of these three gases is present in a flask at 1000 KoC .  The system is known to be at equilibrium.  The concentrations of the gases have been found to be:

[SO2] = 3.77 x 10-3 M

[O2] = 4.30 x 10-3 M

[SO3] = 4.13 x 10-3 M          

            1.  Write the equilibrium constant expression:

 

2.      Determine the equilibrium constant, Kc.

 

 

3.  What is the value for Kp for this reaction at this same temperature?

 

 

2.  Example:  The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M.  Calculate the equilibrium constants Kc and.

1.      Write and balance the equation for the reaction.

CO(g)  +  Cl2(g) ó COCl2(g)

 

2.      Write the equilibrium constant expression:

 

 

3.      Given the concentrations of all the gases at equilibrium, calculate the value for Kc for this reaction at this temperature.

 

 

4.  Calculate the value for Kp for this reaction at this termperature.

 

 

Example:    For the reaction:     2NO2 (g)  ó    2NO (g) + O2 (g)

The equilibrium constant Kp for the reaction is 158 at 1000K.  What is the equilibrium pressure of O2 if at equilibrium, the PNO  = 0.400 atm and

PNO = 0.270 atm?

1.      Write the equilibrium constant expression.

 

2.      Determine the PO2 at equilibrium.

 

What happens to the Equilibrium Constant K, as the chemical equation is manipulated?

A.  If you reverse the reaction, what happens to K? ______________________

            HCO2H(aq) + H2O(g) < =>  HCO2- (aq) + H3O+ (aq)        K = 1.8 x 10 - 8

            HCO2- (aq) + H3O+ (aq) < = > HCO2H(aq) + H2O(g)       K = ___________

 

 

B.  If you add equations, what happens to the K?______________________

            AgCl(s) < => Ag+ (aq) + Cl - (aq)                                        K1 =  1.8 x 10-10

     Ag+ (aq) + 2 NH3(aq)  <=> Ag(NH3)2+ (aq)                              K2 =  1.6 x 107

     AgCl(s) + 2 NH3(aq)   < = > Ag(NH3)2+  (aq) + Cl- (aq)           K3 = ___________

 

 

C.  If you multiply a number times an equation, what happens to K?___________

                                    C(s) + 1/2 O2(g)   <=> CO(g)                       K = 4.6 x 1023

                                 2 C(s) + O2(g)    < = > 2 CO(g)                       K = ____________

 

Again, What does the value of K tell you about a reaction?

            If K is large,_________________________________________________

            If K is small, _________________________________________________

 

In Chem 111 we talk of a reaction “going to completion”.  The information here says that probably no reaction goes absolutely to complete formation of products. 

 

Quiz Yourself    Chemistry 112

                                                                       

1.   Write equilibrium constant expressions for the following reactions.

    (a)   2 SO3(g)  <=>  2 SO2(g) +  O2(g)

 

    (b)   FeO(s) + CO(g) <=> Fe(s) + CO2(g)

 

2. Given:            H2O(g) + CO(g) <=> H2(g) + CO2(g)         K = 1.6

            What is the equilibrium constant for the following reaction: 

            4 H2(g)  +  4 CO2(g)  <=>  4 H2O(g)  +  4CO(g)    Kc = _________

 

3.  Given the reaction:   2 NO(g) <=>  N2O4(g)        K=  170

            A chemist places 0.015 moles of NO gas and 0.25 moles of N2O4 gas in a                             reaction 2.0 liter  container. 

            a.  Is the reaction mixture at equilibrium?______  Explain

 

            b.  If the reaction is not at equilibrium, predict the direction that the

                reaction will go to reach equilibrium.  Circle the direction the reaction

     will move to reach equilibrium.

                                    Forward to make more products?              Or

                                    Backward(reverse) to make more reactants?

 

4  Calculate K for the reaction

SnO2(s) + 2 CO(g) -----> Sn(s) + 2 CO2(g) K = ___________

            given the following information:

SnO2(s) + 2 H2(g)  ---->  Sn(s) + 2 H2O(g)              Kc = 8.12

H2(g) + CO2(g) ---> CO(g) + H2O(g)                                    Kc = 0.771

 

 

5.  Thought question:

What must be true about the equilibrium constant, K, for a reaction to be considered as “going to completion”?  Would it have to be very large, very small, about 1...???

 

 

 

The Reaction Quotient, Q, what is it and how can it be used?

