Consecutive reaction, reversible reaction and parallel reaction

Consecutive reaction

The reactions which proceeds from reactants to products through one or more intermediate stage are known as consecutive reactons.

In these sequence of reactions of varying speed the one that is slowest step will determine the overall rate of reaction.

Let us consider a reaction

A \(\longrightarrow\) B \(\longrightarrow\) C

When, k1> k2the conversion of B to C is slow and determines the rate of overall reaction.

When, K2> k1the conversion of A to B is slow and determines the rate fo overall reaction.

When, k1≅K2, both steps determines the rate of reaction.

Let the initial concentration of A be 'a' moles/l. Let the concentraion of A, B and C at time t be x, y and z moles/l respectively. The concentration of species can be represented as,

A \(\longrightarrow\) B \(\longrightarrow\) C

When, t= 0 a mole/l 0 0

When, t=t x mole/l y mole/l z mole/l

The rate equations in terms of A, B and C are

Rate of disappearance of A,

- \(\frac{dx}{dt}\) = k₁x......(i)

Rate of formation of B

\(\frac{dy}{dt}\) = k1x - k2y.......(ii)

Rate of formation of C,

\(\frac{dz}{dt}\) = k₂y..........(iii)

From equation (i)

-\(\frac{dx}{dt}\) =k₁x

-\(\frac{dx}{x}\) = k₁dt

Integrating,

\(\int\) -\(\frac{dx}{x}\) =k1\(\int\) dt

or, -\(\ln\) x = k1t + c where, c= integration constant

when, t = 0, x = a then,

-\(\ln\) a = c

-\(\ln\) x = k1t - \(\ln\) a

or, \(\ln\) a - \(\ln\) = k₁t

or, \(\ln\) \(\frac{a}{x}\) = k1t

or, \(\frac{a}{x}\) = e k1t

or, \(\frac{x}{a}\) = \(\frac{1}{e^(k_1t)}\)

or, x = ae^-k₁t...............(iv)

Substituting the value of x in equation (ii), we get,

\(\frac{dy}{dt}\) = k₁ae^-k₁t- k₂y

or, \(\frac{dy}{dt}\) + k₂y = k₁ae^-k₁t

Integrating factor = exp \(\int\) k₂dt

= e^k₂t

Multiplying through integration factor,

\(\frac{dy}{dt}\) e^k₂t+ k₂y e^k₂t = k₁ae^-k₁t. e^k₂t

or, \(\frac{d}{dt}\) (ye^k₂t) = k₁ae^(k₂-k₁)t

or, d(ye^k₂t) = k₁ae^(k₂-k₁)tdt

Integrating,

\(\int\) d(ye^k₂t) =k₁a \(\int\)e^(k₂-k₁)t+ c

or, ye^k₂t= \(\frac{k_1a}{k_2 -k_1}\)e^(k₂-k₁)t+ c

When t = 0, y = 0, then

c = -\(\frac{k_1a}{k_2 -k_1}\)

thus, substituting the value of c in the above equation,

ye^k₂t= \(\frac{k_1a}{k_2 -k_1}\)e^(k₂-k₁)t-\(\frac{k_1a}{k_2 -k_1}\) a

or,ye^k₂t= \(\frac{k_1a}{k_2 -k_1}\) [e^(k₂-k₁)t-1]

or, y =\(\frac{k_1a}{k_2 -k_1}\) (e^{-k₁t -}e^-k₂t)

Now,

a = x + y + z

or, z = a - (x+y)

Substituting the value of x and y then,

z = a - {ae^-k₁t+\(\frac{k_1a}{k_2 -k_1}\) (e^{-k₁t -}e^-k₂t)}

or, z = a - [\(\frac{(k_2-k_1)ae^(-k_1 t)+k_1a(e^(-k_1 t) - e^(-k_2 t)}{k_2-k_1}\)]

= a - [\(\frac{k_2ae^(-k_1 t) - k_1a(e^(-k_1 t) + k_1 a(e^(-k_1 t) -k_1 ae^(-k_2 t)}{k_2 -k_1}\)

= a - [\(\frac{k_2 ae^(-k_1 t) - k_1 ae^(-k_2 t)}{k_2 -k_1}\)

= \(\frac{(k_2 - k_1)a - k_2ae^(-k_1 t) + k_1ae^(-k_2 t)}{k_2 -k_1}\)

= \(\frac{a}{k_2 -k_1}\) [(k₂ - k₁) - k₂e^-k₁t+ k₁e^-k₂t]

=\(\frac{a}{k_2 -k_1}\) [(k₂-k₂e^-k₁t) - (k₁- k₁e^-k₂t)............(vi)

The variation of concentration of A, B and C with time when k₁is nearly equal to k₂is shown in the curve

This curve indicates that concentration of A decreases continuously but that of C increases continuously with time. The concentration of B rises first reaches to the peak value and then decreases with time.

