STUDY OF FAST REACTIONS AND RELAXATION METHOD

TEMPERATURE - PRESSURE JUMP METHOD :-

Eigen and his coworkers used relaxation methods to investigate fast reactions in solutions.

In these types of methods, the reacting system is equilibrium is subjected to a sudden variation in some physical parameter on which the equilibrium constant of the reaction depends.

The system will be changes to new state of chemical equilibrium. 

The rate of this change is it's called relaxation is then investigated. 

Relaxation studies are carried and with such parameters as temperature, pressure and electric field.

A HIGHER - VOLTAGE POWER SUPPLY CHARGES A CAPACITOR    C.

                      When a certain voltage is reached, the spark gap G break down, discharging the capacitor and sending a strong current through the cell containing the reactive system at equilibrium in a conducting aqueous solution.

As the current passes, the temperature of the system rises by about 10⁰c in a few microseconds.

In the ensuring time interval the concentration of the reacting species adjust to the new equilibrium. Value appropriate to the temperature jump. 

                                 As a resulting, the intensity of the light beam leaving the cell and entering the detector ( the photomultiplier, PM) is charged.

The output of the photomultiplier tube is displayed on the vertical axis of the oscilloscope. In this way the curve showing the variation of the concentration versus time is displayed on the oscilloscope screen.

Consider a reversible first - order reaction.

Let a be the concentration of A+B and x be the concentration of B at any instant.

Then, 

r =   dx.  = k₁ (a−x) − k₋₁ x  ................. (1)

       dt

If Xe is the equilibrium concentration, then 

∆x = x − xₑ

(or)

X = xₑ + ∆x    ............................... (2)

Since  d(∆x)      =  dx , we have

            dt                 dt

 d(∆x )  = k₁ ( a−xₑ − ∆x) −R₋₁ (xₑ + ∆x).. (3)

   dt

At equilibrium,

 dx  = 0 and x = xₑ

 dt

From equation (1)

k₁ (a−xₑ) − k₋₁ xₑ = 0

k₁ ( a- xₑ ) = k₋₁ xₑ ................... (4)

From equation (3)

 d(∆x ) = k₁ ( a- xₑ ) - k₁ (∆x) - k₋₁ xₑ - k₋₁ 

    dt

( ∆x ) ..................................(5)

Substituting equation (4) in (5)

 d(∆x ) = k₋₁ xₑ - ( k₁ + k-₁ ) ∆x

   dt

 d(∆x ) = - (k₁ + k-₁ ) ∆x

   dt

here, k₁ + k-₁ ₌ kr is the relaxation rate constant.

 d(∆x) = - kᵣ ∆x     .................(6)

   dt

Integerating the above equation ,

∆x = ∆x₀ exp (-krt)      ....................(7)

The reciprocal of kr,

(k₁ + k₋₁ ) ⁻¹ is called the relaxation time, τ,

That is, 

      τ =         1.           [ k₁ + k₋ ₁ ₌ kᵣ]

              k₁ + k₋₁ 

             τ =    1  

                      kᵣ