  

Example:   We are given:                 2 NO(g) <=>  N2O4(g)        Kc=  170

a,   Write the equilibrium constant expression:

 

b.  Suppose 0.015 moles of NO gas and 0.025 moles of N2O4 gas are present in

a 1.00 liter container.  Is the system at equilibrium?  Set up the expression and plug in the values of the concentrations, raise them to the appropriate powers.  The number you get is called the “Reaction Quotient” Q.  Q is calculated with any concentrations that you may have.  You do not know whether it is at equilibrium.  You can compare the value of Q to the equilibrium constant K and tell whether the system is at equilibrium or not.  If the Q does not equal K, the system is not at equilibrium and you can predict in which direction will it move (either forward to make more product or backward to use product to make reactant) to reach equilibrium.

  a.  What are the concentrations of NO and N2O4?

  b.  What is the value of Q, the reaction quotient?

  c.  How does Q compare to K?

            1.  If Q < K , the numerator is “too small, too little product, so the forward                                              reaction will occur, some reactants will be converted to

                                 product until      equilibrium is reached.                                

            2.  If Q = K, the reaction is at equilibrium.

3.      If Q > K,  the numerator is “too big”, too much product, so the reverse

                        reaction will occur, i.e., some product will be converted to

reactant until equilibrium is reached. 

 

 

 

 

 

 

 

Example 2:  Variation on that theme:  Given partial Information about the materials present:

                            2 SO2(g)  +  O2(g) <=>  2 SO3(g)

A student places 1.00 mole of SO2 and 1.00 mole O2 in a 1.00 Liter vessel.  The system is allowed to come to equilibrium.  At equilibrium the number of moles of SO3 gas present is found to be  0.925.  Calculate the Kc for this reaction at this temp.     

            1.  Write the equilibrium constant expression:

            2.  Make an ICE Table

                                                2 SO2(g)   +  O2(g)  <=>   2 SO3(g)

Initial concentrations            _________  _____      __________

Change in concentrations   _________   _____     __________

Equilibrium concentrations_________   ______    __________

 

            3.  Do you know any of the “unknowns” from information given?

4.      Substitute the equilibrium concentrations into the expression and solve

 for K.

 

Another example:  Consider the reaction:  2 SO3(g)  <=>  2 SO2(g) + O2(g)

3.00 moles of pure SO3 are placed in an 8.00-L flask at 1150 K. At equilibrium, 0.58 mole of O2 has formed. Calculate Kc for the reaction at 1150 K. 

1.  WRITE THE EQUILIBRIUM CONSTANT EXPRESSION FIRST.

 

2.  SET-UP AN ICE TABLE FOR YOUR WORK.

                                                            2 SO3(g)    <=>  2 SO2(g)  +   O2(g)

 

Initial concentrations                        _________     ______         ____________

 

Change in concentrations               _________     ______         ____________

 

Equilibrium concentrations            _________     ______         ____________

 

 

 

 

*****

Expanding our capabilities to more complex problems:

Let us use the same ICE system to set up and to work out the following situations:

 

Example 1:     A chemist placed 2.00 moles each of H2 gas and I2 gas into a 1.00 Liter flask at 1000 K.  The following reaction occurs and the system comes to equilibrium.  What is the concentration of each species when the system comes to equilibrium?

 

                                    H2(g)  +  I2(g)  <=>  2 HI(g)      Kc = 33

 

1.  Write the equilibrium constant expression..

2.  Determine the Initial concentrations of each species.

3.  Since you have some of both reactants and products, you must decide which

             way the reaction will proceed to reach equilibrium.  You can do this by

  calculating the Q for the mixture. 

                        If Q < K , the numerator is “too small, too little product, so the

Forward reaction will occur, some reactants will be

converted to product until equilibrium is reached.

If Q = K, the reaction is at equilibrium.

If Q > K,  the numerator is “too big”, too much product, so the

Reverse reaction will occur, i.e., some product will be

converted to reactant until equilibrium is reached. 

 

4.  State whether each concentration will decrease(-)  or increase(+) as the

             reaction occurs.

 

5.Express the Change expected in the concentrations by using the        

 stoichiometry.

                            H2(g)  +  I2(g)  <=>  2 HI(g) 

 

WRITE THE EQUILIBRIUM CONSTANT EXPRESSION FIRST.

 

SET-UP AN ICE TABLE FOR YOUR WORK.

                                                                H2(g)      +     I2(g)       <=>  2 HI(g) 

Initial concentrations                        _________     ______         ____________

 

Change in concentrations               _________     ______         ____________

 

Equilibrium concentrations            _________     ______         ____________

 

 

When you have only one unknown value, use any mathematical technique you can to solve for that unknown!!  Some of those techniques are:

            1.  Use techniques to simplify the math:

a.      Can you take the square root of both sides to give an equation

that is quite simple to solve?

b.      Can you make an estimation that is reasonable and that will

allow for a quick determination of an approximate value for the

unknown.

2.       Use the most efficient method available to solve the algebra.  You may

use the quadratic equation for mathematical equations with no

more that a squared term.  However, it is slow and tedious.  You

need to be able to use your calculators to solve these     equations.