When k₁>> k₂, then the peak of the curve lies above than that of the case k₁≈ k₂

When k₂ >> k₁, then the peak of the curve lies below than that of the casek₁≈ k₂

From the equation (vi), some cases arise

Case I:

When k₁>> k_2,then equation (vi) becomes

z =\(\frac{a}{k_2 -k_1}\) [(k₂-k₂e^-k₁t) - (k₁- k₁e^-k₂t)]

= \(\frac{a}{-k_1}\)× -(k₁-k₁e^-k₂t)

= \(\frac{a}{-k_1}\)× -k₁(1- e^-k₂t)

= a (1 - e^-k₂t)

i,e, the rate of formation of 'C' becomes independent of k₁. In other words the second step is the rate determining step.

Case II

When k₂ >> k₁, then equation (vi) becomes

z = a (1- e^-k₁t)

i.e, the rate of formation of 'c' becomes dependent of k₂. In other words the first step is the rate determining step.

Example:

3NaClO \(\longrightarrow\) 2NaCl + NaClO₃

This reaction proceeds in two steps

1. 2NaClO \(\longrightarrow\) NaCl + NaClO₂
2. NaClO₂+ NaClO \(\longrightarrow\) NaCl + NaClO₃

Experimentallyk₂ >> k₁. So step first is the rate determining step.

Reversible reaction or opposing reaction

Revesible reaction is one in which the products of chemical change react together to form the original products. The reverisble reaction do not proceed to completion. The overall or net rate of reaction is influenced by both forward and backward rates and ultimately becomes zero at equilibrium.

In reversible reaction, the progress of forward and backward reaction be shown as

Here the curve AB represents the variation of rate of forward reaction and the curve CD represents the variation of rate of backward reaction. From these curves it is clear that,

1. Rate of forward reaction gradually decreases.
2. Rate of backward reaction gradually increases.
3. At point E, where curves meet, the rate of reversible reaction becomes equal. This point is called equlibrium point. At equilibrium the net rate of reaction is zero i.e, \(\frac{dx}{dt}\) = 0 (at equilibrium)

Kinetics of reversible/opposing reaction

Let us consider a general reaction, a first order reaction is opposed by reaction by reaction of first order.

where, k₁ = rate constant of first order for forward reaction

k_-1= first order rate constant for backward reaction

Let 'a' moles/l be the initial concentration of reactant A and 'x' moles/l be the amount of A reacted at time ' t' , then concentraion of A left unreacted at time 't' be (a-x) moles/l. At time 't' the concentration of B will be x moles/l

∴rate of forward reaction, (\(\frac{dx}{dt}\))_f= k₁(a-x)

and rate of backward reaction, (\(\frac{dx}{dt}\))_b= k_-1x

The net rate of forward reaction is ,

\(\frac{dx}{dt}\) = (\(\frac{dx}{dt}\))_f- (\(\frac{dx}{dt}\))_b

= k₁(a-x) - k_-1x ..........(i)

This is the differential rate equation of reversible reaction.

At equilibrium state,

\(\frac{dx}{dt}\) = 0

x = x_e

where, x_e = equilibrium concentration of product

The equation (i) becomes,

0 = k₁(a-x_e) - k_-1x_e

or,k_-1x_e=k₁(a-x_e)

or, k_-1= \(\frac{k_1(a-x_e)}{x_e}\)..........(ii)

Putting the value of k_-1in equation (i)

\(\frac{dx}{dt}\) =k₁(a-x) -\(\frac{k_1(a-x_e)}{x_e}\)× x

= k₁a - k₁x - \(\frac{k_1ax - k_1x_ex}{x_e}\)

= \(\frac{k_1ax_e - k_1xx_e - k_1ax + k_1xx_e}{x_e}\)

\(\frac{dx}{dt}\) = \(\frac{k_1a(x_e - x)}{x_e}\)

or, \(\frac{dx}{x_e-x}\) = \(\frac{k_1a}{x_e}\) \(\int\)dt

Integrating,

\(\int\) \(\frac{dx}{x_e-x}\) = \(\frac{k_1a}{x_e}\) + C

where, c= integration constant

when t=0, x = 0 then

-\(\ln\) (x_e - 0) = C

C = -\(\ln\)x_e

Substituting the value of 'C'

-\(\ln\) (x_e - x) = \(\frac{k_1at}{x_e}\) - \(\ln\)x_e

or, \(\lnx_e - \(\ln\)((x_e - x) = \(\frac{k_1at}{x_e}\)

∴k₁= \(\frac{x_e}{at}\) \(\ln\)\(\frac{x_e}{x_e - x}\)............(iii)

From this equation we can find the value of rate constant for forward reaction if the value of equilibrium concenttration of product 'x_e' is known.

References

the goodman group. n.d. <http://www-jmg.ch.cam.ac.uk/tools/magnus/kineticnotes.html>.

Prutton, S. H. Maron and C. Principles of Physical Chemistry . Oxford and IBH Pub. Co, 1992.

Royal Chemical Society. n.d. <http://www.rsc.org/suppdata/dt/c2/c2dt31851b/c2dt31851b.pdf>.