Steady State Approximation

In chemical kinetics, a steady state refers to a condition wherein all the state variables remain constant or change negligibly despite the on-going process trying to change its state. For any reaction to successfully occur, it is imperative to study the involvement of free energy changes in a reaction.  

Subsequently, the rates with which a reaction proceed and rate law are critical aspects of chemical kinetics or physical chemistry. To determine this exact rate of reaction, the study of steady state approximation is integral. 

Reaction Mechanism 

It is an intriguing concept of chemistry which specifically informs about how a chemical reaction occurs, the stages of a reaction, transition state mechanism, break and formation of bonds, etc. Even a simple looking chemical equation is capable of having a complex or multi-stage mechanism. 

For instance, take this reaction into consideration to understand –

2 N2O5 (Di-nitrogen pentoxide) → 4 NO2 (Nitrogen dioxide) + O2 (Oxygen) 

If you look at this chemical reaction, it is impossible to determine its rate of reaction. It depends upon this rate of reactions taking place in each step and mostly on the slowest stage involved in the mechanism of a chemical reaction. 

The three stages in the formation of Di-nitrogen pentoxide are given as follows – 

  1. N2O5 ↔ NO2 + NO3

  2. NO2 + NO3 → NO + NO2 + O2

  3. NO3 + NO → 2 NO

In this above reaction, NO and NO3 are intermediary products which don’t remain as end products when the reaction concludes. 

Few pointers regarding these are – 

  1. Intermediaries formed during any of the elementary stages are consumed in their following stage. 

  2. As per steady state approximation, it is clear that the steady-state variables such as temperature, entropy, pressure, etc. don’t change in intermediate steps. 

  3. Approximation of steady state also deals with the fact that variation in intermediates doesn’t have any impact over the rate of reaction. 

  4. It is to note that a balanced equation differs from the mechanism of a chemical reaction. 

  5. A balanced equation is not helpful while writing rate law. 

  6. A reaction mechanism is helpful while writing rate law. 

  7. Since these intermediates get consumed in the next stage itself and live a short life, they do not have any effect in this rate of a chemical reaction. 

Steady State Theory for Complex Reactions 

A chemical reaction is said to be complex if its mechanism consists of more than one stage. A number of electron transfer steps take place in between the initiation of reaction and formation of an end product. While talking about the mechanism and progress of such chemical reactions, you will also have to study various other integral concepts such as activated complex and reaction coordinate. 

Activated Complex – While the reactants combine to participate in a chemical reaction, an intermediate state is formed. In this state, the point of maximum energy formed for a structure can be defined as the activated complex. 

Reaction Coordinate – It is the measure of progress that comes along its path during the conversion of reactants into products. 

Further, to understand the complex reaction mechanism, you will need to rely on two theories – 

  1. Equilibrium approximation. 

  2. Steady state approximation. 

Study of these two concepts will give you a clearer picture of the reaction rate, reaction law, etc. 

Equilibrium Approximation 

Consider this equation where A and B are reactants which react to form an end product C. In this case, intermediate formed is AB.        

          \[k_{-1}\]

A + B \[\rightleftharpoons\] AB

           \[k_{1}\]


AB \[\overset{k_{2}}{\rightarrow}\] C


Here, the entire sequence of reaction can be defined in one coordinate by labelling the coordinates of individual stages. Amongst these stages, the slowest stage helps determine the rate of reaction and all other stages before it is assumed to remain in the equilibrium. 

Steady State Approximation 

For cases with complex reactions which have multiple stages, determining the rate of a reaction is not easy. The multiple stages will eventually contain multiple intermediates amongst which, one of the intermediate’s concentration remains constant at a certain stage of this reaction. In such a condition, it is assumed to be in steady state. However, its concentration may change with respect to time towards the beginning and end of a chemical reaction. 

Refer to the image shown below to understand this change. 

Image will be uploaded soon

From this, we can infer that the rate of reaction for an intermediate’s formation is equal to the decomposition rate. So, we can write – 

\[\frac{d[INT]}{dt}\] = 0

Here, INT represents the intermediate. You can also write it as –

\[\frac{d|I1|}{dt}\] = \[\frac{d|I1|}{dt}\] = \[\frac{d|I3|}{dt}\] = 0

It is specifically utilised for checking the consistency of chemical reaction rate law. 

Example No. 1 

Derive rate law for the following chemical reaction with the use of steady state approximation. 

