Form 6 Chemistry – PHYSICAL CHEMISTRY 1.4-CHEMICAL EQUILIBRIUM

PREDICTION OF DIRECTION AND EXTENT OF CHEMICAL EQUILIBRIUM

At each point in the progress of a reaction, it is possible to formulate the ratio of concentration having the same form as the equilibrium

constant expression .This generalized ratio is called reaction quotient (Q).

For the reaction; then:- 

Q_C =                                                                                   

Q _p =                                      

Q_C differs from K_C in that the concentration in the expression is not necessarily the equilibrium concentration.

When the values of K_C and Q_C are compared, one can predict the direction of the chemical reaction.

        If Q _c K _c the system is not at equilibrium, the reactants must further be converted to products to achieve equilibrium therefore net reaction proceeds from left to right.

        If Q _c = K _c the system is at equilibrium

        If Q _c > K _c the system is not an equilibrium, the products must converted to reactants to achieve equilibrium therefore a net reaction proceeds from right to left.

Example:

1.Consider the reaction

        

At 250⁰C, K _c = 4.0 x 10^-2. If the concentration of Cl₂ and PCl₃ are both 0.30 M while that of PCl5 is 3.0M, is the system at equilibrium? If not, in which direction does the reaction proceed?

      ^-2

         Q _c =                                                                              

                =                                

  Q _c = 0.03        Q_C ≠ K _c,        Q _c< K _c

The system is not at equilibrium and the reaction proceeds from left to right

2. At 200k, the K _p for the formation of NO is 4 x 10^-4

     N_{2 (g)} + O_{2 (g)}                      2NO _(g)

If at 200k the partial pressure of N₂ is 0.5 atm and that of 0₂ is 0.25 atm, that of NO is 4.2 x 10^-3 atm, decide whether the system is at equilibrium, if not in which direction does the reaction proceed.

  K _p = 4.0 x 10^-4

 Q _p =                                  

       =

  Q _p = 1.4112 x 10^-4           K _p ≠Q _p,      K _p > Q _p

The system is not at equilibrium, therefore the reaction proceeds from left to right

EQUILIBRIUM CONSTANT WITH DEGREE OF DISSOCIATION ()

-It gives to what extent the reactants are converted to products by dissociation.

-This has to be treated similar to moles.

Example1.  0.01 moles of PCl₅was placed in 1L vessel at 210K.It was found to be 52.6% dissociated    into PCl₃ and Cl₂. Calculate the K _c at that temperature.

                          PCl_{5 (g)}                     PCl_{3 (g)} + Cl_{2 (g)}

At start:                      0.01                                0                0

At equilibrium: 0.01-(5.26 x10^-3)            5.26 x 10^-3     5.26×10^-3  

              = 

          = 5.26 x

      PCl₅ = 4.74 x 10^-3 moles

      PCl₃ = 5.26 x 10^-3 moles

       Cl₂   = 5.26 x 10^-3 moles

         K _C =                                     

             =                                         

       K _C = 5.837 x 10^-3 moles

2. At 1 atm and 85^oC, N₂O₄ is 50% dissociated, Calculate the equilibrium constant in terms of pressure and calculate the degree of dissociation of the gas at 10⁰ c and 55⁰c.

                         N₂O₄                          2NO_{2 (g)}      

Start                          1                                        0         

Equilibrium:          1                                     2

  But = 50% = 0.5

                              1-0.5                                   2(0.5)

                               0.5                                      1

n_T = 0.5 + 1

= 1.5 moles

 ₌      x 1                           =    x 1

      = 0.33 moles                            = 0.66moles

   K _{p =}      

           = 

        K _p = 1.32

                        N₂O₄              2NO_{2 (g)} 

Start:                 1                     0

Equilibrium:    1-                  2

      n_T   = (1- ) + 2                     

            =1+

    P_T = 10atm

   =       x 10                        =  x 10

   K _p =                   

       =

    =   x 100 x  x

    =

     1.32 =   

  1.32 – 1.32² = 40²

  1.32 = 40^{2 +} 1.32²

  1.32 = 41.32²

  ² =0.0319

    = 0.1787

Degree of dissociation = 17.87%

 DEGREE OF DISSOCIATION BY DENSITY MEASUREMENT

This method is used for the determination of degree of dissociation of gases in which 1 molecule produces 2 or more molecules.

i.e. PCl_{5 (g)}    PCl_{3 (g)} + Cl_{2 (g)} 

Thus at constant temperature and pressure, the volume increases. The density at constant pressure decreases.

The degree of dissociation can be calculated from the difference in density between the undissociated gas and that of partially dissociated gas at equilibrium.

If we start with 1 mole of the gas (PCl₅) and the degree of dissociation ( ), Then

                                    PCl_{5 (g)}         PCl_{3 (g)} + Cl_{2 (g)}

Moles at equilibrium:  1-                                 

Total moles:   1 –

                      = 1+

Note:

The density of an ideal gas at constant temperature and pressure is inversely proportional to the number of moles for a given weight.