 

 

 

Example 2: 

At 450 °C, 3.60 mol of ammonia is placed in a 2.00-L vessel and allowed to decompose to the elements.  If the experimental value of Kc is 6.3 for this reaction at this temperature, calculate the equilibrium concentration of each reagent.                                2 NH3(g)<=> N2(g) + 3 H2(g)            Kc = 6.3

 

WRITE THE EQUILIBRIUM CONSTANT EXPRESSION FIRST.

 

SET-UP AN ICE TABLE FOR YOUR WORK.

                                                                2 NH3(g) <=>         N2(g)      +    3 H2(g)

Initial concentrations                        _________               ______         _________

 

Change in concentrations               _________                ______         _________

 

Equilibrium concentrations            _________                ______         _________

 

 

SUBSTITUTE THE INFORMATION INTO THE EQUILIBRIUM EXPRESSION AND SOLVE FOR WHAT IS UNKNOWN.

 

Example 3:  At 12800C the equilibrium constant (Kc) for the reaction

                                    Br2(g <+> 2 Br(g)

is 1.1 x 10-3.  If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

1.      Write the equilibrium constant expression.

2.      Set up an ICE table.

3.      Solve for “x”.

4.      Determine the concentrations of each species at equilibrium.

 

 

DISTRUBING EQUILIBRIUM:     Le Chatelier’s Principle

What changes to a system can affect a system at equilibrium?  How can changes affect the system?  What will the changes do to the value of the equilibrium constant, K?

If an external stress is applied to a system at equilibrium, the system

adjusts in such a way that the stress is partially offset as the system

reaches a new equilibrium position = Le Chatelier’s Principle

 

Changes in concentrations, volume, pressure and temperature can SHIFT THE EQUILIBRIUM TO THE RIGHT OR TO THE LEFT.  ONLY CHANGES IN TEMPERATURE WILL CHANGE THE VALUE OF K.

 

 

 

A.  List some changes(stresses) that can affect a system at equilibrium:

1. Increase concentration of one of the reactants: Reaction “shifts to the right”.

    Value of K stays same.

                                    2 NH3(g)<=> N2(g) + 3 H2(g)

 

2. Increase concentration of one of the products:  Reaction “shifts to the left”

    Value of K stays same.

                                    2 NH3(g)<=> N2(g) + 3 H2(g)

 

3.Decrease Volume of the container:  

   Reaction shifts to the side with the fewer  moles of gas.   Value of K stays   

   same

                                    2 NH3(g)<=> N2(g) + 3 H2(g)

 

4.  Increase Volume  of the container :

Reaction shifts to the side with more moles of gas.  Value of K stays

same.

                                    2 NH3(g)<=> N2(g) + 3 H2(g)

 

5.  Raise Temperature: Reaction shifts and value of K changes

Raising the temperature always favors the endothermic direction.

Example:

For an exothermic reaction:  2 NH3(g)<=> N2(g) + 3 H2(g)   DH= -111.3 kJ

            Consider it as:     2 NH3(g)<=> N2(g) + 3 H2(g) + heat

Raising the temperature favors the reverse direction since it is

absorbing heat(endothermic) and causes the reaction to shift left so

that, at equilibrium, there is a net gain in reactants. Therefore, the K value goes down since reactants are the denominator.

 

            For an endothermic reaction: N2O3(g) <=> NO(g) + NO2(g)  DHo =+40.5 kJ

                        Consider it as:    N2O3(g)  + 40.5 kJ<=> NO(g)  +  NO2(g)    

                        Raising the temperature favors the forward direction since it is

absorbing heat(endothermic) and causes the reaction to shift to the

right so that, at equilibrium, there is a net gain in products.

Therefore, the K value goes up since products are numerator.

 

6. Lower Temperature: Reaction shifts and the value of K changes.

Lowering the temperature always favors the exothermic direction.

For an exothermic reaction:  2 NH3(g)<=> N2(g) + 3 H2(g)   DH= -111.3 kJ

            Consider it as:     2 NH3(g)<=> N2(g) + 3 H2(g) + heat

Lowering the temperature favors the forward direction since it is

releasing heat(exothermic) and causes the reaction to shift right so

that, at equilibrium, there is a net gain in products. Therefore, the K value goes up since products are the numerator.

 

           

For an endothermic reaction: N2O3(g) <=> NO(g) + NO2(g)  DHo =+40.5 kJ

                        Consider it as:    N2O3(g)  + 40.5 kJ<=> NO(g)  +  NO2(g)    

                        Lowering the temperature favors the reverse direction since it is

releasing heat(exothermic) and causes the reaction to shift to the

left so that, at equilibrium, there is a net gain in reactants.

Therefore, the K value goes down since reactants are in the

denominator.

 

 

 

 

7.  Raise the pressure by adding a gas that is not part of the reaction?

Does not shift the equilibrium.  Value of K stays the same.