2 N2O5 (Di-nitrogen pentoxide) → 4 NO2 (Nitrogen dioxide) + O2 (Oxygen) 

Solution –

Chemical reaction mentioned here has a three-step mechanism, as shown in the image below. 

            \[k_{f}\] 

\[N_{2}\]\[0_{5}\] \[\rightleftharpoons\] \[NO_{2}\] + \[_{3}\] 

            \[k_{b}\] 


\[NO_{3}\]  + \[NO_{2}\]   \[\overset{k_{2}}{\rightarrow}\] NO + \[NO_{2}\] + \[0_{2}\]

\[NO_{3}\] + NO  \[\overset{k_{3}}{\rightarrow}\]  \[2NO_{2}\]


Clearly, the two intermediates formed during the mechanism reaction are NO and NO3

  • For NO we have, 

[NO] production rate = k2 [NO2] [NO3

[NO] consumption rate = k3 [NO] [NO3

Since the rate of reaction for formation of an intermediate is equal to this rate of reaction for decomposition, we can write –

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

On solving the above equation for [NO], we get 


[NO] = \[\frac{k2[NO_{2}][NO_{3}]}{k3[NO_{3}]}\]   . . . . . . . . (1) [Considering as equation 1]


  • For another intermediate NO3, we have –  

[NO3] production rate = kf [N2O5]

[NO3] consumption rate = k2 [NO2] [NO3] + k3 [NO] [NO3] + Kb [NO2] [NO3]

As per the steady state assumptions, we can write – 

kf [N2O5] = k2 [NO2] [NO3] + k3 [NO] [NO3] + Kb [NO2] [NO3]

On solving the above equation for [NO3], we get 


\[[NO_{3}]\] = \[\frac{kf[N_{2}0_{5}]}{k2[NO_{2}][NO_{3}] + [k_{3}][NO][NO_{3}] + Kb[NO_{2}][NO_{3}]}\]  . . . . . . . . (2) [Considering as equation 2]


Now, considering the three stages of the reaction mechanism, we see that stage 1 can’t provide rate expression as it is in the state of equilibrium. 

In stage 2, an intermediate product is formed, and the reaction is processed further, so that we can write – 


\[\frac{d|0_{2}|}{dt}\] = \[k_{2}\] \[[NO_{2}]\]  \[[NO_{3}]\] . . . . . . . . . . . (3) [Considering as equation 3]


Now, we need to substitute the value of NO from equation 1 in both equation 2 and equation 3. 


\[\frac{d|0_{2}}{dt}\] =  \[\frac{k2[N_{2}0_{5}]}{2k_{2} + kb}\] = k [ \[N_{2}\]\[0_{5}\]]

Here, the value of k can be expressed as \[\frac{k2 kf}{2k2 + kb}\]

Subsequently, the general assumption leads to the scenario that the approximate change in intermediates can be considered as zero. Besides, the system may as well encounter steady state error in case there are variations in reference input or the system has imperfect components. 

Multiple-Choice Questions 

  1. Choose the option which is incorrect for a 2-step reaction. 

  1. Any of the stages can determine the rate of reaction. 

  2. It is possible that the reaction is either exothermic or endothermic. 

  3. This consists of two transition states. 

  4. Two intermediates are formed in a 2-step reaction. 

  1. Choose the incorrect statement regarding the impact of temperature on a chemical reaction. 

  1. Ea varies with the change in temperature. 

  2. Reaction rate varies with changing temperature. 

  3. Rate constant isn’t proportional to temperature. 

  4. Rate constant varies with changing temperature. 

Now that you are familiar with steady state meaning and various other concepts related to it, you will find it easier to answer related questions. Learn these important concepts of Chemistry and improve your knowledge of the subject. These study notes provided by Vedantu will help you score good grades as you rely on quality notes prepared by the best tutors. Download Vedantu official app to have full access to these quality study notes from anywhere and anytime.

FAQ (Frequently Asked Questions)

1. What are the required conditions for steady state approximation?

Ans: This steady state approximation is applicable only on cases wherein the intermediate’s reaction concentration can be constant. A steady state refers to a state of equilibrium where the consumption rate equals the formation rate of an intermediate.

2. What is the application of steady state approximation?

Ans: The method of steady state approximation is utilised for the derivation of rate law. The theory depends upon the assumption that concentration of intermediates remains constant as they are consumed as soon as they are generated.

3. How to differentiate between pre-steady-state and steady-state?

Ans: A pre steady state is concerned about the reactants involved in the reaction, whereas the steady state is concerned about the intermediates and products.

4. What is the difference between steady-state and equilibrium?

Ans: Equilibrium is attained in a reversible chemical reaction wherein the system reaches a point where the concentration of participating reactants and products don’t change at all. However, a steady state is attained when the intermediate’s concentration doesn’t change.