Hence the ratio of density

 

 =

            When = Density of gas mixture of equilibrium

                                                = Density of gas before dissociation

                                              Degree of dissociation

Examples1. When PCl₅ is heated, it gasifies and dissociates into PCl₃ and Cl₂.The density of the gas mixture at 200⁰C is 70.2 Find the degree of dissociation of PCl₅ at 200⁰C

Solution

Observed density, ρ₂ =70.2

                               ρ₁ =?  

                 V.D =  

                    =  

                     =104.25

                

                    =   

              

                     = 48.5%

2. At 90⁰C, the V.D of N₂O₄ is 24.8. Calculate the % dissociation into NO₂ molecules at this temperature.

  N₂O₄        NO₂ + NO₂

Density at equilibrium,   ρ₂ = 24.8

 = V.D =  =   = 46                                        

?

=

    =

     = 85.48%

DETERMINATION OF DEGREE OF DISSOCIATION BY MOLECULAR MASS

Molecular masses are proportional at constant temperature and pressure to the density of their gases; therefore we can substitute the molecular masses for the density in the degree of dissociation.

          

                         

Where:-

 = Molecular mass of undissociated gas   

  = Average Molecular mass of gases at equilibrium    

Example1:1.588g of N₂O₄  gives a total pressure of 1atm when partially dissociated at equilibrium in a 500cm³ glass vessel at 25⁰C.What is the degree of dissociation at this temperature?                    

    N₂O₄         2NO₂

M₁ = (14x 2) + (16 x4)

      = 92gmol^-1

M₂ =?

From PV= nRT, W here n =

M₂ =

      =

     = 77.7gmol^-1

 

       =  

      =    0.184

   

FACTORS AFFECTING EQUILIBRIUM REACTION

These factors are as follows:-

i)  Temperature

ii)  Concentration

iii)  Pressure

        The first three affects both rates and position of chemical equilibrium (i, ii and iii)

        The other three affects the rate of chemical equilibrium

 1) Temperature

a) Increasing the temperature, increases the rate of reaction because usually at high temperature the collision factor increases also the number

of molecules having necessary activation energy is large.

b)  Effect on the position of equilibrium is explained by using Le-Chateliers principle which states that “when a system at equilibrium is

subjected to a change, processes occur which tend to counteract the change” (If a system in equilibrium is disturbed (change in temperature and pressure) the system adjusts itself so as to oppose the disturbance).

Consider the reaction

2SO_{2 (g)} + O_{2 (g)}       2SO_{3 (g)} + Heat (negative)

              

   If temperature is increased in the system, the equilibrium moves in a direction where there is a absorption of heat and if the temperature is decreased  in the system, the equilibrium moves in a direction where there is release of heat.

Effect of temperature, on the position of equilibrium can be explored by Vant Hoff`s law of
mobile chemical equilibrium which states that

“For any system in equilibrium high temperature favours endothermic reactions and low temperature favours exothermic reactions.

The way in which equilibrium constant changes with temperature is found both theoretically and experimentally governed by the following

relationship;

      =                 

 âˆ† Hm = change in molar heat

  K = Equilibrium constant

On intergrating the equation above;

                                           

   Where c = constant

If K₁ and K₂ are equilibrium constants corresponding to T₁ and T_{2 ,}the constant term can be eliminated from the equation above so as to give Vant Hoff`s equation i.e.

lnK₁ =  …………………. (1)

lnK₂ = …………………. (2)

Subtracting equation (1) from (2) i.e.

lnK₂ – lnK₁ =))

=   ……………………1

But 1 ln = 2.303log

 =  ……………………2

But  =

        = ……………………3

  Where 1 = 2 = 3

Example1: For the reaction;

N₂O₄      2NO₂      âˆ†H = 61.5KJ mol^-1

                               K_P = 0.113 at 298K

i) What is the value of K_P at 0⁰C?

ii) At what temperature will K_P =1?

Answers:

i) From  =  

K_P2 = 0.113

K_P1 =?

T ₁ = 273K

T ₂ = 298K

∆Hm = 61.5Kjmol^-1

R=8.314

 =    

 = 3211.9676 (3.0729896 x 10^-4)

   = 0.987

To remove ‘log’ make 10^0.987

 

    

     

ii) From log = 

K_P2 = 0.113                 T₁ =?

K_P1 = 1                         T₂=298K

∆Hm = 61.5KJ mol^-1

R = 8.314

= 

-0.9469 = 3211.9676 

 -2.948 x 10^-4 =

 -0.0878T ₁ = 298 – T ₁

  -0.0878T ₁ + T₁ =298

   0.912T₁ = 298

    T₁ = 326.7k