Adding a gas that is not one of the reactants or one of the products does NOT change the concentration of either a reactant nor a product,

therefore, neither the rate of the forward not the rate of the reverse reaction is disturbed.

 

8.  Adding a catalyst

         Does not shift the equilibrium.  Value of K stays the same.

 

 

 

Quiz Yourself:

NOTE:  The numerical value of K changes only if temperature changes.

           

                        2 NH3(g)<=> N2(g) + 3 H2(g)            DH= -111.3 kJ           Kc = .16

 

Disturbance                What change                        Effect on system        Effect on

occurs as system                  (shift?)                        value of K    returns to equilibrium?

1.  Increase conc of NH3(g)____________           _____________       ________

 

2. Increase conc of H2(g)______________          _____________       ________

 

3.  Decrease Vol. of container__________          _____________       ________

     (Increase pressure of gases)

     

4.  Increase Vol. of container __________            _____________       ________

      (Decrease pressure of gases)

    

5.  Raise Temperature _______________           _____________       ________

 

6.  Lower Temperature _______________          _____________       ________

 

7.  Add an inert gas      ________________        _____________       ________

 

Example 2:                             H2(g)  +  I2(g)  <=>   2 HI(g)

 

Disturbance                           What change             Effect on system        Effect on

                                                occurs as system        (shift)             value of K

                                                returns to equilibrium?

 

1.  Add H2 gas

 

2.  Add HI gas

 

3.  Remove I2 gas

 

4. Reduce volume of container

 

 

 

 

 

 

 

 

 

Example 3:                             2 NO2(g)  <=>  N2O4(g)              DHo =   - 57.2 kJ

 

Disturbance                           What change             Effect on system        Effect on

                                                occurs as system        (shift)             value of K

                                                returns to equilibrium?

1.  Add NO2 gas

 

2.  Remove N2O4 gas

 

3.  Reduce the volume of container

 

4.  cool the system

 

 

Example 4:                             BaCO3(s) <=>  Ba2+(aq)   +  CO32-(aq)

           

Disturbance                           What change             Effect on system        Effect on

                                                occurs as system        (shift)             value of K

                                                returns to equilibrium?

1.  Add  Ba2+ ions

 

2.  Add CO32- ions

 

3.  Add BaCO3 solid

 

 

Example 5:                 N2O3(g)  <=> NO(g)  +  NO2(g)        DHo = + 40.5 kJ

 

Disturbance                           What change             Effect on system        Effect on

                                                occurs as system        (shift)             value of K

                                                returns to equilibrium?

1.Add N2O3 gas

2.Remove NO2 gas

3.Increase the volume of container

4.cool the system

5.Add NO gas

 

 

 

 

 

 

 

 

 

 

Quiz Yourself    Chemistry 112                                                       

 

1.   Write the equilibrium constant expression for each of these reactions:

            a.         SiH4(g)  + 2 O2(g)  <=>  SiO2(g)  +  2 H2O(g)

 

 

            b.         H2O(l)  <=>  H+(aq)  +  OH-(aq)

 

 

            c.         FeO(s)  +  CO(g)  <=>  Fe(s)  +  CO2(g)

 

 

2.  Given the following information:

                        H2O(g)  +  CO(g)  <=>  H2(g)  +  CO2(g)                 K= 1.6

 

                        Fe(s)  +  CO2(g)  <=>  FeO(s)  +  CO(g)                K = 1.5

 

            Find the value of the equilibrium constant, K, for the reaction:

                        2 FeO(s)  +  2 H2(g)  <=> 2 Fe(s)  + 2 H2O(g)       K = ________

 

 

3.  Given the information:  

            H2O(g)  +  CO(g)  <=>  H2(g)  +  CO2(g)                 K= 1.6

 

            A l.00 liter reaction vessel contains the following:  0.10 moles of H20 (g),

            0.10 moles of CO gas, 0 .10 moles of H2 gas and 0.10 moles of CO2 gas.

a.  Is the reaction at equilibrium?_____  On what is your answer based?

 

 

b.  If the reaction is not at equilibrium, which way will it “move/shift” to reach                                   equilibrium?

 

 

a.      Set up an ICE table.

 

 

b.      What is the concentration of each species at equilibrium?

 

 

 

 

 

 

 

 

 

6.                     Butane(g)       <=>       Isobutane(g)         Kc= 2.5  at  25oC

 

How do these changes affect the system and what is the effect on the value of K?

 

Change/Disturbance                                    Effect on system?                 Effect on

                                                                        (shift toward products,          value of K?

                                                                        reactants, or neither?)

1. Add Butane                                               ______________                 __________

2. Add Isobutane                                           ______________                 __________

3. Decrease Vol. of container                      ______________                 __________

4. Lower the Temperature                            ______________                 __________

5. Add some helium gas?                            ______________                 __________