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 –
N2O5 ↔ NO2 + NO3
NO2 + NO3 → NO + NO2 + O2
NO3 + NO → 2 NO2
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 –
Intermediaries formed during any of the elementary stages are consumed in their following stage.
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.
Approximation of steady state also deals with the fact that variation in intermediates doesn’t have any impact over the rate of reaction.
It is to note that a balanced equation differs from the mechanism of a chemical reaction.
A balanced equation is not helpful while writing rate law.
A reaction mechanism is helpful while writing rate law.
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 –
Equilibrium approximation.
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
Choose the option which is incorrect for a 2-step reaction.
Any of the stages can determine the rate of reaction.
It is possible that the reaction is either exothermic or endothermic.
This consists of two transition states.
Two intermediates are formed in a 2-step reaction.
Choose the incorrect statement regarding the impact of temperature on a chemical reaction.
Ea varies with the change in temperature.
Reaction rate varies with changing temperature.
Rate constant isn’t proportional to temperature.
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.
FAQs on Steady State Approximation
1. What is the Steady State Approximation (SSA) in chemical kinetics?
The Steady State Approximation is a method used to simplify the derivation of a rate law for a multi-step reaction mechanism. It is based on the assumption that the concentration of a highly reactive intermediate remains low and nearly constant throughout the reaction. This occurs because the intermediate is consumed almost as quickly as it is formed, meaning its net rate of change is approximately zero (d[Intermediate]/dt ≈ 0).
2. What is the primary application of the Steady State Approximation in chemistry?
The main application of the Steady State Approximation is to derive a plausible rate law for a complex reaction without needing to solve a complicated system of differential equations. By assuming the concentration of the intermediate is constant, we can express it in terms of the concentrations of stable reactants, which simplifies the overall kinetic analysis and allows for the prediction of the reaction's rate equation.
3. What conditions must be met for the Steady State Approximation to be valid?
For the Steady State Approximation to be valid, the following conditions are essential:
- The intermediate species must be highly reactive and have a very short lifetime.
- The concentration of the intermediate must remain very low compared to the concentrations of the reactants and products.
- The rate of formation of the intermediate must be approximately equal to its rate of consumption, leading to a constant, or 'steady state', concentration.
4. How is the Steady State Approximation expressed mathematically for a reaction intermediate?
Mathematically, the Steady State Approximation is applied by setting the rate of change of the intermediate's concentration to zero. For an intermediate 'I', the equation is:
d[I]/dt = (Rate of formation of I) - (Rate of consumption of I) ≈ 0
This simplifies to:
Rate of formation of I ≈ Rate of consumption of I
This algebraic equation allows us to solve for the unknown concentration of the intermediate, [I].
5. Can you provide a simple example of how the Steady State Approximation is used to derive a rate law?
Consider a two-step reaction:
Step 1: A + B → I (rate constant k₁)
Step 2: I → P (rate constant k₂)
Here, 'I' is the intermediate. To find the rate law for the formation of product 'P' (Rate = k₂[I]), we first apply the SSA to I:
d[I]/dt = k₁[A][B] - k₂[I] ≈ 0
Solving for [I], we get: [I] ≈ (k₁/k₂)[A][B].
Now, substitute this into the rate law for P:
Rate = k₂( (k₁/k₂)[A][B] ) = k₁[A][B].
Thus, the SSA helps us express the overall rate in terms of measurable reactants.
6. What is the fundamental difference between a reaction at steady state and a reaction at equilibrium?
The key difference lies in what is being held constant.
- A steady state refers to a condition in a proceeding reaction where the concentration of a transient intermediate is constant. Reactants are continuously being converted to products.
- An equilibrium is a condition where the concentrations of both reactants and products are constant because the forward reaction rate equals the reverse reaction rate. There is no net change in the overall system.
7. How does the Steady State Approximation help simplify the analysis of complex, multi-step reaction mechanisms?
The Steady State Approximation simplifies complex mechanisms by transforming a difficult calculus problem into a simpler algebra problem. Without it, determining the concentration of an intermediate over time requires solving complex differential equations. By assuming the rate of change of the intermediate's concentration is zero, we create a simple algebraic equation. This allows us to easily solve for the intermediate's concentration and substitute it back into the rate equation for the product, yielding a final, manageable rate law that can be tested experimentally.
8. When might the Steady State Approximation fail or lead to an incorrect rate law?
The Steady State Approximation can fail under certain conditions where its core assumption is violated:
- Induction Period: It is not valid at the very beginning of a reaction, as the intermediate's concentration is still building up and is not yet constant.
- Stable Intermediates: If an intermediate is relatively stable and does not react quickly, its concentration can accumulate, meaning its rate of change is not close to zero.
- Explosive Reactions: In some chain reactions, the concentration of radical intermediates grows exponentially, which is the opposite of a steady state.
