REACTION IN SOLUTIONS - EFFECT OF PRESSURE ON REACTION RATE

The rate of a chemical reaction in solution is affected by change in the pressure of the solution or change in the dielectric constant of the solution or change in the ionic strength of the solution.

1) Effect of Pressure on reaction rate :-

  

                       Let us consider a chemical reaction,

A + B             > C    (n=2)

According to Eyring's postulate,

                             The reactants forms a short lived unstable intermediate or transition state via a thermodynamic equilibrium.

A + B ⇔AB ≠ →product.

For the above equilibrium, 

Let k ≠ represents the equilibrium constant. 

According to Eyring,

                           The rate constant of the chemical reaction is directly proportional to    k≠ ,

      κ =   κB    T  k ≠ .................(1)

               h

Isotherm is,

∆G ≠ = − RT lnk ≠

k ≠ =        e   −∆G ≠     .............(2)

                       RT

κ =    κB  T    e   - ∆G ≠    .........(3)

          h                  RT

lnκ = ln  kBT  -   ∆G≠   ..............(4)

                h          RT

Differenting the above equation with respect to pressure under constant temperature.

( ∂lnκ )T = ( ∂ (ln   kBT  ) T +   ∂   (  −∆G≠ )

   ∂p                        ∂p              ∂p       RT

   = 0 +   ∂    (−∆G ≠ )

             ∂p      RT

∴ (   ∂lnk ) =   ∂    ( −∆G ≠  ) ..............(5)

        ∂p      T   ∂p       RT

                     According to thermodynamics, the changes in free energy and the changes in enthalpy is defined as,

∆G ≠ = ∆H≠ − T ∆S≠

∆H ≠ = ∆E ≠ + P ∆V ≠ 

(or)

∆G ≠ = ∆E ≠ + P ∆V ≠ − T ∆S ≠ ........(6)

Substituting equation (6) in (5)

(  ∂lnk   ) T = −   ∂    {  ∆E ≠ + P∆V≠− T∆S≠}

    ∂P                  ∂P               RT

 =  −∂     {  ∆E ≠  +   P∆V ≠  −   T∆S ≠  }

       ∂P       RT          RT              RT

When temperature is constant, ∆E and ∆S≠ are almost constants.

        Differentiation constant will be zero,

        (  ∂ lnk   )T = 0 −  ∆V≠  − 0

             ∂P                     RT

(  ∂ lnk   ) T  =   −∆V ≠   ..........(7)

   ∂P                    RT

Let us consider the following cases.

Cases (1)

    When ∆V ≠ < 0

In a chemical reaction, if the volume of intermediate or the transition state is much lesser than the sum of the total volume of the reactants.

                              Then increase of pressure will increase the rate of chemical reaction.

Plot of ln   κ   vs P will give a straight line 

                   K 

With negative slope.

Slope =    ∆V ≠

                 RT

  ∆V ≠ = slope RT .........(3)