### Steady State

In a given chemical reaction, which involves one or more intermediates and one of the intermediates remains constant at some stage of the reaction. The system is said to reach steady state and name of the technique is called steady state approximation. It is also called as stationary-state approximation where the rate of conversion of intermediate is zero. Whereas, the steady state approximation does not assume that the concentration of intermediate is constant.

The terms like steady and equilibrium are commonly used interchangeably. However, they differ from each other scientifically. Under equilibrium conditions net reaction rate is zero, while no such limitation exists in steady state.

Taking an hypothetical forward reaction as shown in equation 1a

k1            k2

A B C ———- 1a

Where k1 and k2 are two reaction rates for conversion of A and B to C.

Reaction Rate

d[A]/dt=-k1[A] ———- 2a

d[B]/dt= k1[A]-k2[B], ———- 2b

d[C]/dt= k2[B], rate of formation of C ———- 2c

Under steady state condition the concentration of intermediate is set to zero.

\d[B]/dt=0

\k1[A]-k2[B]=0

\k1[A]=k2[B]

\k1[A]/k2=[B]

Substituting in equation 2c.

\d[C]/dt=k1[A], hence [C]= [A] (1-e-k1t)

Derivation of rate constant of reaction using steady state approximation can be understood by taking a simple example as follows: –

If the given reaction of consumption of nitrogen pentoxide to produce nitrogen dioxide and oxygen

2N2O5 4NO2 + O2

can be explained by following three step mechanisms:

kf

i) N2O5 NO2 + NO3

kb

(rate constant k1 divided as kf and kb)

ii) NO3 + NO2 NO + NO2 + O2 (rate constant k2)

iii) NO3 + NO 2NO2 (rate constant k3)

Following is the method to derive the rate constant of the above mentioned three steps using the principle of steady state approximation:

Step1:
Determination of intermediate

It is seen that reaction follows three steps. It can be seen that step 1 is under equilibrium. And there are two major intermediates used in step 2 and 3 viz. NO3 and NO.

Step 2:Developing equation for production and consumption rate for intermediates

Production rate of NO: k2 [NO3] [NO2] ——–(from step ii)

Consumption rate of NO: k3 [NO3] [NO] ———(from step iii)

Production rate of NO3: kf [N2O5] ——–(from step i)

Consumption rate of NO3: k2[NO3] [NO2] + k3[NO3] [NO] + kb[NO3] [NO 2]——–(From step i, ii and iii)

Step 3:
Applying steady state approximation law to equations of intermediates

Under steady state conditions the production and consumption rates are equal.

\ k2[NO3] [NO2]= k3[NO3] [NO]

\ [NO] = k2 [NO3] [NO2]/k3 [NO3] —— (1)

Similarly,

kf[N2O5] = k2[NO3] [NO2] + k3[NO3] [NO] + kb[NO 3] [NO2]

\ [NO3]= kf [N2O5]/ k2 [NO2]+ k3 [NO3] + kb [NO2]——-(2)

Further, the step i of the reaction is under equilibrium condition. But the step ii consumes NO species for carrying out step iii for its completion. This leads to developing rate expression from step ii as

d[O2]/dt= k2[NO3] [NO2] ——-(3)

Step 4: Solving equations 1, 2 and 3 in order to determine final expression for rate of reaction

Substituting value of [NO] from equation 1 in 2 and slowing it we get

\ [NO3]= kf[N2O5]/ k2[NO2] + k2 [NO2] +kb [NO2]

\ [NO3]= kf[N2O5]/ [NO2] [2k2+kb] ——-(4)

Substituting value of [NO3] from equation 4 in 3 and solving it

\d[O2]/dt= k2 kf [N2O5]/ kb + 2k2 = k[N2O5]

Where, k=k2kf/ kb+2k2

This is differential rate constant, which agrees with the experimental results. Moreover, one of the most commonly used examples of steady state approximation is Michaelis-Menten kinetics, which applied, in bio catalytic mechanism.